| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| 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 students to combine fractions, verify indeterminate form, and differentiate twice. While it requires careful algebraic manipulation and multiple applications of the rule, it follows a standard template for such problems with no novel insight needed. Slightly easier than average due to its mechanical nature. |
| Spec | 4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{\sin x} - \frac{1}{x} = \frac{x - \sin x}{x\sin x}\) | M1 | Complete method to write as a quotient |
| \(\dfrac{\frac{d}{dx}(x-\sin x)}{\frac{d}{dx}(x\sin x)} = \dfrac{1\pm\cos x}{\sin x \pm x\cos x}\) | dM1 | Differentiation of both numerator and denominator using product rule; either correct form |
| \(\dfrac{\frac{d}{dx}(x-\sin x)}{\frac{d}{dx}(x\sin x)} = \dfrac{1-\cos x}{\sin x + x\cos x}\) | A1 | Fully correct differentiation of both |
| \(\dfrac{\frac{d}{dx}(1-\cos x)}{\frac{d}{dx}(\sin x + x\cos x)} = \dfrac{\pm\sin x}{\pm\cos x \pm \cos x \pm x\sin x}\) | ddM1 | Recognises need to differentiate again; dependent on previous M |
| \(\lim_{x\to 0}\left(\frac{1}{\sin x}-\frac{1}{x}\right) = \lim_{x\to 0}\left(\frac{\sin x}{2\cos x - x\sin x}\right) = \dfrac{\sin(0)}{2\cos(0)-(0)\sin(0)}\) | M1 | Clear use of L'Hôpital's Rule with substitution of \(x=0\) |
| \(= 0\) | A1* | Must show clear substitution before final answer |
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{\sin x} - \frac{1}{x} = \frac{x - \sin x}{x\sin x}$ | M1 | Complete method to write as a quotient |
| $\dfrac{\frac{d}{dx}(x-\sin x)}{\frac{d}{dx}(x\sin x)} = \dfrac{1\pm\cos x}{\sin x \pm x\cos x}$ | dM1 | Differentiation of both numerator and denominator using product rule; either correct form |
| $\dfrac{\frac{d}{dx}(x-\sin x)}{\frac{d}{dx}(x\sin x)} = \dfrac{1-\cos x}{\sin x + x\cos x}$ | A1 | Fully correct differentiation of both |
| $\dfrac{\frac{d}{dx}(1-\cos x)}{\frac{d}{dx}(\sin x + x\cos x)} = \dfrac{\pm\sin x}{\pm\cos x \pm \cos x \pm x\sin x}$ | ddM1 | Recognises need to differentiate again; dependent on previous M |
| $\lim_{x\to 0}\left(\frac{1}{\sin x}-\frac{1}{x}\right) = \lim_{x\to 0}\left(\frac{\sin x}{2\cos x - x\sin x}\right) = \dfrac{\sin(0)}{2\cos(0)-(0)\sin(0)}$ | M1 | Clear use of L'Hôpital's Rule with substitution of $x=0$ |
| $= 0$ | A1* | Must show clear substitution before final answer |
---\begin{enumerate}
\item Use L'Hospital's rule to show that
\end{enumerate}
$$\lim _ { x \rightarrow 0 } \left( \frac { 1 } { \sin x } - \frac { 1 } { x } \right) = 0$$
(6)
\hfill \mbox{\textit{Edexcel FP1 2024 Q3 [6]}}