| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2020 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Limit using l'Hôpital's rule |
| Difficulty | Standard +0.3 This is a straightforward application of l'Hôpital's rule requiring differentiation of standard functions (exponential, trigonometric) and evaluation at a limit point. While it involves multiple functions and careful algebraic manipulation, it's a direct textbook-style exercise with no conceptual surprises—slightly easier than average for Further Maths FP1. |
| Spec | 4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{\frac{d}{dx}(e^{\sin x} - \cos(3x) - e)}{\frac{d}{dx}(\tan(2x))} = \frac{\pm\cos(x)e^{\sin x} \pm A\sin(3x)}{B\sec^2 2x}\) | M1 | Attempts differentiation of both numerator and denominator, including at least one use of chain rule. Either numerator or denominator of correct form. May be done separately. |
| \(= \frac{\cos(x)e^{\sin x} + 3\sin(3x)}{2\sec^2 2x}\) (numerator correct) | A1 | Numerator correct |
| \(= \frac{\cos(x)e^{\sin x} + 3\sin(3x)}{2\sec^2 2x}\) (denominator correct) | A1 | Denominator correct |
| \(\lim_{x \to \frac{\pi}{2}} \frac{\cos(x)e^{\sin x} + 3\sin(3x)}{2\sec^2 2x} = \frac{\cos\!\left(\frac{\pi}{2}\right)e^{\sin\!\left(\frac{\pi}{2}\right)} + 3\sin\!\left(\frac{3\pi}{2}\right)}{2\sec^2\!\left(\frac{2\pi}{2}\right)}\) or \(= \frac{0 \times e + 3 \times (-1)}{2 \times (-1)^2} = \ldots\) | M1 | Applies l'Hôpital's Rule; must show clear use of substitution of \(x = \frac{\pi}{2}\) into derivatives, not original expression. |
| \(= -\frac{3}{2}\) * | A1\* | Needs correct intermediate line following substitution before reaching printed answer; all aspects of proof clear, no errors seen. |
| (5 marks) |
## Question 1:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{\frac{d}{dx}(e^{\sin x} - \cos(3x) - e)}{\frac{d}{dx}(\tan(2x))} = \frac{\pm\cos(x)e^{\sin x} \pm A\sin(3x)}{B\sec^2 2x}$ | **M1** | Attempts differentiation of both numerator and denominator, including at least one use of chain rule. Either numerator or denominator of correct form. May be done separately. |
| $= \frac{\cos(x)e^{\sin x} + 3\sin(3x)}{2\sec^2 2x}$ (numerator correct) | **A1** | Numerator correct |
| $= \frac{\cos(x)e^{\sin x} + 3\sin(3x)}{2\sec^2 2x}$ (denominator correct) | **A1** | Denominator correct |
| $\lim_{x \to \frac{\pi}{2}} \frac{\cos(x)e^{\sin x} + 3\sin(3x)}{2\sec^2 2x} = \frac{\cos\!\left(\frac{\pi}{2}\right)e^{\sin\!\left(\frac{\pi}{2}\right)} + 3\sin\!\left(\frac{3\pi}{2}\right)}{2\sec^2\!\left(\frac{2\pi}{2}\right)}$ or $= \frac{0 \times e + 3 \times (-1)}{2 \times (-1)^2} = \ldots$ | **M1** | Applies l'Hôpital's Rule; must show clear use of substitution of $x = \frac{\pi}{2}$ into derivatives, not original expression. |
| $= -\frac{3}{2}$ * | **A1\*** | Needs correct intermediate line following substitution before reaching printed answer; all aspects of proof clear, no errors seen. |
| **(5 marks)** | | |
---\begin{enumerate}
\item Use l'Hospital's Rule to show that
\end{enumerate}
$$\lim _ { x \rightarrow \frac { \pi } { 2 } } \frac { \left( e ^ { \sin x } - \cos ( 3 x ) - e \right) } { \tan ( 2 x ) } = - \frac { 3 } { 2 }$$
\hfill \mbox{\textit{Edexcel FP1 2020 Q1 [5]}}