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2011 Paper 1 Q1
D: 1500.0 B: 1479.0

  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 \(a^2-b^2 = \frac12\,\).

Solution:

  1. \(\,\) \begin{align*} && 1 &= \frac{a}{x} + \frac{b}{y} \\ \frac{\d}{\d x}: && 0 &= -\frac{a}{x^2} - \frac{b}{y^2} \frac{\d y}{\d x} \\ \Rightarrow && \frac{\d y}{\d x} &= -\frac{ay^2}{bx^2} \\ \\ (p,q): && -\frac{aq^2}{bp^2} &= -\frac{a}{b} \\ \Rightarrow && p^2 &= q^2 \\ \Rightarrow && p &= \pm q \\ \\ \Rightarrow && ap \pm b p &= 1 \\ \Rightarrow && (a\pm b)p &= 1 \\ \Rightarrow && \frac{a}{p} \pm \frac{b}{p} &= 1 \\ \Rightarrow && (a \pm b)\frac{1}{p} &= 1 \\ \Rightarrow && (a \pm b)^2 &= 1 \end{align*}
  2. \(\,\) \begin{align*} && 1 &= \frac{a}{x} - \frac{b}{y} \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{ay^2}{bx^2} \\ \Rightarrow && \frac{aq^2}{bp^2} &= \frac{b}{a} \\ \Rightarrow && aq &= \pm bp \\ \Rightarrow && 1 &= \frac{a}{p} - \frac{b}{q} \\ &&&= \frac{aq-bp}{pq} \\ \Rightarrow && aq &= -bp \\ \Rightarrow && 1 &= \frac{2aq}{pq} \\ \Rightarrow && p &= 2a \\ \Rightarrow && q &= -2b \\ \Rightarrow && 1 &= 2a^2-2b^2 \\ \Rightarrow && \frac12 &= a^2-b^2 \end{align*}

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2011 Paper 1 Q2
D: 1516.0 B: 1603.0

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. \[ \int_1^{\sqrt2} \frac {\e^{x^2}}x \, \d x \, \]

Solution: \begin{align*} \int_0^1 \frac{x \e^x}{1+x} \, \d x &= \int_0^1 \frac{(x+1-1) \e^x}{1+x} \, \d x \\ &= \int_0^1 \left ( e^x -\frac{\e^x}{1+x} \right )\, \d x \\ &= \e-1-E \end{align*} \begin{align*} \int_0^1 \frac{x^2 \e^x}{1+x} \, \d x &= \int_0^1 \frac{(x^2+x-x) \e^x}{1+x} \, \d x \\ &= \int_0^1 \left ( xe^x -\frac{x\e^x}{1+x} \right )\, \d x \\ &= \left [xe^{x} \right]_0^1 - \int_0^1 e^x \, \d x -(\e-1-E) \\ &= \e-(\e-1)-(\e -1 -E) \\ &= 2-\e + E \end{align*}

  1. Since \(\displaystyle u = \frac{1-x}{1+x},\frac{\d u}{\d x} = \frac{-(1+x)-(1-x)}{(1+x)^2}\), \begin{align*} && \int_0^1 \frac{\e^{\frac{1-x}{1+x}}}{1+x}\, \d x &= \int_{u=1}^{u=0} \frac{e^u}{1+x} \cdot \frac{(1+x)^2}{-2} \d u \\ &&&= \int_0^1 \frac{(1+x) e^u}{2} \d u \\ &&&= \int_0^1 \frac{\left ( 1 + \frac{1-u}{1+u} \right) e^u}{2}\, \d u \\ &&&= \frac12 \int_0^1 \left (e^u + \frac{e^{u}}{1+u} - \frac{ue^u}{1+u} \right) \, \d u \\ &&&= \frac12 \left( \e-1 + E - (\e - 1 - E) \right) \\ &&&= E \end{align*}
  2. Since \(\displaystyle u = x^2-1, \d u = 2x \d x\)\begin{align*} \int_1^{\sqrt2} \frac {\e^{x^2}}x \, \d x &= \int_{u=0}^{u=1} \frac{e^{u+1}}{x} \frac{1}{2x} \d u \\ &= \int_0^1 \frac{e^{u+1}}{2(u+1)} \d u \\ &= \frac{\e}{2} E \\ &= \frac{E\e}{2} \end{align*}

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2011 Paper 1 Q3
D: 1500.0 B: 1500.0

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 \[ \cosec \tfrac19\pi -\cosec \tfrac59\pi +\cosec \tfrac79\pi = 2\sqrt3\,. \]

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2011 Paper 1 Q4
D: 1500.0 B: 1524.2

The distinct points \(P\) and \(Q\), with coordinates \((ap^2,2ap)\) and \((aq^2,2aq)\) respectively, lie on the curve \(y^2=4ax\). The tangents to the curve at \(P\) and \(Q\) meet at the point \(T\). Show that \(T\) has coordinates \(\big(apq, a(p+q)\big)\). You may assume that \(p\ne0\) and \(q\ne0\). The point \(F\) has coordinates \((a,0)\) and \(\phi\) is the angle \(TFP\). Show that \[ \cos\phi = \frac{pq+1}{\sqrt{(p^2+1)(q^2+1)}\ } \] and deduce that the line \(FT\) bisects the angle \(PFQ\).

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2011 Paper 1 Q5
D: 1500.0 B: 1516.7

Given that \(0 < k < 1\), show with the help of a sketch that the equation \[ \sin x = k x \tag{\(*\)}\] has a unique solution in the range \(0 < x < \pi\). Let \[ I= \int_0^\pi \big\vert \sin x -kx\big\vert \, \d x\,. \] Show that \[ I= \frac{\pi^2 \sin\alpha }{2\alpha} -2\cos\alpha - \alpha \sin\alpha\,, \] where \(\alpha\) is the unique solution of \((*)\). Show that \(I\), regarded as a function of \(\alpha\), has a unique stationary value and that this stationary value is a minimum. Deduce that the smallest value of \(I\) is \[ -2 \cos \frac{\pi}{\sqrt2}\, .\]

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2011 Paper 1 Q6
D: 1500.0 B: 1500.0

Use the binomial expansion to show that the coefficient of \(x^r\) in the expansion of \((1-x)^{-3}\) is \(\frac12 (r+1)(r+2)\,\).

  1. Show that the coefficient of \(x^r\) in the expansion of \[ \frac{1-x+2x^2}{(1-x)^3} \] is \(r^2+1\) and hence find the sum of the series \[ 1+\frac22+\frac54+\frac{10}8+\frac{17}{16}+\frac{26}{32}+\frac{37}{64} +\frac{50}{128}+ \cdots \,. \]
  2. Find the sum of the series \[ 1+2+\frac94+2+\frac{25}{16}+\frac{9}{8}+\frac{49}{64} + \cdots \,. \]

Solution: Notice that the coefficient of \(x^r\) is \((-1)^r\frac{(-3) \cdot (-3-1) \cdots (-3-r+1)}{r!} = (-1)^r \frac{(-1)(-2)(-3)(-4) \cdots (-(r+2))}{(-1)(-2)r!} = (-1)^r(-1)^{r+2}\frac{(r+2)!}{2r!} = \frac{(r+2)(r+1)}2\).

  1. The coefficient of \(x^r\) is \begin{align*} && c_r &=\frac{(r+1)(r+2)}{2} - \frac{(r-1+1)(r-1+2)}{2} + 2 \frac{((r-2+1)(r-2+2)}{2} \\ &&&= \frac{r^2+3r+2}{r} - \frac{r^2+r}{2} + \frac{2r^2-2r}{2}\\ &&&= \frac{2r^2+2}{2} = r^2+1 \end{align*} \begin{align*} && S & = 1+\frac22+\frac54+\frac{10}8+\frac{17}{16}+\frac{26}{32}+\frac{37}{64} +\frac{50}{128}+ \cdots \\ &&&= \sum_{r=0}^{\infty} \frac{r^2+1}{2^r} \\ &&&= \frac{1-\tfrac12+2 \cdot \tfrac14}{(1-\tfrac12)^3} \\ &&&= 8 \end{align*}
  2. \(\,\) \begin{align*} && S &= 1+2+\frac94+2+\frac{25}{16}+\frac{9}{8}+\frac{49}{64} + \cdots \\ &&&= \sum_{r=0}^{\infty} \frac{(r+1)^2}{2^r} \\ &&&= 2 \sum_{r=0}^{\infty} \frac{(r+1)^2}{2^{r+1}} \\ &&&= 2 \sum_{r=1}^{\infty} \frac{r^2}{2^{r}} \\ &&&= 2 \left (\sum_{r=0}^{\infty} \frac{r^2+1}{2^{r}} - \sum_{r=0}^{\infty} \frac{1}{2^{r}} \right) \\ &&&= 2 (8 - 1) = 14 \end{align*}

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2011 Paper 1 Q7
D: 1500.0 B: 1500.0

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 \(\alpha\).

Solution:

  1. \begin{align*} \frac{\d h}{\d t} &= \underbrace{c}_{\text{flow in}} - \underbrace{kh}_{\text{flow out}} \end{align*}. We also know that when \(h = \alpha^2 H\), \(\frac{\d h}{\d t} = 0\), ie \(c - k \alpha^2 H = 0\) therefore: \[ \frac{\d h}{\d t} = k(\alpha^2 H - h) \] \begin{align*} && \frac{\d h}{\d t} &= k(\alpha^2 H - h) \\ && \int \frac{1}{\alpha^2 H - h} \d h &= \int k \d t \\ && - \ln |\alpha^2H -h| &= kt + C \\ t = 0, h = H: && -\ln |(1-\alpha^2 )H| &= C \\ \Rightarrow && kt &= \ln \left | \frac{(\alpha^2-1)H}{h-\alpha^2 H }\right | \\ && kT &= \ln \frac{(\alpha^2-1)H}{\alpha H - \alpha^2 H} \\ &&&= \ln \frac{1+\alpha}{\alpha} \\ &&&= \ln \left (1 + \frac1{\alpha} \right) \\ &&&\approx \frac1{\alpha} - \frac12 \frac1{\alpha^2}\\ &&&\approx \alpha^{-1} \end{align*}
  2. \begin{align*} && \frac{\d h}{\d t} &=c(\alpha \sqrt{H} - \sqrt{h}) \\ \Leftrightarrow && c \int_0^{T'} \d t&= \int_{H}^{\alpha H} \frac{1}{\alpha \sqrt{H}-\sqrt{h}} \d h \\ u = \sqrt{h/H}: && cT' &= \int_1^{\sqrt{\alpha}} \frac{1}{\alpha \sqrt{H} - \sqrt{H}u} 2\sqrt{H}u \d u \\ &&&= 2\sqrt{H}\int_1^{\sqrt{\alpha}} \frac{u}{\alpha - u} \d u \\ &&&= 2\sqrt{H}\int_1^{\sqrt{\alpha}} \frac{u - \alpha + \alpha}{\alpha - u} \d u \\ &&&= 2\sqrt{H}\left [-u - \alpha \ln |\alpha - u| \right]_1^{\sqrt{\alpha}} \\ &&&= 2\sqrt{H}\left ( -\sqrt{\alpha} + 1- \alpha \ln (\alpha - \sqrt{\alpha}) + \alpha \ln |\alpha - 1| \right) \\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\alpha-1}{\alpha - \sqrt{\alpha}} \right)\right)\\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\sqrt{\alpha}^2-1}{\sqrt{\alpha}(\sqrt{\alpha}-1)} \right)\right)\\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\sqrt{\alpha}+1}{\sqrt{\alpha}} \right)\right)\\ &&&= \boxed{2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( 1+\frac{1}{\sqrt{\alpha}} \right)\right)}\\ &&&\approx2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \left ( \frac{1}{\sqrt{\alpha}}-\frac12 \frac{1}{\alpha} \right)\right) \\ &&&=2\sqrt{H} \left ( 1 - \frac12 \right) \\ &&&= \sqrt{H} \end{align*} as required.

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2011 Paper 1 Q8
D: 1516.0 B: 1484.0

  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 \[ p^3=q^3+2q^2-1\,. \]

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2011 Paper 1 Q9
D: 1516.0 B: 1484.7

A particle is projected at an angle \(\theta\) above the horizontal from a point on a horizontal plane. The particle just passes over two walls that are at horizontal distances \(d_1\) and \(d_2\) from the point of projection and are of heights \(d_2\) and \(d_1\), respectively. Show that \[ \tan\theta = \frac{d_1^2+d_\subone d_\subtwo +d_2^2}{d_\subone d_\subtwo}\,. \] Find (and simplify) an expression in terms of \(d_1\) and \(d_2\) only for the range of the particle.

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2011 Paper 1 Q10
D: 1516.0 B: 1484.0

A particle, \(A\), is dropped from a point \(P\) which is at a height \(h\) above a horizontal plane. A~second particle, \(B\), is dropped from \(P\) and first collides with \(A\) after \(A\) has bounced on the plane and before \(A\) reaches \(P\) again. The bounce and the collision are both perfectly elastic. Explain why the speeds of \(A\) and \(B\) immediately before the first collision are the same. The masses of \(A\) and \(B\) are \(M\) and \(m\), respectively, where \(M>3m\), and the speed of the particles immediately before the first collision is \(u\). Show that both particles move upwards after their first collision and that the maximum height of \(B\) above the plane after the first collision and before the second collision is \[ h+ \frac{4M(M-m)u^2}{(M+m)^2g}\,. \]

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