A2 June 2023 Paper 1 Q13
13 Use l’Hôpital’s rule to prove that
\[\lim_{x \to \pi}\left(\frac{x\sin 2x}{\cos\left(\frac{x}{2}\right)}\right) = -4\pi\][5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Explains that numerator and denominator are both zero at \(x = \pi\) | E1 | 2.4 |
| Obtains derivatives of numerator and denominator | M1 | 1.1a |
| Correctly evaluates both correct derivatives of numerator and denominator at \(x = \pi\), may be unsimplified | A1 | 1.1b |
| Forms their \(\dfrac{\mathrm{f}^{\prime}(\pi)}{\mathrm{g}^{\prime}(\pi)}\) | M1 | 1.1a |
| Completes a reasoned argument using l’Hôpital’s rule to obtain the required result, including a clear demonstration of the limiting process. Can score E0 M1A1M1R1 | R1 | 2.1 |
| (5 marks) |
Typical solution
Let
\[\mathrm{f}(x) = x\sin 2x\]\[\mathrm{g}(x) = \cos\left(\tfrac{x}{2}\right)\]Then
\[\mathrm{f}(\pi) = 0 \text{ and } \mathrm{g}(\pi) = 0\]So, by l’Hôpital’s rule,
\[\lim_{x \to \pi}\left(\frac{\mathrm{f}(x)}{\mathrm{g}(x)}\right) = \lim_{x \to \pi}\left(\frac{\mathrm{f}^{\prime}(x)}{\mathrm{g}^{\prime}(x)}\right)\]\[\mathrm{f}^{\prime}(x) = 2x\cos 2x + \sin 2x\]\[\mathrm{g}^{\prime}(x) = -\tfrac{1}{2}\sin\left(\tfrac{x}{2}\right)\]\[\mathrm{f}^{\prime}(\pi) = 2\pi\cos 2\pi + \sin 2\pi = 2\pi\]\[\mathrm{g}^{\prime}(\pi) = -\tfrac{1}{2}\sin\left(\tfrac{\pi}{2}\right) = -\tfrac{1}{2}\]\[\lim_{x \to \pi}\left(\frac{\mathrm{f}(x)}{\mathrm{g}(x)}\right) = \lim_{x \to \pi}\left(\frac{\mathrm{f}^{\prime}(x)}{\mathrm{g}^{\prime}(x)}\right) = \frac{2\pi}{-\frac{1}{2}} = -4\pi\]