A2 October 2020 Q1
1. Use l’Hospital’s Rule to show that
\[\lim_{x \to \frac{\pi}{2}}\frac{\left(\mathrm{e}^{\sin x} - \cos(3x) - \mathrm{e}\right)}{\tan(2x)} = -\frac{3}{2}\](5)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{e}^{\sin x} - \cos(3x) - \mathrm{e}\right)}{\frac{\mathrm{d}}{\mathrm{d}x}(\tan(2x))} = \dfrac{\pm\cos(x)\mathrm{e}^{\sin x} \pm A\sin(3x)}{B\sec^2 2x}\) | M1 | 1.1b |
| \(\dfrac{\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{e}^{\sin x} - \cos(3x) - \mathrm{e}\right)}{\frac{\mathrm{d}}{\mathrm{d}x}(\tan(2x))} = \dfrac{\cos(x)\mathrm{e}^{\sin x} + 3\sin(3x)}{2\sec^2 2x}\) | A1 A1 | 1.1b 1.1b |
| \(\lim\limits_{x \to \frac{\pi}{2}}\dfrac{\cos(x)\mathrm{e}^{\sin x} + 3\sin(3x)}{2\sec^2 2x}\) \(= \dfrac{\cos\left(\frac{\pi}{2}\right)\mathrm{e}^{\sin\left(\frac{\pi}{2}\right)} + 3\sin\left(\frac{3\pi}{2}\right)}{2\sec^2\left(\frac{2\pi}{2}\right)}\) or \(= \dfrac{0 \times \mathrm{e} + 3 \times (-1)}{2 \times (-1)^2} = \ldots\) | M1 | 1.2 |
| \(= -\dfrac{3}{2}\ *\) | A1* | 2.1 |
| (5) | ||
| (5 marks) |
Notes
M1: Attempts differentiation of both numerator and denominator, including at least one use of the chain rule. Either numerator or denominator of the correct form. May be done separately.
A1: Numerator correct
A1: Denominator correct
M1: Applies l’Hospital’s Rule, must see clear use of a substitution of \(x = \dfrac{\pi}{2}\) into their derivatives, not the original expression. (no need to see check that limits of numerator and denominator are non-zero).
A1*: Needs to be a correct intermediate line following substitution before reaching the printed answer with use of some limit notation. All aspects of the proof should be clear for this mark to be awarded and no errors seen.