PDF generation for Dimensional Analysis has been started.
PDF generation for Approximating Binomial to Normal Distribution has been started.

Problems

Add Problem
Filters
Clear Filters
2006 Paper 1 Q1
D: 1500.0 B: 1516.0

Find the integer, \(n\), that satisfies \(n^2 < 33\,127< (n+1)^2\). Find also a small integer \(m\) such that \((n+m)^2-33\,127\) is a perfect square. Hence express \(33\,127\) in the form \(pq\), where \(p\) and \(q\) are integers greater than \(1\). By considering the possible factorisations of \(33\, 127\), show that there are exactly two values of \(m\) for which \((n+m)^2 -33\,127\) is a perfect square, and find the other value.

Solution: \begin{align*} 180^2 &= 32400 \\ 181^2 &= 32761 \\ 182^2 &= 33124 \\ 183^2 &= 33489 \\ 184^2 &= 33856 \end{align*} Therefore \(182^2 < 33\,127 < (182+1)^2\). and \((182+2)^2 - 33\,127 = 729 = 27^2\). Therefore \(33\,127 = 184^2 - 27^2 = 211 \times 157\). (Note both of these numbers are prime). Suppose \((n+m)^2 - 33\,127 = k^2\) then \(33\,127 = (n+m)^2-k^2 = (n+m-k)(n+m+k)\). Since there are only two factorisations of \(33\,127\) into positive integer factors with one factor larger than the other, the other factorisation must be: \(n+m+k = 33\,127, n+m-k = 1 \Rightarrow k = \frac{33\, 126}{2} = 16563\), ie \(16564^2 - 33\,127 = 16563^2\)

Edit
2006 Paper 1 Q2
D: 1516.0 B: 1500.0

A small goat is tethered by a rope to a point at ground level on a side of a square barn which stands in a large horizontal field of grass. The sides of the barn are of length \(2a\) and the rope is of length \(4a\). Let \(A\) be the area of the grass that the goat can graze. Prove that \(A\le14\pi a^2\) and determine the minimum value of \(A\).

Solution:

TikZ diagram
The areas are \(8\pi a^2 + \frac14 \pi (4a-x)^2 + \frac14 \pi (2a-x)^2 + \frac14\pi(2a+x)^2+\frac14 \pi x^2\) ie \begin{align*} A &= \frac{\pi}{4} \left ( x^2 \left (1 + 1 + 1 + 1 \right) + x \left (4a-4a-8a \right)+\left (32a^2+16a^2+4a^2+4a^2 \right)\right) \\ &= \frac{\pi}{4} \left (4x^2-8ax+56a^2 \right) \\ &= \pi(x^2-2ax+14a^2) \\ &= \pi ((x-a)^2+13a^2) \end{align*} Since \(x \in [0, 2a]\) we have \(13\pi a^2 \leq A \leq 14 \pi a^2\)

Edit
2006 Paper 1 Q3
D: 1500.0 B: 1500.0

In this question \(b\), \(c\), \(p\) and \(q\) are real numbers.

  1. By considering the graph \(y=x^2 + bx + c\) show that \(c < 0\) is a sufficient condition for the equation \(\displaystyle x^2 + bx + c = 0\) to have distinct real roots. Determine whether \(c < 0\) is a necessary condition for the equation to have distinct real roots.
  2. Determine necessary and sufficient conditions for the equation \(\displaystyle x^2 + bx + c = 0\) to have distinct positive real roots.
  3. What can be deduced about the number and the nature of the roots of the equation \(x^3 + px + q = 0\) if \(p>0\) and \(q<0\)? What can be deduced if \(p<0\,\) and \(q<0\)? You should consider the different cases that arise according to the value of \(4p^3+ 27q^2\,\).

Solution:

  1. TikZ diagram
    Since \(y(0) < 0\) and \(y(\pm \infty) > 0\) we must cross the axis twice. Therefore there are two distinct real roots. It is not necessary, for example \((x-2)(x-3)\) has distinct real roots by the constant term is \(6 > 0\)
  2. For \(x^2+bx+c=0\) to have distinct, positive real roots we need \(\Delta > 0\) and \(\frac{-b -\sqrt{\Delta}}{2a} > 0\) where \(\Delta = b^2-4ac\), ie \(b < 0\) and \(b^2 > \Delta = b^2-4ac\) or \(4ac > 0\). Therefore we need \(b^2-4ac > 0, b < 0, 4ac > 0\)
  3. Since \(q < 0\) at least one of the roots is positive. The gradient is \(3x^2+p > 0\) therefore there is exactly one positive root. If \(p < 0\) then there are turning points when \(3x^2+p = 0\) ie \(x = \pm \sqrt{\frac{-p}{3}}\). If the first turning point is above the \(x\)-axis then there will be 3 roots. If it is on the \(x\)-axis then 2, otherwise only 1. \begin{align*} y &= \left (-\sqrt{\frac{-p}{3}}\right)^3 + p\left (-\sqrt{\frac{-p}{3}}\right)+q \\ &= \sqrt{\frac{-p}{3}} \left (p - \frac{p}{3} \right) + q \\ &= \frac{2}{3} \sqrt{\frac{-p}{3}}p +q \\ \end{align*} Therefore it is positive if \(-\frac{4}{27}p^3 >q^2\) ie if \(4p^3+27q^2 < 0\)

Edit
2006 Paper 1 Q4
D: 1500.0 B: 1514.2

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 \(x\).
A regular polygon has \(n\) sides, and perimeter \(P\). Show that the area of the polygon is \[ \displaystyle \frac{P^2} { {4n \tan \l\dfrac{ \pi} { n} \r}} \;. \] Show by differentiation (treating \(n\) as a continuous variable) that the area of the polygon increases as \(n\) increases with \(P\) fixed. Show also that, for large \(n\), the ratio of the area of the polygon to the area of the smallest circle which can be drawn around the polygon is approximately \(1\).

Edit
2006 Paper 1 Q5
D: 1500.0 B: 1500.0

  1. Use the substitution \(u^2=2x+1\) to show that, for \(x>4\), \[ \int \frac{3} { ( x-4 ) \sqrt {2x+1}} \; \d x = \ln \l \frac{\sqrt{2x+1}-3} {\sqrt{2x+1}+3} \r + K\,, \] where \(K\) is a constant.
  2. Show that $ \displaystyle \int_{\ln 3}^{\ln 8} \frac{2} { \e^x \sqrt{ \e^x + 1}}\; \mathrm{d}x\, = \frac 7{12} + \ln \frac23 $ .

Solution:

  1. \begin{align*} && I &= \int \frac{3}{(x-4)\sqrt{2x+1}}\, \d x \\ u^2 =2x+1, 2u \frac{\d u}{\d x}=2: && &= \int \frac{3}{\left(\frac{u^2-1}{2}-4\right)u} u \d u \\ &&&= \int \frac{6}{u^2-9} \d u \\ &&&= \int \frac{6}{(u-3)(u+3)} \d u\\ &&&= \int \left ( \frac{1}{u-3} - \frac{1}{u+3} \right )\d u \\ &&&= \ln (u-3) - \ln (u+3) + C \\ &&&= \ln \frac{u-3}{u+3} + C \\ &&&= \ln \left (\frac{\sqrt{2x+1}-3}{\sqrt{2x+1}+3} \right) + C \end{align*}
  2. \begin{align*} && I &= \int_{\ln 3}^{\ln 8} \frac{2}{e^x\sqrt{e^x+1}} \d x \\ u = e^x, \frac{\d u}{\d x} = e^x: &&&= \int_{u=3}^{u=8} \frac{2}{u\sqrt{u+1}} \frac{1}{u} \d u \\ v^2=u+1, 2v \frac{\d v}{\d u} = 1: &&&= \int_{v=2}^{v=3} \frac{2}{v(v^2-1)^2} \d v \\ &&&= \int_2^3 \left ( \frac{2}{v} - \frac{}{v-1} - \frac{1}{2(v-1)^2} - \frac{1}{v+1} - \frac{1}{2(v+1)^2}\right) \d v \\ &&&= \left [2\ln v - \ln(v^2-1)+\frac12(v-1)^{-1}+\frac12(v+1)^{-1} \right]_2^3 \\ &&&= \left ( 2\ln3-\ln 8+\frac14+\frac18\right)-\left ( 2\ln2-\ln 3+\frac12+\frac16\right) \\ &&&= \end{align*}

Edit
2006 Paper 1 Q6
D: 1500.0 B: 1606.3

  1. Show that, if \(\l a \, , b\r\) is any point on the curve \(x^2 - 2y^2 = 1\), then \(\l 3a + 4b \, , 2a + 3b \r\,\) also lies on the curve.
  2. Determine the smallest positive integers \(M\) and \(N\) such that, if \(\l a \,, b\r\) is any point on the curve \(Mx^2 - Ny^2 = 1\), then \((5a+6b\,, 4a+5b)\) also lies on the curve.
  3. Given that the point \(\l a \, , b\r\) lies on the curve \(x^2 - 3y^2 = 1\,\), find positive integers \(P\), \(Q\), \(R\) and \(S\) such that the point \((P a +Q b\,, R a + Sb)\) also lies on the curve.

Solution:

  1. Suppose \(a^2-2b^2=1\) then \begin{align*} (3a+4b)^2-2(2a+3b)^2 &= 9a^2+24ab+16b^2-2\cdot(4a^2+12ab+9b^2) \\ &=a^2-2b^2 \\ &= 1 \end{align*} Therefore \((3a+4b,2a+3b)\) also lies on the curve.
  2. Suppose \(Ma^2-Nb^2 = 1\) then \begin{align*} M(5a+6b)^2-N(4a+5b)^2 &= M\cdot(25a^2+60ab+36b^2) - N\cdot(16a^2+40ab+25b^2) \\ &= (25M-16N)a^2+20\cdot(3M-2N)ab+(36M-25N)b^2 \end{align*} Therefore we need \(3M = 2N\) so the smallest possible value would have to be \(M = 2, N = 3\), which does work
  3. Consider \(x + \sqrt{3}y\), then consider \((x+\sqrt{3}y)(2+\sqrt{3}) = (2x+3y)+(x+2y)\sqrt{3}\). Notice that \((x+\sqrt{3}y)(x-\sqrt{3}y) = 1\) and \((2+\sqrt{3})(2-\sqrt{3}) = 1\) so \(((2x+3y)+(x+2y)\sqrt{3})((2x+3y)-(x+2y)\sqrt{3}) = 1\), so we can take \(P=2,Q=3,R=1,S=2\)

Edit
2006 Paper 1 Q7
D: 1516.0 B: 1503.2

  1. Sketch on the same axes the functions \({\rm cosec}\, x\) and \(2x/ \pi\), for \(0 < x < \pi\,\). Deduce that the equation \(x\sin x = \pi/2 \) has exactly two roots in the interval \(0 < x < \pi\,\). Show that \[ \displaystyle \int_{\pi/2}^{\pi} \left \vert x\sin x - \frac{\pi} { 2} \right \vert \; \mathrm{d}x = 2\sin\alpha +\frac{3\pi^2} 4 - \alpha \pi -\pi -2\alpha \cos\alpha -1 \] where \(\alpha\) is the larger of the roots referred to above.
  2. Show that the region bounded by the positive \(x\)-axis, the \(y\)-axis and the curve \[y = \Bigl| \vert \e^x - 1 \vert - 1 \Bigr|\] has area \(\ln 4-1\).

Edit
2006 Paper 1 Q8
D: 1500.0 B: 1484.0

{\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 %\(\displaystyle \tfrac12 \sqrt{ \vphantom{\dot A } a^2b^2 +b^2c^2 + c^2 a^2 \;} \;\).
Hence show that \(d\), the perpendicular distance of the origin from the triangle \(ABC\), satisfies \[ \frac 1{d^2} = \frac 1 {a^2} + \frac 1 {b^2} + \frac 1 {c^2} \,. \]

Edit
2006 Paper 1 Q9
D: 1500.0 B: 1484.0

A block of mass \(4\,\)kg is at rest on a smooth, horizontal table. A smooth pulley \(P\) is fixed to one edge of the table and a smooth pulley \(Q\) is fixed to the opposite edge. The two pulleys and the block lie in a straight line. Two horizontal strings are attached to the block. One string runs over pulley \(P\); a particle of mass \(x\,\)kg hangs at the end of this string. The other string runs over pulley \(Q\); a particle of mass \(y\,\)kg hangs at the end of this string, where \(x > y\) and \(x + y = 6\,\). The system is released from rest with the strings taut. When the \(4\,\)kg block has moved a distance \(d\), the string connecting it to the particle of mass \(x\,\)kg is cut. Show that the time taken by the block from the start of the motion until it first returns to rest (assuming that it does not reach the edge of the table) is \(\sqrt{d/(5g)\,} \,\f(y)\), where \[ \f(y)= \frac{10}{ \sqrt{6-2y}}+ \left(1 + \frac{4}{ y} \right) \sqrt {6 -2y}. \] Calculate the value of \(y\) for which \(\f'(y)=0\).

Edit
2006 Paper 1 Q10
D: 1484.0 B: 1500.0

A particle \(P\) is projected in the \(x\)-\(y\) plane, where the \(y\)-axis is vertical and the \(x\)-axis is horizontal. The particle is projected with speed \(V\) from the origin at an angle of \(45 ^\circ\) above the positive \(x\)-axis. Determine the equation of the trajectory of \(P\). The point of projection (the origin) is on the floor of a barn. The roof of the barn is given by the equation \(y= x \tan \alpha +b\,\), where \(b>0\) and \(\alpha\) is an acute angle. Show that, if the particle just touches the roof, then \(V(-1+ \tan\alpha) =-2 \sqrt{bg}\); you should justify the choice of the negative root. If this condition is satisfied, find, in terms of \(\alpha\), \(V\) and \(g\), the time after projection at which touching takes place. A particle \(Q\) can slide along a smooth rail fixed, in the \(x\)-\(y\) plane, to the under-side of the roof. It is projected from the point \((0,b)\) with speed \(U\) at the same time as \(P\) is projected from the origin. Given that the particles just touch in the course of their motions, show that \[ 2 \sqrt 2 \, U \cos \alpha = V \big(2 + \sin\alpha\cos\alpha -\sin^2\alpha) . \]

Edit