Problems

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Solution:

Note that the line \(y = x\) is the tangent \((0,0)\) and \(y = \sin x \) is always below it. For any other line through the origin with gradient \(0 < k < 1\) it must start below \(y = \sin x\), but finish above it at \(x = \pi\). It also can only cross once due the the convexity of \(\sin\) in this interval. \begin{align*} && I &= \int_0^\pi | \sin x -kx | \d x \\ &&&= \int_0^\alpha (\sin x -k x) \d x + \int_{\alpha}^\pi (kx - \sin x) \d x \\ &&&= \left [ -\cos x - \frac{kx^2}{2} \right]_0^{\alpha} + \left [ \cos x +\frac{kx^2}{2} \right]_{\alpha}^\pi \\ &&&= -\cos \alpha - \frac{k \alpha^2}{2} +1+(-1)+\frac{k\pi^2}{2} - \cos \alpha - \frac{k\alpha^2}{2} \\ &&&= -2\cos \alpha - k\left (\alpha^2 - \frac{\pi^2}{2} \right) \\ &&&= -2\cos \alpha - \frac{\sin \alpha}{\alpha}\left (\alpha^2 - \frac{\pi^2}{2} \right) \\ &&&= \frac{\pi^2 \sin \alpha}{\alpha} - \alpha \sin \alpha - 2\cos \alpha \end{align*} \begin{align*} && \frac{\d I}{\d \alpha} &= \frac{\pi^2(\alpha \cos \alpha - \sin \alpha)}{2\alpha^2} + 2 \sin \alpha - \sin \alpha - \alpha \cos \alpha \\ &&&= \frac{-2\alpha^3 \cos \alpha + 2\alpha \sin\alpha + \pi^2 \alpha \cos \alpha - \pi^2 \sin \alpha}{2\alpha^2} \\ \\ &&&= \left ( \alpha \cos \alpha - \sin \alpha\right) \left ( \frac{\pi^2}{2\alpha^2}-1 \right)\ \end{align*} Therefore \(I' = 0\) if \(\tan \alpha = \alpha\) or \(\alpha = \frac{\pi}{\sqrt{2}}\). Since \(\tan \alpha = \alpha\) only at \(\alpha = 0\) (between \(0 \leq \alpha < \pi\) (by considering the tangent), we must have a unique turning point when \(\alpha = \frac{\pi}{\sqrt{2}}\). Note that \(I(\frac{\pi}{\sqrt{2}}) = \frac{\pi^2 \sqrt{2} \sin \alpha}{2\pi} - \frac{\pi}{\sqrt{2}} \sin \alpha - 2\cos \frac{\pi}{\sqrt{2}}=- 2\cos \frac{\pi}{\sqrt{2}}\). Notice that \(I(0) = \frac{\pi^2}2 - 2 > 2\) and \(I(\pi) = 2\), but \(-2\cos \frac{\pi}{\sqrt{2}} < 2\) so we must be a at a minimum