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2005 Paper 2 Q3
D: 1600.0 B: 1469.5
integration integration by parts curve sketching trigonometric inequality product of trig and polynomial definite integral bounds

Give a sketch, for \(0 \le x \le \frac{1}{2}\pi\), of the curve $$ y = (\sin x - x\cos x)\;, $$ and show that \(0\le y \le 1\,\). Show that:

  1. \(\displaystyle \int_0^{\frac{1}{2}\pi}\,y\;\d x = 2 -\frac \pi 2 \)
  2. \(\displaystyle \int_0^{\frac{1}{2}\pi}\,y^2\,\d x = \frac{\pi^3}{48}-\frac \pi 8 \)
Deduce that \(\pi^3 +18 \pi< 96\,\).

Solution:

TikZ diagram
Since \(y' = \cos x - \cos x + x \sin x = x \sin x > 0\) which is positive on \((0, \frac{\pi}{2})\), so \(y\) is increasing, and therefore will achieve it's highest value at \(\frac{\pi}{2}\) which is \(y(\frac{\pi}{2}) = 1\) and it's smallest value at \(y(0) = 0\). Therefore \(0 \leq y \leq 1\)
  1. \(\,\) \begin{align*} \int_0^{\frac{1}{2}\pi}\,y\;\d x &= \int_0^{\frac{1}{2}\pi} (\sin x - x \cos x) \d x \\ &= \left [-\cos x \right]_0^{\frac{1}{2}\pi} +\left [ -x \sin x \right]_0^{\frac{1}{2}\pi} + \int_0^{\frac{1}{2}\pi} \sin x \d x \\ &= 1-\frac{\pi}{2} + 1 = 2 - \frac{\pi}{2} \end{align*}
  2. \(\,\) \begin{align*} \int_0^{\frac{1}{2}\pi}y^2\d x &= \int_0^{\frac{1}{2}\pi} (\sin x - x \cos x)^2 \d x \\ &= \int_0^{\frac{1}{2}\pi} (\sin^2x - 2x\sin x \cos x+x^2\cos^2 x) \d x\\ &= \int_0^{\frac{1}{2}\pi} (\sin^2x -x \sin 2x+\tfrac12x^2(\cos 2 x + 1)) \d x\\ &= \frac{\pi}{4} + \frac{\pi^3}{48} + \int_0^{\frac{1}{2}\pi} (-x \sin 2x+\tfrac12x^2\cos 2 x) \d x \\ &= \frac{\pi}{4} + \frac{\pi^3}{48} + \left [\frac12 x \cos 2x +\frac14 x^2 \sin2x\right]_0^{\frac{1}{2}\pi}-\int_0^{\frac{1}{2}\pi}(\tfrac12 \cos 2x +\tfrac12 x \sin 2x) \d x\\ &= \frac{\pi}{4} + \frac{\pi^3}{48} - \frac{\pi}{4} - \left [ \frac14 \sin 2x \right]_0^{\frac{1}{2}\pi} - \int_0^{\frac{1}{2}\pi} \tfrac12 x \sin 2x \d x\\ &= \frac{\pi^3}{48} - \left( \left[ -\frac14 x \cos 2x \right]_0^{\frac{1}{2}\pi} - \int_0^{\frac{1}{2}\pi} -\frac14 \cos 2x \d x \right)\\ &= \frac{\pi^3}{48} - \left( \frac{\pi}{8} + \left[ \frac18 \sin 2x \right]_0^{\frac{1}{2}\pi} \right)\\ &= \frac{\pi^3}{48} - \frac{\pi}{8} \end{align*}
Since \(y^2 < y\) on this interval, we must have \( \frac{\pi^3}{48} - \frac{\pi}{8} < 2 - \frac{\pi}{2} \Rightarrow \pi^3 +18\pi < 96\) as required.

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2004 Paper 3 Q3
D: 1700.0 B: 1516.0
sequences series inequality integration convexity telescoping curve sketching summation reciprocal squares bounds

Given that \(\f''(x) > 0\) when \(a \le x \le b\,\), explain with the aid of a sketch why \[ (b-a) \, \f \Big( {a+b \over 2} \Big) < \int^b_a \f(x) \, \mathrm{d}x < (b-a) \, \displaystyle \frac{\f(a) + \f(b)}{2} \;. \] By choosing suitable \(a\), \(b\) and \(\f(x)\,\), show that \[ {4 \over (2n-1)^2} < {1 \over n-1} - {1 \over n} < {1 \over 2} \l {1 \over n^2} + {1 \over (n-1)^2}\r \,, \] where \(n\) is an integer greater than 1. Deduce that \[ 4 \l {1 \over 3^2} +{1 \over 5^2} + {1 \over 7^2} + \cdots \r < 1 < {1 \over 2} + \left( {1 \over 2^2} +{1 \over 3^2} + {1 \over 4^2} + \cdots \right)\,. \] Show that \[ {1 \over 2} \l {1 \over 3^2} + {1 \over 4^2} + {1 \over 5^2} + \frac 1 {6^2} + \cdots \r < {1 \over 3^2} +{1 \over 5^2} + {1 \over 7^2} + \cdots \] and hence show that \[ {3 \over 2} \displaystyle < \sum_{n=1}^\infty {1 \over n^2} <{7 \over 4}\;. \]

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1997 Paper 2 Q11
D: 1600.0 B: 1484.0
projectiles quadratic equations trigonometry inequality tennis serve trajectory bounds optimisation

A tennis player serves from height \(H\) above horizontal ground, hitting the ball downwards with speed \(v\) at an angle \(\alpha\) below the horizontal. The ball just clears the net of height \(h\) at horizontal distance \(a\) from the server and hits the ground a further horizontal distance \(b\) beyond the net. Show that $$v^2 = \frac{ g(a+b)^2(1+\tan^2\alpha)}{ 2[H-(a+b)\tan\alpha]}$$ and $$\tan\alpha = \frac{2a+b }{ a(a+b)}H - \frac{a+b }{ ab}h \,.$$ By considering the signs of \(v^2\) and \(\tan\alpha\), find upper and lower bounds on \(H\) for such a serve to be possible.

Solution: We have \begin{align*} \rightarrow: && a &= v\cos \alpha t_{net} \\ \Rightarrow && t_{net} &= \frac{a}{v \cos \alpha} \\ \downarrow: && H-h &= v\sin \alpha t_{net} + \frac12 g t_{net}^2 \\ &&&= a \tan \alpha + \frac12 g \frac{a^2}{v^2} \sec^2 \alpha \\ &&&= a \tan \alpha + \frac{a^2g}{2v^2}(1 + \tan^2 \alpha) \tag{*}\\ \\ \rightarrow: && a+b &= v \cos \alpha t_{ground} \\ && t_{ground} &= \frac{a+b}{v \cos \alpha}\\ \downarrow: && H &= v\sin \alpha t_{ground} + \frac12 g t_{ground}^2 \\ &&&= (a+b)\tan \alpha + \frac{(a+b)^2g}{2v^2}(1+\tan^2\alpha) \tag{**} \\ \\ (**): && v^2 &= \frac{g(a+b)^2(1+\tan^2\alpha)}{2[H-(a+b)\tan \alpha]} \\ (a+b)^2(*) - a^2(**): && (a+b)^2(H-h) -a^2H &= [(a+b)^2a - a^2(a+b)]\tan \alpha \\ \Rightarrow && (2ab+b^2)H - (a+b)^2h &= ab(a+b) \tan \alpha \\ \Rightarrow && \tan \alpha &= \frac{2a+b}{a(a+b)}H - \frac{a+b}{ab} h \end{align*} Noting that \(v^2 \geq 0\) and the numerator is positive, we must have \begin{align*} && H &> (a+b)\tan \alpha \\ &&&= \frac{2a+b}{a}H - \frac{(a+b)^2}{ab} h \\ \Rightarrow && \frac{a+b}{a}H &< \frac{(a+b)^2}{ab} h \\ \Rightarrow && H &< \frac{a+b}{b} h \end{align*} Noting that \(\tan \alpha > 0\) we must have \begin{align*} && \frac{2a+b}{a(a+b)} H & > \frac{a+b}{ab} h \\ \Rightarrow && H &> \frac{(a+b)^2}{b(2a+b)}h \end{align*}

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