Problems

Add Problem
Filters
Clear Filters
2010 Paper 3 Q1
D: 1700.0 B: 1500.8

Let \(x_{\low1}\), \(x_{\low2}\), \ldots, \(x_n\) and \(x_{\vphantom {\dot A} n+1}\) be any fixed real numbers. The numbers \(A\) and \(B\) are defined by \[ A = \frac 1 n \sum_{k=1}^n x_{ \low k} \,, \ \ \ B= \frac 1 n \sum_{k=1}^n (x_{\low k}-A)^2 \,, \ \ \ \] and the numbers \(C\) and \(D\) are defined by \[ C = \frac 1 {n+1} \sum\limits_{k=1}^{n+1} x_{\low k} \,, \ \ \ D = \frac1{n+1} \sum_{k=1}^{n+1} (x_{\low k}-C)^2 \,. \]

  1. Express \( C\) in terms of \(A\), \(x_{\low n+1}\) and \(n\).
  2. Show that $ \displaystyle B= \frac 1 n \sum_{k=1}^n x_{\low k}^2 - A^2\,\(.
  3. Express \)D \( in terms of \)B\(, \)A\(, \)x_{\low n+1}\( and \)n$. Hence show that \((n + 1)D \ge nB\) for all values of \(x_{\low n+1}\), but that \(D < B\) if and only if \[ A-\sqrt{\frac{(n+1)B}{n}}

Edit
2010 Paper 3 Q2
D: 1700.0 B: 1485.5

In this question, \(a\) is a positive constant.

  1. Express \(\cosh a\) in terms of exponentials. By using partial fractions, prove that \[ \int_0^1 \frac 1{ x^2 +2x\cosh a +1} \, \d x = \frac a {2\sinh a}\,. \]
  2. Find, expressing your answers in terms of hyperbolic functions, \[ \int_1^\infty \frac 1 {x^2 +2x \sinh a -1} \,\d x \, \] and \[ \int_0^\infty \frac 1 {x^4 +2x^2\cosh a +1} \,\d x \,.\]

Solution:

  1. \(\cosh a = \frac12 (e^a + e^{-a})\) \begin{align*} \int_0^1 \frac 1{ x^2 +2x\cosh a +1} \, \d x &= \int_0^1 \frac{1}{x^2+(e^a+e^{-a})x+e^ae^{-a}} \d x \\ &= \int_0^1 \frac{1}{e^a-e^{-a}}\left (\frac{1}{x+e^{-a}}-\frac{1}{x+e^a} \right)\d x \\ &= \frac{1}{2 \sinh a} \int_0^1 \left (\frac{1}{x+e^{-a}}-\frac{1}{x+e^a} \right)\d x \\ &= \frac{1}{2 \sinh a}\left [\ln(x+e^{-a})-\ln(x+e^a) \right]_0^1 \\ &= \frac{1}{2 \sinh a} \left (\ln(1+e^a)-\ln(1+e^{-a}) - (\ln e^{-a}-\ln e^a) \right) \\ &= \frac{1}{2\sinh a}\left (2a + \ln \frac{1+e^a}{1+e^{-a}}\right) \\ &= \frac1{2\sinh a} \left ( 2a -a \right) \\ &= \frac{a}{2 \sinh a} \end{align*}
  2. \begin{align*} \int_1^\infty \frac 1 {x^2 +2x \sinh a -1} \,\d x &= \int_1^{\infty} \frac{1}{(x+e^a)(x-e^{-a})} \d x \\ &= \int_1^{\infty} \frac{1}{e^a+e^{-a}} \left ( \frac{1}{x-e^{-a}} - \frac{1}{x+e^{a}} \right)\d x \\ &= \frac{1}{2\cosh a} \int_1^{\infty} \left ( \frac{1}{x-e^{-a}} - \frac{1}{x+e^{a}} \right)\d x \\ &= \frac{1}{2\cosh a} \left [\ln(x-e^{-a}) - \ln (x + e^{a} ) \right]_1^{\infty} \\ &= \frac1{2\cosh a} \left [ \ln \frac{x-e^{-a}}{x+e^{a}} \right]_1^{\infty} \\ &= \frac{1}{2\cosh a} \left ( 0 - \ln \frac{1-e^{-a}}{1+e^a}{}\right) \\ &= \frac{1}{2\cosh a} \ln \frac{1+e^a}{1-e^{-a}}\\ &= \frac{1}{2\cosh a} \left ( a + \ln \coth \frac{a}{2} \right) \end{align*} and \begin{align*} \int_0^\infty \frac 1 {x^4 +2x^2\cosh a +1} \,\d x &= \int_0^\infty\frac{1}{(x^2+e^a)(x^2+e^{-a})} \d x \\ &= \int_0^\infty \frac{1}{e^a-e^{-a}} \left ( \frac{1}{x^2+e^{-a}} - \frac{1}{x^2+e^{a}} \right) \d x \\ &= \frac{1}{2\sinh a} \left [ \frac{1}{e^{-a/2}} \tan^{-1} \frac{x}{e^{-a/2}} - \frac{1}{e^{a/2}}\tan^{-1} \frac{x}{e^{a/2}} \right]_0^{\infty} \\ &= \frac{1}{2\sinh a} \left (e^{a/2}\frac{\pi}{2}-e^{-a/2}\frac{\pi}{2} - 0 \right) \\ &= \frac{1}{2\sinh a} \pi \sinh \frac{a}{2} \\ &= \frac{\pi \sinh \tfrac{a}{2}}{2\sinh a} \\ &= \frac{\pi \sinh \tfrac{a}{2}}{4\sinh \tfrac{a}{2} \cosh \tfrac{a}{2}} \\ &= \frac{\pi}{4\cosh \tfrac{a}{2}} \end{align*}

Edit
2010 Paper 3 Q3
D: 1700.0 B: 1545.2

For any given positive integer \(n\), a number \(a\) (which may be complex) is said to be a primitive \(n\)th root of unity if \(a^n=1\) and there is no integer \(m\) such that \(0 < m < n\) and \(a^m = 1\). Write down the two primitive 4th roots of unity. Let \({\rm C}_n(x)\) be the polynomial such that the roots of the equation \({\rm C}_n(x)=0\) are the primitive \(n\)th roots of unity, the coefficient of the highest power of \(x\) is one and the equation has no repeated roots. Show that \({\rm C}_4(x) = x^2+1\,\).

  1. Find \({\rm C}_1(x)\), \({\rm C}_2(x)\), \({\rm C}_3(x)\), \({\rm C}_5(x)\) and \({\rm C}_6(x)\), giving your answers as unfactorised polynomials.
  2. Find the value of \(n\) for which \({\rm C}_n(x) = x^4 + 1\).
  3. Given that \(p\) is prime, find an expression for \({\rm C}_p(x)\), giving your answer as an unfactorised polynomial.
  4. Prove that there are no positive integers \(q\), \(r\) and \(s\) such that \({\rm C}_q(x) \equiv {\rm C}_r(x) {\rm C}_s(x)\,\).

Solution: The primitive 4th roots of unity are \(i\) and \(-i\). (Since the other two roots of \(x^4-1\) are also roots of \(x^2-1\) \({\rm C}_4(x) = (x-i)(x+i) = x^2+1\) as required.

  1. \(\,\) \begin{align*} && {\rm C}_1 (x) &= x-1 \\ && {\rm C}_2 (x) &= x+1 \\ && {\rm C}_3 (x) &= x^2+x+1 \\ && {\rm C}_5 (x) &= x^4+x^3+x^2+x+1 \\ && {\rm C}_6 (x) &= x^2-x+1 \\ \end{align*}
  2. Since \((x^4+1)(x^4-1) = x^8-1\) we must have \(n \mid 8\). But \(n \neq 1,2,4\) so \(n = 8\).
  3. \({\rm C}_p(x) = x^{p-1} +x^{p-2}+\cdots+x+1\)
  4. Suppose \({\rm C_q}(x) \equiv {\rm C}_r(x){\rm C}_s(x)\), then if \(\omega\) is a primitive \(q\)th root of unity we must \({\rm C}_q(\omega) = 0\), but that means that one of \({\rm C}_r(\omega)\), \({\rm C}_s(\omega)\) is \(0\). But that's only possible if \(r\) or \(s\) \(=q\). If this were the case, then what would the other value be? There are no possible values, hence it's not possible.

Edit
2010 Paper 3 Q4
D: 1700.0 B: 1516.0

  1. The number \(\alpha\) is a common root of the equations \(x^2 +ax +b=0\) and \(x^2+cx+d=0\) (that is, \(\alpha\) satisfies both equations). Given that \(a\ne c\), show that \[ \alpha =- \frac{b-d}{a-c}\,. \] Hence, or otherwise, show that the equations have at least one common root if and only if \[ (b-d)^2 -a(b-d)(a-c) + b(a-c)^2 =0\,. \] Does this result still hold if the condition \(a\ne c\) is not imposed?
  2. Show that the equations \(x^2+ax+b=0\) and \(x^3+(a+1)x^2+qx+r=0\) have at least one common root if and only if \[ (b-r)^2-a(b-r)(a+b-q) +b(a+b-q)^2=0\,. \] Hence, or otherwise, find the values of \(b\) for which the equations \(2x^2+5 x+2 b=0\) and \(2x^3+7x^2+5x+1=0\) have at least one common root.

Solution:

  1. \begin{align*} && 0 &= \alpha^2 + a \alpha + b \tag{1} \\ && 0 &= \alpha^2 + c \alpha + d \tag{2} \\ \\ (1) - (2): && 0 & =\alpha ( a-c) + (b-d) \\ \Rightarrow && \alpha &= - \frac{b-d}{a-c} \tag{\(a\neq c\)} \end{align*} (\(\Rightarrow\)) Suppose they have a common root, then given we know it's form, we must have: \begin{align*} && 0 &= \left ( - \frac{b-d}{a-c} \right)^2 +a\left ( - \frac{b-d}{a-c} \right) + b \\ \Rightarrow && 0 &= (b-d)^2 - a(b-d)(a-c) + b(a-c)^2 \end{align*} (\(\Leftarrow\)) Suppose the equation holds, then \begin{align*} && 0 &= (b-d)^2 - a(b-d)(a-c) + b(a-c)^2 \\ \Rightarrow && 0 &= \left ( - \frac{b-d}{a-c} \right)^2 +a\left ( - \frac{b-d}{a-c} \right) + b \\ \end{align*} So \(\alpha\) is a root of the first equation. Considering \((1) - (2)\) we must have that \(\alpha(a-c) +(b-d) = t\) (whatever the second equation is), but that value is clearly \(0\), therefore \(\alpha\) is a root of both equations. If \(a = c\) then the equation becomes \(0 = (b-d)^2\), ie the two equations are the same, therefore they must have common roots!
  2. \begin{align*} && 0 &= x^2+ax+b \tag{1} \\ && 0 &= x^3+(a+1)x^2+qx+r \tag{2} \\ \\ (2) - x(1) && 0 &= x^2 + (q-b)x + r \tag{3} \end{align*} Therefore if the equations have a common root, \((1)\) and \((3)\) have a common root, ie \((b-r)^2-a(b-r)(a-(q-b))+b(a-(q-b))^2 = 0\) which is exactly our condition. \(a = \frac52, q = \frac52, r = \frac12\) \begin{align*} && 0 &= \left (b-\frac12 \right)^2 - \frac52\left (b-\frac12\right) b + b^3 \\ &&&= b^2 -b + \frac14 - \frac52 b^2+\frac54b + b^3 \\ &&&= b^3 -\frac32 b^2 +\frac14 b + \frac14 \\\Rightarrow && 0 &= 4b^3 - 6b^2+b + 1 \\ &&&= (b-1)(4b^2-2b-1) \\ \Rightarrow && b &= 1, \frac{1 \pm \sqrt{5}}{4}\end{align*}

Edit
2010 Paper 3 Q5
D: 1700.0 B: 1486.7

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.]

Edit
2010 Paper 3 Q6
D: 1700.0 B: 1484.0

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.

Edit
2010 Paper 3 Q7
D: 1700.0 B: 1516.0

Given that \(y = \cos(m \arcsin x)\), for \(\vert x \vert <1\), prove that \[ (1-x^2) \frac {\d^2 y}{\d x^2} -x \frac {\d y}{\d x} +m^2y=0\,. \] Obtain a similar equation relating \(\dfrac{\d^3y}{\d x^3}\,\),\; \(\dfrac{\d^2y}{\d x^2}\, \) and \(\, \dfrac{\d y}{\d x}\,\), and a similar equation relating \(\dfrac{\d^4y}{\d x^4}\,\),~~\(\dfrac{\d^3y}{\d x^3}\,\) and \(\,\dfrac{\d^2 y}{\d x^2}\,\). Conjecture and prove a relation between \(\dfrac{\d^{n+2}y}{\d x^{n+2}}\,\), \ \(\dfrac{\d^{n+1}y}{\d x^{n+1}}\;\) and \(\;\dfrac{\d^n y}{\d x^n}\,\). Obtain the first three non-zero terms of the Maclaurin series for \(y\). Show that, if \(m\) is an even integer, \(\cos m\theta\) may be written as a polynomial in \(\sin\theta\) beginning \[ 1 - \frac{m^2\sin^2\theta}{2!}+ \frac{m^2(m^2-2^2)\sin^4\theta}{4!} -\cdots \,. \, \tag{\(\vert\theta\vert < \tfrac12 \pi\)} \] State the degree of the polynomial.

Edit
2010 Paper 3 Q8
D: 1700.0 B: 1531.5

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 \[ (1+\cos x +2 \sin x) \frac {\d y}{\d x} +(\sin x -2 \cos x)y = 5 - 3 \cos x + 4 \sin x\,. \]

Edit
2010 Paper 3 Q9
D: 1700.0 B: 1468.9

\(\,\)

\newrgbcolor{wwwwww}{0.4 0.4 0.4} \psset{xunit=1.0cm,yunit=1.0cm,algebraic=true,dotstyle=o,dotsize=3pt 0,linewidth=0.3pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(1.92,2.15)(7.25,6.21) \pspolygon[linecolor=wwwwww,fillcolor=wwwwww,fillstyle=solid,opacity=0.75](2.27,2.85)(2.27,2.52)(6.48,2.52)(6.48,2.85) \psline[linecolor=wwwwww](2.27,2.85)(2.27,2.52) \psline[linecolor=wwwwww](2.27,2.52)(6.48,2.52) \psline[linecolor=wwwwww](6.48,2.52)(6.48,2.85) \psline[linewidth=1.2pt,linecolor=wwwwww](6.48,2.85)(2.27,2.85) \psline(6.48,2.85)(6.48,5.88) \pscustom[linewidth=0.5pt]{\parametricplot{1.5707963267948966}{3.141592653589793}{1*3.02*cos(t)+0*3.02*sin(t)+6.48|0*3.02*cos(t)+1*3.02*sin(t)+2.85}\lineto(6.48,2.85)\closepath} \parametricplot{1.5450288353258959}{2.629484171074415}{1*3.1*cos(t)+0*3.1*sin(t)+6.48|0*3.1*cos(t)+1*3.1*sin(t)+2.85} \psline(6.56,3.94)(6.56,5.96) \psline[linestyle=dashed,dash=1pt 1pt](3.78,4.37)(6.48,2.85) \rput[tl](5.07,4.02){\(a\)} \rput[tl](5.76,3.17){\(\theta \)} \parametricplot{2.629484171074415}{3.141592653589793}{1*0.87*cos(t)+0*0.87*sin(t)+6.48|0*0.87*cos(t)+1*0.87*sin(t)+2.85} \rput[tl](6.61,3.14){\(O\)} \rput[tl](3.31,4.89){\(P\)} \rput[tl](6.85,4.39){\(Q\)} \begin{scriptsize} \psdots[dotsize=6pt 0,dotstyle=*](3.78,4.37) \psdots[dotsize=6pt 0,dotstyle=*](6.56,3.94) \end{scriptsize} \end{pspicture*}
The diagram shows two particles, \(P\) and \(Q\), connected by a light inextensible string which passes over a smooth block fixed to a horizontal table. The cross-section of the block is a quarter circle with centre \(O\), which is at the edge of the table, and radius \(a\). The angle between \(OP\) and the table is \(\theta\). The masses of \(P\) and \(Q\) are \(m\) and \(M \), respectively, where \(m < M\). Initially, \(P\) is held at rest on the table and in contact with the block, \(Q\) is vertically above \(O\), and the string is taut. Then \(P\) is released. Given that, in the subsequent motion, \(P\) remains in contact with the block as \(\theta\) increases from \(0\) to \(\frac12\pi\), find an expression, in terms of \(m\), \(M\), \(\theta\) and \(g\), for the normal reaction of the block on \(P\) and show that \[ \frac{m}{M} \ge \frac{\pi-1}3\,. \]

Edit
2010 Paper 3 Q10
D: 1700.0 B: 1500.0

A small bead \(B\), of mass \(m\), slides without friction on a fixed horizontal ring of radius \(a\). The centre of the ring is at \(O\). The bead is attached by a light elastic string to a fixed point \(P\) in the plane of the ring such that \(OP = b\), where \(b > a\). The natural length of the elastic string is \(c\), where \(c < b - a\), and its modulus of elasticity is \(\lambda\). Show that the equation of motion of the bead is \[ ma\ddot \phi = -\lambda\left( \frac{a\sin\phi}{c\sin\theta}-1\right)\sin(\theta+\phi) \,, \] where \(\theta=\angle BPO\) and \(\phi=\angle BOP\). Given that \(\theta\) and \(\phi\) are small, show that $a(\theta+\phi)\approx b\theta$. Hence find the period of small oscillations about the equilibrium position \(\theta=\phi =0\).

Edit