LFM Pure

Let \(f(x)\) be defined and positive for \(x > 0\). Let \(a\) and \(b\) be real numbers with \(0 < a < b\) and define the points \(A = (a, f(a))\) and \(B = (b, -f(b))\). Let \(X = (m,0)\) be the point of intersection of line \(AB\) with the \(x\)-axis.

1. Find an expression for \(m\) in terms of \(a\), \(b\), \(f(a)\) and \(f(b)\).
2. Show that, if \(f(x) = \sqrt{x}\), then \(m = \sqrt{ab}\). Find, in terms of \(n\), \(a\) function \(f(x)\) such that \(m = \frac{a^{n+1} + b^{n+1}}{a^n + b^n}\).
3. Let \(g_1(x)\) and \(g_2(x)\) be defined and positive for \(x > 0\). Let \(m = M_1\) when \(f(x) = g_1(x)\) and let \(m = M_2\) when \(f(x) = g_2(x)\). Show that if \(\frac{g_1(x)}{g_2(x)}\) is a decreasing function then \(M_1 > M_2\). Hence show that $$\frac{a+b}{2} > \sqrt{ab} > \frac{2ab}{a+b}$$
4. Let \(p\) and \(c\) be chosen so that the curve \(y = p(c-x)^3\) passes through both \(A\) and \(B\). Show that $$\frac{c-a}{b-c} = \left(\frac{f(a)}{f(b)}\right)^{1/3}$$ and hence determine \(c\) in terms of \(a\), \(b\), \(f(a)\) and \(f(b)\). Show that if \(f\) is a decreasing function, then \(c < m\).

1. \noindent\vspace{-4cm} %%%%%%%The diagram requires scale of 1 unit = 15cm and 11 pt font.
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   \vspace{-6cm} The diagram shows three touching circles \(A\), \(B\) and \(C\), with a common tangent \(PQR\). The radii of the circles are \(a\), \(b\) and \(c\), respectively. Show that \[ \frac 1 {\sqrt b} = \frac 1 {\sqrt{a}} + \frac1{\sqrt{c}} \tag{\(*\)} \] and deduce that \[ 2\left(\frac1{a^2} + \frac1 {b^2} + \frac1 {c^2} \right) = \left(\frac1 a + \frac1 {b} + \frac1 {c} \right)^{\!2} . \tag{\(**\)} \]
2. Instead, let \(a\), \(b\) and \(c\) be positive numbers, with \(b

A circle \(C\) is said to be {\em bisected} by a curve \(X\) if \(X\) meets \(C\) in exactly two points and these points are diametrically opposite each other on \(C\).

1. Let \(C\) be the circle of radius \(a\) in the \(x\)-\(y\) plane with centre at the origin. Show, by giving its equation, that it is possible to find a circle of given radius \(r\) that bisects \(C\) provided \(r>a\). Show that no circle of radius \(r\) bisects \(C\) if \(r\le a\,\).
2. Let \(C_1\) and \(C_2\) be circles with centres at \((-d,0)\) and \((d,0)\) and radii \(a_1\) and \(a_2\), respectively, where \(d>a_1\) and \(d>a_2\). Let \(D\) be a circle of radius~\(r\) that bisects both \(C_1\) and \(C_2\). Show that the \(x\)-coordinate of the centre of \(D\) is \(\dfrac{a_2^2 - a_1^2}{4d}\). Obtain an expression in terms of \(d\), \(r\), \(a_1\) and \(a_2\) for the \(y\)-coordinate of the centre of~\(D\), and deduce that \(r\) must satisfy \[ 16r^2d^2 \ge \big (4d^2 +(a_2-a_1)^2\big) \, \big (4d^2 +(a_2+a_1)^2\big) \,. \]

The line passing through the point \((a,0)\) with gradient \(b\) intersects the circle of unit radius centred at the origin at \(P\) and \(Q\), and \(M\) is the midpoint of the chord \(PQ\). Find the coordinates of \(M\) in terms of \(a\) and \(b\).

1. Suppose \(b\) is fixed and positive. As \(a\) varies, \(M\) traces out a curve (the locus of \(M\)). Show that \(x=- by\) on this curve. Given that \(a\) varies with \(-1\le a \le 1\), show that the locus is a line segment of length \(2b/(1+b^2)^\frac12\). Give a sketch showing the locus and the unit circle.
2. Find the locus of \(M\) in the following cases, giving in each case its cartesian equation, describing it geometrically and sketching it in relation to the unit circle:
   • \(a\) is fixed with \(0 < a < 1\), and \(b\) varies with \(-\infty < b < \infty\);
   • \(ab=1\), and \(b\) varies with \(0< b\le1\).

The angle \(A\) of triangle \(ABC\) is a right angle and the sides \(BC\), \(CA\) and \(AB\) are of lengths \(a\), \(b\) and \(c\), respectively. Each side of the triangle is tangent to the circle \(S_1\) which is of radius \(r\). Show that \(2r = b+c-a\). Each vertex of the triangle lies on the circle~\(S_2\). The ratio of the area of the region between~\(S_1\) and the triangle to the area of \(S_2\) is denoted by \(R\,\). Show that $$ \pi R = -(\pi-1)q^2 + 2\pi q -(\pi+1) \;, $$ where \(q=\dfrac{b+c}a\,\). Deduce that $$ R\le \frac1 {\pi( \pi - 1)} \;. $$

1. The point \(A\) has coordinates \(\l 5 \, , 16 \r\) and the point \(B\) has coordinates \(\l -4 \, , 4 \r\). The variable point \(P\) has coordinates \(\l x \, , y \r\,\) and moves on a path such that \(AP=2BP\). Show that the Cartesian equation of the path of \(P\) is \[ \displaystyle \l x+7 \r^2 + y^2 =100 \;. \]
2. The point \(C\) has coordinates \(\l a \, , 0 \r\) and the point \(D\) has coordinates \(\l b \, , 0 \r\), where \(a\ne b\). The variable point \(Q\) moves on a path such that \[ QC = k \times QD\;, \] where \(k>1\,\). Given that the path of \(Q\) is the same as the path of \(P\), show that \[ \frac{a+7}{b+7}=\frac{a^2+51}{b^2+51}\;. \] Show further that \((a+7)(b+7)=100\,\).

The three points \(A\), \(B\) and \(C\) have coordinates \(\l p_1 \, , \; q_1 \r\), \(\l p_2 \, , \; q_2 \r\) and \(\l p_3 \, , \; q_3 \r\,\), respectively. Find the point of intersection of the line joining \(A\) to the midpoint of \(BC\), and the line joining~\(B\) to the midpoint of \(AC\). Verify that this point lies on the line joining \(C\) to the midpoint of~\(AB\). The point \(H\) has coordinates \(\l p_1 + p_2 + p_3 \, , \; q_1 + q_2 + q_3 \r\,\). Show that if the line \(AH\) intersects the line \(BC\) at right angles, then \(p_2^2 + q_2^2 = p_3^2 + q_3^2\,\), and write down a similar result if the line \(BH\) intersects the line \(AC\) at right angles. Deduce that if \(AH\) is perpendicular to \(BC\) and also \(BH\) is perpendicular to \(AC\), then \(CH\) is perpendicular to \(AB\).

The line \(y=d\,\), where \(d>0\,\), intersects the circle \(x^2+y^2=R^2\) at \(G\) and \(H\). Show that the area of the minor segment \(GH\) is equal to \begin{equation} R^2\arccos \left({d \over R}\right) -d\sqrt{R^2 - d^2}\;. \tag {\(*\)} \end{equation} In the following cases, the given line intersects the given circle. Determine how, in each case, the expression \((*)\) should be modified to give the area of the minor segment.

1. Line: \(y=c\,\); \ \ \ circle: \((x-a)^2+(y-b)^2=R^2\,\).
2. Line: \(y=mx+c\, \); \ \ \ circle: \(x^2+y^2=R^2\,\).
3. Line: \(y=mx+c\,\); \ \ \ circle: \((x-a)^2+(y-b)^2=R^2\,\).

A pyramid stands on horizontal ground. Its base is an equilateral triangle with sides of length~\(a\), the other three sides of the pyramid are of length \(b\) and its volume is \(V\). Given that the formula for the volume of any pyramid is $ \textstyle \frac13 \times \mbox{area of base} \times \mbox {height} \,, $ show that \[ V= \frac1{12} {a^2(3b^2-a^2)}^{\frac12}\;. \] The pyramid is then placed so that a non-equilateral face lies on the ground. Show that the new height, \(h\), of the pyramid is given by \[ h^2 = \frac{a^2(3b^2-a^2)}{4b^2-a^2}\;. \] Find, in terms of \(a\) and \(b\,\), the angle between the equilateral triangle and the horizontal.

Show that the equation of any circle passing through the points of intersection of the ellipse \((x+2)^2 +2y^2 =18\) and the ellipse \(9(x-1)^2 +16y^2 = 25\) can be written in the form \[ x^2-2ax +y^2 =5-4a\;. \]

The points \(A\), \(B\) and \(C\) lie on the sides of a square of side 1 cm and no two points lie on the same side. Show that the length of at least one side of the triangle \(ABC\) must be less than or equal to \((\sqrt6 -\sqrt2)\) cm.

Sketch on the same axes the two curves \(C_1\) and \(C_2\), given by

Show Solution

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\(Q = (-a,-b)\) \begin{align*} && \frac{\d y}{\d x} &= -\frac{1}{x^2} \\ \Rightarrow && \frac{y-b}{x-a} &= -\frac{1}{a^2} \\ \Rightarrow && 0 &= a^2y+x-a^2b-a \\ &&&= a^2y+x - 2a\\ \\ && 2x - 2y \frac{\d y}{\d x} &= 0 \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{x}{y} \\ \Rightarrow && \frac{y+b}{x+a} &= \frac{a}{b} \\ \Rightarrow && 0 &= by-ax+b^2 - a^2 \\ &&&= by - ax -2 \end{align*} Notice that \((-b,a)\) is on both lines, therefore it is their point of intersection. The coordinates of \(N\) will be \((a,-b)\). We can see this is a square by noting each point is a rotation (centre the origin) of \(90^\circ\) of each other.

A point moves in the \(x\)-\(y\) plane so that the sum of the squares of its distances from the three fixed points \((x_{1},y_{1})\), \((x_{2},y_{2})\), and \((x_{3},y_{3})\) is always \(a^{2}\). Find the equation of the locus of the point and interpret it geometrically. Explain why \(a^2\) cannot be less than the sum of the squares of the distances of the three points from their centroid. [The centroid has coordinates \((\bar x, \bar y)\) where \(3\bar x = x_1+x_2+x_3,\) $3\bar y =y_1+y_2+y_3. $]

Consider a fixed square \(ABCD\) and a variable point \(P\) in the plane of the square. We write the perpendicular distance from \(P\) to \(AB\) as \(p\), from \(P\) to \(BC\) as \(q\), from \(P\) to \(CD\) as \(r\) and from \(P\) to \(DA\) as \(s\). (Remember that distance is never negative, so \(p,q,r,s\geqslant 0\).) If \(pr=qs\), show that the only possible positions of \(P\) lie on two straight lines and a circle and that every point on these two lines and a circle is indeed a possible position of \(P\).

The diagram shows a circle, of radius \(r\) and centre \(I\), touching the three sides of a triangle \(ABC\). We write \(a\) for the length of \(BC\) and \(\alpha\) for the angle \(\angle BAC\) and so on. Let \(s=\frac{1}{2}\left(a+b+c\right)\) and let \(\triangle\) be the area of the triangle.

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1. By considering the area of the triangles \(AIB,\) \(BIC\) and \(CIA\), or otherwise, show that \(\Delta=rs\).
2. By using the formula \(\Delta=\frac{1}{2}bc\sin\alpha\), show that \[ \Delta^{2}=\tfrac{1}{16}[4b^{2}c^{2}-\left(2bc\cos\alpha\right)^{2}]. \] Now use the formula \(a^{2}=b^{2}+c^{2}-2bc\cos\alpha\) to show that \[ \Delta^{2}=\tfrac{1}{16}[(a^{2}-\left(b-c\right)^{2})(\left(b+c\right)^{2}-a^{2})] \] and deduce that \[ \Delta=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}. \]
3. A hole in the shape of the triangle \(ABC\) is cut in the top of a level table. A sphere of radius \(R\) rests in the hole. Find the height of the centre of the sphere above the level of the table top, expressing your answer in terms of \(a,b,c,s\) and \(R\).

The famous film star Birkhoff Maclane is sunning herself by the side of her enormous circular swimming pool (with centre \(O\)) at a point \(A\) on its circumference. She wants a drink from a small jug of iced tea placed at the diametrically opposite point \(B\). She has three choices:

1. to swim directly to \(B\).
2. to choose \(\theta\) with \(0<\theta<\pi,\) to run round the pool to a point \(X\) with \(\angle AOX=\theta\) and then to swim directly from \(X\) to \(B\).
3. to run round the pool from \(A\) to \(B\).

Show that the equation \[ ax^{2}+ay^{2}+2gx+2fy+c=0 \] where \(a>0\) and \(f^{2}+g^{2}>ac\) represents a circle in Cartesian coordinates and find its centre. The smooth and level parade ground of the First Ruritanian Infantry Division is ornamented by two tall vertical flagpoles of heights \(h_{1}\) and \(h_{2}\) a distance \(d\) apart. As part of an initiative test a soldier has to march in such a way that he keeps the angles of elevation of the tops of the two flagpoles equal to one another. Show that if the two flagpoles are of different heights he will march in a circle. What happens if the two flagpoles have the same height? To celebrate the King's birthday a third flagpole is added. Soldiers are then assigned to each of the three different pairs of flagpoles and are told to march in such a way that they always keep the tops of their two assigned flagpoles at equal angles of elevation to one another. Show that, if the three flagpoles have different heights \(h_{1},h_{2}\) and \(h_{3}\) and the circles in which the soldiers march have centres of \((x_{ij},y_{ij})\) (for the flagpoles of height \(h_{i}\) and \(h_{j}\)) relative to Cartesian coordinates fixed in the parade ground, then the \(x_{ij}\) satisfy \[ h_{3}^{2}\left(h_{1}^{2}-h_{2}^{2}\right)x_{12}+h_{1}^{2}\left(h_{2}^{2}-h_{3}^{2}\right)x_{23}+h_{2}^{2}\left(h_{3}^{2}-h_{1}^{2}\right)x_{31}=0, \] and the same equation connects the \(y_{ij}\). Deduce that the three centres lie in a straight line.

My house has an attic consisting of a horizontal rectangular base of length \(2q\) and breadth \(2p\) (where \(p < q\)) and four plane roof sections each at angle \(\theta\) to the horizontal. Show that the length of the roof ridge is independent of \(\theta\) and find the volume of the attic and the surface area of the roof.

Show Solution

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The distance to the top of the house (viewed from above) from the long side will be \(p\). The distance from the short side will also be the same, since the roof sections are climbing at the same angle, so they will take just as far to reach the top. Therefore the length of the roof ridge will be \(2q - 2p\) which is independent of \(\theta\). \vspace{1em} The height of the roof will be \(h = p \tan \theta\). The attic can be split into a prism (along the roof ridge) and a pyramid (along the sloping sides). The pyramid will have volume \(\frac13 p \tan\theta (2p)^2 = \frac83 \tan\theta p^3\). The prism will have volume \(2(q-p)p^2 \tan\theta\). Therefore the total volume will be \(\l \frac{2}{3}p + 2q \r p^2\tan\theta \) The distance (along the plane) to the roof of the house will be \(\frac{p}{\cos \theta}\) and therefore the two end roof-sections will be triangles of area \(\frac{p^2}{\cos \theta}\). The two side roof-sections will be trapiziums will area \(\frac{1}{2} \l 2q + 2(q-p) \r \frac{p}{\cos \theta}\) Therefore the total area will be \(\frac{1}{\cos \theta} \l 2p^2 + 4pq - 2p^2 \r = \frac{4pq}{\cos \theta}\)

1. Prove that the intersection of the surface of a sphere with a plane is always a circle, a point or the empty set. Prove that the intersection of the surfaces of two spheres with distinct centres is always a circle, a point or the empty set. {[}If you use coordinate geometry, a careful choice of origin and axes may help.{]}
2. The parish council of Little Fitton have just bought a modern sculpture entitled `Truth, Love and Justice pouring forth their blessings on Little Fitton.' It consists of three vertical poles \(AD,BE\) and \(CF\) of heights 2 metres, 3 metres and 4 metres respectively. Show that \(\angle DEF=\cos^{-1}\frac{1}{5}.\) Vandals now shift the pole \(AD\) so that \(A\) is unchanged and the pole is still straight but \(D\) is vertically above \(AB\) with \(\angle BAD=\frac{1}{4}\pi\) (in radians). Find the new angle \(\angle DEF\) in radians correct to four figures.

Prove that the area of the zone of the surface of a sphere between two parallel planes cutting the sphere is given by \[ 2\pi\times(\mbox{radius of sphere})\times(\mbox{perpendicular distance between the planes}). \] A tangent from the origin \(O\) to the curve with cartesian equation \[ (x-c)^{2}+y^{2}=a^{2}, \] where \(a\) and \(c\) are positive constants with \(c>a,\) touches the curve at \(P\). The \(x\)-axis cuts the curve at \(Q\) and \(R\), the points lying in the order \(OQR\) on the axis. The line \(OP\) and the arc \(PR\) are rotated through \(2\pi\) radians about the line \(OQR\) to form a surface. Find the area of this surface.

Show Solution

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We can choose a coordinate frame where the parallel planes are parallel to the \(y-z\) axis. Then we can compute the surface area as an integral of the surface of revolution for \(x^2 + y^2 = r^2\). Using \(y = r \sin t, x = r \cos t\) we have: \begin{align*} S &= 2\pi\int_{\cos^{-1}a}^{\cos^{-1}b}y \sqrt{\left ( \frac{\d x}{\d t} \right)^2+\left ( \frac{\d y}{\d t} \right)^2} \d t \\ &=2\pi\int_{\cos^{-1}a}^{\cos^{-1}b} r^2 \sin t \d t \\ &= 2\pi \cdot r^2 \cdot (a - b) \\ &= 2 \pi \cdot r \cdot (ra-rb) \\ &= 2\pi\times(\mbox{radius of sphere})\times(\mbox{perpendicular distance between the planes}). \end{align*}

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We can view this surface as a sphere missing a cap of height \(XQ\) and adding a cone of slant height \(OP\) and radius \(PX\) The centre of the circle is at \((c,0)\) and \(OP^2 + a^2 = c^2 \Rightarrow OP = \sqrt{c^2-a^2}\) Since \(OPC \sim OXP\) we must have that \(\frac{OX}{OP} = \frac{OP}{OC} \Rightarrow OX = \frac{c^2-a^2}{c}\) and \(\frac{PX}{OP} = \frac{CP}{OC} \Rightarrow PX = \frac{a}{c}\sqrt{c^2-a^2}\) \(QX = OX - OQ = \frac{c^2-a^2}{c}-(c-a) = \frac{ac-a^2}{c}\) Therefore the surface area is: \begin{align*} S &= 4 \pi a^2 - 2\pi \cdot a \cdot QX+ \pi PX \cdot OP \\ &= 4 \pi a^2 - 2\pi a \cdot \frac{ac-a^2}{c}+\pi \frac{a}{c}\sqrt{c^2-a^2}\cdot \sqrt{c^2-a^2} \\ &= 4\pi a^2 -2\pi \frac{a^2c-a^3}{c}+\pi \frac{ac^2-a^3}{c} \\ &= \pi a \frac{(a+c)^2}{c} \end{align*}

Two points \(P\) and \(Q\) lie within, or on the boundary of, a square of side 1cm, one corner of which is the point \(O\). Show that the length of at least one of the lines \(OP,PQ\) and \(QO\) must be less than or equal to \((\sqrt{6}-\sqrt{2})\) cm.

1. In the cubic equation \(x^3-3pqx+pq(p+q)=0\,\), where \(p\) and \(q\) are distinct real numbers, use the substitution \[ x=\frac{pz+q}{z+1} \] to show that the equation reduces to \(az^3+b = 0\,\), where \(a\) and \(b\) are to be expressed in terms of \(p\) and \(q\).
2. Show further that the equation \(x^3 - 3cx + d = 0\,\), where \(c\) and \(d\) are non-zero real numbers, can be written in the form \(x^3-3pqx+pq(p+q)=0\,\), where \(p\) and \(q\) are distinct real numbers, provided \(d^2 > 4c^3\,\).
3. Find the real root of the cubic equation \(x^3+6x-2=0\,\).
4. Find the roots of the equation \(x^3 - 3p^2x +2p^3=0\,\), and hence show how the equation \(x^3 - 3cx + d = 0\) can be solved in the case \(d^2 = 4c^3\,\).

1. The function f satisfies, for all \(x\), the equation \[ \f(x) + (1- x)\f(-x) = x^2\, . \] Show that \(\f(-x) + (1 + x)\f(x) = x^2\,\). Hence find \(\f(x)\) in terms of \(x\). You should verify that your function satisfies the original equation.
2. The function \({\rm K}\) is defined, for \(x\ne 1\), by \[{\rm K}(x) = \dfrac{x+1}{x-1}\,.\] Show that, for \(x\ne1\), \({\rm K(K(}x)) =x\,\). The function g satisfies the equation \[ \g(x)+ x\, \g\Big(\frac{ x+1 }{x-1}\Big) = x \ \ \ \ \ \ \ \ \ \ \ ( x\ne 1) \,. \] Show that, for \(x\ne1\), \(\g(x)= \dfrac{2x}{x^2+1}\,\).
3. Find \(\h(x)\), for \(x\ne0\), \(x\ne1\), given that \[ \h(x)+ \h\Big(\frac 1 {1-x}\Big)= 1-x -\frac1{1-x} \ \ \ \ \ \ ( x\ne0, \ \ x\ne1 ) \,. \]

It is given that the two curves \[ y=4-x^2 \text{ and } m x = k-y^2\,, \] where \(m > 0\), touch exactly once.

1. In each of the following four cases, sketch the two curves on a single diagram, noting the coordinates of any intersections with the axes:
   1. \(k < 0\, \);
   2. \(0 < k < 16\), \(k/m < 2\,\);
   3. \(k > 16\), \(k/m > 2\,\);
   4. \(k > 16\), \(k/m < 2\,\).
2. Now set \(m=12\). Show that the \(x\)-coordinate of any point at which the two curves meet satisfies \[ x^4-8x^2 +12x +16-k=0\,. \] Let \(a\) be the value of \(x\) at the point where the curves touch. Show that \(a\) satisfies \[ a^3 -4a +3 =0 \] and hence find the three possible values of \(a\). Derive also the equation \[ k= -4a^2 +9a +16\,. \] Which of the four sketches in part (i) arise?

Given that \[ 5x^{2}+2y^{2}-6xy+4x-4y\equiv a\left(x-y+2\right)^{2} +b\left(cx+y\right)^{2}+d\,, \] find the values of the constants \(a\), \(b\), \(c\) and \(d\). Solve the simultaneous equations \begin{align*} 5x^{2}+2y^{2}-6xy+4x-4y&=9\,, \\ 6x^{2}+3y^{2}-8xy+8x-8y&=14\,. \end{align*}

Two curves have equations \(\; x^4+y^4=u\;\) and \(\; xy = v\;\), where \(u\) and \(v\) are positive constants. State the equations of the lines of symmetry of each curve. The curves intersect at the distinct points \(A\), \(B\), \(C\) and \(D\) (taken anticlockwise from \(A\)). The coordinates of \(A\) are \((\alpha,\beta)\), where \(\alpha > \beta > 0\). Write down, in terms of \(\alpha\) and \(\beta\), the coordinates of \(B\), \(C\) and \(D\). Show that the quadrilateral \(ABCD\) is a rectangle and find its area in terms of \(u\) and \(v\) only. Verify that, for the case \(u=81\) and \(v=4\), the area is \(14\).

Show Solution

The curve \(x^4 + y^4 = u\) has lines of symmetry:

• \(y = 0\)
• \(x = 0\)
• \(y = x\)
• \(y = -x\)

The curve \(xy = v\) has lines of symmetry:

• \(y = x\)
• \(y = -x\)

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The points are \(A = (\alpha, \beta), B = (\beta, \alpha), C = (-\alpha, -\beta), D = (-\beta, -\alpha)\) \(AD\) has gradient \(\frac{\beta+\alpha}{\alpha+\beta} = 1\), \(BC\) has the same gradient. \(AB\) has gradient \(\frac{\alpha-\beta}{\beta-\alpha} = -1\), as does \(CD\). Therefore it has two sets of perpendicular and parallel sides, hence a rectangle. The area is \(|AD||AB| = \sqrt{2(\alpha+\beta)^2}\sqrt{2(\alpha-\beta)^2} = 2(\alpha^2-\beta^2)\) The squared area is \(4(\alpha^4+\beta^4 - 2 \alpha^2\beta^2) = 4(u - 2v^2)\) ie the area is \(2\sqrt{u-2v^2}\) When \(u = 81, v = 4\) we have the area is \(2 \sqrt{81 - 2 \cdot 16} = 14\) as required.

A transformation \(T\) of the real numbers is defined by \[ y=T(x)=\frac{ax-b}{cx-d}\,, \] where \(a,b,c\), \(d\) are real numbers such that \(ad\neq bc\). Find all numbers \(x\) such that \(T(x)=x.\) Show that the inverse operation, \(x=T^{-1}(y)\) expressing \(x\) in terms of \(y\) is of the same form as \(T\) and find corresponding numbers \(a',b',c'\),\(d'\). Let \(S_{r}\) denote the set of all real numbers excluding \(r\). Show that, if \(c\neq0,\) there is a value of \(r\) such that \(T\) is defined for all \(x\in S_{r}\) and find the image \(T(S_{r}).\) What is the corresponding result if \(c=0\)? If \(T_{1},\) given by numbers \(a_{1},b_{1},c_{1},d_{1},\) and \(T_{2},\) given by numbers \(a_{2},b_{2},c_{2},d_{2}\) are two such transformations, show that their composition \(T_{3},\) defined by \(T_{3}(x)=T_{2}(T_{1}(x)),\) is of the same form. Find necessary and sufficient conditions on the numbers \(a,b,c,d\) for \(T^{2}\), the composition of \(T\) with itself, to be the identity. Hence, or otherwise, find transformations \(T_{1},T_{2}\) and their composition \(T_{3}\) such that \(T_{1}^{2}\) and \(T_{2}^{2}\) are each the identity but \(T_{3}^{2}\) is not.

It is given that \(x,y\) and \(z\) are distinct and non-zero, and that they satisfy \[ x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}. \] Show that \(x^{2}y^{2}z^{2}=1\) and that the value of \(x+\dfrac{1}{y}\) is either \(+1\) or \(-1\).

Let \(\mathrm{h}(x)=ax^{2}+bx+c,\) where \(a,b\) and \(c\) are constants, and \(a\neq0\). Give a condition which \(a,b\) and \(c\) must satisfy in order that \(\mathrm{h}(x)\) can be written in the form \[ a(x+k)^{2},\tag{*} \] where \(k\) is a constant. If \(\mathrm{f}(x)=3x^{2}+4x\) and \(\mathrm{g}(x)=x^{2}-2\), find the two constant values of \(\lambda\) such that \(\mathrm{f}(x)+\lambda\mathrm{g}(x)\) can be written in the form \((*)\). Hence, or otherwise, find constants \(A,B,C,D,m\) and \(n\) such that \begin{alignat*}{1} \mathrm{f}(x) & =A(x+m)^{2}+B(x+n)^{2}\\ \mathrm{g}(x) & =C(x+m)^{2}+D(x+n)^{2}. \end{alignat*} If \(\mathrm{f}(x)=3x^{2}+4x\) and \(\mathrm{g}(x)=x^{2}+\alpha\) and it is given by that there is only one value of \(\lambda\) for which \(\mathrm{f}(x)+\lambda\mathrm{g}(x)\) can be written in the form \((*)\), find \(\alpha\).

The numbers \(x,y\) and \(z\) are non-zero, and satisfy \[ 2a-3y=\frac{\left(z-x\right)^{2}}{y}\quad\mbox{ and }\quad2a-3z=\frac{\left(x-y\right)^{2}}{z}, \] for some number \(a\). If \(y\neq z\), prove that \[ x+y+z=a, \] and that \[ 2a-3x=\frac{\left(y-z\right)^{2}}{x}. \] Determine whether this last equation holds only if \(y\neq z\).

If we split a set \(S\) of integers into two subsets \(A\) and \(B\) whose intersection is empty and whose union is the whole of \(S\), and such that

• the sum of the elements of \(A\) is equal to the sum of the elements of \(B\)
• and the sum of the squares of the elements of \(A\) is equal to the sum of the squares of the elements of \(B\),

1. Find a balanced partition of the set \(\{1, 2, 3, 4, 5, 6, 7, 8\}\) into two subsets \(A\) and \(B\), each of size 4.
2. Given that \(a_1, a_2, \ldots, a_m\) and \(b_1, b_2, \ldots, b_m\) are sequences with \[\sum_{k=1}^m a_k = \sum_{k=1}^m b_k \quad \text{and} \quad \sum_{k=1}^m a_k^2 = \sum_{k=1}^m b_k^2,\] show that \[\sum_{k=1}^m a_k^3 + \sum_{k=1}^m (c + b_k)^3 = \sum_{k=1}^m b_k^3 + \sum_{k=1}^m (c + a_k)^3\] for any real number \(c\).
3. Find, with justification, a balanced partition of the set \(\{1, 2, 3, \ldots, 16\}\) into two subsets \(A\) and \(B\), each of size 8, which also has the property that
   • the sum of the cubes of the elements of \(A\) is equal to the sum of the cubes of the elements of \(B\).
4. You are given that the sets \(A = \{1, 3, 4, 5, 9, 11\}\) and \(B = \{2, 6, 7, 8, 10\}\) form a balanced partition of the set \(\{1, 2, 3, \ldots, 11\}\). Let \(S = \{n^2, (n+1)^2, (n+2)^2, \ldots, (n+11)^2\}\), where \(n\) is any positive integer. Find, with justification, two subsets \(C\) and \(D\) of \(S\) whose intersection is empty and whose union is the whole of \(S\), and such that
   • the sum of the elements of \(C\) is equal to the sum of the elements of \(D\).

The sequence \(u_0, u_1, \ldots\) is said to be a constant sequence if \(u_n = u_{n+1}\) for \(n = 0, 1, 2, \ldots\). The sequence is said to be a sequence of period 2 if \(u_n = u_{n+2}\) for \(n = 0, 1, 2, \ldots\) and the sequence is not constant.

1. A sequence of real numbers is defined by \(u_0 = a\) and \(u_{n+1} = f(u_n)\) for \(n = 0, 1, 2, \ldots\), where $$f(x) = p + (x - p)x,$$ and \(p\) is a given real number. Find the values of \(a\) for which the sequence is constant. Show that the sequence has period 2 for some value of \(a\) if and only if \(p > 3\) or \(p < -1\).
2. A sequence of real numbers is defined by \(u_0 = a\) and \(u_{n+1} = f(u_n)\) for \(n = 0, 1, 2, \ldots\), where $$f(x) = q + (x - p)x,$$ and \(p\) and \(q\) are given real numbers. Show that there is no value of \(a\) for which the sequence is constant if and only if \(f(x) > x\) for all \(x\). Deduce that, if there is no value of \(a\) for which the sequence is constant, then there is no value of \(a\) for which the sequence has period 2. Is it true that, if there is no value of \(a\) for which the sequence has period 2, then there is no value of \(a\) for which the sequence is constant?

Consider the following steps in a proof that \(\sqrt{2} + \sqrt{3}\) is irrational.

1. If an integer \(a\) is not divisible by 3, then \(a = 3k \pm 1\), for some integer \(k\). In both cases, \(a^2\) is one more than a multiple of 3.
2. Suppose that \(\sqrt{2} + \sqrt{3}\) is rational, and equal to \(\frac{a}{b}\), where \(a\) and \(b\) are positive integers with no common factor greater than one.
3. Then \(a^4 + b^4 = 10a^2b^2\).
4. So if \(a\) is divisible by 3, then \(b\) is divisible by 3.
5. Hence \(\sqrt{2} + \sqrt{3}\) is irrational.

1. Show clearly that steps 1, 3 and 4 are all valid and that the conclusion 5 follows from the previous steps of the argument.
2. Prove, by means of a similar method but using divisibility by 5 instead of 3, that \(\sqrt{6} + \sqrt{7}\) is irrational. Why can divisibility by 3 not be used in this case?

Find the set of positive integers \(n\) for which \(n\) does not divide \((n-1)!.\) Justify your answer. [Note that small values of \(n\) may require special consideration.]

1. Find all pairs of positive integers \((n,p)\), where \(p\) is a prime number, that satisfy \[ n!+ 5 =p \,. \]
2. In this part of the question you may use the following two theorems:
   1. For \(n\ge 7\), \(1! \times 3! \times \cdots \times (2n-1)! > (4n)!\,\).
   2. For every positive integer \(n\), there is a prime number between \(2n\) and \(4n\).
   Find all pairs of positive integers \((n,m)\) that satisfy \[ 1! \times 3! \times \cdots \times (2n-1)! = m! \,. \]

The sequence of numbers \(x_0\), \(x_1\), \(x_2\), \(\ldots\) satisfies \[ x_{n+1} = \frac{ax_n-1}{x_n+b} \,. \] (You may assume that \(a\), \(b\) and \(x_0\) are such that \(x_n+b\ne0\,\).) Find an expression for \(x_{n+2}\) in terms of \(a\), \(b\) and \(x_n\).

1. Show that \(a+b=0\) is a necessary condition for the sequence to be periodic with period 2. {\bf Note: } The sequence is said to be periodic with period \(k\) if \(x_{n+k} = x_n\) for all \(n\), and there is no integer \(m\) with \(0 < m < k\) such that \(x_{n+m} = x_n\) for all \(n\).
2. Find necessary and sufficient conditions for the sequence to have period 4.

The set \(S\) % = \{1, 5, 9, 13, \,\ldots \}$ consists of all the positive integers that leave a remainder of 1 upon division by 4. The set \(T\) % = \{1, 5, 9, 13, \,\ldots \}$ consists of all the positive integers that leave a remainder of 3 upon division by 4.

1. Describe in words the sets \(S \cup T\) and \(S \cap T\).
2. Prove that the product of any two integers in \(S\) is also in \(S\). Determine whether the product of any two integers in \(T\) is also in \(T\).
3. Given an integer in \(T\) that is not a prime number, prove that at least one of its prime factors is in \(T\).
4. For any set \(X\) of positive integers, an integer in \(X\) (other than 1) is said to be {\em \(X\)-prime} if it cannot be expressed as the product of two or more integers {\em all in \(X\)} (and all different from 1).
   • [\bf (a)] Show that every integer in \(T\) is either \(T\)-prime or is the product of an odd number of \(T\)-prime integers.
   • [\bf (b)] Find an example of an integer in \(S\) that can be expressed as the product of \hbox{\(S\)-prime} integers in two distinct ways. [Note: \(s_1s_2\) and \(s_2s_1\) are not counted as distinct ways of expressing the product of \(s_1\) and \(s_2\).]

An operator \(\rm D\) is defined, for any function \(\f\), by \[ {\rm D}\f(x) = x\frac{\d\f(x)}{\d x} .\] The notation \({\rm D}^n\) means that \(\rm D\) is applied \(n\) times; for example \[ \displaystyle {\rm D}^2\f(x) = x\frac{\d\ }{\d x}\left( x\frac{\d\f(x)}{\d x} \right) \,. \] Show that, for any constant \(a\), \({\rm D}^2 x^a = a^2 x^a\,\).

1. Show that if \(\P(x)\) is a polynomial of degree \(r\) (where \(r\ge1\)) then, for any positive integer \(n\), \({\rm D}^n\P(x)\) is also a polynomial of degree \(r\).
2. Show that if \(n\) and \(m\) are positive integers with \(n < m\), then \({\rm D}^n(1-x)^m\) is divisible by \((1-x)^{m-n}\).
3. Deduce that, if \(m\) and \(n\) are positive integers with \(n < m\), then \[ \sum_{r=0}^m (-1)^r \binom m r r^n =0 \, . \]
4. [Not on original paper] Let \(\f_n(x) = D^n(1-x)^n\,\), where \(n\) is a positive integer. Prove that \(\f_n(1)=(-1)^nn!\, \).

1. In the following argument to show that \(\sqrt2\) is irrational, give proofs appropriate for steps 3, 5 and 6.
   1. Assume that \(\sqrt2\) is rational.
   2. Define the set \(S\) to be the set of positive integers with the following property:
      \(n\) is in \(S\) if and only if \(n \sqrt2\) is an integer.
   3. Show that the set \(S\) contains at least one positive integer.
   4. Define the integer \(k\) to be the smallest positive integer in \(S\).
   5. Show that \((\sqrt2-1)k\) is in \(S\).
   6. Show that steps 4 and 5 are contradictory and hence that \(\sqrt2\) is irrational.
2. Prove that \(2^{\frac13} \) is rational if and only if \(2^{\frac23}\) is rational. Use an argument similar to that of part (i) to prove that \(2^{\frac13}\) and \(2^{\frac23}\) are irrational.

If \(s_1\), \(s_2\), \(s_3\), \(\ldots\) and \(t_1\), \(t_2\), \(t_3\), \(\ldots\) are sequences of positive numbers, we write \[ (s_n)\le (t_n) \] to mean

1. \((1000n) \le (n^2)\,\).
2. If it is not the case that \((s_n)\le (t_n)\), then it is the case that \((t_n)\le (s_n)\,\).
3. If \((s_n)\le (t_n)\) and \((t_n) \le (u_n)\), then \((s_n)\le (u_n)\,\).
4. \((n^2)\le (2^n)\,\).

Show that:

1. \(1+2+3+ \cdots + n = \frac12 n(n+1)\);
2. if \(N\) is a positive integer, \(m\) is a non-negative integer and \(k\) is a positive odd integer, then \((N-m)^k +m^k\) is divisible by \(N\).

1. Let \(\.f(x) = (x+2a)^3 -27 a^2 x\), where \(a\ge 0\). By sketching \(\.f(x)\), show that \(\.f(x)\ge 0\) for~\(x \ge0\).
2. Use part (i) to find the greatest value of \(xy^2\) in the region of the \(x\)-\(y\) plane given by \(x\ge0\), \(y\ge0\) and \(x+2y\le 3\,\). For what values of \(x\) and \(y\) is this greatest value achieved?
3. Use part (i) to show that \((p+q+r)^3 \ge 27pqr\) for any non-negative numbers \(p\), \(q\) and~\(r\). If \((p+q+r)^3 = 27pqr\), what relationship must \(p\), \(q\) and \(r\) satisfy?

All numbers referred to in this question are non-negative integers.

1. Express each of the numbers 3, 5, 8, 12 and 16 as the difference of two non-zero squares.
2. Prove that any odd number can be written as the difference of two squares.
3. Prove that all numbers of the form \(4k\), where \(k\) is a non-negative integer, can be written as the difference of two squares.
4. Prove that no number of the form \(4k+2\), where \(k\) is a non-negative integer, can be written as the difference of two squares.
5. Prove that any number of the form \(pq\), where \(p\) and \(q\) are prime numbers greater than 2, can be written as the difference of two squares in exactly two distinct ways. Does this result hold if \(p\) is a prime greater than 2 and \(q=2\)?
6. Determine the number of distinct ways in which 675 can be written as the difference of two squares.

In this question, you may assume that, if \(a\), \(b\) and \(c\) are positive integers such that \(a\) and \(b\) are coprime and \(a\) divides \(bc\), then \(a\) divides \(c\). (Two positive integers are said to be coprime if their highest common factor is 1.)

1. Suppose that there are positive integers \(p\), \(q\), \(n\) and \(N\) such that \(p\) and \(q\) are coprime and \(q^nN=p^n\). Show that \(N=kp^n\) for some positive integer \(k\) and deduce the value of \(q\). Hence prove that, for any positive integers \(n\) and \(N\), \(\sqrt[n]N\) is either a positive integer or irrational.
2. Suppose that there are positive integers \(a\), \(b\), \(c\) and \(d\) such that \(a\) and \(b\) are coprime and \(c\) and \(d\) are coprime, and \(a^ad^b = b^a c^b \,\). Prove that \(d^b = b^a\) and deduce that, if \(p\) is a prime factor of \(d\), then \(p\) is also a prime factor of \(b\). If \(p^m\) and \(p^n\) are the highest powers of the prime number \(p\) that divide \(d\) and \(b\), respectively, express \(b\) in terms of \(a\), \(m\) and \(n\) and hence show that \(p^n\le n\). Deduce the value of \(b\). (You may assume that if \(x > 0\) and \(y\ge2\) then \(y^x > x\).) Hence prove that, if \(r\) is a positive rational number such that \(r^r\) is rational, then \(r\) is a positive integer.

1. Write down a solution of the equation \[ x^2-2y^2 =1\,, \tag{\(*\)} \] for which \(x\) and \(y\) are non-negative integers. Show that, if \(x=p\), \(y=q\) is a solution of (\(*\)), then so also is \(x=3p+4q\), \(y=2p+3q\). Hence find two solutions of \((*)\) for which \(x\) is a positive odd integer and \(y\) is a positive even integer.
2. Show that, if \(x\) is an odd integer and \(y\) is an even integer, \((*)\) can be written in the form \[ n^2 = \tfrac12 m(m+1)\,, \] where \(m\) and \(n\) are integers.
3. The positive integers \(a\), \(b\) and \(c\) satisfy \[ b^3=c^4-a^2\,, \] where \(b\) is a prime number. Express \(a\) and \(c^2\) in terms of \(b\) in the two cases that arise. Find a solution of \(a^2+b^3=c^4\), where \(a\), \(b\) and \(c\) are positive integers but \(b\) is not prime.

1. The point with coordinates \((a, b)\), where \(a\) and \(b\) are rational numbers, is called: \newline \hspace*{1cm} an {\em integer rational point} if both \(a\) and \(b\) are integers; \newline\hspace*{1cm} a {\em non-integer rational point} if neither \(a\) nor \(b\) is an integer.
   • [\bf (a)] Write down an integer rational point and a non-integer rational point on the circle \(x^2+y^2 =1\).
   • [\bf (b)] Write down an integer rational point on the circle \(x^2+y^2=2\). Simplify \[ (\cos\theta + \sqrt m \sin\theta)^2 + (\sin\theta - \sqrt m \cos\theta)^2 \, \] and hence obtain a non-integer rational point on the circle \(x^2+y^2=2\,\).
2. The point with coordinates \((p+\sqrt 2 \, q\,,\, r+\sqrt 2 \, s)\), where \(p\), \(q\), \(r\) and \(s\) are rational numbers, is called: \newline\hspace*{1cm} an {\em integer \(2\)-rational point} if all of \(p\), \(q\), \(r\) and \(s\) are integers; \newline\hspace*{1cm} a {\em non-integer \(2\)-rational point} if none of \(p\), \(q\), \(r\) and \(s\) is an integer.
   • [\bf (a)] Write down an integer \(2\)-rational point, and obtain a non-integer \(2\)-rational point, on the circle \(x^2+y^2=3\,\).
   • [\bf(b)] Obtain a non-integer \(2\)-rational point on the circle \(x^2+y^2=11\,\).
   • [\bf(c)]Obtain a non-integer \(2\)-rational point on the hyperbola \(x^2-y^2 =7 \).

Write down the cubes of the integers \(1, 2, \ldots , 10\). The positive integers \(x\), \(y\) and \(z\), where \(x < y\), satisfy \[ x^3+y^3 = kz^3\,, \tag{\(*\)} \] where \(k\) is a given positive integer.

1. In the case \(x+y =k\), show that \[ z^3 = k^2 -3kx+3x^2\,. \] Deduce that \((4z^3 - k^2)/3\) is a perfect square and that \(\frac14 {k^2} \le z^3 < k^2\,\). Use these results to find a solution of \((*)\) when \(k=20\).
2. By considering the case \(x+y = z^2\), find two solutions of \((*)\) when \(k=19\).

1. The numbers \(m\) and \(n\) satisfy \[ m^3=n^3+n^2+1\,. \tag{\(*\)} \]
   • Show that \(m > n\). Show also that \(m < n+1\) if and only if \(2n^2+3n > 0\,\). Deduce that \(n < m < n+1\) unless \(-\frac32 \le n \le 0\,\).
   • Hence show that the only solutions of \((*)\) for which both \(m\) and \(n\) are integers are \((m,n) = (1,0)\) and \((m,n)= (1,-1)\).
2. Find all integer solutions of the equation \[ p^3=q^3+2q^2-1\,. \]

The vertices \(A\), \(B\), \(C\) and \(D\) of a square have coordinates \((0,0)\), \((a,0)\), \((a,a)\) and \((0,a)\), respectively. The points \(P\) and \(Q\) have coordinates \((an,0)\) and \((0,am)\) respectively, where \(0 < m < n < 1\). The line \(CP\) produced meets \(DA\) produced at \(R\) and the line \(CQ\) produced meets \(BA\) produced at \(S\). The line \(PQ\) produced meets the line \(RS\) produced at \(T\). Show that \(TA\) is perpendicular to \(AC\). Explain how, given a square of area \(a^2\), a square of area \(2a^2\) may be constructed using only a straight-edge. [{\bf Note}: a straight-edge is a ruler with no markings on it; no measurements (and no use of compasses) are allowed in the construction.]

1. Suppose that \(a\), \(b\) and \(c\) are integers that satisfy the equation \[ a^{3}+3b^{3}=9c^{3}. \] Explain why \(a\) must be divisible by 3, and show further that both \(b\) and \(c\) must also be divisible by 3. Hence show that the only integer solution is \(a=b=c=0\,\).
2. Suppose that \(p\), \(q\) and \(r\) are integers that satisfy the equation \[ p^4 +2q^4 = 5r^4 \,.\] By considering the possible final digit of each term, or otherwise, show that \(p\) and \(q\) are divisible by 5. Hence show that the only integer solution is \(p=q=r=0\,\).

A {\em proper factor} of an integer \(N\) is a positive integer, not \(1\) or \(N\), that divides \(N\).

1. Show that \(3^2\times 5^3\) has exactly \(10\) proper factors. Determine how many other integers of the form \(3^m\times5^n\) (where \(m\) and \(n\) are integers) have exactly 10 proper factors.
2. Let \(N\) be the smallest positive integer that has exactly \(426\) proper factors. Determine \(N\), giving your answer in terms of its prime factors.

A sequence of points \((x_1,y_1)\), \((x_2,y_2)\), \(\ldots\) in the cartesian plane is generated by first choosing \((x_1,y_1)\) then applying the rule, for \(n=1\), \(2\), \(\ldots\), \[ (x_{n+1}, y_{n+1}) = (x_n^2-y_n^2 +a, \; 2x_ny_n+b+2)\,, \] where \(a\) and \(b\) are given real constants.

1. In the case \(a=1\) and \(b=-1\), find the values of \((x_1,y_1)\) for which the sequence is constant.
2. Given that \((x_1,y_1) = (-1,1)\), find the values of \(a\) and \(b\) for which the sequence has period 2.

The polynomial \(\p(x)\) is given by \[ \ds \p(x)= x^n +\sum\limits_{r=0}^{n-1}a_rx^r\,, \] where \(a_0\), \(a_1\), \(\ldots\) , \(a_{n-1}\) are fixed real numbers and \(n\ge1\). Let \(M\) be the greatest value of \(\big\vert \p(x) \big\vert\) for $\vert x \vert\le 1\(. Then Chebyshev's theorem states that \)M\ge 2^{1-n}$.

1. Prove Chebyshev's theorem in the case \(n=1\) and verify that Chebyshev's theorem holds in the following cases:
   1. \( \p(x) = x^2 - \frac12\,\);
   2. \(\p(x) = x^3 - x \,\).
2. Use Chebyshev's theorem to show that the curve $ \ y= 64x^5+25x^4-66x^3-24x^2+3x+1 \ $ has at least one turning point in the interval \(-1\le x \le 1\).

Prove that, if \(c\ge a\) and \(d\ge b\), then \[ ab+cd\ge bc+ad\,. \tag{\(*\)} \]

1. If \(x\ge y\), use \((*)\) to show that \(x^2+y^2\ge 2xy\,\). If, further, \(x\ge z\) and \(y\ge z\), use \((*)\) to show that \(z^2+xy\ge xz+yz\) and deduce that \(x^2+y^2+z^2\ge xy+yz+zx\,\). Prove that the inequality \(x^2+y^2+z^2\ge xy+yz+zx\,\) holds for all \(x\), \(y\) and \(z\).
2. Show similarly that the inequality \[\frac st +\frac tr +\frac rs +\frac ts +\frac rt +\frac sr \ge 6\] holds for all positive \(r\), \(s\) and \(t\).

What does it mean to say that a number \(x\) is irrational? Prove by contradiction statements A and B below, where \(p\) and \(q\) are real numbers.

• A: If \(pq\) is irrational, then at least one of \(p\) and \(q\) is irrational.
• B: If \(p+q\) is irrational, then at least one of \(p\) and \(q\) is irrational.

• C: If \(p\) and \(q\) are irrational, then \(p+q\) is irrational.

Let \(\lfloor x \rfloor\) denote the largest integer that satisfies \(\lfloor x \rfloor \leq x\). For example, if \(x = -4.2\), then \(\lfloor x \rfloor = -5\).

1. Show that, if \(n\) is an integer, then \(\lfloor x + n \rfloor = \lfloor x \rfloor + n\).
2. Let \(n\) be a positive integer and define function \(f_n\) by \[f_n(x) = \lfloor x \rfloor + \left\lfloor x + \frac{1}{n} \right\rfloor + \left\lfloor x + \frac{2}{n} \right\rfloor + \ldots + \left\lfloor x + \frac{n-1}{n} \right\rfloor - \lfloor nx \rfloor\]
   1. Show that \(f_n\left(x + \frac{1}{n}\right) = f_n(x)\).
   2. Evaluate \(f_n(t)\) for \(0 \leq t < \frac{1}{n}\).
   3. Hence show that \(f_n(x) \equiv 0\).
3. 1. Show that \(\left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x+1}{2} \right\rfloor = \lfloor x \rfloor\).
   2. Hence, or otherwise, simplify \[\left\lfloor \frac{x+1}{2} \right\rfloor + \left\lfloor \frac{x+2}{2^2} \right\rfloor + \ldots + \left\lfloor \frac{x+2^k}{2^{k+1}} \right\rfloor + \ldots\]

The Bernoulli polynomials \(P_{n}(x)\), where \(n\) is a non-negative integer, are defined by \(P_{0}(x)=1\) and, for \(n\geqslant1\), \[ \frac{\mathrm{d}P_{n}}{\mathrm{d}x}=nP_{n-1}(x),\qquad\int_{0}^{1}P_{n}(x)\,\mathrm{d}x=0 \] Show by induction or otherwise, that \[ P_{n}(x+1)-P_{n}(x)=nx^{n-1},\quad\mbox{ for }n\geqslant1. \] Deduce that \[ n\sum_{m=0}^{k}m^{n-1}=P_{n}(k+1)-P_{n}(0) \] Hence show that \({\displaystyle \sum_{m=0}^{1000}m^{3}=(500500)^{2}}\)

Let \[ S_n = \sum_{r=1}^n \frac 1 {\sqrt r \ } \,, \] where \(n\) is a positive integer.

1. Prove by induction that \[ S_n \le 2\sqrt n -1\, . \]
2. Show that \((4k+1)\sqrt{k+1} > (4k+3)\sqrt k\,\) for \(k\ge0\,\). Determine the smallest number \(C\) such that \[ S_n \ge 2\sqrt n + \frac 1 {2\sqrt n} -C \,.\]

Two sequences are defined by \(a_1 = 1\) and \(b_1 = 2\) and, for \(n \ge 1\), \begin{equation*} \begin{split} a_{n+1} & = a_n+ 2b_n \,, \\ b_{n+1} & = 2a_n + 5b_n \,. \end{split} \end{equation*} Prove by induction that, for all \(n \ge 1\), \[ a_n^2+2a_nb_n - b_n^2 = 1 \,. \tag{\(*\)}\]

1. Let \(c_n = \dfrac{a_n}{b_n}\)\,. Show that \(b_n \ge 2 \times 5^{n-1}\) and use \((*)\) to show that \[ c_n \to \sqrt 2 -1 \text{ as } n\to\infty\,. \]
2. Show also that \(c_n > \sqrt2 -1\) and hence that $\dfrac2 {c_n+1}<\sqrt2

Let \[ T _n = \left( \sqrt{a+1} + \sqrt a\right)^n\,, \] where \(n\) is a positive integer and \(a\) is any given positive integer.

1. In the case when \(n\) is even, show by induction that \(T_n\) can be written in the form \[ A_n +B_n \sqrt{a(a+1)}\,, \] where \(A_n\) and \(B_n\) are integers (depending on \(a\) and \(n\)) and \(A_n^2 =a(a+1)B_n^2 +1\).
2. In the case when \(n\) is odd, show by considering \((\sqrt{a+1} +\sqrt a)T_m\) where \(m\) is even, or otherwise, that \(T_n\) can be written in the form \[ C_n \sqrt {a+1} + D_n \sqrt a \,, \] where \(C_n\) and \(D_n\) are integers (depending on \(a\) and \(n\)) and \( (a+1)C_n^2 = a D_n^2 +1\,\).
3. Deduce that, for each \(n\), \(T_n\) can be written as the sum of the square roots of two consecutive integers.

The functions \({\rm T}_n(x)\), for \(n=0\), 1, 2, \(\ldots\,\), satisfy the recurrence relation \[ {\rm T}_{n+1}(x) -2x {\rm T}_n(x) + {\rm T}_{n-1}(x) =0\, \ \ \ \ \ \ \ (n\ge1). \tag{\(*\)} \] Show by induction that \[ \left({\rm T}_n(x)\right)^2 - {\rm T}_{n-1}(x) {\rm T}_{n+1}(x) = \f(x)\,, \] where \(\f(x) = \left({\rm T}_1(x)\right)^2 - {\rm T}_0(x){\rm T}_2(x)\,\). In the case \(\f(x)\equiv 0\), determine (with proof) an expression for \({\rm T}_n(x)\) in terms of \({\rm T}_0(x)\) (assumed to be non-zero) and \({\rm r}(x)\), where \({\rm r}(x) = {\rm T}_1(x)/ {\rm T}_0(x)\). Find the two possible expressions for \({\rm r}(x)\) in terms of \(x\). %Conjecture (without proof) the general form of the solution of \((*)\).

A sequence of numbers, \(F_1\), \(F_2\), \(\ldots\), is defined by $ F_1=1\(, \)F_2=1$, and \[ F_n=F_{n-1}+F_{n-2}\, \text{ \ \ \ for \(n\ge 3\)}. \]

1. Write down the values of \(F_3\), \(F_4\), \(\ldots\) , \(F_8\).
2. Prove that $F^{\vphantom{2}}_{2k+3}F^{\vphantom{2}}_{2k+1} -F_{2k+2}^2 = -F^{\vphantom{2}}_{2k+2}F^{\vphantom{2}}_{2k}+F_{2k+1}^2\,$.
3. Prove by induction or otherwise that $ F^{\vphantom{2}}_{2n+1} F^{\vphantom{2}}_{2n-1}- F^2_{2n}=1\,$ and deduce that \(F^2_{2n}+1\,\) is divisible by \(F^{\vphantom{2}}_{2n+1}\,.\)
4. Prove that \(F^2_{2n-1}+1\,\) is divisible by \(F^{\vphantom{2}}_{2n+1}\,.\)

It is given that \(\sum\limits_{r=-1}^ {n} r^2\) can be written in the form \(pn^3 +qn^2+rn+s\,\), where \(p\,\), \(q\,\), \(r\,\) and \(s\) are numbers. By setting \(n=-1\), \(0\), \(1\) and \(2\), obtain four equations that must be satisfied by \(p\,\), \(q\,\), \(r\,\) and \(s\) and hence show that \[ { \sum\limits_{r=0} ^n} r^2= {\textstyle \frac16} n(n+1)(2n+1)\;. \] Given that \(\sum\limits_{r=-2}^ nr^3\) can be written in the form \(an^4 +bn^3+cn^2+dn +e\,\), show similarly that \[ { \sum\limits_{r=0} ^n} r^3= {\textstyle \frac14} n^2(n+1)^2\;. \]

The \(n\)th Fermat number, \(F_n\), is defined by \[ F_n = 2^{2^n} +1\, , \ \ \ \ \ \ \ n=0, \ 1, \ 2, \ \ldots \ , \] where \(\ds 2^{2^n}\) means \(2\) raised to the power \(2^n\,\). Calculate \(F_0\), \(F_1\), \(F_2\) and \(F_3\,\). Show that, for \(k=1\), \(k=2\) and \(k=3\,\), $$ F_0F_1 \ldots F_{k-1} = F_k-2 \;. \tag{*} $$ Prove, by induction, or otherwise, that \((*)\) holds for all \(k\ge1\). Deduce that no two Fermat numbers have a common factor greater than \(1\). Hence show that there are infinitely many prime numbers.

Let $$ {\rm S}_n(x)=\mathrm{e}^{x^3}{{\d^n}\over{\d x^n}}{(\mathrm{e}^{-x^3})}. $$ Show that \({\rm S}_2(x)=9x^4-6x\) and find \({\rm S}_3(x)\). Prove by induction on \(n\) that \({\rm S}_n(x)\) is a polynomial. By means of your induction argument, determine the order of this polynomial and the coefficient of the highest power of~\(x\). Show also that if \ \(\displaystyle \frac{\d S_n}{\d x}=0\) \ for some value \(a\) of \(x\), then \( \ S_n(a)S_{n+1}(a)\le0\).

Suppose that $$3=\frac{2}{ x_1}=x_1+\frac{2}{ x_2} =x_2+\frac{2}{ x_3}=x_3+\frac{2}{ x_4}=\cdots.$$ Guess an expression, in terms of \(n\), for \(x_n\). Then, by induction or otherwise, prove the correctness of your guess.

The Fibonacci numbers \(F_{n}\) are defined by the conditions \(F_{0}=0\), \(F_{1}=1\) and \[F_{n+1}=F_{n}+F_{n-1}\] for all \(n\geqslant 1\). Show that \(F_{2}=1\), \(F_{3}=2\), \(F_{4}=3\) and compute \(F_{5}\), \(F_{6}\) and~\(F_{7}\). Compute \(F_{n+1}F_{n-1}-F_{n}^{2}\) for a few values of \(n\); guess a general formula and prove it by induction, or otherwise. By induction on \(k\), or otherwise, show that \[F_{n+k}=F_{k}F_{n+1}+F_{k-1}F_{n}\] for all positive integers \(n\) and \(k\).

I have \(n\) fence posts placed in a line and, as part of my spouse's birthday celebrations, I wish to paint them using three different colours red, white and blue in such a way that no adjacent fence posts have the same colours. (This allows the possibility of using fewer than three colours as well as exactly three.) Let \(r_{n}\) be the number of ways (possibly zero) that I can paint them if I paint the first and the last post red and let \(s_{n}\) be the number of ways that I can paint them if I paint the first post red but the last post either of the other two colours. Explain why \(r_{n+1}=s_{n}\) and find \(r_{n}+s_{n}.\) Hence find the value of \(r_{n+1}+r_{n}\) for all \(n\geqslant1.\) Prove, by induction, that \[ r_{n}=\frac{2^{n-1}+2(-1)^{n-1}}{3}. \] Find the number of ways of painting \(n\) fence posts (where \(n\geqslant3\)) placed in a circle using three different colours in such a way that no adjacent fence posts have the same colours.

Suppose that \(a_{i}>0\) for all \(i>0\). Show that \[ a_{1}a_{2}\leqslant\left(\frac{a_{1}+a_{2}}{2}\right)^{2}. \] Prove by induction that for all positive integers \(m\) \[ a_{1}\cdots a_{2^{m}}\leqslant\left(\frac{a_{1}+\cdots+a_{2^{m}}}{2^{m}}\right)^{2^{m}}.\tag{\ensuremath{*}} \] If \(n<2^{m}\), put \(b_{1}=a_{2},\) \(b_{2}=a_{2},\cdots,b_{n}=a_{n}\) and \(b_{n+1}=\cdots=b_{2^{m}}=A\), where \[ A=\frac{a_{1}+\cdots+a_{n}}{n}. \] By applying \((*)\) to the \(b_{i},\) show that \[ a_{1}\cdots a_{n}A^{(2^{m}-n)}\leqslant A^{2^{m}} \] (notice that \(b_{1}+\cdots+b_{n}=nA).\) Deduce the (arithmetic mean)/(geometric mean) inequality \[ \left(a_{1}\cdots a_{n}\right)^{1/n}\leqslant\frac{a_{1}+\cdots+a_{n}}{n}. \]

Let \(\mathrm{g}(x)=ax+b.\) Show that, if \(\mathrm{g}(0)\) and \(\mathrm{g}(1)\) are integers, then \(\mathrm{g}(n)\) is an integer for all integers \(n\). Let \(\mathrm{f}(x)=Ax^{2}+Bx+C.\) Show that, if \(\mathrm{f}(-1),\mathrm{f}(0)\) and \(\mathrm{f}(1)\) are integers, then \(\mathrm{f}(n)\) is an integer for all integers \(n\). Show also that, if \(\alpha\) is any real number and \(\mathrm{f}(\alpha-1),\) \(\mathrm{f}(\alpha)\) and \(\mathrm{f}(\alpha+1)\) are integers, then \(\mathrm{f}(\alpha+n)\) is an integer for all integers \(n\).

A plane contains \(n\) distinct given lines, no two of which are parallel, and no three of which intersect at a point. By first considering the cases \(n=1,2,3\) and \(4\), provide and justify, by induction or otherwise, a formula for the number of line segments (including the infinite segments). Prove also that the plane is divided into \(\frac{1}{2}(n^{2}+n+2)\) regions (including those extending to infinity).

In both parts of this question, \(x\) is real and \(0 < \theta < \pi\).

1. By completing the square, find in terms of \(\theta\) the minimum value as \(x\) varies of $$9x^2 - 12x \cos \theta + 4.$$ Find also the maximum value as \(x\) varies of \(12x^2 \sin \theta - 9x^4\). Hence determine the values of \(x\) and \(\theta\) that satisfy the equation $$9x^4 + (9 - 12 \sin \theta)x^2 - 12x \cos \theta + 4 = 0.$$
2. Sketch the curve $$y = \frac{x^2}{x - \theta},$$ where \(\theta\) is a constant. Deduce that either \(\frac{x^2}{x - \theta} \leq 0\) or \(\frac{x^2}{x - \theta} \geq 4\theta\). By considering the numerator and denominator separately, or otherwise, show that $$\frac{\sin^2 \theta \cos^2 x}{1 + \cos^2 \theta \sin^2 x} \leq 1.$$ Hence determine the values of \(x\) and \(\theta\) that satisfy the equation $$\frac{x^2}{4\theta(x - \theta)} = \frac{\sin^2 \theta \cos^2 x}{1 + \cos^2 \theta \sin^2 x}.$$

The triangle \(ABC\) has side lengths \(\left| BC \right| = a\), \(\left| CA \right| = b\) and \(\left| AB \right| = c\). Equilateral triangles \(BXC\), \; \(CY\!A\) \hspace{0.0mm} and \(AZB\) are erected on the sides of the triangle \(ABC\), with~\(X\) on the other side of \(BC\) from \(A\), and similarly for \(Y\) and \(Z\). Points \(L\), \(M\) and \(N\) are the centres of rotational symmetry of triangles \(BXC\), \(CY\!A\) and \(AZB\) respectively.

1. Show that \(| CM| = \dfrac {\ b} {\sqrt3} \,\) and write down the corresponding expression for \(| CL|\).
2. Use the cosine rule to show that \[ 6 \left| LM \right|^2 = a^2+b^2+c^2 + 4\sqrt3 \, \Delta \,, \] where \(\Delta\) is the area of triangle \(ABC\). Deduce that \(LMN\) is an equilateral triangle. Show further that the areas of triangles \(LMN\) and \(ABC\) are equal if and only if \[ a^2+b^2 +c^2 = 4\sqrt3 \, \Delta \,. \]
3. Show that the conditions \[ (a -b)^2 = -2ab \big( 1 -\cos(C-60^\circ)\big) \,\] and \[ a^2+b^2 +c^2 = 4\sqrt3 \, \Delta \] are equivalent. Deduce that the areas of triangles \(LMN\) and \(ABC\) are equal if and only if \(ABC\) is equilateral.

In the triangle \(ABC\), angle \(BAC = \alpha\) and angle \(CBA= 2\alpha\), where \(2\alpha\) is acute, and \(BC= x\). Show that \(AB = (3-4 \sin^2\alpha)x\). The point \(D\) is the midpoint of \(AB\) and the point \(E\) is the foot of the perpendicular from \(C\) to \(AB\). Find an expression for \(DE\) in terms of \(x\). The point \(F\) lies on the perpendicular bisector of \(AB\) and is a distance \(x\) from \(C\). The points \(F\) and \(B\) lie on the same side of the line through \(A\) and \(C\). Show that the line \(FC\) trisects the angle \(ACB\).

A prison consists of a square courtyard of side \(b\) bounded by a perimeter wall and a square building of side \(a\) placed centrally within the courtyard. The sides of the building are parallel to the perimeter walls. Guards can stand either at the middle of a perimeter wall or in a corner of the courtyard. If the guards wish to see as great a length of the perimeter wall as possible, determine which of these positions is preferable. You should consider separately the cases \(b<3a\) and \(b>3a\,\).

In the triangle \(ABC\), the base \(AB\) is of length 1 unit and the angles at~\(A\) and~\(B\) are \(\alpha\) and~\(\beta\) respectively, where \(0<\alpha\le\beta\). The points \(P\) and~\(Q\) lie on the sides \(AC\) and \(BC\) respectively, with \(AP=PQ=QB=x\). The line \(PQ\) makes an angle of~\(\theta\) with the line through~\(P\) parallel to~\(AB\).

1. Show that \(x\cos\theta = 1- x\cos\alpha - x\cos\beta\), and obtain an expression for \(x\sin\theta\) in terms of \(x\), \(\alpha\) and~\(\beta\). Hence show that \begin{equation} \label{eq:2*} \bigl(1+2\cos(\alpha+\beta)\bigr)x^2 - 2(\cos\alpha + \cos\beta)x + 1 = 0\,. \tag{\(*\)} \end{equation} Show that \((*)\) is also satisfied if \(P\) and \(Q\) lie on \(AC\) produced and \(BC\) produced, respectively. [By definition, \(P\) lies on \(AC\) produced if \(P\) lies on the line through \(A\) and~\(C\) and the points are in the order \(A\), \(C\), \(P\)\,.]
2. State the condition on \(\alpha\) and \(\beta\) for \((*)\) to be linear in \(x\). If this condition does not hold (but the condition \(0<\alpha \le \beta\) still holds), show that \((*)\) has distinct real roots.
3. Find the possible values of~\(x\) in the two cases (a) \(\alpha = \beta = 45^\circ\) and (b) \(\alpha = 30^\circ\), \(\beta = 90^\circ\), and illustrate each case with a sketch.

A cyclic quadrilateral \(ABCD\) has sides \(AB\), \(BC\), \(CD\) and \(DA\) of lengths \(a\), \(b\), \(c\) and \(d\), respectively. The area of the quadrilateral is \(Q\), and angle \(DAB\) is \(\theta\). Find an expression for \(\cos\theta\) in terms of \(a\), \(b\), \(c\) and \(d\), and an expression for \(\sin\theta\) in terms of \(a\), \(b\), \(c\), \(d\) and \(Q\). Hence show that \[ 16Q^2 = 4(ad+bc)^2 - (a^2+d^2-b^2-c^2)^2 \,, \] and deduce that \[ Q^2 = (s-a)(s-b)(s-c)(s-d)\,, \] where \(s= \frac12(a+b+c+d)\). Deduce a formula for the area of a triangle with sides of length \(a\), \(b\) and \(c\).

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\begin{align*} && BD^2 &= a^2+d^2 - 2ad \cos \theta \\ && BD^2 &= b^2+c^2-2bc \cos (\pi - \theta) \\ \Rightarrow && a^2+d^2 - 2ad \cos \theta &= b^2+c^2+2bc \cos \theta \\ \Rightarrow && 2(ad+bc)\cos \theta &= a^2+d^2-b^2-c^2 \\ \Rightarrow && \cos \theta &= \frac{a^2+d^2-b^2-c^2}{2(ad+bc)} \\ \\ && Q &= \frac12 ad \sin \theta + \frac12 bc \sin (\pi - \theta) \\ &&&= \frac12 (ad+bc) \sin \theta \\ \Rightarrow && \sin \theta &= \frac{2Q}{ad+bc} \\ \\ && 1 &= \sin^2 \theta + \cos^2 \theta \\ &&&= \frac{4Q^2}{(ad+bc)^2} + \frac{(a^2+d^2-b^2-c^2)^2}{4(ad+bc)^2} \\ \Rightarrow && 4(ad+bc)^2 &= 16Q^2 + (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= 4(ad+bc)^2- (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= (2ad+2bc - a^2-d^2+b^2+c^2)(2ad+2bc+a^2+d^2-b^2-c^2) \\ &&&= ((b+c)^2-(a-d)^2)((a+d)^2-(b-c)^2) \\ &&&= (b+c-a+d)(b+c+a-d)(a+d+b-c)(a+d-b+c) \\ \Rightarrow && Q^2 &= (s-a)(s-b)(s-c)(s-d) \end{align*} Since all triangles are cyclic, we can place \(D\) at the same point as \(A\) to obtain Heron's formula \(A = \sqrt{s(s-a)(s-b)(s-c)}\) where \(s = \frac12(a+b+c)\)

Each edge of the tetrahedron \(ABCD\) has unit length. The face \(ABC\) is horizontal, and \(P\) is the point in \(ABC\) that is vertically below \(D\).

1. Find the length of \(PD\).
2. Show that the cosine of the angle between adjacent faces of the tetrahedron is \(1/3\).
3. Find the radius of the largest sphere that can fit inside the tetrahedron.

Prove that \[ \tan \left ( \tfrac14 \pi -\tfrac12 x \right)\equiv \sec x -\tan x\,. \tag{\(*\)} \]

1. Use \((*)\) to find the value of \(\tan\frac18\pi\,\). Hence show that \[ \tan \tfrac{11}{24} \pi = \frac{\sqrt3 + \sqrt2 -1}{\sqrt3 -\sqrt6+1}\;. \]
2. Show that \[ \frac{\sqrt3 + \sqrt2 -1}{\sqrt3 -\sqrt6+1}= 2+\sqrt2+\sqrt3+\sqrt6\,. \]
3. Use \((*)\) to show that \[ \tan \tfrac1{48}\pi = \sqrt{16+10\sqrt2+8\sqrt3 +6\sqrt6 \ }-2-\sqrt2-\sqrt3-\sqrt6\,. \]

1. The equation of the circle \(C\) is \[ (x-2t)^2 +(y-t)^2 =t^2, \] where \(t\) is a positive number. Show that \(C\) touches the line \(y=0\,\). Let \(\alpha\) be the acute angle between the \(x\)-axis and the line joining the origin to the centre of \(C\). Show that $\tan2\alpha =\frac43\( and deduce that \)C\( touches the line \)3y=4x\,$.
2. Find the equation of the incircle of the triangle formed by the lines \(y=0\), \(3y=4x\) and \(4y+3x=15\,\). {\bf Note:} The {\em incircle} of a triangle is the circle, lying totally inside the triangle, that touches all three sides.

The sides of a triangle have lengths \(p-q\), \(p\) and \(p+q\), where \(p>q> 0\,\). The largest and smallest angles of the triangle are \(\alpha\) and \(\beta\), respectively. Show by means of the cosine rule that \[ 4(1-\cos\alpha)(1-\cos\beta) = \cos\alpha + \cos\beta \,. \] In the case \(\alpha = 2\beta\), show that \(\cos\beta=\frac34\) and hence find the ratio of the lengths of the sides of the triangle.

{\it Note that the volume of a tetrahedron is equal to \(\frac1 3\) \(\times\) the area of the base \(\times\) the height.} The points \(O\), \(A\), \(B\) and \(C\) have coordinates \((0,0,0)\), \((a,0,0)\), \((0,b,0)\) and \((0,0,c)\), respectively, where \(a\), \(b\) and \(c\) are positive.

1. Find, in terms of \(a\), \(b\) and \(c\), the volume of the tetrahedron \(OABC\).
2. Let angle \(ACB = \theta\). Show that \[ \cos\theta = \frac {c^2} { { \sqrt{\vphantom{ \dot b} (a^2+c^2)(b^2+c^2)} } ^{\vphantom A} \ } \] and find, in terms of \(a\), \(b\) and \(c\), the area of triangle \(ABC\). % is %\(\displaystyle \tfrac12 \sqrt{ \vphantom{\dot A } a^2b^2 +b^2c^2 + c^2 a^2 \;} \;\).

Given that \(\alpha\) and \(\beta\) are acute angles, show that \(\alpha + \beta = \tfrac{1}{2}\pi\) if and only if \(\cos^2 \alpha + \cos^2 \beta = 1\). In the \(x\)--\(y\) plane, the point \(A\) has coordinates \((0,s)\) and the point \(C\) has coordinates \((s,0)\), where \(s>0\). The point \(B\) lies in the first quadrant (\(x>0\), \(y>0\)). The lengths of \(AB\), \(OB\) and \(CB\) are respectively \(a\), \(b\) and \(c\). Show that \[ (s^2 +b^2 - a^2)^2 + (s^2 +b^2 -c^2)^2 = 4s^2b^2 \] and hence that \[ (2s^2 -a^2-c^2)^2 + (2b^2 -a^2-c^2)^2 =4a^2c^2\;. \] Deduce that $$ \l a - c \r^2 \le 2b^2 \le \l a + c \r^2\;. $$ %Show, %by considering the case \(a=1+\surd2\,\), \(b=c=1\,\), % that the condition \(\l \ast \r\,\) %is not sufficient to ensure that \(B\) lies in the first quadrant.

Arthur and Bertha stand at a point \(O\) on an inclined plane. The steepest line in the plane through \(O\) makes an angle \(\theta\) with the horizontal. Arthur walks uphill at a steady pace in a straight line which makes an angle \(\alpha\) with the steepest line. Bertha walks uphill at the same speed in a straight line which makes an angle \(\beta\) with the steepest line (and is on the same side of the steepest line as Arthur). Show that, when Arthur has walked a distance \(d\), the distance between Arthur and Bertha is \(2d \vert\sin\frac12(\alpha-\beta)\vert\). Show also that, if \(\alpha\ne\beta\), the line joining Arthur and Bertha makes an angle \(\phi\) with the vertical, where \[ \cos\phi = \sin\theta \sin \frac12(\alpha+\beta). \]

Let \(f(x) = 7 - 2|x|\). A sequence \(u_0, u_1, u_2, \ldots\) is defined by \(u_0 = a\) and \(u_n = f(u_{n-1})\) for \(n > 0\).

1. 1. Sketch, on the same axes, the graphs with equations \(y = f(x)\) and \(y = f(f(x))\).
   2. Find all solutions of the equation \(f(f(x)) = x\).
   3. Find the values of \(a\) for which the sequence \(u_0, u_1, u_2, \ldots\) has period 2.
   4. Show that, if \(a = \frac{28}{5}\), then the sequence \(u_2, u_3, u_4, \ldots\) has period 2, but neither \(u_0\) or \(u_1\) is equal to either of \(u_2\) or \(u_3\).
2. 1. Sketch, on the same axes, the graphs with equations \(y = f(x)\) and \(y = f(f(f(x)))\).
   2. Consider the sequence \(u_0, u_1, u_2, \ldots\) in the cases \(a = 1\) and \(a = -\tfrac79\). Hence find all the solutions of the equation \(f(f(f(x))) = x\).
   3. Find a value of \(a\) such that the sequence \(u_3, u_4, u_5, \ldots\) has period 3, but where none of \(u_0, u_1\) or \(u_2\) is equal to any of \(u_3, u_4\) or \(u_5\).

1. The function \(\f\) is defined by \(\f(x)= |x-a| + |x-b| \), where \(a < b\). Sketch the graph of \(\f(x)\), giving the gradient in each of the regions \(x < a\), \(a < x < b\) and \(x > b\). Sketch on the same diagram the graph of \(\g(x)\), where \(\g(x)= |2x-a-b|\). What shape is the quadrilateral with vertices \((a,0)\), \((b,0)\), \((b,\f(b))\) and \((a, \f(a))\)?
2. Show graphically that the equation \[ |x-a| + |x-b| = |x-c|\,, \] where \(a < b\), has \(0\), \(1\) or \(2\) solutions, stating the relationship of \(c\) to \(a\) and \(b\) in each case.
3. For the equation \[ |x-a| + |x-b| = |x-c|+|x-d|\,, \] where \(a < b\), \(c < d\) and \(d-c < b-a\), determine the number of solutions in the various cases that arise, stating the relationship between \(a\), \(b\), \(c\) and \(d\) in each case.

Sketch the following subsets of the \(x\)-\(y\) plane:

1. \(|x|+|y|\le 1\) ;
2. \(|x-1|+|y-1|\le 1 \) ;
3. \(|x-1|-|y+1|\le 1 \) ;
4. \(|x|\, |y-2|\le 1\) .

Find all the solutions of the equation \[|x+1|-|x|+3|x-1|-2|x-2|=x+2.\]

The function \(\mathrm{f}\) is defined for \(x<2\) by \[ \mathrm{f}(x)=2| x^{2}-x|+|x^{2}-1|-2|x^{2}+x|. \] Find the maximum and minimum points and the points of inflection of the graph of \(\mathrm{f}\) and sketch this graph. Is \(\mathrm{f}\) continuous everywhere? Is \(\mathrm{f}\) differentiable everywhere? Find the inverse of the function \(\mathrm{f}\), i.e. expressions for \(\mathrm{f}^{-1}(x),\) defined in the various appropriate intervals.

Show Solution

\[ f(x) = 2|x(x-1)| + |(x-1)(x+1)|-2|x(x+1)| \] Therefore the absolute value terms will change behaviour at \(x = -1, 0, 1\). Then \begin{align*} f(x) &= \begin{cases} 2(x^2-x)+(x^2-1)-2(x^2+x) & x \leq -1 \\ 2(x^2-x)-(x^2-1)+2(x^2+x) & -1 < x \leq 0 \\ -2(x^2-x)-(x^2-1)-2(x^2+x) & 0 < x \leq 1 \\ 2(x^2-x)+(x^2-1)-2(x^2+x) & 1 < x\end{cases} \\ &= \begin{cases} x^2-4x-1 & x \leq -1 \\ 3x^2+1& -1 < x \leq 0 \\ -5x^2+1& 0 < x \leq 1 \\ x^2-4x-1 & 1 < x\end{cases} \\ \\ f'(x) &= \begin{cases} 2x-4 & x <-1 \\ 6x & -1 < x < 0 \\ -10x & 0 < x < 1 \\ 2x-4 & 1 < x\end{cases} \\ \end{align*} Therefore \(f'(x) = 0 \Rightarrow x = 0, 2\) and so we should check all the turning points. Therefore the minimum is \(x = 2, y = -5\), maximum is \(x = -2, y = 11\) (assuming the range is actually \(|x| < 2\). There is a point of inflection at \(x = 0, y = 1\).

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\(f\) is continuous everywhere as a sum of continuous functions. \(f\) is not differentiable at \(x = -1, 1\) Suppose \begin{align*} &&y &=x^2-4x-1 \\ &&&= (x-2)^2 -5 \\ \Rightarrow &&x &= 2\pm \sqrt{y+5} \\ \\ && y &= 3x^2+1 \\ \Rightarrow && x &= \pm \sqrt{\frac{y-1}{3}} \\ \\ && y &= -5x^2+1 \\ \Rightarrow && x &=\pm \sqrt{\frac{1-y}{5}} \\ \\ \Rightarrow && f^{-1}(y) &= \begin{cases} 2 - \sqrt{y+5} & y > 4 \\ -\sqrt{\frac{y-1}{3}} & 1 < y < 4 \\ \sqrt{\frac{1-y}{5}} & -4 < y < 1 \\ 2 + \sqrt{y+5} & y < -4 \end{cases} \end{align*}

The domain of the function f is the set of all \(2 \times 2\) matrices and its range is the set of real numbers. Thus, if \(M\) is a \(2 \times 2\) matrix, then \(f(M) \in \mathbb{R}\). The function f has the property that \(f(MN) = f(M)f(N)\) for any \(2 \times 2\) matrices \(M\) and \(N\).

1. You are given that there is a matrix \(M\) such that \(f(M) \neq 0\). Let \(I\) be the \(2 \times 2\) identity matrix. By considering \(f(MI)\), show that \(f(I) = 1\).
2. Let \(J = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\). You are given that \(f(J) \neq 1\). By considering \(J^2\), evaluate \(f(J)\). Using \(J\), show that, for any real numbers \(a\), \(b\), \(c\) and \(d\), $$.f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) = -f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right) = f\left(\begin{pmatrix} d & c \\ b & a \end{pmatrix}\right)$$
3. Let \(K = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\) where \(k \in \mathbb{R}\). Use \(K\) to show that, if the second row of the matrix \(A\) is a multiple of the first row, then \(f(A) = 0\).
4. Let \(P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\). By considering the matrices \(P^2\), \(P^{-1}\), and \(K^{-1}PK\) for suitable values of \(k\), evaluate \(f(P)\).

For any square matrix \(\mathbf{A}\) such that \(\mathbf{I-A}\) is non-singular (where \(\mathbf{I}\) is the unit matrix), the matrix \(\mathbf{B}\) is defined by \[ \mathbf{B}=(\mathbf{I+A})(\mathbf{I-A})^{-1}. \] Prove that \(\mathbf{B}^{\mathrm{T}}\mathbf{B}=\mathbf{I}\) if and only if \(\mathbf{A+A}^{\mathrm{T}}=\mathbf{O}\) (where \(\mathbf{O}\) is the zero matrix), explaining clearly each step of your proof. {[}You may quote standard results about matrices without proof.{]}

For any two points \(X\) and \(Y\), with position vectors \(\bf x\) and \(\bf y\) respectively, \(X*Y\) is defined to be the point with position vector \(\lambda {\bf x}+ (1-\lambda){\bf y}\), where \(\lambda\) is a fixed number.

1. If \(X\) and \(Y\) are distinct, show that \(X*Y\) and \(Y*X\) are distinct unless \(\lambda\) takes a certain value (which you should state).
2. Under what conditions are \((X*Y)*Z\) and \(X*(Y*Z)\,\) distinct?
3. Show that, for any points \(X\), \(Y\) and \(Z\), \[ (X*Y)*Z = (X*Z)*(Y*Z)\, \] and obtain the corresponding result for \(X*(Y*Z)\).
4. The points \(P_1\), \(P_2\), \(\ldots\) are defined by \( P_1 = X*Y\) and, for \(n \ge2\), \(P_n= P_{n-1}*Y\,.\) Given that \(X\) and \(Y\) are distinct and that \(0<\lambda<1\), find the ratio in which \(P_n\) divides the line segment \(XY\).

Verify that if \[ \mathbf{P}=\begin{pmatrix}1 & 2\\ 2 & -1 \end{pmatrix}\qquad\mbox{ and }\qquad\mathbf{A}=\begin{pmatrix}-1 & 8\\ 8 & 11 \end{pmatrix} \] then \(\mathbf{PAP}\) is a diagonal matrix. Put $\mathbf{x}=\begin{pmatrix}x\\ y \end{pmatrix}\( and \)\mathbf{x}_{1}=\begin{pmatrix}x_{1}\\ y_{1} \end{pmatrix}.$ By writing \[ \mathbf{x}=\mathbf{P}\mathbf{x}_{1}+\mathbf{a} \] for a suitable vector \(\mathbf{a},\) show that the equation \[ \mathbf{x}^{\mathrm{T}}\mathbf{Ax}+\mathbf{b}^{\mathrm{T}}\mathbf{x}-11=0, \] where $\mathbf{b}=\begin{pmatrix}18\\ 6 \end{pmatrix}\( and \) \mathbf{x}^{\mathrm{T}} \( is the transpose of \)\mathbf{x},$ becomes \[ 3x_{1}^{2}-y_{1}^{2}=c \] for some constant \(c\) (which you should find).

In this question, \(\mathbf{A,\mathbf{B\) }}and \(\mathbf{X\) are non-zero \(2\times2\) real matrices.} Are the following assertions true or false? You must provide a proof or a counterexample in each case.

1. If \(\mathbf{AB=0}\) then \(\mathbf{BA=0}.\)
2. \((\mathbf{A-B)(A+B)=}\mathbf{A}^{2}-\mathbf{B}^{2}.\)
3. The equation \(\mathbf{AX=0}\) has a non-zero solution \(\mathbf{X}\) if and only if \(\det\mathbf{A}=0.\)
4. For any \(\mathbf{A}\) and \(\mathbf{B}\) there are at most two matrices \(\mathbf{X}\) such that \(\mathbf{X}^{2}+\mathbf{AX}+\mathbf{B}=\mathbf{0}.\)

The matrices \(\mathbf{I}\) and \(\mathbf{J}\) are \[ \mathbf{I}=\begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}\quad\mbox{ and }\quad\mathbf{J}=\begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix} \] respectively and \(\mathbf{A}=\mathbf{I}+a\mathbf{J},\) where \(a\) is a non-zero real constant. Prove that \[ \mathbf{A}^{2}=\mathbf{I}+\tfrac{1}{2}[(1+2a)^{2}-1]\mathbf{J}\quad\mbox{ and }\quad\mathbf{A}^{3}=\mathbf{I}+\tfrac{1}{2}[(1+2a)^{3}-1]\mathbf{J} \] and obtain a similar form for \(\mathbf{A}^{4}.\) If \(\mathbf{A}^{k}=\mathbf{I}+p_{k}\mathbf{J},\) suggest a suitable form for \(p_{k}\) and prove that it is correct by induction, or otherwise.

Two square matrices \(\mathbf{A}\) and \(\mathbf{B}\) satisfies \(\mathbf{AB=0}.\) Show that either \(\det\mathbf{A}=0\) or \(\det\mathbf{B}=0\) or \(\det\mathbf{A}=\det\mathbf{B}=0\). If \(\det\mathbf{B}\neq0\), what must \(\mathbf{A}\) be? Give an example to show that the condition \(\det\mathbf{A}=\det\mathbf{B}=0\) is not sufficient for the equation \(\mathbf{AB=0}\) to hold. Find real numbers \(p,q\) and \(r\) such that \[ \mathbf{M}^{3}+2\mathbf{M}^{2}-5\mathbf{M}-6\mathbf{I}=(\mathbf{M}+p\mathbf{I})(\mathbf{M}+q\mathbf{I})(\mathbf{M}+r\mathbf{I}), \] where \(\mathbf{M}\) is any square matrix and \(\mathbf{I}\) is the appropriate identity matrix. Hence, or otherwise, find all matrices \(\mathbf{M}\) of the form $\begin{pmatrix}a & c\\ 0 & b \end{pmatrix}$ which satisfy the equation \[ \mathbf{M}^{3}+2\mathbf{M}^{2}-5\mathbf{M}-6\mathbf{I}=\mathbf{0}. \]

In a crude model of population dynamics of a community of aardvarks and buffaloes, it is assumed that, if the numbers of aardvarks and buffaloes in any year are \(A\) and \(B\) respectively, then the numbers in the following year at \(\frac{1}{4}A+\frac{3}{4}B\) and \(\frac{3}{2}B-\frac{1}{2}A\) respectively. It does not matter if the model predicts fractions of animals, but a non-positive number of buffaloes means that the species has become extinct, and the model ceases to apply. Using matrices or otherwise, show that the ratio of the number of aardvarks to the number of buffaloes can remain the same each year, provided it takes one of two possible values. Let these two possible values be \(x\) and \(y\), and let the numbers of aardvarks and buffaloes in a given year be \(a\) and \(b\) respectively. By writing the vector \((a,b)\) as a linear combination of the vectors \((x,1)\) and \((y,1),\) or otherwise, show how the numbers of aardvarks and buffaloes in subsequent years may be found. On a sketch of the \(a\)-\(b\) plane, mark the regions which correspond to the following situations

1. an equilibrium population is reached as time \(t\rightarrow\infty\);
2. buffaloes become extinct after a finite time;
3. buffaloes approach extinction as \(t\rightarrow\infty.\)

Show Solution

If the population in a given year is \(\mathbf{v} = \begin{pmatrix}A \\ B \end{pmatrix}\) then the population the next year is \(\mathbf{Mv}\) where \(\mathbf{M} = \begin{pmatrix} \frac14 & \frac34 \\ -\frac12 &\frac32 \end{pmatrix}\) The ratio is the same if \(\mathbf{Mv} = \lambda \mathbf{v}\) ie if \(\mathbf{v}\) is an eigenvector of \(\mathbf{M}\). The eigenvalues will be \(1\) and \(\frac38\) (by inspection) so we should be able to solve for the eigenvectors: \(\lambda = 1\) we have \(\frac14A + \frac34B = A \Rightarrow A = B\) a ratio of \(1\). \(\lambda = \frac38\) we have \(\frac14A + \frac34B = \frac38A \Rightarrow \frac34B = \frac18A \Rightarrow A = 6B\) a ratio of \(6\). If we write \(\begin{pmatrix} a \\ b \end{pmatrix}\) as \(x_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + x_6 \begin{pmatrix} 6 \\ 1 \end{pmatrix}\) we find that after \(n\) years, we have: \(x_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + \l \frac38 \r^n x_6 \begin{pmatrix} 6 \\ 1 \end{pmatrix}\) for the populations. Therefore if \(x_1\) is \(< 0\) then in finite time we will end up with one population being 0. If \(x_1 > 0\) are positive we tend to a finite population and if \(x_1 = 0\) then over time the population will tend to \(0\) at infinity. In our diagram these areas correspond to (red) - die out in finite time, (green) population stable and the thick black line where the population goes extinct as \(t \to \infty\)

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1. \(x_2\) and \(y_2\) are defined in terms of \(x_1\) and \(y_1\) by the equation $$\begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$$ \(G_1\) is the graph with equation $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$ and \(G_2\) is the graph with equation $$\frac{\left(\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}\right)^2}{9} + \frac{\left(-\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}\right)^2}{4} = 1$$ Show that, if \((x_1, y_1)\) is a point on \(G_1\), then \((x_2, y_2)\) is a point on \(G_2\). Show that \(G_2\) is an anti-clockwise rotation of \(G_1\) through \(45°\) about the origin.
2. 1. The matrix $$\begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix}$$ represents a reflection. Find the line of invariant points of this matrix.
   2. Sketch, on the same axes, the graphs with equations $$y = 2^x \text{ and } 0.8x + 0.6y = 2^{-0.6x+0.8y}$$
3. Sketch, on the same axes, for \(0 \leq x \leq 2\pi\), the graphs with equations $$y = \sin x \text{ and } y = \sin(x - 2y)$$ You should determine the exact co-ordinates of the points on the graph with equation \(y = \sin(x - 2y)\) where the tangent is horizontal and those where it is vertical.

The matrix A is given by $$\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.$$

1. You are given that the transformation represented by A has a line \(L_1\) of invariant points (so that each point on \(L_1\) is transformed to itself). Let \((x, y)\) be a point on \(L_1\). Show that \(((a - 1)(d - 1) - bc)xy = 0\). Show further that \((a - 1)(d - 1) = bc\). What can be said about A if \(L_1\) does not pass through the origin?
2. By considering the cases \(b \neq 0\) and \(b = 0\) separately, show that if \((a - 1)(d - 1) = bc\) then the transformation represented by A has a line of invariant points. You should identify the line in the different cases that arise.
3. You are given instead that the transformation represented by A has an invariant line \(L_2\) (so that each point on \(L_2\) is transformed to a point on \(L_2\)) and that \(L_2\) does not pass through the origin. If \(L_2\) has the form \(y = mx + k\), show that \((a - 1)(d - 1) = bc\).

Show that, if the lengths of the diagonals of a parallelogram are specified, then the parallogram has maximum area when the diagonals are perpendicular. Show also that the area of a parallelogram is less than or equal to half the square of the length of its longer diagonal. The set \(A\) of points \((x,y)\) is given by \begin{alignat*}{1} \left|a_{1}x+b_{1}y-c_{1}\right| & \leqslant\delta,\\ \left|a_{2}x+b_{2}y-c_{2}\right| & \leqslant\delta, \end{alignat*} with \(a_{1}b_{2}\neq a_{2}b_{1}.\) Sketch this set and show that it is possible to find \((x_{1},y_{1}),(x_{2},y_{2})\in A\) with \[ (x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}\geqslant\frac{8\delta^{2}}{\left|a_{1}b_{2}-a_{2}b_{1}\right|}. \]

Show Solution

In a parallelogram the diagonals meet at their mid points. Fixing one diagonal, we can look at the two triangles formed by the other diagonal. Suppose the angle between them is \(\theta\). Then the area of the triangles will be \(\frac12 \frac{l_1}{2} \frac{l_2}2 \sin \theta+\frac12 \frac{l_1}{2} \frac{l_2}2 \sin (\pi -\theta) = \frac{l_1l_2}{4} \sin \theta\). This will be true on both sides. Therefore we can maximise this area by setting \(\theta = \frac{\pi}{2}\).

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Consider the (darker) shaded area. This is our set \(A\). The area of the set is indifferent to a parallel shift in the lines, so without loss of generality, we can consider \(c_1 = 0, c_2 = 0\), so our lines meet at the origin. Now also consider the linear transformation \(\begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix}^{-1}\) which takes the coordinate axes to these lines. This will take the square \([-\delta, \delta] \times [-\delta, \delta]\) which has area \(4\delta^2\) to our square, which will have area \(\frac{4 \delta^2}{ |a_1b_2 - a_2b_1|}\). If we consider the length of the two diagonals of this area, \(l_1, l_2\) we know that \(\frac{l_1l_2}2 \sin \theta = \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}\), if we consider the larger of \(l_1\) and \(l_2\) (wlog \(l_1\)) we must have that \(\frac{l_1^2}{2} \geq \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}\) and so points on opposite ends of the diagonal will satisfy the inequality in the question.

Let \(R_{\alpha}\) be the \(2\times2\) matrix that represents a rotation through the angle \(\alpha\) and let $$A=\begin{pmatrix}a&b\\b&c\end{pmatrix}.$$

1. Find in terms of \(a\), \(b\) and \(c\) an angle \(\alpha\) such that \(R_{-\alpha}AR_{\alpha}\) is a diagonal matrix (i.e. has the value zero in top-right and bottom-left positions).
2. Find values of \(a\), \(b\) and \(c\) such that the equation of the ellipse \[x^2+(y+2x\cot2\theta)^2=1\qquad(0 < \theta < \tfrac{1}{4}\pi)\] can be expressed in the form \[\begin{pmatrix}x&y\end{pmatrix}A\begin{pmatrix}x\\y\end{pmatrix}=1.\] Show that, for this \(A\), \(R_{-\alpha}AR_{\alpha}\) is diagonal if \(\alpha=\theta\). Express the non--zero elements of this matrix in terms of \(\theta\).
3. Deduce, or show otherwise, that the minimum and maximum distances from the centre to the circumference of this ellipse are \(\tan\theta\) and \(\cot\theta\).

The transformation \(T\) of the point \(P\) in the \(x\),\(y\) plane to the point \(P'\) is constructed as follows: \hfil\break Lines are drawn through \(P\) parallel to the lines \(y=mx\) and \(y=-mx\) to cut the line \(y=kx\) at \(Q\) and \(R\) respectively, \(m\) and \(k\) being given constants. \(P'\) is the fourth vertex of the parallelogram \(PQP'R\). Show that if \(P\) is \((x_1,y_1)\) then \(Q\) is $$ \left( {mx_1-y_1 \over m-k}, {k(mx_1-y_1)\over m-k}\right). $$ Obtain the coordinates of \(P'\) in terms of \(x_1\), \(y_1\), \(m\) and \(k\), and express \(T\) as a matrix transformation. Show that areas are transformed under \(T\) into areas of the same magnitude.

The transformation \(T\) from \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} X \\ Y \end{pmatrix}\) is given by \[ \begin{pmatrix}X\\ Y \end{pmatrix}=\frac{2}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix}. \] Show that \(T\) leaves the vector \(\begin{pmatrix} 1\\ 2 \end{pmatrix}\) unchanged in direction but multiplied by a scalar, and that \(\begin{pmatrix} 2\\ -1 \end{pmatrix}\) is similarly transformed. The circle \(C\) whose equation is \(x^{2}+y^{2}=1\) transforms under \(T\) to a curve \(E\). Show that \(E\) has equation \[ 8X^{2}+12XY+17Y^{2}=80, \] and state the area of the region bounded by \(E\). Show also that the greatest value of \(X\) on \(E\) is \(2\sqrt{17/5}.\) Find the equation of the tangent to \(E\) at the point which corresponds to the point \(\frac{1}{5}(3,4)\) on \(C\).

The linear transformation \(\mathrm{T}\) is a shear which transforms a point \(P\) to the point \(P'\) defined by

1. \(\overrightarrow{PP'}\) makes an acute angle \(\alpha\) (anticlockwise) with the \(x\)-axis,
2. \(\angle POP'\) is clockwise (i.e. the rotation from \(OP\) to \(OP'\) clockwise is less than \(\pi),\)
3. \(PP'=k\times PN,\) where \(PN\) is the perpendicular onto the line \(y=x\tan\alpha,\) where \(k\) is a given non-zero constant.

Show Solution

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We can see that \(\mathbf{M}\) sends \(\begin{pmatrix} 1 \\ \tan \alpha \end{pmatrix}\) to itself, and \(\begin{pmatrix} -\tan \alpha \\ 1 \end{pmatrix}\) to \(\begin{pmatrix} -\tan \alpha \\ 1 \end{pmatrix} + k \begin{pmatrix} 1 \\ \tan \alpha \end{pmatrix}\) Therefore, we have: \begin{align*} && \mathbf{M} \begin{pmatrix} 1 & -\tan \alpha \\ \tan \alpha & 1 \end{pmatrix} &= \begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix} \\ && \sec \alpha \mathbf{M} \begin{pmatrix} \cos \alpha & -\sin\alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} &= \begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix} \\ \Rightarrow && \mathbf{M} &= \cos \alpha\begin{pmatrix} 1 & k - \tan \alpha \\ \tan \alpha & 1 + k\tan \alpha \end{pmatrix}\begin{pmatrix} \cos \alpha & \sin\alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \\ &&&= \cos\alpha \begin{pmatrix}\cos \alpha -k\sin\alpha + \frac{\sin^2 \alpha}{\cos \alpha} & \sin \alpha + k \cos \alpha - \sin \alpha \\ \sin \alpha - \sin \alpha - k\frac{\sin^2 \alpha}{\cos \alpha} & \frac{\sin^2 \alpha}{\cos \alpha} + \cos\alpha + k \sin \alpha \end{pmatrix} \\ &&&= \begin{pmatrix}1-k\sin\alpha\cos\alpha & k\cos^{2}\alpha\\ -k\sin^{2}\alpha & 1+k\sin\alpha\cos\alpha \end{pmatrix} \end{align*} Suppose $\begin{pmatrix}p & q\\ r & t \end{pmatrix} \mathbf{M} = \mathbf{M} \begin{pmatrix}p & q\\ r & t \end{pmatrix}$ then, \begin{align*} && \begin{pmatrix}p & q\\ r & t \end{pmatrix} \mathbf{M} &= \mathbf{M} \begin{pmatrix}p & q\\ r & t \end{pmatrix} \\ \Leftrightarrow && \small \begin{pmatrix} p(1-k\sin\alpha\cos\alpha) + q(-k\sin^{2}\alpha) & pk\cos^{2}\alpha + q(1+k\sin\alpha\cos\alpha)\\ r(1-k\sin\alpha\cos\alpha) + t(-k\sin^{2}\alpha) & rk\cos^{2}\alpha + t(1+k\sin\alpha\cos\alpha)\end{pmatrix} &= \\ && \qquad \small \begin{pmatrix} p(1-k\sin\alpha\cos\alpha) + rk\cos^{2}\alpha & q(1-k\sin\alpha\cos\alpha) + tk\cos^{2}\alpha \\ -pk\sin^{2}\alpha + r(1+k\sin\alpha\cos\alpha) & -qk\sin^{2}\alpha+t (1+k\sin\alpha\cos\alpha) \end{pmatrix} \\ \Leftrightarrow && \begin{cases} p(1-k\sin\alpha\cos\alpha) + q(-k\sin^{2}\alpha) &= p(1-k\sin\alpha\cos\alpha) + rk\cos^{2}\alpha \\ pk\cos^{2}\alpha + q(1+k\sin\alpha\cos\alpha) &=q(1-k\sin\alpha\cos\alpha) + tk\cos^{2}\alpha \\ r(1-k\sin\alpha\cos\alpha) + t(-k\sin^{2}\alpha) &=-pk\sin^{2}\alpha + r(1+k\sin\alpha\cos\alpha) \\ rk\cos^{2}\alpha + t(1+k\sin\alpha\cos\alpha) &= -qk\sin^{2}\alpha+t (1+k\sin\alpha\cos\alpha) \end{cases} \\ \Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\ p\cos^{2}\alpha + q\sin\alpha\cos\alpha &=-q\sin\alpha\cos\alpha + t\cos^{2}\alpha \\ -r\sin\alpha\cos\alpha + -t\sin^{2}\alpha &=-p\sin^{2}\alpha + r\sin\alpha\cos\alpha \\ r &= -q\tan^{2}\alpha \end{cases} \\ \Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\ 2q\sin\alpha\cos\alpha &=(t-p)\cos^{2}\alpha \\ (p-t)\sin^{2}\alpha &=2r\sin\alpha\cos\alpha \end{cases} \\ \Leftrightarrow && \begin{cases} -q\tan^{2}\alpha &= r \\ 2q\tan \alpha &=(t-p) \end{cases} \\ \end{align*} as required

A pyramid has a horizontal rectangular base \(ABCD\) and its vertex \(V\) is vertically above the centre of the base. The acute angle between the face \(AVB\) and the base is \(\alpha\), the acute angle between the face \(BVC\) and the base is \(\beta\) and the obtuse angle between the faces \(AVB\) and \(BVC\) is \(\pi - \theta\).

1. The edges \(AB\) and \(BC\) are parallel to the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\), respectively, and the unit vector \(\mathbf{k}\) is vertical. Find a unit vector that is perpendicular to the face \(AVB\). Show that $$\cos \theta = \cos \alpha \cos \beta.$$
2. The edge \(BV\) makes an angle \(\phi\) with the base. Show that $$\cot^2 \phi = \cot^2 \alpha + \cot^2 \beta.$$ Show also that $$\cos^2 \phi = \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1 - \cos^2 \theta} \geq \frac{2 \cos \theta - 2 \cos^2 \theta}{1 - \cos^2 \theta}$$ and deduce that \(\phi < \theta\).

You are not required to consider issues of convergence in this question. For any sequence of numbers \(a_1, a_2, \ldots, a_m, \ldots, a_n\), the notation \(\prod_{i=m}^{n} a_i\) denotes the product \(a_m a_{m+1} \cdots a_n\).

1. Use the identity \(2 \cos x \sin x = \sin(2x)\) to evaluate the product \(\cos(\frac{\pi}{9}) \cos(\frac{2\pi}{9}) \cos(\frac{4\pi}{9})\).
2. Simplify the expression $$\prod_{k=0}^{n} \cos\left(\frac{x}{2^k}\right) \quad (0 < x < \frac{1}{2}\pi).$$ Using differentiation, or otherwise, show that, for \(0 < x < \frac{1}{2}\pi\), $$\sum_{k=0}^{n} \frac{1}{2^k} \tan\left(\frac{x}{2^k}\right) = \frac{1}{2^n} \cot\left(\frac{x}{2^n}\right) - 2 \cot(2x).$$
3. Using the results \(\lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1\) and \(\lim_{\theta\to 0} \frac{\tan \theta}{\theta} = 1\), show that $$\prod_{k=1}^{\infty} \cos\left(\frac{x}{2^k}\right) = \frac{\sin x}{x}$$ and evaluate $$\sum_{j=2}^{\infty} \frac{1}{2^{j-2}} \tan\left(\frac{\pi}{2^j}\right).$$

Show that if at least one of the four angles \(A\pm B\pm C\) is a multiple of \(\pi\), then \begin{alignat*}{1} \sin^{4}A+\sin^{4}B+\sin^{4}C & -2\sin^{2}B\sin^{2}C-2\sin^{2}C\sin^{2}A\\ & -2\sin^{2}A\sin^{2}B+4\sin^{2}A\sin^{2}B\sin^{2}C=0. \end{alignat*}

In this question, you may use the following identity without proof: \[ \cos A + \cos B = 2\cos\tfrac12(A+B) \, \cos \tfrac12(A-B) \;. \]

1. Given that \(0\le x \le 2\pi\), find all the values of \(x\) that satisfy the equation \[ \cos x + 3\cos 2x + 3\cos 3 x + \cos 4x= 0 \,. \]
2. Given that \(0\le x \le \pi\) and \(0\le y \le \pi\) and that \[ \cos (x+y) + \cos (x-y) -\cos2x = 1 \,, \] show that either \(x=y\) or \(x\) takes one specific value which you should find.
3. Given that \(0\le x \le \pi\) and \(0\le y \le \pi\,\), find the values of \(x\) and \(y\) that satisfy the equation \[ \cos x + \cos y -\cos (x+y) = \tfrac32 \,. \]

A function \(\f(x)\) is said to be {\em concave} for \(a< x < b\) if \[ \ t\,\f(x_1) +(1-t)\,\f(x_2) \le \f\big(tx_1+ (1-t)x_2\big) \, ,\] for \(a< x_1 < b\,\), \ \(a< x_2< b\) and \(0\le t \le 1\,\). Illustrate this definition by means of a sketch, showing the chord joining the points \(\big(x_1, \f(x_1)\big) \) and \(\big(x_2, \f(x_2)\big) \), in the case \(x_1

Use the identity \[ 2 \sin P\,\sin Q = \cos(Q-P)-\cos(Q+P)\, \] to show that \[ 2\sin\theta \,\big (\sin\theta + \sin 3\theta + \cdots + \sin (2n-1)\theta\,\big ) = 1-\cos 2n\theta \,. \]

1. Let \(A_n\) be the approximation to the area under the curve \(y=\sin x\) from \(x=0\) to \(x=\pi\), using \(n\) rectangular strips each of width \(\frac{{\displaystyle \pi}}{\displaystyle n}\), such that the midpoint of the top of each strip lies on the curve. Show that \[ A_n \sin \left( \frac{\pi}{2n} \right) = \frac \pi n\,. \]
2. Let \(B_n\) be the approximation to the area under the curve \(y=\sin x\) from \(x=0\) to \(x=\pi\,\), using the trapezium rule with \(n\) strips each of width \(\frac{\displaystyle \pi}{ \displaystyle n}\). Show that \[B_n \sin \left( \frac{\pi}{2n} \right) = \frac{\pi}{n} \cos \left( \frac{\pi}{2n} \right) . \]
3. Show that \[ \frac{1}{2}(A_n + B_n) = B_{2n}\,, \] and that \[ A_n B_{2n} = A^2_{2n}\, . \]

The points \(R\) and \(S\) have coordinates \((-a,\, 0)\) and \((2a,\, 0)\), respectively, where \(a > 0\,\). The point \(P\) has coordinates \((x,\, y)\) where \(y > 0\) and \(x < 2a\). Let \(\angle PRS = \alpha \) and \(\angle PSR = \beta\,\).

1. Show that, if \(\beta = 2 \alpha\,\), then \(P\) lies on the curve \(y^2=3(x^2-a^2)\,\).
2. Find the possible relationships between \(\alpha\) and \(\beta\) when \(0 < \alpha < \pi\,\) and \(P\) lies on the curve \(y^2=3(x^2-a^2)\,\).

Show Solution

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1. \begin{align*} &&\tan \beta &= \frac{y}{2a - x} \\ &&\tan \alpha &= \frac{y}{x+a} \\ && \tan \beta &= \tan 2 \alpha \\ && &= \frac{\tan \alpha}{1 - \tan^2 (2\alpha)} \\ \Leftrightarrow && \frac{y}{2a-x}&= \frac{\l \frac{y}{x+a} \r}{1 - \l \frac{y}{x+a} \r^2} \\ && &= \frac{2y(x+a)}{(x+a)^2 - y^2} \\ \Leftrightarrow && (x+a)^2 - y^2 &= 2(x+a)(2a-x) \tag{\(y \neq 0\)} \\ \Leftrightarrow && x^2 + 2ax + a^2 - y^2 &= -2x^2 + 2ax - 4a^2 \\ \Leftrightarrow && y^2 &= 3(x^2-a^2) \end{align*}
2. Therefore if \(y^2 = 3(x^2-a^2)\) we know that \(\tan \beta = \tan 2\alpha\), so \(2\alpha = \beta + n \pi\). Since \(0 < \alpha + \beta < \pi\) (since they are angles in a triangle we must have that \(0 < \alpha + 2\alpha - n \pi = 3\alpha - n\pi < \pi\), so \(0 < \alpha - \frac{n\pi}{3} < \frac{\pi}3\), therefore we have \(3\) cases:

In this question, the \(\mathrm{arctan}\) function satisfies \(0\le \arctan x <\frac12 \pi\) for \(x\ge0\,\).

1. Let \[ S_n= \sum_{m=1}^n \arctan \left(\frac1 {2m^2}\right) \,, \] for \(n=1, 2, 3, \ldots\) . Prove by induction that \[ \tan S_n = \frac n {n+1} \,. \] Prove also that \[ S_n = \arctan \frac n {n+1} \,. \]
2. In a triangle \(ABC\), the lengths of the sides \(AB\) and \(BC\) are \(4n^2\) and \(4n^4-1\), respectively, and the angle at \(B\) is a right angle. Let \(\angle BCA = 2\alpha_n\). Show that \[ \sum_{n=1}^\infty \alpha_n = \tfrac14\pi \,. \]

1. The continuous function \(\f\) is defined by \[ \tan \f(x) = x \ \ \ \ \ (-\infty < x <\infty) \] and \(\f(0)=\pi\). Sketch the curve \(y=\f(x)\)\,.
2. The continuous function \(\g\) is defined by \[ \tan \g(x) = \frac x {1+x^2} \ \ \ \ \ \ (-\infty < x <\infty) \] and \(\g(0)=\pi\). Sketch the curves \(y= \dfrac x {1+x^2} \ \) and \(y=\g(x)\)\,.
3. The continuous function \(\h \) is defined by \(\h (0)=\pi\) and \[ \tan \h (x)= \frac x {1-x^2}\, \ \ \ \ \ (x \ne \pm 1) \,. \] (The values of \(\h (x)\) at \(x=\pm1\) are such that \(\h (x)\) is continuous at these points.) Sketch the curves \(y= \dfrac x {1-x^2} \ \) and \(y=\h (x)\). %
4. The continuous functions \(\h_1\) and \(\h_2\) are % defined by: \(\h_1(0)=\h_2(0)=\pi \), %\[ %\tan \h_1(x) = \frac {x+x^4} {1+x^2+x^4} %\ \ \ \ \ \text{and} \ \ \ \ \ \ %\tan \h_2(x) = \frac {4x-x^3} {1-x^4} %\,. %\] %for values of \(x\) at which the right hand sides are defined. %Find \(\lim\limits_{x\to\infty}\h_1(x)\) and \(\lim\limits_{x\to\infty}\h_2(x)\,\).

1. Show that \(\cos 15^\circ = \dfrac{\sqrt3 +1}{2\sqrt2}\) and find a similar expression for \(\sin 15^\circ\).
2. Show that \(\cos \alpha\) is a root of the equation \[ 4x^3-3 x -\cos 3\alpha =0\,, \] and find the other two roots in terms of \(\cos\alpha\) and \(\sin\alpha\).
3. Use parts (i) and (ii) to solve the equation \(y^3-3y -\sqrt2 =0\,\), giving your answers in surd form.

1. The sequence of numbers \(u_0, u_1, \ldots \) is given by \(u_0=u\) and, for \(n\ge 0\), \begin{equation} u_{n+1} =4u_n(1- u_n)\,. \tag{\(*\)} \end{equation} In the case \(u= \sin^2\theta\) for some given angle \(\theta\), write down and simplify expressions for \(u_1\) and \(u_2\) in terms of \(\theta\). Conjecture an expression for \(u_n\) and prove your conjecture.
2. The sequence of numbers \(v_0, v_1, \ldots\) is given by $v_0= v \text{ and, for }n\ge 0$, \[ v_{n+1} = -pv_n^2 +qv_n +r\,, \] where \(p\), \(q\) and \(r\) are given numbers, with \(p\ne0\). Show that a substitution of the form \(v_n =\alpha u_n +\beta\), where \(\alpha\) and \(\beta\) are suitably chosen, results in the sequence \((*)\) provided that \[ 4pr = 8 +2q -q^2 \,. \] Hence obtain the sequence satisfying \(v_0=1\) and, for \(n\ge0\), \(v_{n+1} = -v_n^2 +2 v_n +2 \,\).

A thin circular path with diameter \(AB\) is laid on horizontal ground. A vertical flagpole is erected with its base at a point \(D\) on the diameter \(AB\). The angles of elevation of the top of the flagpole from \(A\) and \(B\) are \(\alpha\) and \(\beta\) respectively (both are acute). The point \(C\) lies on the circular path with \(DC\) perpendicular to \(AB\) and the angle of elevation of the top of the flagpole from \(C\) is \(\phi\). Show that \(\cot\alpha\cot \beta = \cot^2\phi\). Show that, for any \(p\) and \(q\), \[ \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) = \tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q) .\] Deduce that, if \(p\) and \(q\) are positive and \( p+q \le \tfrac12 \pi\), then \[ \cot p\cot q\, \ge \cot^2 \tfrac12(p+q) \, \] and hence show that \(\phi \le \tfrac12(\alpha+\beta)\) when \( \alpha +\beta \le \tfrac12 \pi\,\).

Show Solution

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\begin{align*} && \cot \alpha &= \frac{AD}{h} \\ && \cot \beta &= \frac{BD}{h} \\ && \cot \phi &= \frac{DC}h \\ && CD^2 &= AB \cdot BD \tag{intersecting chords} \\ \Rightarrow && \cot^2 \phi &= \cot \alpha \cot \beta \end{align*} \begin{align*} && LHS &= \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) \\ &&&= \cos p \cos q \left ( \frac{1-\cos(p+q)}{2} \right) - \sin p\sin q \left (\frac{1+\cos(p+q)}{2} \right) \\ &&&= \frac12 \left (\cos p \cos q(1-\cos(p+q)) - \sin p\sin q (1+\cos(p+q)) \right) \\ &&&= \frac12 \left ((\cos p \cos q- \sin p\sin q) - (\cos p \cos q+ \sin p\sin q)\cos(p+q) \right) \\ &&&= \frac12 \left (\cos(p+q) - \cos (p-q)\cos(p+q) \right) \\ &&&= RHS \end{align*} Therefore \begin{align*} \cot p \cot q -\cot^2 \tfrac12 (p+q) &= \frac{\tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q)}{\sin p \sin q \sin^2 \tfrac12(p+q)} \\ &=\frac{\cos(p+q)(1-\cos(p-q))}{\sin p \sin q \sin^2 \tfrac12(p+q)} \end{align*} Since \(p,q\) are acute, the denominator is positive. Since \(p+q \leq \frac{\pi}{2}\), we have \(\cos(p+q) \geq 0\). Also \((1-\cos(p-q)) \geq 0\). Thus, the expression is \(\geq 0\). So we must have \begin{align*} && \cot^2 \phi &= \cot \alpha \cot \beta \\ &&&\geq \cot^2 \tfrac12(\alpha+\beta) \end{align*} Since \(\cot\) is decreasing on \((0, \tfrac12 \pi)\) we can deduce \(\phi \leq \tfrac12 (\alpha+\beta)\)

1. Find all the values of \(\theta\), in the range \(0^\circ <\theta<180^\circ\), for which \(\cos\theta=\sin 4\theta\). Hence show that \[ \sin 18^\circ = \frac14\left( \sqrt 5 -1\right). \]
2. Given that \[ 4\sin^2 x + 1 = 4\sin^2 2x \,, \] find all possible values of \(\sin x\,\), giving your answers in the form \(p+q\sqrt5\) where \(p\) and \(q\) are rational numbers.
3. Hence find two values of \(\alpha\) with \(0^\circ <\alpha<90^\circ\) for which \[ \sin^23\alpha + \sin^25\alpha = \sin^2 6\alpha\,. \]

Prove the identity \[ 4\sin\theta \sin(\tfrac13\pi-\theta) \sin (\tfrac13\pi+\theta)= \sin 3\theta\, . \tag{\(*\)}\]

1. By differentiating \((*)\), or otherwise, show that \[ \cot \tfrac19\pi - \cot \tfrac29\pi + \cot \tfrac49\pi = \sqrt3\,. \]
2. By setting \(\theta = \frac16\pi-\phi\) in \((*)\), or otherwise, obtain a similar identity for \(\cos3\theta\) and deduce that \[ \cot \theta \cot (\tfrac13\pi-\theta) \cot (\tfrac13\pi+\theta) =\cot3\theta\,. \] Show that \[ \cosec \tfrac19\pi -\cosec \tfrac59\pi +\cosec \tfrac79\pi = 2\sqrt3\,. \]

The points \(P\), \(Q\) and \(R\) lie on a sphere of unit radius centred at the origin, \(O\), which is fixed. Initially, \(P\) is at \(P_0(1, 0, 0)\), \(Q\) is at \(Q_0(0, 1, 0)\) and \(R\) is at \(R_0(0, 0, 1)\).

1. The sphere is then rotated about the \(z\)-axis, so that the line \(OP\) turns directly towards the positive \(y\)-axis through an angle \(\phi\). The position of \(P\) after this rotation is denoted by \(P_1\). Write down the coordinates of \(P_1\).
2. The sphere is now rotated about the line in the \(x\)-\(y\) plane perpendicular to \(OP_1\), so that the line \(OP\) turns directly towards the positive \(z\)-axis through an angle \(\lambda\). The position of \(P\) after this rotation is denoted by \(P_2\). Find the coordinates of \(P_2\). Find also the coordinates of the points \(Q_2\) and \(R_2\), which are the positions of \(Q\) and \(R\) after the two rotations.
3. The sphere is now rotated for a third time, so that \(P\) returns from \(P_2\) to its original position~\(P_0\). During the rotation, \(P\) remains in the plane containing \(P_0\), \(P_2\) and \(O\). Show that the angle of this rotation, \(\theta\), satisfies \[ \cos\theta = \cos\phi \cos\lambda\,, \] and find a vector in the direction of the axis about which this rotation takes place.

Show that \[ \sin(x+y) -\sin(x-y) = 2 \cos x \, \sin y \] and deduce that \[ \sin A - \sin B = 2 \cos \tfrac12 (A+B) \, \sin\tfrac12 (A-B) \,. \] Show also that \[ \cos A - \cos B = -2 \sin \tfrac12(A+B) \, \sin\tfrac12(A-B)\,. \] The points \(P\), \(Q\), \(R\) and \(S\) have coordinates \(\left(a\cos p,b\sin p\right)\), \(\left(a\cos q,b\sin q\right)\), \(\left(a\cos r,b\sin r\right)\) and \(\left(a\cos s,b\sin s\right)\) respectively, where \(0\le p < q < r < s <2\pi\), and \(a\) and \(b\) are positive. Given that neither of the lines \(PQ\) and \(SR\) is vertical, show that these lines are parallel if and only if \[ r+s-p-q = 2\pi\,. \]

A curve has the equation \(y=\f(x)\), where \[ \f(x) = \cos \Big( 2x+ \frac \pi 3\Big) + \sin \Big ( \frac{3x}2 - \frac \pi 4\Big). \]

1. Find the period of \(\f(x)\).
2. Determine all values of \(x\) in the interval \(-\pi\le x \le \pi\) for which \(\f(x)=0\). Find a value of \(x\) in this interval at which the curve touches the \(x\)-axis without crossing it.
3. Find the value or values of \(x\) in the interval \(0\le x \le 2\pi\) for which \(\f(x)=2\,\).

The point \(P\) has coordinates \((x,y)\) with respect to the origin \(O\). By writing \(x=r\cos\theta\) and \(y=r\sin\theta\), or otherwise, show that, if the line \(OP\) is rotated by \(60^\circ\) clockwise about \(O\), the new \(y\)-coordinate of \(P\) is \(\frac12(y-\sqrt3\,x)\). What is the new \(y\)-coordinate in the case of an anti-clockwise rotation by \(60^\circ\,\)? An equilateral triangle \(OBC\) has vertices at \(O\), \((1,0)\) and \((\frac12,\frac12 \sqrt3)\), respectively. The point \(P\) has coordinates \((x,y)\). The perpendicular distance from \(P\) to the line through \(C\) and \(O\) is \(h_1\); the perpendicular distance from \(P\) to the line through \(O\) and \(B\) is \(h_2\); and the perpendicular distance from \(P\) to the line through \(B\) and \(C\) is \(h_3\). Show that \(h_1=\frac12 \big\vert y-\sqrt3\,x\big\vert\) and find expressions for \(h_2\) and \(h_3\). Show that \(h_1+h_2+h_3=\frac12 \sqrt3\) if and only if \(P\) lies on or in the triangle \(OBC\).

In this question, \(\f^2(x)\) denotes \(\f(\f(x))\), \(\f^3(x)\) denotes \(\f( \f (\f(x)))\,\), and so on.

1. The function \(\f\) is defined, for \(x\ne \pm 1/ \sqrt3\,\), by $$ \f(x) = \ds \frac{x+\sqrt3} {1-\sqrt3\, x }\,. $$ Find by direct calculation \(\f^2(x) \) and \(\f^3(x)\), and determine \(\f^{2007}(x)\,\).
2. Show that \(\f^n(x) = \tan(\theta + \frac 13 n\pi)\), where \(x=\tan\theta\) and \(n\) is any positive integer.
3. The function \(\g(t)\) is defined, for \(\vert t\vert\le1\) by \(\g(t) = \frac {\sqrt3}2 t + \frac 12 \sqrt {1-t^2}\,\). Find an expression for \(\g^n(t)\) for any positive integer \(n\).

Given that \(\cos A\), \(\cos B\) and \(\beta\) are non-zero, show that the equation \[ \alpha \sin(A-B) + \beta \cos(A+B) = \gamma \sin(A+B) \] reduces to the form \[ (\tan A-m)(\tan B-n)=0\,, \] where \(m\) and \(n\) are independent of \(A\) and \(B\), if and only if \(\alpha^2=\beta^2+\gamma^2\). Determine all values of \(x\), in the range \(0\le x <2\pi\), for which:

1. $2\sin(x-\frac14\pi) + \sqrt 3 \cos(x+\frac14\pi) = \sin(x+\frac14\pi)\, \(;
2. \)2\sin(x-\frac16\pi) + \sqrt 3 \cos(x+\frac16\pi) = \sin(x+\frac16\pi)\, \(;
3. \)2\sin(x+\frac13\pi) + \sqrt 3 \cos(3x) = \sin(3x)\, $.

1. Given that \(A = \arctan \frac12\) and that \(B = \arctan\frac13\,\) (where \(A\) and \(B\) are acute) show, by considering \(\tan \left( A + B \right)\), that \(A + B = {\frac{1}{4}\pi }\). The non-zero integers \(p\) and \(q\) satisfy \[ \displaystyle \arctan {\frac1 p} + \arctan {\frac1 q} = {\frac\pi 4}\,. \] Show that \( \left ( p-1 \right) \left(q-1 \right) = 2\) and hence determine \(p\) and \(q\).
2. Let \(r\), \(s\) and \(t\) be positive integers such that the highest common factor of \(s\) and \(t\) is \(1\). Show that, if \[ \arctan {\frac1 r} + \arctan \frac s {s+t} = {\frac\pi 4}\,, \] then there are only two possible values for \(t\), and give \(r\) in terms of \(s\) in each case.

Show that \(\sin A = \cos B\) if and only if \(A = (4n+1)\frac{\pi}{2}\pm B\) for some integer \(n\). Show also that \(\big\vert\sin x \pm \cos x \big\vert \le \sqrt{2}\) for all values of \(x\) and deduce that there are no solutions to the equation \(\sin\left( \sin x \right) = \cos \left( \cos x \right)\). Sketch, on the same axes, the graphs of \(y= \sin \left( \sin x \right)\) and \(y = \cos \left( \cos x \right)\). Sketch, not on the previous axes, the graph of \(y= \sin \left(2 \sin x \right)\).

Show Solution

\begin{align*} && \sin A &= \cos B \\ \Leftrightarrow && 0 &= \sin A - \cos B \\ &&&= \sin A - \sin ( \frac{\pi}{2} - B) \\ &&&= 2 \sin \left ( \frac{A + B - \frac{\pi}{2}}{2} \right) \cos \left (\frac{A - B + \frac\pi2}{2} \right) \\ \Leftrightarrow && n \pi &= \frac{A+B - \frac{\pi}{2}}{2}, n\pi + \frac{\pi}{2} = \frac{A-B+\frac{\pi}{2}}{2} \\ \Leftrightarrow && A \pm B &= 2n\pi + \frac{\pi}{2} \\ &&&= (4n+1) \frac{\pi}{2} \end{align*} \begin{align*} |\sin x \pm \cos x| &= | \sqrt{2} \sin(x \pm \frac{\pi}{4} )| \\ & \leq \sqrt{2} \end{align*} Therefore if \(\sin(\sin x) = \cos (\cos x)\) we must have that \(|\sin x \pm \cos x| = |(4n+1) \frac{\pi}{2}| \geq \frac{\pi}{2} > 1.5 > \sqrt{2}\) contradiction.

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The positive numbers \(a\), \(b\) and \(c\) satisfy \(bc=a^2+1\). Prove that $$ \arctan\left(\frac1 {a+b}\right)+ \arctan\left(\frac1 {a+c}\right)= \arctan\left(\frac1 a \right). $$ The positive numbers \(p\), \(q\), \(r\), \(s\), \(t\), \(u\) and \(v\) satisfy $$ st = (p+q)^2 + 1 \;, \ \ \ \ \ \ uv=(p+r)^2 + 1 \;, \ \ \ \ \ \ qr = p^2+1\;. $$ Prove that $$ \arctan \! \!\left(\!\frac1 {p+q+s}\!\right) + \arctan \! \!\left(\!\frac 1{p+q+t}\!\right) + \arctan \! \!\left(\!\frac 1 {p+r+u}\!\right) + \arctan \! \!\left(\!\frac1 {p+r+v}\!\right) =\arctan \! \!\left( \! \frac1 p \! \right) . $$ Hence show that $$ \arctan\left(\frac1 {13}\right) +\arctan\left(\frac1 {21}\right) +\arctan\left(\frac1 {82}\right) +\arctan\left(\frac1 {187}\right) =\arctan\left(\frac1 {7}\right). $$ [\,Note that \(\arctan x\) is another notation for \( \tan^{-1}x \,.\,\)]

The notation \(\displaystyle \prod^n_{r=1} \f (r)\) denotes the product $\f (1) \times \f (2) \times \f(3) \times \cdots \times \f(n)$. %For example, \(\displaystyle \prod_{r=1}^4 r = 24\). %Simplify \(\displaystyle \prod^n_{r=1} \frac{\g (r) }{ \g (r-1) }\). %You may assume that \(\g (r) \neq 0\) for any integer \(0 \le r \le n \). Simplify the following products as far as possible:

1. \(\displaystyle \prod^n_{r=1} \l \frac{r+ 1 }{ r } \r\,\);
2. \(\displaystyle \prod^n_{r=2} \l \frac{r^2 -1}{r^2 } \r\,\);
3. $\displaystyle \prod^n_{r=1} \l {\cos \frac{2\pi }{ n} + \sin \frac{2\pi}{ n} \cot \frac{\l 2r-1 \r \pi }{ n} }\r\,$, where \(n\) is even.

1. Given that \(\displaystyle \cos \theta = \frac35\) and that \(\displaystyle \frac{3\pi }{ 2} \le \theta \le 2\pi\), show that \(\displaystyle \sin 2 \theta = -\frac{24}{25}\), and evaluate \(\cos 3 \theta\).
2. Prove the identity \(\displaystyle \tan 3\theta \equiv \frac {3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}\). Hence evaluate \(\tan \theta\), given that \(\displaystyle \tan 3\theta = \frac{11}{ 2}\) and that \(\displaystyle \frac{\pi}{ 4} \le \theta \le \frac{\pi}{2}\).

The curve \(C\) has equation \[ y= a^{\sin (\pi \e^ x)}\,, \] where \(a>1\).

1. Find the coordinates of the stationary points on \(C\).
2. Use the approximations \(\e^t \approx 1+t\) and \(\sin t \approx t\) (both valid for small values of \(t\)) to show that \[ y\approx 1-\pi x \ln a \; \] for small values of \(x\).
3. Sketch \(C\).
4. By approximating \(C\) by means of straight lines joining consecutive stationary points, show that the area between \(C\) and the \(x\)-axis between the \(k\)th and \((k+1)\)th maxima is approximately \[ \Big( \frac {a^2+1}{2a} \Big) \ln \Big ( 1+ \big( k-\tfrac34)^{-1} \Big)\,. \]

The lengths of the sides \(BC\), \(CA\), \(AB\) of the triangle \(ABC\) are denoted by \(a\), \(b\), \(c\), respectively. Given that $$ b = 8+{\epsilon}_1, \, c=3+{\epsilon}_2,\, A=\tfrac{1}{3}\pi + {\epsilon}_3, $$ where \({\epsilon}_1\), \({\epsilon}_2\), and \( {\epsilon}_3\) are small, show that \(a \approx 7 + {\eta}\), where ${\eta}= {\left(13 \, {{\epsilon}_1}-2\,{\epsilon}_2 + 24{\sqrt 3} \;{{\epsilon}_3}\right)}/14$. Given now that $$ {\vert {\epsilon}_1} \vert \le 2 \times 10^{-3}, \ \ \ {\vert {\epsilon}_2} \vert \le 4\cdot 9\times 10^{-2}, \ \ \ {\vert {\epsilon}_3} \vert \le \sqrt3 \times 10^{-3}, $$ find the range of possible values of \({\eta}\).

For this question, you may use the following approximations, valid if \(\theta \) is small: \ \(\sin\theta \approx \theta\) and \(\cos\theta \approx 1-\theta^2/2\,\). A satellite \(X\) is directly above the point \(Y\) on the Earth's surface and can just be seen (on the horizon) from another point \(Z\) on the Earth's surface. The radius of the Earth is \(R\) and the height of the satellite above the Earth is \(h\).

1. Find the distance \(d\) of \(Z\) from \(Y\) along the Earth's surface.
2. If the satellite is in low orbit (so that \(h\) is small compared with \(R\)), show that $$d \approx k(Rh)^{1/2},$$ where \(k\) is to be found.
3. If the satellite is very distant from the Earth (so that \(R\) is small compared with \(h\)), show that $$d\approx aR+b(R^2/h),$$ where \(a\) and \(b\) are to be found.

Find the limit, as \(n\rightarrow\infty,\) of each of the following. You should explain your reasoning briefly. \begin{alignat*}{4} \mathbf{(i)\ \ } & \dfrac{n}{n+1}, & \qquad & \mathbf{(ii)\ \ } & \dfrac{5n+1}{n^{2}-3n+4}, & \qquad & \mathbf{(iii)\ \ } & \dfrac{\sin n}{n},\\ \\ \mathbf{(iv)\ \ } & \dfrac{\sin(1/n)}{(1/n)}, & & \mathbf{(v)}\ \ & (\arctan n)^{-1}, & & \mathbf{(vi)\ \ } & \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}. \end{alignat*}

Explain briefly, by means of a diagram, or otherwise, why \[ \mathrm{f}(\theta+\delta\theta)\approx\mathrm{f}(\theta)+\mathrm{f}'(\theta)\delta\theta, \] when \(\delta\theta\) is small. Two powerful telescopes are placed at points \(A\) and \(B\) which are a distance \(a\) apart. A very distant point \(C\) is such that \(AC\) makes an angle \(\theta\) with \(AB\) and \(BC\) makes an angle \(\theta+\phi\) with \(AB\) produced. (A sketch of the arrangement is given in the diagram.) \noindent

By considering the graphs of \(y=kx\) and \(y=\sin x,\) show that the equation \(kx=\sin x,\) where \(k>0,\) may have \(0,1,2\) or \(3\) roots in the interval \((4n+1)\frac{\pi}{2} < x < (4n+5)\frac{\pi}{2},\) where \(n\) is a positive integer. For a certain given value of \(n\), the equation has exactly one root in this interval. Show that \(k\) lies in an interval which may be written \(\sin\delta < k < \dfrac{2}{(4n+1)\pi},\) where \(0 < \delta < \frac{1}{2}\pi\) and \[ \cos\delta=\left((4n+5)\frac{\pi}{2}-\delta\right)\sin\delta. \] Show that, if \(n\) is large, then \(\delta\approx\dfrac{2}{(4n+5)\pi}\) and obtain a second, improved, approximation.

Show Solution

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Clearly we can achieve \(0\), \(1\), and \(2\) intersections by never entering the range, entering too flat, or entering before hitting the second branch. To achieve \(3\) we can go at a flat enough slope that we hit somewhere near the top of the second branch, and since the gradient there will be \(\approx 0\), and our gradient is positive, we must intersect before that point as well, ie \(3\) intersections. Clearly we cannot intersect the second branch \(3\) times or the first branch twice, therefore there are at most \(3\) intersections. To intersect the graph only once, we need to:

• be below \(((4n+1)\tfrac{\pi}{2}, 1)\) and
• not touch the second gradient

The first condition means that \(k (4n+1)\tfrac{\pi}{2} < 1 \Rightarrow k < \frac{2}{(4n+1)\pi}\). For the second condition, consider a point on the curve \(\sin x\) whose tangent line goes through the origin, ie \(\frac{y - \sin t}{x - t} = \cos t \Rightarrow y = (\cos t)x - t \cos t+\sin t\) ie \(\sin t = t \cos t\). For this point \(t\) to be in the required interval, we need \((4n+5) \tfrac{\pi}{2} -t \in (0, \frac{\pi}{2})\), so let's call this value \(\delta\). Then our result is: The gradient needs to be steeper than \(\cos t = \cos \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) = \sin \delta\) and \(\cos \delta =\left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \sin \delta \). If \(n\) is large, then, \begin{align*} && 1 &\approx \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \delta \\ \Rightarrow && 1 &\approx (4n+5) \tfrac{\pi}{2} \delta \\ \Rightarrow && \delta &\approx \frac{2}{(4n+5)\pi} \end{align*}. To higher order: \begin{align*} && 1-\frac12 \delta^2 &\approx \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \delta \\ \Rightarrow && 1-\frac12 \delta^2 &\approx (4n+5) \tfrac{\pi}{2} \delta - \delta^2 \\ \Rightarrow && 0 &\approx 1 - (4n + 5)\tfrac{\pi}{2} \delta + \frac12 \delta^2 \\ \Rightarrow && \delta &\approx (4n+5) \tfrac{\pi}{2} - \sqrt{(4n+5)^2 \frac{\pi^2}{4} - 2} \\ &&&= \frac{2}{(4n+5) \tfrac{\pi}{2} + \sqrt{(4n+5)^2 \frac{\pi^2}{4} - 2}} \end{align*}.

Show that \[ \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)=\frac{\sin\alpha}{4\sin\left(\dfrac{\alpha}{4}\right)}\,, \] where \(\alpha\neq k\pi\), \(k\) is an integer. Prove that, for such \(\alpha\), \[ \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)=\frac{\sin\alpha}{2^{n}\sin\left(\dfrac{\alpha}{2^{n}}\right)}\,, \] where \(n\) is a positive integer. Deduce that \[ \alpha=\frac{\sin\alpha}{\cos\left(\dfrac{\alpha}{2}\right)\cos\left(\dfrac{\alpha}{4}\right)\cos\left(\dfrac{\alpha}{8}\right)\cdots}\,, \] and hence that \[ \frac{\pi}{2}=\frac{1}{\sqrt{\frac{1}{2}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}\cdots}\,. \]

1. The circle \(x^2 + (y-a)^2 = r^2\) touches the parabola \(2ky = x^2\), where \(k > 0\), tangentially at two points. Show that \(r^2 = k(2a - k)\). Show further that if \(r^2 = k(2a - k)\) and \(a > k > 0\), then the circle \(x^2 + (y-a)^2 = r^2\) touches the parabola \(2ky = x^2\) tangentially at two points.
2. The lines \(y = c \pm x\) are tangents to the circle \(x^2 + (y-a)^2 = r^2\). Find \(r^2\), and the coordinates of the points of contact, in terms of \(a\) and \(c\).
3. \(C_1\) and \(C_2\) are circles with equations \(x^2 + (y-a_1)^2 = r_1^2\) and \(x^2 + (y-a_2)^2 = r_2^2\) respectively, where \(a_1 \neq a_2\) and \(r_1 \neq r_2\). Each circle touches the parabola \(2ky = x^2\) tangentially at two points and the lines \(y = c \pm x\) are tangents to both circles.
   1. Show that \(a_1 + a_2 = 2c + 4k\) and that \(a_1^2 + a_2^2 = 2c^2 + 16kc + 12k^2\).
   2. The circle \(x^2 + (y-d)^2 = p^2\) passes through the four points of tangency of the lines \(y = c \pm x\) to the two circles, \(C_1\) and \(C_2\). Find \(d\) and \(p^2\) in terms of \(k\) and \(c\).
   3. Show that the circle \(x^2 + (y-d)^2 = p^2\) also touches the parabola \(2ky = x^2\) tangentially at two points.

1. Sketch a graph of \(y = \frac{\ln x}{x}\) for \(x > 0\).
2. Use your graph to show the following.
   1. \(3^{\pi} > \pi^3\)
   2. \(\left(\frac{9}{4}\right)^{\sqrt{5}} > \sqrt{5}^{\frac{9}{4}}\)
3. Given that \(1 < x < 2\), decide, with justification, which is the larger of \(x^{x+2}\) or \((x+2)^x\).
4. Show that the inequalities \(9^{\sqrt{2}} > \sqrt{2}^9\) and \(3^{2\sqrt{2}} > (2\sqrt{2})^3\) are equivalent. Given that \(e^2 < 8\), decide, with justification, which is the larger of \(9^{\sqrt{2}}\) and \(\sqrt{2}^9\).
5. Decide, with justification, which is the larger of \(8^{\sqrt[4]{3}}\) and \(\sqrt[3]{8}\).

The function f satisfies \(f(0) = 0\) and \(f'(t) > 0\) for \(t > 0\). Show by means of a sketch that, for \(x > 0\), $$\int_0^x f(t) \, dt + \int_0^{f(x)} f^{-1}(y) \, dy = xf(x).$$

1. The (real) function g is defined, for all \(t\), by $$(g(t))^3 + g(t) = t.$$ Prove that \(g(0) = 0\), and that \(g'(t) > 0\) for all \(t\). Evaluate \(\int_0^2 g(t) \, dt\).
2. The (real) function h is defined, for all \(t\), by $$(h(t))^3 + h(t) = t + 2.$$ Evaluate \(\int_0^8 h(t) \, dt\).

Show Solution

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Notice the total area is \(xf(x)\) and it is made up of the sum of the two integrals.

1. Suppose \((g(t))^3 + g(t) = t\). Notice that \((g(0))^3 + g(0) =0 \Rightarrow g(0)((g(0))^2 + 1) = 0 \Rightarrow g(0) = 0\). \begin{align*} && t &= (g(t))^3 + g(t) \\ \Rightarrow && 1 &= 3(g(t))^2 g'(t) + g'(t) \\ \Rightarrow && g'(t) &= \frac{1}{1 + 3(g(t))^2} > 0 \end{align*}
   

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   From our sketch, we can see we are interested in: \begin{align*} && \int_0^2 g(t) \d t &= 2 - \int_0^1 (x^3 + x) \d x \\ &&&= 2 - \frac14 - \frac12 = \frac54 \end{align*}
2. \(\,\)
   

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   From our second sketch, we can see that: \begin{align*} && \int_0^8 h(t) \d t &= 16 - \int_1^2 (x^3+x-2) \d x \\ &&&= 16 - \left ( \frac{8}{4} + \frac{2^2}{2} - 2 \cdot 2 \right)+ \left ( \frac{1}{4} + \frac{1}{2} - 2 \right) \\ &&&= \frac{59}{4} \end{align*}

A straight line passes through the fixed point \((1 , k)\) and has gradient \(- \tan \theta\), where \(k > 0\) and \(0 < \theta < \frac{1}{2}\pi\). Find, in terms of \(\theta\) and \(k\), the coordinates of the points \(X\) and \(Y\) where the line meets the \(x\)-axis and the \(y\)-axis respectively.

1. Find an expression for the area \(A\) of triangle \(OXY\) in terms of \(k\) and \(\theta\). (The point \(O\) is the origin.) You are given that, as \(\theta\) varies, \(A\) has a minimum value. Find an expression in terms of \(k\) for this minimum value.
2. Show that the length \(L\) of the perimeter of triangle \(OXY\) is given by $$L = 1 + \tan \theta + \sec \theta + k(1 + \cot \theta + \cosec \theta).$$ You are given that, as \(\theta\) varies, \(L\) has a minimum value. Show that this minimum value occurs when \(\theta = \alpha\) where $$\frac{1 - \cos \alpha}{1 - \sin \alpha} = k.$$ Find and simplify an expression for the minimum value of \(L\) in terms of \(\alpha\).

A woman stands in a field at a distance of \(a\,\mathrm{m}\) from the straight bank of a river which flows with negligible speed. She sees her frightened child clinging to a tree stump standing in the river \(b\,\mathrm{m}\) downstream from where she stands and \(c\,\mathrm{m}\) from the bank. She runs at a speed of \(u\,\mathrm{ms}^{-1}\) and swims at \(v\,\mathrm{ms}^{-1}\) in straight lines. Find an equation to be satisfied by \(x,\) where \(x\,\mathrm{m}\) is the distance upstream from the stump at which she should enter the river if she is to reach the child in the shortest possible time. Suppose now that the river flows with speed \(v\) ms\(^{-1}\) and the stump remains fixed. Show that, in this case, \(x\) must satisfy the equation \[ 2vx^{2}(b-x)=u(x^{2}-c^{2})[a^{2}+(b-x)^{2}]^{\frac{1}{2}}. \] For this second case, draw sketches of the woman's path for the three possibilities \(b>c,\) \(b=c\) and \(b< c\).

Show Solution

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The distance to where she enters the water is \(\sqrt{a^2+(b-x)^2}\) and the distance through the water is \(\sqrt{x^2+c^2}\). The total time will be \(\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{\sqrt{x^2+c^2}}{v}\). To minimise this, we can differentiate. \begin{align*} \frac{\d}{\d x}: && \frac{-(b-x)}{u\sqrt{a^2+(b-x)^2}} + \frac{x}{v \sqrt{x^2+c^2}} &= 0 \\ \Rightarrow && v(b-x)(x^2+c^2)^{\frac12} &= xu(a^2+(b-x)^2)^{\frac12} \end{align*} When she is in the water, she can will move with velocity \(\begin{pmatrix} v \cos \theta \\ v \sin \theta -v \end{pmatrix}\). She needs to travel a distance \(\begin{pmatrix} c \\ -x \end{pmatrix}\), so we must have that \begin{align*} && \frac{x}{c} &= \frac{1-\sin \theta}{\cos \theta} \\ \Rightarrow && \sec \theta - \tan \theta &= \frac{x}{c} \\ \Rightarrow && \sec \theta &= \tan \theta + \frac{x}{c} \\ \Rightarrow && \sec^2 \theta &= \tan^2 \theta + 2 \tan \theta \frac{x}{c} + \frac{x^2}{c^2} \\ \Rightarrow && 1 + \tan^2 \theta &= \tan^2 \theta + 2 \tan \theta \frac{x}{c} + \frac{x^2}{c^2} \\ \Rightarrow && \tan \theta &=\frac{c^2-x^2}{2xc} \\ \Rightarrow && \sin \theta &= \frac{c^2-x^2}{c^2+x^2} \\ && \cos \theta &= \frac{2xc}{c^2+x^2} \\ \end{align*} (where we have taken the positive value for \(\cos \theta\) since we must be heading towards the child). Since \(v \cos \theta t = c\) the time taken to reach the child in the water is \(\frac{c}{v} \frac{c^2+x^2}{2xc} = \frac{c^2+x^2}{2xv}\). So the total time is: \(\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{c^2+x^2}{2xv}\). To minimise this, we can differentiate. \begin{align*} \frac{\d}{\d x}: && \frac{-(b-x)}{u\sqrt{a^2+(b-x)^2}} -\frac{c^2}{2vx^2} + \frac{x^2}{2vx^2}&= 0 \\ \Rightarrow && u(x^2-c^2)\sqrt{a^2+(b-x)^2}&= 2vx^2(b-x) \end{align*} as required. When \(b = c\), the shortest path will be running directly to the bank (there's no quicker way to get to the bank) then swimming directly out (and letting the current take you downstream exactly as far as you need)). Therefore the path will be:

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If \(b > c\) then she should run a little downstream first.

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and if \(c > b\) she should actually run a little upstream to take advantage of the current:

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Find the stationary points of the function \(\mathrm{f}\) given by \[ \mathrm{f}(x)=\mathrm{e}^{ax}\cos bx,\mbox{ }(a>0,b>0). \] Show that the values of \(\mathrm{f}\) at the stationary points with \(x>0\) form a geometric progression with common ratio \(-\mathrm{e}^{a\pi/b}\). Give a rough sketch of the graph of \(\mathrm{f}\).

Show Solution

Let \(f(x) = e^{ax} \cos bx\) then, \(f'(x) = ae^{ax} \cos bx - be^{ax} \sin bx = e^{ax} \l a\cos bx - b \sin bx \r\). Therefore the stationary points are where \(f'(x) = 0 \Leftrightarrow \tan bx = \frac{b}a\), ie \(x = \tan^{-1} \frac{a}{b} + \frac{n}{b} \pi, n \in \mathbb{Z}\). \begin{align*} f(\tan^{-1} \frac{a}{b} + \frac{n}{b} \pi) &= e^{a \tan^{-1} \frac{a}{b} + \frac{an}{b} \pi} \cos \l b \tan^{-1} \frac{a}{b} +n \pi\r \\ &= e^{a \tan^{-1} \frac{a}{b}} \cos \l b \tan^{-1} \frac{a}{b}\r \cdot e^{\frac{an}{b} \pi}(-1)^n \\ &= e^{a \tan^{-1} \frac{a}{b}} \cos \l b \tan^{-1} \frac{a}{b}\r \cdot (-e^{\frac{a}{b} \pi})^n \\ \end{align*} showing the form the desired geometric progression.

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The sequence of functions \(y_0\), \(y_1\), \(y_2\), \(\ldots\,\) is defined by \(y_0=1\) and, for \(n\ge1\,\), \[ y_n = (-1)^n \frac {1}{z} \, \frac{\d^{n} z}{\d x^n} \,, \] where \(z= \e^{-x^2}\!\).

1. Show that \(\dfrac{\d y_n}{\hspace{-4.7pt}\d x} = 2x y_n -y_{n+1}\,\) for \(n\ge1\,\).
2. Prove by induction that, for \(n\ge1\,\), \[ y_{n+1} = 2x y_n -2ny_{n-1} \,. \] Deduce that, for \(n\ge1\,\), \[ y_{n+1}^2 - {y}_n {y}_{n+2} = 2n (y_n^2 - y_{n-1}y_{n+1}) + 2 y_n^2 \,. \]
3. Hence show that $y_{n}^2 - y^2_{n-1} y^2_{n+1} > 0\( for \)n \ge 1$.

%In this question, %the definition of \(a^b\) (for \(a>0\)) is %$ %a^b = \e^{b \ln a} \,. %$ %\\ The functions \(\f\) and \(\g\) are defined, for \(x>0\), by \[ \f(x) = x^x\,, \ \ \ \ \ \g(x) = x^{\f(x)}\,. \]

1. By taking logarithms, or otherwise, show that \(\f(x)> x\) for \(01 \,\).
2. Find the value of \(x\) for which \(\f'(x)=0\,\).
3. Use the result \(x\ln x \to 0\) as \(x\to 0\) to find \(\lim\limits_{x\to0}\f(x)\), and write down \(\lim\limits_{x\to0}\g(x)\,\).
4. Show that \( x^{-1}+\, \ln x \ge 1\,\) for \(x>0\). \\[5pt] Using this result, or otherwise, show that~\(\g'(x) >0\,\).

A circle of radius \(a\) is centred at the origin \(O\). A rectangle \(PQRS\) lies in the minor sector \(OMN\) of this circle where \(M\) is \((a,0)\) and \(N\) is \((a \cos \beta, a \sin \beta)\), and \(\beta\) is a constant with \(0 < \beta < \frac{\pi}{2}\,\). Vertex \(P\) lies on the positive \(x\)-axis at \((x,0)\); vertex \(Q\) lies on \(ON\); vertex \(R\) lies on the arc of the circle between \(M\) and \(N\); and vertex \(S\) lies on the positive \(x\)-axis at \((s,0)\). Show that the area \(A\) of the rectangle can be written in the form \[ A= x(s-x)\tan\beta \,. \] Obtain an expression for \(s\) in terms of \(a\), \(x\) and \(\beta\), and use it to show that \[ \frac{\d A}{\d x} = (s-2x) \tan \beta - \frac {x^2} s \tan^3\beta \,. \] Deduce that the greatest possible area of rectangle \(PQRS\) occurs when \(s= x(1+\sec\beta)\) and show that this greatest area is \(\tfrac12 a^2 \tan \frac12 \beta\,\). Show also that this greatest area occurs when \(\angle ROS = \frac12\beta\,\).

For each non-negative integer \(n\), the polynomial \(\f_n\) is defined by \[ \f_n(x) = 1 + x + \frac{x^2}{2!} + \frac {x^3}{3!} + \cdots + \frac{x^n}{n!} \]

1. Show that \(\f'_{n}(x) = \f_{n-1}(x)\,\) (for \(n\ge1\)).
2. Show that, if \(a\) is a real root of the equation \[\f_n(x)=0\,,\tag{\(*\)}\] then \(a<0\).
3. Let \(a\) and \(b\) be distinct real roots of \((*)\), for \(n\ge2\). Show that \(\f_n'(a)\, \f_n'(b)>0\,\) and use a sketch to deduce that \(\f_n(c)=0\) for some number \(c\) between \(a\) and \(b\). Deduce that \((*)\) has at most one real root. How many real roots does \((*)\) have if \(n\) is odd? How many real roots does \((*)\) have if \(n\) is even?

Differentiate, with respect to \(x\), \[ (ax^2+bx+c)\,\ln \big( x+\sqrt{1+x^2}\big) +\big(dx+e\big)\sqrt{1+x^2} \,, \] where \(a\), \(b\), \(c\), \(d\) and \(e\) are constants. You should simplify your answer as far as possible. Hence integrate:

1. \( \ln \big( x+\sqrt{1+x^2}\,\big) \,;\)
2. \(\sqrt{1+x^2} \,; \)
3. \( x\ln \big( x+\sqrt{1+x^2}\,\big) \,.\)

By simplifying \(\sin(r+\frac12)x - \sin(r-\frac12)x\) or otherwise show that, for \(\sin\frac12 x \ne0\), \[ \cos x + \cos 2x +\cdots + \cos nx = \frac{\sin(n+\frac12)x - \sin\frac12 x}{2\sin\frac12x}\,. \] The functions \(\.S_n\), for \(n=1\), \(2\), \dots, are defined by \[ \.S_n(x) = \sum_{r=1}^n \frac 1 r \sin rx \qquad (0\le x \le \pi). \]

1. Find the stationary points of \(\.S_2(x)\) for \(0\le x\le\pi\), and sketch this function.
2. Show that if \(\.S_n(x)\) has a stationary point at \(x=x_0\), where \(0< x_0 < \pi\), then \[ \sin nx_0 = (1-\cos nx_0) \tan\tfrac12 x_0 \] and hence that \(\.S_n(x_0) \ge \.S_{n-1}(x_0)\). Deduce that if \(\.S_{n-1}(x)>0\) for all \(x\) in the interval \(00\) for all \(x\) in this interval.
3. Prove that \(\.S_n(x)\ge0\) for \(n\ge1\) and \(0\le x\le\pi\).

An accurate clock has an hour hand of length \(a\) and a minute hand of length \(b\) (where \(b>a\)), both measured from the pivot at the centre of the clock face. Let \(x\) be the distance between the ends of the hands when the angle between the hands is \(\theta\), where \(0\le\theta < \pi\). Show that the rate of increase of \(x\) is greatest when \(x=(b^2-a^2)^\frac12\). In the case when \(b=2a\) and the clock starts at mid-day (with both hands pointing vertically upwards), show that this occurs for the first time a little less than 11 minutes later.

1. Find the value of \(m\) for which the line \(y = mx\) touches the curve \(y = \ln x\,\). If instead the line intersects the curve when \(x = a\) and \(x = b\), where \(a < b\), show that \(a^b = b^a\). Show by means of a sketch that \(a < \e < b\).
2. The line \(y=mx+c\), where \(c>0\), intersects the curve \(y=\ln x\) when \(x=p\) and \(x=q\), where \(p < q\). Show by means of a sketch, or otherwise, that \(p^q > q^p\).
3. Show by means of a sketch that the straight line through the points \((p, \ln p)\) and \((q, \ln q)\), where \(\e\le p < q\,\), intersects the \(y\)-axis at a positive value of \(y\). Which is greater, \(\pi^\e\) or \(\e^\pi\)?
4. Show, using a sketch or otherwise, that if \(0 < p < q\) and \(\dfrac{\ln q - \ln p}{q-p} = \e^{-1}\), then \(q^p > p^q\).

1. Sketch the curve \(y=\f(x)\), where \[ \f(x) = \frac 1 {(x-a)^2 -1} \hspace{2cm}(x\ne a\pm1), \] and \(a\) is a constant.
2. The function \(\g(x)\) is defined by \[ \g(x) = \frac 1 {\big( (x-a)^2-1 \big) \big( (x-b)^2 -1\big)} \hspace{1cm}(x\ne a\pm1, \ x\ne b\pm1), \] where \(a\) and \(b\) are constants, and \(b>a\). Sketch the curves \(y=\g(x)\) in the two cases \(b>a+2\) and \(b=a+2\), finding the values of \(x\) at the stationary points.

The line \(L\) has equation \(y=c-mx\), with \(m>0\) and \(c>0\). It passes through the point \(R(a,b)\) and cuts the axes at the points \(P(p,0)\) and \(Q(0,q)\), where \(a\), \(b\), \(p\) and \(q\) are all positive. Find \(p\) and \(q\) in terms of \(a\), \(b\) and \(m\). As \(L\) varies with \(R\) remaining fixed, show that the minimum value of the sum of the distances of \(P\) and \(Q\) from the origin is \((a^{\frac12} + b^{\frac12})^2\), and find in a similar form the minimum distance between \(P\) and \(Q\). (You may assume that any stationary values of these distances are minima.)

In this question, you may assume without proof that any function \(\f\) for which \(\f'(x)\ge 0\) is increasing; that is, \(\f(x_2)\ge \f(x_1)\) if \(x_2\ge x_1\,\).

1. 1. Let \(\f(x) =\sin x -x\cos x\). Show that \(\f(x)\) is increasing for \(0\le x \le \frac12\pi\,\) and deduce that \(\f(x)\ge 0\,\) for \(0\le x \le \frac12\pi\,\).
   2. Given that \(\dfrac{\d}{\d x} (\arcsin x) \ge1\) for \(0\le x< 1\), show that \[ \arcsin x\ge x \quad (0\le x < 1). \]
   3. Let \(\g(x)= x\cosec x\, \text{ for }0< x < \frac12\pi\). Show that \(\g\) is increasing and deduce that \[ ({\arcsin x})\, x^{-1} \ge x\,{\cosec x} \quad (0 < x < 1). \]
2. Given that $\dfrac{\d}{\d x} (\arctan x)\le 1\text{ for }x\ge 0$, show by considering the function \(x^{-1} \tan x\) that \[ (\tan x)( \arctan x) \ge x^2 \quad (0< x < \tfrac12\pi). \]

The number \(E\) is defined by $\displaystyle E= \int_0^1 \frac{\e^x}{1+x} \, \d x\,.$ Show that \[ \int_0^1 \frac{x \e^x}{1+x} \, \d x = \e -1 -E\, ,\] and evaluate \(\ds \int_0^1 \frac{x^2\e^x}{1+x} \, \d x\) in terms of \(\e\) and \(E\). Evaluate also, in terms of \(E\) and \(\rm e\) as appropriate:

1. \[ \int_0^1 \frac{\e^{\frac{1-x}{1+x}}}{1+x}\, \d x\,\]
2. \[ \int_1^{\sqrt2} \frac {\e^{x^2}}x \, \d x \, \]

Let \(P\) be a given point on a given curve \(C\). The {\em osculating circle} to \(C\) at \(P\) is defined to be the circle that satisfies the following two conditions at \(P\): it touches \(C\); and the rate of change of its gradient is equal to the rate of change of the gradient of \(C\). Find the centre and radius of the osculating circle to the curve \(y=1-x+\tan x\) at the point on the curve with \(x\)-coordinate \(\frac14 \pi\).

The curve \(\displaystyle y=\Bigl(\frac{x-a}{x-b}\Bigr)\e^{x}\), where \(a\) and \(b\) are constants, has two stationary points. Show that \[ a-b<0 \ \ \ \text{or} \ \ \ a-b>4 \,. \]

1. [(i)] Show that, in the case \(a=0\) and \(b= \frac12\), there is one stationary point on either side of the curve's vertical asymptote, and sketch the curve.
2. [(ii)] Sketch the curve in the case \( a=\tfrac{9}{2}\) and \(b=0\,\).

1. The gradient \(y'\) of a curve at a point \((x,y)\) satisfies \[ (y')^2 -xy'+y=0\,. \tag{\(*\)} \] By differentiating \((*)\) with respect to \(x\), show that either \(y''=0\) or \(2y'=x\,\). Hence show that the curve is either a straight line of the form \(y=mx+c\), where \(c=-m^2\), or the parabola \(4y=x^2\).
2. The gradient \(y'\) of a curve at a point \((x,y)\) satisfies \[ (x^2-1)(y')^2 -2xyy'+y^2-1=0\,. \] Show that the curve is either a straight line, the form of which you should specify, or a circle, the equation of which you should determine.

A function \(\f(x)\) is said to be convex in the interval \(a < x < b\) if \(\f''(x)\ge0\) for all \(x\) in this interval.

1. Sketch on the same axes the graphs of \(y= \frac23 \cos^2 x\) and \(y=\sin x\) in the interval \(0\le x \le 2\pi\). The function \(\f(x)\) is defined for \(0 < x < 2\pi\) by \[\f(x) = \e^{\frac23 \sin x}. \] Determine the intervals in which \(\f(x)\) is convex.
2. The function \(\g(x)\) is defined for \(0 < x < \frac12\pi\) by \[\g(x) = \e^{-k \tan x}. \] If \(k=\sin 2 \alpha\) and \(0 < \alpha < \frac{1}{4}\pi\), show that \(\g(x)\) is convex in the interval \(0 < x < \alpha\), and give one other interval in which \(\g(x)\) is convex.

Using the series \[ \e^x = 1 + x +\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}+\cdots\,, \] show that \(\e>\frac83\). Show that \(n!>2^n\) for \(n\ge4\) and hence show that \(\e<\frac {67}{24}\). Show that the curve with equation \[ y= 3\e^{2x} +14 \ln (\tfrac43-x)\,, \qquad {x<\tfrac43} \] has a minimum turning point between \(x=\frac12\) and \(x=1\) and give a sketch to show the shape of the curve.

Show Solution

\begin{align*} && e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\ &&&> 1 + 1+ \frac12 + \frac16 \\ &&&= \frac{12+3+1}{6} = \frac83 \end{align*} \(4! = 24 > 16 = 2^4\), notice that \(n! = \underbrace{n \cdot (n-1) \cdots 5}_{>2^{n-4}} \cdot \underbrace{4!}_{>2^4} >2^n\). \begin{align*} && e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\ &&&< \frac83 + \frac{1}{2^4} + \frac{1}{2^5} + \cdots \\ &&&= \frac83 + \frac{1}{2^4} \frac{1}{1-\tfrac12} \\ &&&= \frac83 + \frac1{8} \\ &&&= \frac{67}{24} \end{align*} \begin{align*} && y &= 3e^{2x} +14 \ln(\tfrac43-x) \\ && y' &= 6e^{2x} - \frac{14}{\tfrac43-x} \\ && y'(\tfrac12) &= 6e - \frac{14}{\tfrac43-\tfrac12} \\ &&&= 6e -\tfrac{84}{5} = 6(e-\tfrac{14}5) < 0 \\ && y'(1) &= 6e^2 - \frac{14}{\tfrac43-1} \\ &&&= 6e^2 - 42 = 6(e^2-7) \\ &&&> 6(\tfrac{64}{9} - 7) > 0 \end{align*} Therefore \(y'\) changes from negative (decreasing) to positive (increasing) in our range, and therefore there is a minima in this range.

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By sketching on the same axes the graphs of \(y=\sin x\) and \(y=x\), show that, for \(x>0\):

1. \(x>\sin x\,\);
2. \(\dfrac {\sin x} {x} \approx 1\) for small \(x\).

Find the three values of \(x\) for which the derivative of \(x^2 \e^{-x^2}\) is zero. Given that \(a\) and \(b\) are distinct positive numbers, find a polynomial \(\P(x)\) such that the derivative of \(\P(x)\e^{-x^2}\) is zero for \(x=0\), \(x=\pm a\) and \(x=\pm b\,\), but for no other values of \(x\).

You need not consider the convergence of the improper integrals in this question. For \(p, q > 0\), define $$b(p,q) = \int_0^1 x^{p-1}(1-x)^{q-1} \, dx$$

1. Show that \(b(p,q) = b(q,p)\).
2. Show that \(b(p+1,q) = b(p,q) - b(p,q+1)\) and hence that \(b(p+1,p) = \frac{1}{2}b(p,p)\).
3. Show that $$b(p,q) = 2\int_0^{\pi/2} (\sin\theta)^{2p-1}(\cos\theta)^{2q-1} \, d\theta$$ Hence show that \(b(p,p) = \frac{1}{2^{2p-1}}b(p,\frac{1}{2})\).
4. Show that $$b(p,q) = \int_0^\infty \frac{t^{p-1}}{(1+t)^{p+q}} \, dt$$
5. Evaluate $$\int_0^\infty \frac{t^{3/2}}{(1+t)^6} \, dt$$

You need not consider the convergence of the improper integrals in this question.

1. Use the substitution \(x = u^{-1}\) to show that \[\int_0^{\infty} \frac{\sqrt{x}-1}{\sqrt{x(x^3+1)}} \, dx = 0.\]
2. Use the substitution \(x = u^{-2}\) to show that \[\int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx = 2\int_0^{\infty} \frac{1}{\sqrt{x^6+1}} \, dx.\]
3. Find, in terms of \(p\) and \(s\), a value of \(r\) for which \[\int_0^{\infty} \frac{x^r - 1}{\sqrt{x^s(x^p+1)}} \, dx = 0,\] given that \(p\) and \(s\) are fixed values for which the required integrals converge.
4. Show that, for any positive value of \(k\), it is possible to find values of \(p\) and \(q\) for which \[\int_0^{\infty} \frac{1}{\sqrt{x^p+1}} \, dx = k\int_0^{\infty} \frac{1}{\sqrt{x^q+1}} \, dx.\]

1. Let $$f(x) = \frac{x}{\sqrt{x^2 + p}},$$ where \(p\) is a non-zero constant. Sketch the curve \(y = f(x)\) for \(x \geq 0\) in the case \(p > 0\).
2. Let $$I = \int \frac{1}{(b^2 - y^2)\sqrt{c^2 - y^2}} \, dy,$$ where \(b\) and \(c\) are positive constants. Use the substitution \(y = \frac{cx}{\sqrt{x^2 + p}}\), where \(p\) is a suitably chosen constant, to show that $$I = \int \frac{1}{b^2 + (b^2 - c^2)x^2} \, dx.$$ Evaluate $$\int_1^{\sqrt{2}} \frac{1}{(3 - y^2)\sqrt{2 - y^2}} \, dy.$$ [ Note: \(\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \frac{x}{a} + \text{constant.}\) ] Hence evaluate $$\int_{\frac{1}{\sqrt{2}}}^1 \frac{y}{(3y^2 - 1)\sqrt{2y^2 - 1}} \, dy.$$
3. By means of a suitable substitution, evaluate $$\int_{\frac{1}{\sqrt{2}}}^1 \frac{1}{(3y^2 - 1)\sqrt{2y^2 - 1}} \, dy.$$

By first multiplying the numerator and the denominator of the integrand by \((1 - \sin x)\), evaluate $$\int_0^{\frac{1}{4}\pi} \frac{1}{1 + \sin x} dx.$$ Evaluate also: $$\int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \frac{1}{1 + \sec x} dx \quad \text{and} \quad \int_0^{\frac{1}{3}\pi} \frac{1}{(1 + \sin x)^2} dx.$$

The function \(f\) is defined, for \(x > 1\), by $$f(x) = \int_1^x \sqrt{\frac{t-1}{t+1}} dt.$$ Do not attempt to evaluate this integral.

1. Show that, for \(x > 2\), $$\int_2^x \sqrt{\frac{u-2}{u+2}} du = 2f\left(\frac{1}{2}x\right).$$
2. Evaluate in terms of \(f\), for \(x > 0\), $$\int_0^x \sqrt{\frac{u}{u+4}} du.$$
3. Evaluate in terms of \(f\), for \(x > 5\), $$\int_5^x \sqrt{\frac{u-5}{u+1}} du.$$
4. Evaluate in terms of \(f\) $$\int_1^2 \frac{u^2}{\sqrt{u^2+4}} du.$$

A definite integral can be evaluated approximately by means of the Trapezium rule: \[ \int_{x_{0}}^{x_{N}}\mathrm{f}(x)\,\mathrm{d}x\approx\tfrac{1}{2}h\left\{ \mathrm{f}\left(x_{0}\right)+2\mathrm{f}\left(x_{1}\right)+\ldots+2\mathrm{f}\left(x_{N-1}\right)+\mathrm{f}\left(x_{N}\right)\right\} , \] where the interval length \(h\) is given by \(Nh=x_{N}-x_{0}\), and \(x_{r}=x_{0}+rh\). Justify briefly this approximation. Use the Trapezium rule with intervals of unit length to evaluate approximately the integral \[ \int_{1}^{n}\ln x\,\mathrm{d}x, \] where \(n(>2)\) is an integer. Deduce that \(n!\approx\mathrm{g}(n)\), where \[ \mathrm{g}(n)=n^{n+\frac{1}{2}}\mathrm{e}^{1-n}, \] and show by means of a sketch, or otherwise, that \[ n!<\mathrm{g}(n). \] By using the Trapezium rule on the above integral with intervals of width \(k^{-1}\), where \(k\) is a positive integer, show that \[ \left(kn\right)!\approx k!n^{kn+\frac{1}{2}}\left(\frac{\mathrm{e}}{k}\right)^{k\left(1-n\right)}. \] Determine whether this approximation or \(\mathrm{g}(kn)\) is closer to \(\left(kn\right)!\).

Show Solution

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We can approximate the integral by \(N\) trapeziums, each with height \(x_{i+1}-x_{i} = \frac{x_N-x_0}{N} = \frac{h}{N}\). The will have area \(\frac{(f(x_i)+f(x_{i+1}))h}{2}\) and summing all these areas we will get: \[\frac12 h \l f(x_0) + f(x_1) + f(x_1)+f(x_2) + \cdots + f(x_{N-1})+f(x_N) \r = \frac12 h \l f(x_0) +2 f(x_1) + + \cdots +2f(x_{N-1})+f(x_N) \r\] But this is approximately the integral \(\displaystyle \int_{x_0}^{x_N} f(x) \d x\) \begin{align*} && \int_1^n \ln x \d x &= [x \ln x]_1^n - \int_1^n x \cdot \frac{1}{x} \d x \\ &&&= n \ln n - n+1 \\ &&&\approx \frac12 \l \ln 1 + 2\sum_{k=2}^{n-1} \ln k + \ln n \r \\ &&&= \ln (n!) - \frac12 \ln n \\ \Rightarrow && \ln (n!) &\approx n \ln n + \frac12 \ln n - n + 1 \\ \Rightarrow && n! &\approx \exp(n \ln n + \frac12 \ln n - n + 1) \\ &&&=n^{n+\frac12}e^{1-n} \end{align*} Since \(\ln x\) is a concave function, we should expect all the trapeziums to all lie under the curve, therefore this is always an underestimate for the integral, ie \(n! < g(n)\)

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\begin{align*} && \int_1^n \ln x \d x &= n \ln n - n+1 \\ &&&\approx \frac12 k^{-1} \l \ln 1 + 2\sum_{r=1}^{k(n-1)-1} \ln \l 1+\frac{r}{k} \r + \ln n \r \\ &&&=\frac{1}{2k} \l 2\sum_{r=1}^{k(n-1)-1} \l \ln(k+r) - \ln k)\r + \ln n\r \\ &&&=\frac1{k} \l \ln ((k+k(n-1)-1)!) - \ln(k!) - (k(n-1)-1) \ln k+\frac12\ln n \r \\ &&&=\frac1{k} \l \ln ((kn-1)!) - \ln(k!) - (k(n-1)-1) \ln k+\frac12 \ln n \r \\ &&&=\frac1{k} \l \ln ((kn)! ) -\ln k -\ln n - \ln(k!) - (k(n-1)-1) \ln k+\frac12\ln n \r \\ &&&= \frac1{k} \l \ln ((kn)! ) - \ln(k!) - (k(n-1)) \ln k - \frac12 \ln n\r \\ \Rightarrow && \ln ((kn)!) &\approx kn \ln n - kn + k + \ln(k!) + (k(n-1)) \ln k + \frac12 \ln n\\ \Rightarrow && (kn)! &\approx n^{kn+\frac12}e^{-k(n-1)}k!k^{k(n-1)} \\ &&&= n^{kn+\frac12} k! \l \frac{e}{k} \r^{k(1-n)} \end{align*} I would expect this approximation to be a better approximation for \((kn)!\) since it is created using a finer mesh.

Let \[ I=\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\frac{\cos^{2}\theta}{1-\sin\theta\sin2\alpha}\,\mathrm{d}\theta\, , \] where \(0<\alpha<\frac{1}{4}\pi\). Show that \[ I=\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\frac{\cos^{2}\theta}{1+\sin\theta\sin2\alpha}\,\mathrm{d}\theta\, , \] and hence that \[ I=\frac{\pi}{\sin^{2}2\alpha}-\cot^{2}2\alpha\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\frac{\sec^{2}\theta}{1+\cos^{2}2\alpha\tan^{2}\theta}\,\mathrm{d}\theta. \] Show that \(I=\frac{1}{2}\pi\sec^{2}\alpha\), and state the value of \(I\) if \(\frac{1}{4}\pi<\alpha<\frac{1}{2}\pi\).

Explain why the use of the substitution \(x=\dfrac{1}{t}\) does not demonstrate that the integrals \[ \int_{-1}^{1}\frac{1}{(1+x^{2})^{2}}\,\mathrm{d}x\quad\mbox{ and }\quad\int_{-1}^{1}\frac{-t^{2}}{(1+t^{2})^{2}}\,\mathrm{d}t \] are equal. Evaluate both integrals correctly.

Let \(y=\mathrm{f}(x)\), \((0\leqslant x\leqslant a)\), be a continuous curve lying in the first quadrant and passing through the origin. Suppose that, for each non-negative value of \(y\) with \(0\leqslant y\leqslant\mathrm{f}(a)\), there is exactly one value of \(x\) such that \(\mathrm{f}(x)=y\); thus we may write \(x=\mathrm{g}(y)\), for a suitable function \(\mathrm{g}.\) For \(0\leqslant s\leqslant a,\) \(0\leqslant t\leqslant \mathrm{f}(a)\), define \[ \mathrm{F}(s)=\int_{0}^{s}\mathrm{f}(x)\,\mathrm{d}x,\qquad\mathrm{G}(t)=\int_{0}^{t}\mathrm{g}(y)\,\mathrm{d}y. \] By a geometrical argument, show that \[ \mathrm{F}(s)+\mathrm{G}(t)\geqslant st.\tag{*} \] When does equality occur in \((*)\)? Suppose that \(y=\sin x\) and that the ranges of \(x,y,s,t\) are restricted to \(0\leqslant x\leqslant s\leqslant\frac{1}{2}\pi,\) \(0\leqslant y\leqslant t\leqslant1\). By considering \(s\) such that the equality holds in \((*)\), show that \[ \int_{0}^{t}\sin^{-1}y\,\mathrm{d}y=t\sin^{-1}t-\left(1-\cos(\sin^{-1}t)\right). \] Check this result by differentiating both sides with respect to \(t\).

Show Solution

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The blue area is \(F(s)\) the red area is \(G(t)\), the dashed rectangle (which is a subset of the red and blue areas) has area \(st\) therefore \(F(s) + G(t) \geq st\). Equality holds if \(f(s) = t\). \begin{align*} && \int_0^t \sin^{-1} y \d y + \int_0^{\sin^{-1} t} \sin x \d x &= t \sin^{-1} t \\ \Rightarrow && \int_0^t \sin^{-1} y \d y &= t \sin^{-1} t - \left [ -\cos (x) \right]_0^{\sin^{-1} t} \\ &&&= t \sin^{-1} t - (1- \cos (\sin^{-1} t)) \end{align*} Let \(y = t \sin^{-1} t - (1- \cos (\sin^{-1} t))\) then, \begin{align*} \frac{\d y}{\d t} &= \sin^{-1} t +t \frac{\d}{\d t} \l \sin^{-1} (t) \r - \sin ( \sin^{-1} t) \frac{\d}{\d t} \l \sin^{-1} (t) \r \\ &= \sin^{-1} t \end{align*} as required

Using the substitution \(x=\alpha\cos^{2}\theta+\beta\sin^{2}\theta,\) show that, if \(\alpha<\beta\), \[ \int_{\alpha}^{\beta}\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x=\pi. \] What is the value of the above integral if \(\alpha>\beta\)? Show also that, if \(0<\alpha<\beta\), \[ \int_{\alpha}^{\beta}\frac{1}{x\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x=\frac{\pi}{\sqrt{\alpha\beta}}. \]

1. Let \[ \f(x) = \frac 1 {1+\tan x} \] for \(0\le x < \frac12\pi\,\). Show that \(\f'(x)= -\dfrac{1}{1+\sin 2x}\) and hence find the range of \(\f'(x)\). Sketch the curve \(y=\f(x)\).
2. The function \(\g(x)\) is continuous for \(-1\le x \le 1\,\). Show that the curve \(y=\g(x)\) has rotational symmetry of order 2 about the point \((a,b)\) on the curve if and only if \[ \g(x) + \g(2a-x) = 2b\,. \] Given that the curve \(y=\g(x)\) passes through the origin and has rotational symmetry of order 2 about the origin, write down the value of \[\displaystyle \int_{-1}^1 \g(x)\,\d x\,. \]
3. Show that the curve \(y=\dfrac{1}{1+\tan^kx}\,\), where \(k\) is a positive constant and \(0 < x < \frac12\pi\,\), \\[3mm] has rotational symmetry of order~2 about a certain point (which you should specify) and evaluate \[ \int_{\frac16\pi}^{\frac13\pi} \frac 1 {1+\tan^kx} \, \d x \,. \]

The functions \(\s\) and \(\c\) satisfy \(\s(0)= 0\,\), \(\c(0)=1\,\) and \[ \s'(x) = \c(x)^2 ,\] \[ \c'(x)=-\s(x)^2. \] You may assume that \(\s\) and \(\c\) are uniquely defined by these conditions.

1. Show that \(\s(x)^3+\c(x)^3\) is constant, and deduce that \(\s(x)^3+\c(x)^3=1\,\).
2. Show that \[ \frac{\d }{\d x} \, \Big( \s(x) \c(x) \Big) = 2\c(x)^3-1 \] and find (and simplify) an expression in terms of \(\c(x)\) for $\dfrac{\d }{\d x} \left( \dfrac{\s(x)}{\c(x)} \right) $.
3. Find the integrals \[ \int \s(x)^2 \, \d x \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \int \s(x)^5 \, \d x \,. \]
4. Given that \(\s\) has an inverse function, \(\s^{-1}\), use the substitution \(u = \s(x)\) to show that \[ \int \frac{1}{(1-u^3)^{\frac{2}{3}}} \, \d u = \s^{-1}(u) \, + \text{constant}. \]
5. Find, in terms of \(u\), the integrals \[ \int \frac{1}{{(1-u^3)}^{\frac{4}{3}}} \, \d u \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \int {(1-u^3)}^{\frac{1}{3}} \, \d u \,. \]

The function \(\f\) is defined by \[ \phantom{\ \ \ \ \ \ \ \ \ \ \ \ (x>0, \ \ x\ne1)} \f(x) = \frac{1}{x\ln x} \left(1 - (\ln x)^2 \right)^2 \ \ \ \ \ \ \ \ \ \ \ \ (x>0, \ \ x\ne1) \,.\] Show that,\, when \(( \ln x )^2 = 1\,\),\, both \(\f(x)=0\) and \(\f'(x)=0\,\). The function \(F\) is defined by \begin{align*} F(x) = \begin{cases} \displaystyle \int_{ 1/\text{e}}^x \f(t) \; \mathrm{d}t & \text{ for } 0 < x < 1\,, \\[7mm] \displaystyle \int_{\text{e}}^x \f(t) \; \mathrm{d}t & \text{ for } x > 1\,. \\ \end{cases} \end{align*}

1. Find \(F(x)\) explicitly and hence show that \(F(x^{-1})=F(x)\,\).
2. Sketch the curve with equation \(y=F(x)\,\). %You may assume that \(\dfrac{ (\ln x)^k} x\to 0\) as \(x\to\infty\) for %any constant \(k\).

In this question, you are not permitted to use any properties of trigonometric functions or inverse trigonometric functions. The function \(\T\) is defined for \(x>0\) by \[ \T(x) = \int_0^x \! \frac 1 {1+u^2} \, \d u\,, \] and $\displaystyle T_\infty = \int_0^\infty \!\! \frac 1 {1+u^2} \, \d u\,$ (which has a finite value).

1. By making an appropriate substitution in the integral for \(\T(x)\), show that \[\T(x) = \T_\infty - \T(x^{-1})\,.\]
2. Let \(v= \dfrac{u+a}{1-au}\), where \(a\) is a constant. Verify that, for \(u\ne a^{-1}\), \[ \frac{\d v}{\d u} = \frac{1+v^2}{1+u^2} \,. \] Hence show that, for \(a>0\) and \(x< \dfrac1a\,\), \[ \T(x) = \T\left(\frac{x+a}{1-ax}\right) -\T(a) \,. \] Deduce that \[ \T(x^{-1}) = 2\T_\infty -\T\left(\frac{x+a}{1-ax}\right) -\T(a^{-1}) \] and hence that, for \(b>0\) and \(y>\dfrac1b\,\), \[ \T(y) =2\T_\infty - \T\left(\frac{y+b}{by-1}\right) - \T(b) \,. \]
3. Use the above results to show that \(\T(\sqrt3)= \tfrac23 \T_\infty \,\) and \(\T(\sqrt2 -1)= \frac14 \T_\infty\,\).

For any function \(\f\) satisfying \(\f(x) > 0\), we define the {\em geometric mean}, F, by \[ \F(y) \; = \mbox{ \fontsize{12}{15.6}\selectfont \(\e\)} \mbox{ \fontsize{14}{15.6}\selectfont $ ^{\! \raisemath {3pt} {\frac{1}{y} \! \int_{\raisemath{-1pt}{0}}^{\raisemath{1pt}{y}} \ln \f(x) \, \d x} } $ } \ \ \ \ \ \ (y>0)\,. \]

1. The function f satisfies \(\f(x) > 0\) and \(a\) is a positive number with \(a\ne1\). Prove that \[ \F(y) = \mbox{ \fontsize{12}{15.6}\selectfont \(a\)} \mbox{ \fontsize{14}{15.6}\selectfont $ ^{ \! \raisemath {3pt} {\frac{1}{y} \! \int_{\raisemath{-1pt}{0}}^{\raisemath{1pt}{y}} \log_a \f(x) \, \d x} } $ } . \]
2. The functions f and g satisfy \(\f(x) > 0\) and \(\g(x) > 0\), and the function \(\h\) is defined by \(\h(x) = \f(x)\g(x)\). Their geometric means are F, G and H, respectively. Show that \(\H(y)= \F(y) \G(y)\,\).
3. Prove that, for any positive number \(b\), the geometric mean of \(b^x\) is \(\sqrt{b^y}\,\).
4. Prove that, if \(\f(x)>0\) and the geometric mean of \(\f(x)\) is \(\sqrt{\f(y)}\,\), then \(\f(x) = b^x\) for some positive number \(b\).

The Schwarz inequality is \[ \left( \int_a^b \f(x)\, \g(x)\,\d x\right)^{\!\!2} \le \left( \int_a^b \big( \f(x)\big)^2 \d x \right) \left( \int_a^b \big( \g(x)\big)^2 \d x \right) . \tag{\(*\)} \]

1. By setting \( \f(x)=1\) in \((*)\), and choosing \(\g(x)\), \(a\) and \(b\) suitably, show that for \(t> 0\,\), \[ \frac {\e^t -1}{\e^t+1} \le \frac t 2 \,. \]
2. By setting \( \f(x)= x\) in \((*)\), and choosing \( \g(x)\) suitably, show that \[ \int_0^1\e^{-\frac12 x^2}\d x \ge 12 \big(1-\e^{-\frac14})^2 \,. \]
3. Use \((*)\) to show that \[ \frac {64}{25\pi} \le \int_0^{\frac12\pi} \!\! {\textstyle \sqrt{\, \sin x\, } } \, \d x \le \sqrt{\frac \pi 2 } \,. \]

Note: In this question you may use without proof the result \( \dfrac{\d \ }{\d x}\big(\!\arctan x \big) = \dfrac 1 {1+x^2}\,\). Let \[ I_n = \int_0^1 x^n \arctan x \, \d x \;, \] where \(n=0\), 1, 2, 3, \(\ldots\) .

1. Show that, for \(n\ge0\,\), \[ (n+1) I_n = \frac \pi 4 - \int _0^1 \frac {x^{n+1}}{1+x^2} \, \d x \, \] and evaluate \(I_0\).
2. Find an expression, in terms of \(n\), for \((n+3)I_{n+2}+(n+1)I_{n}\,\). Use this result to evaluate \(I_4\).
3. Prove by induction that, for \(n\ge1\), \[ (4n+1) I_{4n} =A - \frac12 \sum_{r=1}^{2n} (-1)^r \frac 1 {r} \,, \] where \(A\) is a constant to be determined.

1. The inequality \(\dfrac 1 t \le 1\) holds for \(t\ge1\). By integrating both sides of this inequality over the interval \(1\le t \le x\), show that \[ \ln x \le x-1 \tag{\(*\)} \] for \(x \ge 1\). Show similarly that \((*)\) also holds for \(0 < x \le 1\).
2. Starting from the inequality \(\dfrac{1}{t^2} \le \dfrac1 t \) for \(t \ge 1\), show that \[ \ln x \ge 1-\frac{1}{x} \tag{\(**\)} \] for \(x > 0\).
3. Show, by integrating (\(*\)) and (\(**\)), that \[ \frac{2}{ y+1} \le \frac{\ln y}{ y-1} \le \frac{ y+1}{2 y} \] for \( y > 0\) and \( y\ne1\).

1. Use the substitution \(u= x\sin x +\cos x\) to find \[ \int \frac{x }{x\tan x +1 } \, \d x \,. \] Find by means of a similar substitution, or otherwise, \[ \int \frac{x }{x\cot x -1 } \, \d x \,. \]
2. Use a substitution to find \[ \int \frac{x\sec^2 x \, \tan x}{x\sec^2 x -\tan x} \,\d x \, \] and \[ \int \frac{x\sin x \cos x}{(x-\sin x \cos x)^2} \, \d x \,. \]

1. Given that \[ \int \frac {x^3-2}{(x+1)^2}\, \e ^x \d x = \frac{\P(x)}{Q(x)}\,\e^x + \text{constant} \,, \] where \(\P(x)\)and \(Q(x)\) are polynomials, show that \(Q(x)\) has a factor of \(x + 1\). Show also that the degree of \(\P(x)\) is exactly one more than the degree of \(Q(x)\), and find \(\P(x)\) in the case \(Q(x) =x+1\).
2. Show that there are no polynomials \(\P(x)\) and \(Q(x)\) such that \[ \int \frac 1 {x+1} \, \, \e^x \d x = \frac{\P(x)}{Q(x)}\,\e^x +\text{constant} \,. \] You need consider only the case when \(\P(x)\) and \(Q(x)\) have no common factors.

Show that \[ \int_0^a \f(x) \d x= \int _0^a \f(a-x) \d x\,, \tag{\(*\)} \] where f is any function for which the integrals exist.

1. Use (\(*\)) to evaluate \[ \int_0^{\frac12\pi} \frac{\sin x}{\cos x + \sin x} \, \d x \,. \]
2. Evaluate \[ \int_0^{\frac14\pi} \frac{\sin x}{\cos x + \sin x} \, \d x \,. \]
3. Evaluate \[ \int_0^{\frac14\pi} \ln (1+\tan x) \, \d x \,. \]
4. Evaluate \[ \int_0^{\frac14 \pi} \frac x {\cos x \, (\cos x + \sin x)}\, \d x \,. \]

1. Show that \[ \mathrm{sec}^2\left(\tfrac14\pi-\tfrac12 x\right)=\frac{2}{1+\sin x} \,. \] Hence integrate \(\dfrac{1}{1+\sin x}\) with respect to \(x\).
2. By means of the substitution \(y=\pi -x\), show that \[ \int_0^\pi x \f (\sin x)\, \d x = \frac \pi 2 \int_0^\pi \f(\sin x) \, \d x ,\] where \(\mathrm{f}\) is any function for which these integrals exist. Hence evaluate \[ \int_0^\pi \frac x {1+\sin x} \, \d x \,. \]
3. Evaluate \[ \int_0^\pi\frac{ 2x^3 -3\pi x^2}{(1+\sin x)^2}\, \d x .\]

1. The function \(\f\) is defined, for \(x>0\), by \[ \f(x) =\int_{1}^3 (t-1)^{x-1} \, \d t \,. \] By evaluating the integral, sketch the curve \(y=\f(x)\).
2. The function \(\g\) is defined, for \(-\infty < x < \infty\), by \[ \g(x)= \int_{-1}^1 \frac 1 {\sqrt{1-2xt +x^2} \ }\, \d t \,.\] By evaluating the integral, sketch the curve \(y=\g(x)\).

1. Let \[ I = \int_0^1 \bigl((y')^2 -y^2\bigr)\d x \qquad\text{and}\qquad I_1=\int_0^1 (y'+y\tan x)^2 \d x \,, \] where \(y\) is a given function of \(x\) satisfying \(y=0\) at \(x=1\). Show that \(I-I_1=0\) and deduce that \(I\ge0\). Show further that \(I=0\) only if \(y=0\) for all \(x\) (\(0\le x \le 1\)).
2. Let \[ J = \int_0^1 \bigl((y')^2 -a^2y^2\bigr)\d x \,, \] where \(a\) is a given positive constant and \(y\) is a given function of \(x\), not identically zero, satisfying \(y=0\) at \(x=1\). By considering an integral of the form \[ \int_0^1 (y'+ay\tan bx)^2 \d x \,, \] where \(b\) is suitably chosen, show that \(J\ge0\). You should state the range of values of \(a\), in the form \(a < k\), for which your proof is valid. In the case \(a=k\), find a function \(y\) (not everywhere zero) such that \(J=0\).

1. By using the substitution \(u=1/x\), show that for \(b>0\) \[ \int_{1/b}^b \frac{x \ln x}{(a^2+x^2)(a^2x^2+1)} \ud x =0 \,. \]
2. By using the substitution \(u=1/x\), show that for \(b>0\), \[ \int_{1/b}^b \frac{\arctan x}{x} \ud x = \frac{\pi \ln b} 2\,. \]
3. By using the result \( \displaystyle \int_0^\infty \frac 1 {a^2+x^2} \ud x = \frac {\pi}{2 a} \) \ (where \(a>0\)), and a substitution of the \\[5pt] form \(u=k/x\), for suitable \(k\), show that \[ \int_0^\infty \frac 1 {(a^2+x^2)^2} \ud x = \frac {\pi}{4a^3 } \, \ \ \ \ \ \ (a>0). \]

The curve \(C\) is given parametrically by the equations \(x = 3t^2\), \(y = 2t^3\). Show that the equation of the tangent to \(C\) at the point \((3p^2 , 2p^3)\) is \(y = px - p^3\). Find the point of intersection of the tangents to \(C\) at the distinct points \((3p^2 , 2p^3)\) and \((3q^2 , 2q^3)\). Hence show that, if these two tangents are perpendicular, their point of intersection is \((u^2 + 1 , -u)\), where \(u = p + q\). The curve \(\tilde{C}\) is given parametrically by the equations \(x = u^2 + 1\), \(y = -u\). Find the coordinates of the points that lie on both \(C\) and \(\tilde{C}\). Sketch \(C\) and \(\tilde{C}\) on the same axes.

Show Solution

\begin{align*} && \frac{\d y}{\d x} &= \frac{\frac{\d y}{\d t}}{\frac{\d x}{\d t}} \\ &&&= \frac{6t^2}{6t} = t \\ \Rightarrow && \frac{y-2p^3}{x - 3p^2} &= p \\ \Rightarrow && y &= px-3p^3+2p^3 \\ && y &= px - p^3 \end{align*} The two lines will be \begin{align*} && y &= px - p^3 \\ && y &= qx - q^3 \\ \Rightarrow && p^3-q^3 &= (p-q)x \\ \Rightarrow && x &= p^2+pq+q^2 \\ && y &= p(p^2+pq+q^2)-p^3 \\ &&&= pq(p+q) \\ && (x,y) &= (p^2+pq+q^2,pq(p+q)) \\ \end{align*} If the tangents are \(\perp\) then \(pq=-1\), so we have \begin{align*} && (x,y) &= (p^2+2pq+q^2-pq, pq(p+q)) \\ &&&= ((p+q)^2-1, -(p+q)) \\ &&&= (u^2-1, -u) \end{align*} We have \(x = y^2+1\) and \(\left ( \frac{x}{3} \right)^3 = \left ( \frac{y}{2}\right)^2 \Rightarrow y^2 = \frac{4}{27}x^3\) so \begin{align*} && 0 &= \frac{4}{27}x^3-x+1 \\ &&0&=4x^3-27x+27 \\ &&&= (x+3)(2x-3)^2 \end{align*} So we have the points \((x,y) = \left (\frac32, \pm\frac{1}{\sqrt{2}}\right)\)

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1. Sketch, on \(x\)-\(y\) axes, the set of all points satisfying \(\sin y = \sin x\), for \(-\pi \le x \le \pi\) and \(-\pi \le y \le \pi\). You should give the equations of all the lines on your sketch.
2. Given that \[ \sin y = \tfrac12 \sin x \] obtain an expression, in terms of \(x\), for \(y'\) when \(0\le x \le \frac12 \pi\) and \(0\le y \le \frac12 \pi\), and show that \[ y'' = - \frac {3\sin x}{(4-\sin^2 x)^{\frac32}} \;. \] Use these results to sketch the set of all points satisfying \(\sin y = \tfrac12 \sin x\) for \(0 \le x \le \frac12 \pi\) and \(0 \le y \le \frac12 \pi\). Hence sketch the set of all points satisfying \(\sin y = \tfrac12 \sin x\) for \(-\pi\! \le \! x \! \le \! \pi\) and \mbox{\( -\pi \, \le\, y\, \le\, \pi\,\)}.
3. Without further calculation, sketch the set of all points satisfying \(\cos y = \tfrac12 \sin x\) for \(- \pi \le x \le \pi\) and \( -\pi \le y \le \pi\).

1. Show, geometrically or otherwise, that the shortest distance between the origin and the line \(y= mx+c\), where \(c\ge0\), is \(c(m^2+1)^{-\frac12}\).
2. The curve \(C\) lies in the \(x\)-\(y\) plane. Let the line \(L\) be tangent to~\(C\) at a point~\(P\) on~\(C\), and let \(a\)~be the shortest distance between the origin and \(L\). The curve~\(C\) has the property that the distance~\(a\) is the same for all points~\(P\) on~\(C\). Let \(P\) be the point on \(C\) with coordinates \((x,y(x))\). Given that the tangent to \(C\) at \(P\) is not vertical, show that \begin{equation} \label{eq:8*} (y-xy')^2 = a^2\big (1+(y')^2 \big) \,. \tag{\(*\)} \end{equation} By first differentiating \((*)\) with respect to \(x\), show that either \(y= mx \pm a(1+m^2)^{\frac12}\) for some \(m\) or \(x^2+y^2 =a^2\).
3. Now suppose that \(C\) (as defined above) is a continuous curve for \(-\infty < x < \infty\), consisting of the arc of a circle and two straight lines. Sketch an example of such a curve which has a non-vertical tangent at each point.

1. Show that the gradient of the curve \(\; \dfrac a x + \dfrac by =1\), where \(b\ne0\), is \(\; -\dfrac{ay^2}{bx^2}\,\). The point \((p,q)\) lies on both the straight line \(ax+by=1\) and the curve \(\dfrac a x + \dfrac by =1\,\), where \(ab\ne0\). Given that, at this point, the line and the curve have the same gradient, show that \( p=\pm q\,\). Show further that either \((a-b)^2 =1\,\) or \((a+b)^2 =1\,\).
2. Show that if the straight line \(ax+by=1\), where \(ab\ne0\), is a normal to the curve \(\dfrac a x - \dfrac by =1\), then \(a^2-b^2 = \frac12\,\).

A right circular cone has base radius \(r\), height \(h\) and slant length \(\ell\). Its volume \(V\), and the area~\(A\) of its curved surface, are given by \[ V= \tfrac13 \pi r^2 h \,, \ \ \ \ \ \ \ A = \pi r\ell\,. \] \vspace*{-1cm}

1. Given that \(A\) is fixed and \(r\) is chosen so that \(V\) is at its stationary value, show that \(A^2 = 3\pi^2r^4\) and that $ \ell =\sqrt3\,r\(.
2. Given, instead, that \)V\( is fixed and \)r$ is chosen so that \(A\) is at its stationary value, find~\(h\) in terms of \(r\).

A curve has the equation \[ y^3 = x^3 +a^3+b^3\,, \] where \(a\) and \(b\) are positive constants. Show that the tangent to the curve at the point \((-a,b)\) is \[ b^2y-a^2x = a^3+b^3\,. \] In the case \(a=1\) and \(b=2\), show that the \(x\)-coordinates of the points where the tangent meets the curve satisfy \[ 7x^3 -3x^2 -27x-17 =0\,. \] Hence find positive integers \(p\), \(q\), \(r\) and \(s\) such that \[ p^3 = q^3 +r^3 +s^3\,. \]

A curve is given by \[x^2+y^2 +2axy = 1,\] where \(a\) is a constant satisfying \(0 < a < 1\). Show that the gradient of the curve at the point~\(P\) with coordinates \((x,y)\) is \[\displaystyle - \frac {x+ay}{ax+y}\,,\] provided \(ax+y \ne0\). Show that \(\theta\), the acute angle between \(OP\) and the normal to the curve at \(P\), satisfies \[ \tan\theta = a\vert y^2-x^2\vert\;. \] Show further that, if \(\ \displaystyle \frac{\d \theta}{\d x}=0\) at \(P\), then:

1. \(a(x^2+y^2)+2xy=0\,\);
2. \((1+a)(x^2+y^2+2xy)=1\,\);
3. \(\displaystyle \tan\theta = \frac a{\sqrt{1-a^2}}\,\).

The variables \(t\) and \(x\) are related by \(t=x+ \sqrt{x^2+2bx+c\;} \,\), where \(b\) and \(c\) are constants and \(b^2 < c\). Show that \[ \frac{\d x}{\d t} = \frac{t-x}{t+b}\;, \] and hence integrate \(\displaystyle \frac1 {\sqrt{x^2+2bx+c}}\,\). Verify by direct integration that your result holds also in the case \(b^2=c\) if \(x+b > 0\) but that your result does not hold in the case \(b^2=c\) if \(x+b < 0\,\).

1. If \[{\mathrm f}(x)=\tan^{-1}x+\tan^{-1}\left(\frac{1-x}{1+x}\right),\] find \({\mathrm f}'(x)\). Hence, or otherwise, find a simple expression for \({\mathrm f}(x)\).
2. Suppose that \(y\) is a function of \(x\) with \(0 < y < (\pi/2)^{1/2}\) and \[x=y\sin y^{2}\] for \(0 < x < (\pi/2)^{1/2}\). Show that (for this range of \(x\)) \[\frac{{\mathrm d}y}{{\mathrm d}x}= \frac{y}{x+2y^2\sqrt{y^{2}-x^{2}}}.\]

The function \(\mathrm{g}\) satisfies, for all positive \(x\) and \(y\), \[ \mathrm{g}(x)+\mathrm{g}(y)=\mathrm{g}(z),\tag{\ensuremath{*}} \] where \(z=xy/(x+y+1).\) By treating \(y\) as a constant, show that \[ \mathrm{g}'(x)=\frac{y^{2}+y}{(x+y+1)^{2}}\mathrm{g}'(z)=\frac{z(z+1)}{x(x+1)}\mathrm{g}'(z), \] and deduce that \(2\mathrm{g}'(1)=(u^{2}+u)\mathrm{g}'(u)\) for all \(u\) satisfying \(0 < u < 1.\) Now by treating \(u\) as a variable, show that \[ \mathrm{g}(u)=A\ln\left(\frac{u}{u+1}\right)+B, \] where \(A\) and \(B\) are constants. Verify that \(\mathrm{g}\) satisfies \((*)\) for a suitable value of \(B\). Can \(A\) be determined from \((*)\)? The function \(\mathrm{f}\) satisfies, for all positive \(x\) and \(y\), \[ \mathrm{f}(x)+\mathrm{f}(y)=\mathrm{f}(z) \] where \(z=xy.\) Show that \(\mathrm{f}(x)=C\ln x\) where \(C\) is a constant.

The equation of a hyperbola (with respect to axes which are displaced and rotated with respect to the standard axes) is \[ 3y^{2}-10xy+3x^{2}+16y-16x+15=0.\tag{\(\dagger\)} \] By differentiating \((\dagger)\), or otherwise, show that the equation of the tangent through the point \((s,t)\) on the curve is \[ y=\left(\frac{5t-3s+8}{3t-5s+8}\right)x-\left(\frac{8t-8s+15}{3t-5s+8}\right). \] Show that the equations of the asymptote (the limiting tangents as \(s\rightarrow\infty\)) are \[ y=3x-4\qquad\mbox{ and }\qquad3y=x-4. \] {[}Hint: You will need to find a relationship between \(s\) and \(t\) which is valid in the limit as \(s\rightarrow\infty.\){]} Show that the angle between one asymptote and the \(x\)-axis is the same as the angle between the other asymptote and the \(y\)-axis. Deduce the slopes of the lines that bisect the angles between the asymptotes and find the equations of the axes of the hyperbola.

Sketch the curve \(y^{2}=1-\left|x\right|\). A rectangle, with sides parallel to the axes, is inscribed within this curve. Show that the largest possible area of the rectangle is \(8/\sqrt{27}\). Find the maximum area of a rectangle similarly inscribed within the curve given by \(y^{2m}=\left(1-\left|x\right|\right)^{n}\), where \(m\) and \(n\) are positive integers, with \(n\) odd.

Show Solution

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Suppose one of the \(x\) coordinates is \(t > 0\), then the coordinates are \(y = \pm \sqrt{1-t}, x = \pm t\). The area will be \(A = 2t \cdot 2 \sqrt{1-t}\). To maximise this, \begin{align*} && \frac{\d A}{\d t} &= 4 \sqrt{1-t} - 2t(1-t)^{-\frac12} \\ &&&= \frac{4(1-t) - 2t}{\sqrt{1-t}} \\ &&&= \frac{4-6t}{\sqrt{1-t}} \end{align*} Therefore there is a stationary point at \(t = \frac23\). Since we know the area is \(0\) when \(t = 0, 1\) we can see this must be a maximum for the area. Therefore the area is \(\displaystyle 4 \frac23 \sqrt{1-\frac23} = \frac{8}{3\sqrt{3}} = \frac{8}{\sqrt{27}}\). For this similar problem, using a similar approach we find \(y = \pm (1- t)^{n/2m}, x = \pm t\) and so the area is \(A = 4 t \cdot (1-t)^{n/2m}\). \begin{align*} && \frac{\d A}{\d t} &= 4(1-t)^{n/2m} - 4t \frac{n}{2m} (1-t)^{\frac{n}{2m} - 1} \\ &&&= (1-t)^{\frac{n}{2m}-1} \left ( 4(1-t) - \frac{2n}{m} t\right) \\ &&&= (1-t)^{\frac{n}{2m}-1} \left ( 4 - (4 + \frac{2n}{m})t \right) \\ \end{align*} Therefore \(\displaystyle t = \frac{2m}{2m+n}\) and \(\displaystyle A = 4\cdot \frac{2m}{2m+n} \cdot (1 - \frac{2m}{2m+n})^{n/2m} = \frac{8m}{2m+n} \cdot \left ( \frac{n}{2m+n}\right)^{n/2m}\)

Note: You may assume that if the functions \(y_1(x)\) and \(y_2(x)\) both satisfy one of the differential equations in this question, then the curves \(y = y_1(x)\) and \(y = y_2(x)\) do not intersect.

1. Find the solution of the differential equation $$\frac{dy}{dx} = y + x + 1$$ that has the form \(y = mx + c\), where \(m\) and \(c\) are constants. Let \(y_3(x)\) be the solution of this differential equation with \(y_3(0) = k\). Show that any stationary point on the curve \(y = y_3(x)\) lies on the line \(y = -x - 1\). Deduce that solution curves with \(k < -2\) cannot have any stationary points. Show further that any stationary point on the solution curve is a local minimum. Use the substitution \(Y = y + x\) to solve the differential equation, and sketch, on the same axes, the solutions with \(k = 0\), \(k = -2\) and \(k = -3\).
2. Find the two solutions of the differential equation $$\frac{dy}{dx} = x^2 + y^2 - 2xy - 4x + 4y + 3$$ that have the form \(y = mx + c\). Let \(y_4(x)\) be the solution of this differential equation with \(y_4(0) = -2\). (Do not attempt to find this solution.) Show that any stationary point on the curve \(y = y_4(x)\) lies on one of two lines that you should identify. What can be said about the gradient of the curve at points between these lines? Sketch the curve \(y = y_4(x)\). You should include on your sketch the two straight line solutions and the two lines of stationary points.

This question concerns solutions of the differential equation \[ (1-x^2) \left(\frac{\d y}{\d x}\right)^2 + k^2 y^2 = k^2\, \tag{\(*\)} \] where \(k\) is a positive integer. For each value of \(k\), let \(y_k(x)\) be the solution of \((*)\) that satisfies \(y_k(1)=1\); you may assume that there is only one such solution for each value of \(k\).

1. Write down the differential equation satisfied by \(y_1(x)\) and verify that \(y_1(x) = x\,\).
2. Write down the differential equation satisfied by \(y_2(x)\) and verify that \(y_2(x) = 2x^2-1\,\).
3. Let \(z(x) = 2\big(y_n(x)\big)^2 -1\). Show that \[ (1-x^2) \left(\frac{\d z}{\d x}\right)^2 +4n^2 z^2 = 4n^2\, \] and hence obtain an expression for \(y_{2n}(x)\) in terms of \(y_n(x)\).
4. Let \(v(x) = y_n\big(y_m(x)\big)\,\). Show that \(v(x) = y_{mn}(x)\,\).

1. Differentiate $\displaystyle \; \frac z {(1+z^2)^{\frac12}} \;$ with respect to \(z\).
2. The {\em signed curvature} \(\kappa\) of the curve \(y=\f(x)\) is defined by \[ \kappa = \frac {\f''(x)}{\big({1+ (\f'(x))^2\big)^{\frac32}}} \,.\] Use this definition to determine all curves for which the signed curvature is a non-zero constant. For these curves, what is the geometrical significance of \(\kappa\)?

1. Use the substitution \(y=ux\), where \(u\) is a function of \(x\), to show that the solution of the differential equation \[ \frac{\d y}{\d x} = \frac x y + \frac y x \quad \quad (x > 0, y> 0) \] that satisfies \(y=2\) when \(x=1\) is \[ y= x\, \sqrt{4+2\ln x \, } ( x > \e^{-2}). \]
2. Use a substitution to find the solution of the differential equation \[ \frac{\d y}{\d x} = \frac x y + \frac {2y} x \quad \quad (x > 0, y > 0) \] that satisfies \(y=2\) when \(x=1\).
3. Find the solution of the differential equation \[ \frac{\d y}{\d x} = \frac {x^2} y + \frac {2y} x \quad \quad (x> 0, \ y> 0) \] that satisfies \(y=2\) when \(x=1\).

In this question, you may assume that \(\ln (1+x) \approx x -\frac12 x^2\) when \(\vert x \vert \) is small. The height of the water in a tank at time \(t\) is \(h\). The initial height of the water is \(H\) and water flows into the tank at a constant rate. The cross-sectional area of the tank is constant.

1. Suppose that water leaks out at a rate proportional to the height of the water in the tank, and that when the height reaches \(\alpha^2 H\), where \(\alpha\) is a constant greater than 1, the height remains constant. Show that \[ \frac {\d h}{\d t } = k( \alpha^2 H -h)\,, \] for some positive constant \(k\). Deduce that the time \(T\) taken for the water to reach height \(\alpha H\) is given by \[ kT = \ln \left(1+\frac1\alpha\right)\,, \] and that \(kT\approx \alpha^{-1}\) for large values of \(\alpha\).
2. Suppose that the rate at which water leaks out of the tank is proportional to \(\sqrt h\) (instead of \(h\)), and that when the height reaches \(\alpha^2 H\), where \(\alpha\) is a constant greater than 1, the height remains constant. Show that the time \(T'\) taken for the water to reach height \(\alpha H\) is given by \[ cT'=2\sqrt H \left( 1 - \sqrt\alpha +\alpha \ln \left(1+\frac1 {\sqrt\alpha} \right)\right)\, \] for some positive constant \(c\), and that \(cT'\approx \sqrt H\) for large values of \(\alpha\).

Given that \({\rm P} (x) = {\rm Q} (x){\rm R}'(x) - {\rm Q}'(x){\rm R}(x)\), write down an expression for \[ \int \frac{{\rm P} ( x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x\, . \]

1. By choosing the function \({\rm R}(x)\) to be of the form \(a +bx+c x^2\), find \[ \int \frac{5x^2 - 4x - 3} {(1 + 2x + 3x^2 )^2 } \, \d x \,.\] Show that the choice of \({\rm R}(x)\) is not unique and, by comparing the two functions \({\rm R}(x)\) corresponding to two different values of \(a\), explain how the different choices are related.
2. Find the general solution of \[ (1+\cos x +2 \sin x) \frac {\d y}{\d x} +(\sin x -2 \cos x)y = 5 - 3 \cos x + 4 \sin x\,. \]

1. By writing \(y=u{(1+x^2)\vphantom{\dot A}}^{\frac12}\), where \(u\) is a function of \(x\), find the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = xy + \frac x {1+x^2} \] for which \(y=1\) when \(x=0\).
2. Find the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = x^2y + \frac {x^2 } {1+x^3} \] for which \(y=1\) when \(x=0\).
3. Give, without proof, a conjecture for the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = x^{n-1}y + \frac {x^{n-1} } {1+x^n} \] for which \(y=1\) when \(x=0\), where \(n\) is an integer greater than 1.

1. Differentiate \(\ln\big (x+\sqrt{3+x^2}\,\big)\) and \(x\sqrt{3+x^2}\) and simplify your answers. Hence find \(\int \! \sqrt{3+x^2}\, \d x\).
2. Find the two solutions of the differential equation \[ 3\left(\frac{\d y}{\d x}\right)^{\!2} + 2 x \frac {\d y}{\d x} =1 \] that satisfy \(y=0\) when \(x=1\).

Find the general solution of the differential equation \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{xy}{x^2+a^2}\;\), where \(a\ne0\,\), and show that it can be written in the form \(\displaystyle y^2(x^2+a^2)= c^2\,\), where \(c\) is an arbitrary constant. Sketch this curve. Find an expression for \(\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} (x^2+y^2)\) and show that \[ \frac{\mathrm{d^2}}{\mathrm{d}x^2} (x^2+y^2) = 2\left(1 -\frac {c^2}{(x^2+a^2)^2} \right) + \frac{8c^2x^2}{(x^2+a^2)^3}\;. \]

1. Show that, if \(0 < c < a^2\), the points on the curve whose distance from the origin is least are \(\displaystyle \l 0\,,\;\pm \frac{c}{a}\r\).
2. If \(c > a^2\), determine the points on the curve whose distance from the origin is least.

Show Solution

\begin{align*} && \frac{\d y}{\d x} &= - \frac{xy}{x^2+a^2} \\ \Rightarrow && \int \frac{1}{y} \d y &= \int -\frac{x}{x^2+a^2} \d x \\ && \ln |y| &= -\frac12 \ln |x^2 + a^2| + C \\ \Rightarrow && C' &= \ln y^2 + \ln (x^2+a^2) \\ \Rightarrow && c^2 &= y^2(x^2+a^2) \end{align*} (where the final constant \(c^2\) can be taken as a square since it is clearly positive).

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\begin{align*} && \frac{\d }{\d x} \left (x^2 + y^2 \right) &= 2x - \frac{2xy^2}{x^2+a^2} \\ &&&=2x - \frac{2x c^2}{(x^2+a^2)^2} \\ &&&= 2x \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) \\ \\ && \frac{\d ^2}{\d x^2} \left (x^2 + y^2 \right) &= \frac{\d }{\d x} \left (2x \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) \right) \\ &&&= 2 \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) + 2x \left (\frac{2c^2 \cdot 2x}{(x^2+a^2)^3} \right) \\ &&&= 2 \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) + \frac{8x^2c^2 }{(x^2+a^2)^3} \\ \end{align*}

1. The shortest distance from the origin will have the first derivative as \(0\), ie \(x = 0\) or \(x^2 + a^2 = c\), but if \(c < a^2\) this can only occur for \(x = 0\), so the closest to the origin is \((0, \pm \frac{c}{a})\)
2. If \(c > a^2\) then we can have \(x = 0\) or \(x = \pm \sqrt{c-a^2}\). Looking at the second derivative, when \(x = 0\) we have \(2(1- \frac{c^2}{a^4}) < 0\) which is a local maximum. When \(x = \pm\sqrt{c-a^2}\) we have \(8(c-a^2)c^2/c^3 > 0\) which is the minimum, therefore the points are \((\pm \sqrt{c-a^2}, c)\)

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For \(x \ge 0\) the curve \(C\) is defined by $$ {\frac{\d y} {\d x}} = \frac{x^3y^2}{(1 + x^2)^{5/2}} $$ with \(y = 1\) when \(x=0\,\). Show that \[ \frac 1 y = \frac {2+3x^2}{3(1+x^2)^{3/2}} +\frac13 \] and hence that for large positive \(x\) $$ y \approx 3 - \frac 9 x\;. $$ Draw a sketch of \(C\). On a separate diagram, draw a sketch of the two curves defined for \(x \ge 0\) by $$ \frac {\d z} {\d x} = \frac{x^3z^3}{2(1 + x^2)^{5/2}} $$ with \(z = 1\) at \(x=0\) on one curve, and \(z = -1\) at \(x=0\) on the other.

Show Solution

\begin{align*} && {\frac{\d y} {\d x}} &= \frac{x^3y^2}{(1 + x^2)^{5/2}} \\ \Rightarrow &&\int \frac{1}{y^2} \d y &= \int \frac{x^3}{(1+x^2)^{5/2}} \d x \\ \Rightarrow && -\frac1y &= \int \frac{x^3+x-x}{(1+x^2)^{5/2}} \d x \\ &&&= \int \left ( \frac{x}{(1+x^2)^{3/2}}-\frac{x}{(1+x^2)^{5/2}} \right) \d x \\ &&&= \frac{-1}{(1+x^2)^{1/2}} + \frac{1}{3(1+x^2)^{3/2}} + C \\ &&&= \frac{1-3(1+x^2)}{3(1+x^2)^{3/2}} + C \\ &&&= \frac{-3x^2-2}{3(1+x^2)^{3/2}} + C \\ (x,y) = (0,1): &&-1 &= -\frac23 + C \\ \Rightarrow && C &= -\frac13 \\ \Rightarrow && \frac1y &= \frac{3x^2+2}{3(1+x^2)^{3/2}} + \frac13 \end{align*} \begin{align*} y &= \frac{1}{\frac13 +\frac{3x^2+2}{3(1+x^2)^{3/2}} } \\ &= \frac{3}{1+ \frac{3x^2+2}{3(1+x^2)^{3/2}}} \\ &= \frac{3}{1+ \frac{3}{x} + \cdots} \\ &\approx 3 - \frac{9}{x} \end{align*}

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\begin{align*} && \frac {\d z} {\d x} &= \frac{x^3z^3}{2(1 + x^2)^{5/2}} \\ \Rightarrow && \int \frac{1}{z^3} \d z &= \int \frac{x^3}{2(1+x^2)^{5/2}} \\ && -\frac{1}{2z^2} &= -\frac{3x^2+2}{3(1+x^2)^{3/2}} - C \\ (x,z) = (0, \pm 1): && \frac{1}{2} &= \frac{2}{3} + C \\ \Rightarrow && C &= -\frac16 \\ \Rightarrow && \frac{1}{z^2} &= \frac{6x^2+4}{3(1+x^2)^{3/2}} - \frac13 \end{align*} So as \(x \to \infty\) \(z \sim \pm (3 + \frac{2}{x} + \cdots)\) and so:

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Show that, if \(y^2 = x^k \f(x)\), then $\displaystyle 2xy \frac{\mathrm{d}y }{ \mathrm{d}x} = ky^2 + x^{k+1} \frac{\mathrm{d}\f }{ \mathrm{d}x}$\,.

1. By setting \(k=1\) in this result, find the solution of the differential equation \[ \displaystyle 2xy \frac{\mathrm{d}y }{ \mathrm{d}x} = y^2 + x^2 - 1 \] for which \(y=2\) when \(x=1\). Describe geometrically this solution.
2. Find the solution of the differential equation \[ 2x^2y\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = 2 \ln(x) - xy^2 \] for which \(y=1\) when \(x=1\,\).

Let \(x\) satisfy the differential equation $$ \frac {\d x}{\d t} = {\big( 1-x^n\big)\vphantom{\Big)}}^{\!1/n} $$ and the condition \(x=0\) when \(t=0 \,\).

1. Solve the equation in the case \(n=1\) and sketch the graph of the solution for \(t > 0 \,\).
2. Prove that \(1-x < (1-x^2)^{1/2} \) for \(0 < x < 1 \,\). Use this result to sketch the graph of the solution in the case \(n=2\) for \(0 < t < \frac12 \pi \,\), using the same axes as your previous sketch. By setting \(x=\sin y\,\), solve the equation in this case.
3. Use the result (which you need not prove) \[ (1-x^2)^{1/2} < (1-x^3)^{1/3} \text{ \ \ for \ \ } 0 < x < 1 \;, \] to sketch, without solving the equation, the graph of the solution of the equation in the case \(n=3\) using the same axes as your previous sketches. Use your sketch to show that \(x=1\) at a value of \(t\) less than \(\frac12 \pi \,\).

Evaluate \(\int_0^{{\pi}} x \sin x\,\d x\) and \(\int_0^{{\pi}} x \cos x\,\d x\;\). The function \(\f\) satisfies the equation \begin{equation} \f(t)=t + \int_0^{{\pi}} \f(x)\sin(x+t)\,\d x\;. \tag{\(*\)} \end{equation} Show that \[ \f(t)=t + A\sin t + B\cos t\;, \] where \(A= \int_0^{{\pi}}\,\f(x)\cos x\,\d x\;\) and \(B= \int_0^{{\pi}}\,\f(x)\sin x\,\d x\;\). Find \(A\) and \(B\) by substituting for \(\f(t)\) and \(\f(x)\) in \((*)\) and equating coefficients of \(\sin t\) and \(\cos t\,\).

It is given that \(y\) satisfies $$ {{\d y} \over { \d t}} + k\left({{t^2-3t+2} \over {t+1}}\right)y = 0\;, $$ where \(k\) is a constant, and \(y=A \) when \(t=0\,\), where \(A\) is a positive constant. Find \(y\) in terms of \(t\,\), \(k\) and \(A\,\). Show that \(y\) has two stationary values whose ratio is \((3/2)^{6k}\e^{-5{k}/2}.\) Describe the behaviour of \(y\) as \(t \to +\infty\) for the case where \(k> 0\) and for the case where \(k<0\,.\) In separate diagrams, sketch the graph of \(y\) for \(t>0\) for each of these cases.

Show Solution

\begin{align*} && \frac{\d y}{\d t} &= - k \left (\frac{t^2-3t+2}{t+1} \right) y \\ \Rightarrow && \int \frac1y \d y &= -k\int \left (t-4 + \frac{6}{t+1}\right) \d t \\ \Rightarrow && \ln y &= -k \left ( \frac12 t^2 -4t + 6\ln (t+1) \right) + C \\ (t,y) = (0,A): && \ln A &=C \\ \Rightarrow && \ln y &= -k \left ( \frac12 t^2 -4t + 6\ln (t+1) \right) + \ln A \\ && \ln \left ( \frac{y}{A}(t+1)^{6k} \right) &= -k \l \frac12 t^2 - 4t \r \\ \Rightarrow && y &= A\frac{\exp \l -k(\frac12 t^2-4t)\r}{(t+1)^{6k}} \end{align*} \(y\) wil have stationary values when \(\frac{\d y}{\d t} = 0\), ie \begin{align*} k \left (\frac{t^2-3t+2}{t+1} \right) y &= 0 \\ k \left ( \frac{(t-2)(t-1)}{t+1} \right) y &= 0 \end{align*} ie when \(y = 0, t = 1, t =2\). Clearly \(y = 0\) is not a solution, so \(y\) has the values: \begin{align*} t = 1: && y &= A\frac{\exp \l -k(\frac12 -4)\r}{(2)^{6k}} \\ &&&= A \frac{e^{7/2 k}}{2^{6k}} \\ t = 2: && y &= A\frac{\exp \l -k(2 -8)\r}{(3)^{6k}} \\ &&&= A \frac{e^{6 k}}{3^{6k}} \\ \text{ratio}: && \frac{e^{7/2k}}{2^{6k}} \cdot \frac{3^{6k}}{e^{6k}} &= (3/2)^{6k} e^{-5k/2} \end{align*} If \(k > 0\) as \(t \to \infty\) \(y \to 0\) since the \(e^{-kt^2/2}\) term dominates everything. If \(k < 0\) as \(t \to \infty\) \(y \to \infty\) as since the \(e^{-kt^2}\) term also dominates but now it growing to infinity faster than everything else.

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A liquid of fixed volume \(V\) is made up of two chemicals \(A\) and \(B\,\). A reaction takes place in which \(A\) converts to \(B\,\). The volume of \(A\) at time \(t\) is \(xV\) and the volume of \(B\) at time \(t\) is \(yV\) where \(x\) and \(y\) depend on \(t\) and \(x+y=1\,\). The rate at which \(A\) converts into \(B\) is given by \(kVxy\,\), where \(k\) is a positive constant. Show that if both \(x\) and \(y\) are strictly positive at the start, then at time \(t\) \[ y= \frac {D\e^{kt}}{1+D \e^{kt}} \;, \] where \(D\) is a constant. Does \(A\) ever completely convert to \(B\,\)? Justify your answer.

Find all the solution curves of the differential equation \[ y^4 \l {\mathrm{d}y \over \mathrm{d}x }\r^{\! \! 4} = \l y^2 - 1 \r^2 \] that pass through either of the points

1. \(\l 0, \, \frac{1}{2}\sqrt3 \r\),
2. \(\l 0, \, \frac{1}{2}\sqrt5 \r\).

Show Solution

\begin{align*} && y^4 \left (\frac{\d y}{\d x} \right)^4 &= (y^2 - 1)^2 \\ \Rightarrow && y^2 \left (\frac{\d y}{\d x} \right)^2 &= |y^2 - 1| \\ && y \left (\frac{\d y}{\d x} \right) &= \pm \sqrt{|y^2-1|} \\ \Rightarrow &&\int \frac{y}{\sqrt{|y^2-1|}} \d y &= \int \pm 1 \d x \\ \Rightarrow && \pm \sqrt{|y^2-1|} &= \pm x + C \\ \end{align*}

1. Since \(y^2 < 1\), our solution curve should be of the from \(-\sqrt{1-y^2} = \pm x + C\) Plugging in \((0, \tfrac12 \sqrt{3})\), we obtain \(-\tfrac12 = C\), therefore our solution curves are \(\pm x = \frac12 - \sqrt{1-y^2}\)
2. Since \(y^2 > 1\), our solution curve should be of the from \(\sqrt{y^2-1} = \pm x + C\) Plugging in \((0, \tfrac12 \sqrt{5})\), we obtain \(\tfrac12 = C\), therefore our solution curves are \(\pm x = \sqrt{y^2-1}-\frac12\)

Clearly if \(y = \pm 1\), \(y'=0\) and the equation is satisfied.

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Find \(y\) in terms of \(x\), given that: \begin{eqnarray*} \mbox{for \(x < 0\,\)}, && \frac{\d y}{\d x} = -y \mbox{ \ \ and \ \ } y = a \mbox{ when } x = -1\;; \\ \mbox{for \(x > 0\,\)}, && \frac{\d y}{\d x} = y \mbox{ \ \ \ \ and \ \ } y = b \ \mbox{ when } x = 1\;. \end{eqnarray*} Sketch a solution curve. Determine the condition on \(a\) and \(b\) for the solution curve to be continuous (that is, for there to be no `jump' in the value of \(y\)) at \(x = 0\). Solve the differential equation \[ \frac{\d y}{\d x} = \left\vert \e^x-1\right\vert y \] given that \(y=\e^{\e}\) when \(x=1\) and that \(y\) is continuous at \(x=0\,\). Write down the following limits: \ \[ \text{(i)} \ \ \lim_ {x \to +\infty} y\exp(-\e^x)\;; \ \ \ \ \ \ \ \ \ \text{(ii)} \ \ \lim_{x \to -\infty}y \e^{-x}\,. \]

Sketch the graph of the function \(\ln x - {1 \over 2} x^2\). Show that the differential equation \[ {\mathrm{d} y \over \mathrm{d} x} = {2xy \over x^2 - 1} \] describes a family of parabolas each of which passes through the points \((1,0)\) and \((-1,0)\) and has its vertex on the \(y\)-axis. Hence find the equation of the curve that passes through the point \((1,1)\) and intersects each of the above parabolas orthogonally. Sketch this curve. [Two curves intersect orthogonally if their tangents at the point of intersection are perpendicular.]

Show Solution

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\begin{align*} && y' &= \frac{2xy}{x^2-1} \\ \Rightarrow && \int \frac{1}{y} \d y &= \int \frac{2x}{x^2-1} \d x \\ \Rightarrow && \ln |y| &= \ln |x^2-1| + C \\ \Rightarrow && y &= A(x^2-1) \end{align*} which is a family of parabolas each passing through \((\pm1, 0)\) and with a vertex on the \(y\)-axis. The curve we seek must satisfy \begin{align*} && y' &= \frac{1-x^2}{2xy} \\ \Rightarrow && \int2 y \d y &= \int \left ( \frac{1}{x} - x \right) \d x \\ \Rightarrow && y^2 &= \ln x - \tfrac12 x^2 + C \\ (1,1): && 1 &= -\tfrac12+C \\ \Rightarrow && C &= \frac32 \\ \Rightarrow && y^2 &= \tfrac32 + \ln x - \tfrac12 x^2 \end{align*}

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The function \(\f\) satisfies \(\f(x+1)= \f(x)\) and \(\f(x)>0\) for all \(x\).

1. Give an example of such a function.
2. The function \(\F\) satisfies \[ \frac{\d \F}{\d x} =\f(x) \] and \(\F(0)=0\). Show that \(\F(n) = n\F(1)\), for any positive integer \(n\).
3. Let \(y\) be the solution of the differential equation \[ \frac{\d y}{\d x} +\f(x) y=0 \] that satisfies \(y=1\) when \(x=0\). Show that \(y(n) \to 0\) as \(n\to\infty\), where \(n= 1,\,2,\, 3,\, \ldots\)

The curve \(C_1\) passes through the origin in the \(x\)--\(y\) plane and its gradient is given by $$ \frac{\d y}{\d x} =x(1-x^2)\e^{-x^2}. $$ Show that \(C_1\) has a minimum point at the origin and a maximum point at \(\left(1,{\frac12\, \e^{-1}} \right)\). Find the coordinates of the other stationary point. Give a rough sketch of \(C_1\). The curve \(C_2\) passes through the origin and its gradient is given by $$ \frac{\d y}{\d x}= x(1-x^2)\e^{-x^3}. $$ Show that \(C_2\) has a minimum point at the origin and a maximum point at \((1,k)\), where \phantom{} \(k > \frac12 \,\e^{-1}.\) (You need not find \(k\).)

In a cosmological model, the radius \(\rm R\) of the universe is a function of the age \(t\) of the universe. The function \(\rm R\) satisfies the three conditions: $$ \mbox{\({\rm R}(0)=0\)}, \ \ \ \ \ \ \ \ \ \mbox{\({\rm R'}(t)>0\) for \(t>0\)}, \ \ \ \ \ \ \ \ \ \ \mbox{\({\rm R''}(t)<0\) for \(t>0\)}, \tag{*} $$ where \({\rm R''}\) denotes the second derivative of \(\rm R\). The function \({\rm H}\) is defined by \[ {\rm H} (t)= \frac{{\rm R}'(t)}{{\rm R}( t)}\;. \]

1. Sketch a graph of \({\rm R} (t)\). By considering a tangent to the graph, show that \(t<1/{\rm H}(t)\).
2. Observations reveal that \({\rm H}(t) = a/t\), where \(a\) is constant. Derive an expression for \({\rm R}(t)\). What range of values of \(a\) is consistent with the three conditions \((*)\)?
3. Suppose, instead, that observations reveal that \({\rm H}(t)= b t^{-2}\), where \(b\) is constant. Show that this is not consistent with conditions \((*)\) for any value of \(b\).

Fluid flows steadily under a constant pressure gradient along a straight tube of circular cross-section of radius \(a\). The velocity \(v\) of a particle of the fluid is parallel to the axis of the tube and depends only on the distance \(r\) from the axis. The equation satisfied by \(v\) is \[\frac{1}{r}\frac{{\mathrm d}\ }{{\mathrm d}r} \left(r\frac{{\mathrm d}v}{{\mathrm d}r}\right) =-k,\] where \(k\) is constant. Find the general solution for \(v\). Show that \(|v|\rightarrow\infty\) as \(r\rightarrow 0\) unless one of the constants in your solution is chosen to be~\(0\). Suppose that this constant is, in fact, \(0\) and that \(v=0\) when \(r=a\). Find \(v\) in terms of \(k\), \(a\) and \(r\). The volume \(F\) flowing through the tube per unit time is given by \[F=2\pi\int_{0}^{a}rv\,{\mathrm d}r. \] Find \(F\).

1. At time \(t=0\) a tank contains one unit of water. Water flows out of the tank at a rate proportional to the amount of water in the tank. The amount of water in the tank at time \(t\) is \(y\). Show that there is a constant \(b<1\) such that \(y=b^{t}.\)
2. Suppose instead that the tank contains one unit of water at time \(t=0,\) but that in addition to water flowing out as described, water is added at a steady rate \(a>0.\) Show that \[ \frac{\mathrm{d}y}{\mathrm{d}t}-y\ln b=a, \] and hence find \(y\) in terms of \(a,b\) and \(t\).

1. In the differential equation \[ \frac{1}{y^{2}}\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{1}{y}=\mathrm{e}^{2x} \] make the substitution \(u=1/y,\) and hence show that the general solution of the original equation is \[ y=\frac{1}{A\mathrm{e}^{x}-\mathrm{e}^{2x}}\,. \]
2. Use a similar method to solve the equation \[ \frac{1}{y^{3}}\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{1}{y^{2}}=\mathrm{e}^{2x}. \]

Find the two solutions of the differential equation \[ \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}=4y \] which pass through the point \((a,b^{2}),\) where \(b\neq0.\) Find two distinct points \((a_{1},1)\) and \((a_{2},1)\) such that one of the solutions through each of them also passes through the origin. Show that the graphs of these two solutions coincide and sketch their common graph, together with the other solutions through \((a_{1},1)\) and \((a_{2},1)\). Now sketch sufficient members of the family of solutions (for varying \(a\) and \(b\)) to indicate the general behaviour. Use your sketch to identify a common tangent, and comment briefly on its relevance to the differential equation.

The vectors \({\bf a}\) and \({\bf b}\) lie in the plane \(\Pi\,\). Given that \(\vert {\bf a} \vert= 1\) and \({\bf a}.{\bf b} = 3,\) find, in terms of \({\bf a}\) and \({\bf b}\,\), a vector \({\bf p}\) parallel to \({\bf a}\) and a vector \({\bf q}\) perpendicular to \({\bf a}\,\), both lying in the plane \(\Pi\,\), such that $${\bf p}+{\bf q}={\bf a}+{\bf b}\;.$$ The vector \({\bf c}\) is not parallel to the plane \(\Pi\) and is such that \({\bf a}.{\bf c} = -2\) and \({\bf b}.{\bf c} = 2\,\). Given that \(\vert {\bf b} \vert = 5\,\), find, in terms of \({\bf a}, {\bf b}\) and \({\bf c},\) vectors \({\bf P}\), \({\bf Q}\) and \({\bf R}\) such that \({\bf P}\) and \({\bf Q}\) are parallel to \({\bf p}\) and \({\bf q},\) respectively, \({\bf R}\) is perpendicular to the plane \(\Pi\) and $${\bf P} + {\bf Q} + {\bf R} = {\bf a}+{\bf b}+{\bf c}\;.$$

The position vectors of the points \(A\,\), \(B\,\) and \(P\) with respect to an origin \(O\) are \(a{\bf i}\,\), \(b{\bf j}\,\) and \(l{\bf i}+m{\bf j}+n{\bf k}\,\), respectively, where \(a\), \(b\), and \(n\) are all non-zero. The points \(E\), \(F\), \(G\) and~\(H\) are the midpoints of \(OA\), \(BP\), \(OB\) and \(AP\), respectively. Show that the lines \(EF\) and \(GH\) intersect. Let \(D\) be the point with position vector \(d{\bf k}\), where \(d\) is non-zero, and let \(S\) be the point of intersection of \(EF\) and \(GH.\) The point \(T\) is such that the mid-point of \(DT\) is \(S\). Find the position vector of \(T\) and hence find \(d\) in terms of \(n\) if \(T\) lies in the plane \(OAB\).

Consider the equations \begin{alignat*}{2} ax-&y- \ z && =3 \;,\\ 2ax -&y -3z && = 7 \;,\\ 3ax-&y-5z && =b \;, \end{alignat*} where \(a\) and \(b\) are given constants.

1. In the case \(a=0\,\), show that the equations have a solution if and only if \(b=11\,\).
2. In the case \(a\ne0\) and \(b=11\,\) show that the equations have a solution with \(z=\lambda\) for any given number \(\lambda\,\).
3. In the case \(a=2\) and \(b=11\,\) find the solution for which \(x^2+y^2+z^2\) is least.
4. Find a value for \(a\) for which there is a solution such that \(x>10^6\) and \(y^2+z^2<1\,\).

For all values of \(a\) and \(b,\) either solve the simultaneous equations \begin{alignat*}{1} x+y+az & =2\\ x+ay+z & =2\\ 2x+y+z & =2b \end{alignat*} or prove that they have no solution.

Consider the system of equations \begin{alignat*}{1} 2yz+zx-5xy & =2\\ yz-zx+2xy & =1\\ yz-2zx+6xy & =3 \end{alignat*} Show that \[xyz=\pm 6\] and find the possible values of \(x\), \(y\) and \(z\).

Find the simultaneous solutions of the three linear equations \begin{alignat*}{1} a^{2}x+ay+z & =a^{2}\\ ax+y+bz & =1\\ a^{2}bx+y+bz & =b \end{alignat*} for all possible real values of \(a\) and \(b\).

The matrices \(\mathbf{A},\mathbf{B}\) and \(\mathbf{M}\) are given by \[ \mathbf{A}=\begin{pmatrix}a & 0 & 0\\ b & c & 0\\ d & e & f \end{pmatrix},\quad\mathbf{B}=\begin{pmatrix}1 & p & q\\ 0 & 1 & r\\ 0 & 0 & 1 \end{pmatrix},\quad\mathbf{M}=\begin{pmatrix}1 & 3 & 2\\ 4 & 13 & 5\\ 3 & 8 & 7 \end{pmatrix}, \] where \(a,b,\ldots,r\) are real numbers. Given that \(\mathbf{M=AB},\) show that \(a=1,b=4,c=1,d=3,e=1,f=-2,p=3,q=2\) and \(r=-3\) gives the unique solution for \(\mathbf{A}\) and \(\mathbf{B}.\) Evaluate \(\mathbf{A}^{-1}\) and \(\mathbf{B}^{-1},\) Hence, or otherwise, solve the simultaneous equations \begin{alignat*}{1} x+3y+2z & =7\\ 4x+13y+5z & =18\\ 3x+8y+7z & =25. \end{alignat*}

The point \(P\) moves on a straight line in three-dimensional space. The position of \(P\) is observed from the points \(O_{1}(0,0,0)\) and \(O_{2}(8a,0,0).\) At times \(t=t_{1}\) and \(t=t_{1}'\), the lines of sight from \(O_{1}\) are along the lines \[ \frac{x}{2}=\frac{z}{3},y=0\quad\mbox{ and }\quad x=0,\frac{y}{3}=\frac{z}{4} \] respectively. At times \(t=t_{2}\) and \(t=t_{2}'\), the lines of sight from \(O_{2}\) are \[ \frac{x-8a}{-3}=\frac{y}{1}=\frac{z}{3}\quad\mbox{ and }\quad\frac{x-8a}{-4}=\frac{y}{2}=\frac{z}{5} \] respectively. Find an equation or equations for the path of \(P\).

The distinct points \(P_{1},P_{2},P_{3},Q_{1},Q_{2}\) and \(Q_{3}\) in the Argand diagram are represented by the complex numbers \(z_{1},z_{2},z_{3},w_{1},w_{2}\) and \(w_{3}\) respectively. Show that the triangles \(P_{1}P_{2}P_{3}\) and \(Q_{1}Q_{2}Q_{3}\) are similar, with \(P_{i}\) corresponding to \(Q_{i}\) (\(i=1,2,3\)) and the rotation from \(1\) to \(2\) to \(3\) being in the same sense for both triangles, if and only if \[ \frac{z_{1}-z_{2}}{z_{2}-z_{3}}=\frac{w_{1}-w_{2}}{w_{1}-w_{3}}. \] Verify that this condition may be written \[ \det\begin{pmatrix}z_{1} & z_{2} & z_{3}\\ w_{1} & w_{2} & w_{3}\\ 1 & 1 & 1 \end{pmatrix}=0. \]

1. Show that if \(w_{i}=z_{i}^{2}\) (\(i=1,2,3\)) then triangle \(P_{1}P_{2}P_{3}\) is not similar to triangle \(Q_{1}Q_{2}Q_{3}.\)
2. Show that if \(w_{i}=z_{i}^{3}\) (\(i=1,2,3\)) then triangle \(P_{1}P_{2}P_{3}\) is similar to triangle \(Q_{1}Q_{2}Q_{3}\) if and only if the centroid of triangle \(P_{1}P_{2}P_{3}\) is the origin. {[}The centroid of triangle \(P_{1}P_{2}P_{3}\) is represented by the complex number \(\frac{1}{3}(z_{1}+z_{2}+z_{3})\).{]}
3. Show that the triangle \(P_{1}P_{2}P_{3}\) is equilateral if and only if \[ z_{2}z_{3}+z_{3}z_{1}+z_{1}z_{2}=z_{1}^{2}+z_{2}^{2}+z_{3}^{2}. \]

State carefully the conditions which the fixed vectors \(\mathbf{a,b,u}\) and \(\mathbf{v}\) must satisfy in order to ensure that the line \(\mathbf{r=a+}\lambda\mathbf{u}\) intersects the line \(\mathbf{r=b+\mu}\mathbf{v}\) in exactly one point. Find the two values of the fixed scalar \(b\) for which the planes with equations \[ \left.\begin{array}{c} x+y+bz=b+2\\ bx+by+z=2b+1 \end{array}\right\} \tag{*} \] do not intersect in a line. For other values of \(b\), express the line of intersection of the two planes in the form \(\mathbf{r=a}+\lambda\mathbf{u},\) where \(\mathbf{a\cdot u}=0\). Find the conditions which \(b\) and the fixed scalars \(c\) and \(d\) must satisfy to ensure that there is exactly one point on the line \[ \mathbf{r=}\left(\begin{array}{c} 0\\ 0\\ c \end{array}\right)+\mu\left(\begin{array}{c} 1\\ d\\ 0 \end{array}\right) \] whose coordinates satisfy both equations \((*)\).