| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Exponential or trigonometric base functions |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring chain rule differentiation of a composite exponential function, followed by routine substitution into the Maclaurin series formula. While it involves multiple applications of the chain rule and some trigonometric manipulation, the techniques are standard and the question structure is typical for F2 Taylor series problems. |
| Spec | 1.07q Product and quotient rules: differentiation4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = -2\sin x \cos x e^{\cos^2 x} = -\sin 2x e^{\cos^2 x}\) | M1 | Differentiates using chain rule to obtain form \(\alpha \sin x \cos x e^{\cos^2 x}\) or \(\beta \sin 2x e^{\cos^2 x}\) |
| Correct derivative | A1 | Note candidates may use \(\frac{1}{2}(1+\cos 2x)\) instead of \(\cos^2 x\) |
| \(\frac{d^2y}{dx^2} = -\sin 2x(-2\sin x \cos x e^{\cos^2 x}) - 2\cos 2x e^{\cos^2 x}\) | dM1 | Correct use of Product Rule on first derivative. Dependent on first method mark |
| \(\frac{d^2y}{dx^2} = e^{\cos^2 x}(\sin^2 2x - 2\cos 2x)\)* | A1* | Achieves printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = e^{\cos^2 x} \Rightarrow \ln y = \cos^2 x \Rightarrow \frac{1}{y}\frac{dy}{dx} = -2\sin x\cos x\) | M1 | \(\frac{1}{y}\frac{dy}{dx} = k\sin x\cos x\) or \(k\sin 2x\) |
| \(\frac{1}{y}\frac{dy}{dx} = -2\sin x\cos x\) | A1 | |
| \(\frac{dy}{dx} = -y\sin 2x \Rightarrow \frac{d^2y}{dx^2} = -\frac{dy}{dx}\sin 2x - 2y\cos 2x\) | M1 | Correct use of product rule |
| \(\frac{d^2y}{dx^2} = e^{\cos^2 x}(\sin^2 2x - 2\cos 2x)\)* | A1 | Achieves printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 0\), \(\frac{d^2y}{dx^2} = -2e\) | B1 | Both seen, can be implied by subsequent work |
| \(f(x) = f(0) + xf'(0) + \frac{x^2}{2}f''(0) + ...\) \(= e^{\cos^2 0} - \sin 0 e^{\cos^2 0}x + \frac{1}{2}e^{\cos^2 0}(\sin^2 0 - 2\cos 0)x^2 + ...\) | M1 | Applies correct Maclaurin expansion; \(\frac{1}{2}\) is required; must be no \(x\)'s in derivatives |
| \(\left(e^{\cos^2 x}\right) = e(1 - x^2 + ...)\) | A1 | Or exact equivalent e.g. \(e - ex^2\); all trig evaluated |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -2\sin x \cos x e^{\cos^2 x} = -\sin 2x e^{\cos^2 x}$ | M1 | Differentiates using chain rule to obtain form $\alpha \sin x \cos x e^{\cos^2 x}$ or $\beta \sin 2x e^{\cos^2 x}$ |
| Correct derivative | A1 | Note candidates may use $\frac{1}{2}(1+\cos 2x)$ instead of $\cos^2 x$ |
| $\frac{d^2y}{dx^2} = -\sin 2x(-2\sin x \cos x e^{\cos^2 x}) - 2\cos 2x e^{\cos^2 x}$ | dM1 | Correct use of Product Rule on first derivative. Dependent on first method mark |
| $\frac{d^2y}{dx^2} = e^{\cos^2 x}(\sin^2 2x - 2\cos 2x)$* | A1* | Achieves printed answer with no errors |
**Alternative using logs:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = e^{\cos^2 x} \Rightarrow \ln y = \cos^2 x \Rightarrow \frac{1}{y}\frac{dy}{dx} = -2\sin x\cos x$ | M1 | $\frac{1}{y}\frac{dy}{dx} = k\sin x\cos x$ or $k\sin 2x$ |
| $\frac{1}{y}\frac{dy}{dx} = -2\sin x\cos x$ | A1 | |
| $\frac{dy}{dx} = -y\sin 2x \Rightarrow \frac{d^2y}{dx^2} = -\frac{dy}{dx}\sin 2x - 2y\cos 2x$ | M1 | Correct use of product rule |
| $\frac{d^2y}{dx^2} = e^{\cos^2 x}(\sin^2 2x - 2\cos 2x)$* | A1 | Achieves printed answer with no errors |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 0$, $\frac{d^2y}{dx^2} = -2e$ | B1 | Both seen, can be implied by subsequent work |
| $f(x) = f(0) + xf'(0) + \frac{x^2}{2}f''(0) + ...$ $= e^{\cos^2 0} - \sin 0 e^{\cos^2 0}x + \frac{1}{2}e^{\cos^2 0}(\sin^2 0 - 2\cos 0)x^2 + ...$ | M1 | Applies correct Maclaurin expansion; $\frac{1}{2}$ is required; must be no $x$'s in derivatives |
| $\left(e^{\cos^2 x}\right) = e(1 - x^2 + ...)$ | A1 | Or exact equivalent e.g. $e - ex^2$; all trig evaluated |
---5.
$$y = \mathrm { e } ^ { \cos ^ { 2 } x }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { \cos ^ { 2 } x } \left( \sin ^ { 2 } 2 x - 2 \cos 2 x \right)$$
\item Hence find the Maclaurin series expansion of $\mathrm { e } ^ { \cos ^ { 2 } x }$ up to and including the term in $x ^ { 2 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2017 Q5 [7]}}