| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Exponential or trigonometric base functions |
| Difficulty | Standard +0.8 This is a Further Maths FP2 question requiring multiple derivatives of a product of exponential and trigonometric functions, establishing recurrence relations, and constructing a Maclaurin series. While systematic, it demands careful differentiation, algebraic manipulation to verify the given relation, and pattern recognition across multiple derivatives—more demanding than standard A-level calculus but follows a structured path once the technique is understood. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07q Product and quotient rules: differentiation4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f'(x) = -\sin x \cdot e^{-x} + \cos x \cdot e^{-x}\) | M1 | Differentiation using product correctly |
| \(\Rightarrow f'(0) = 1\) | A1 | |
| \(f(0) = 0\) | A1 | For both values |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f'(x) = \cos x \cdot e^{-x} - \sin x \cdot e^{-x} = \cos x \cdot e^{-x} - f(x)\) | M1 | Differentiate |
| \(f''(x) = -f'(x) - \cos x \cdot e^{-x} - f(x)\) | ||
| \(= -f'(x) - f'(x) - f(x) - f(x)\) | ||
| \(f''(x) = -2f'(x) - 2f(x)\) OR \(-2\cos x \cdot e^{-x}\) | A1 | |
| Showing the two equal | A1 | |
| \(f''(0) = -2\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f''(x) = -2f'(x) - 2f(x)\) | ||
| \(\Rightarrow f'''(x) = -2f''(x) - 2f'(x)\) | B1 | Not involving trig or exp functions; \(= -f''+2f\) or \(2f'+4f\) |
| \(\Rightarrow f'''(0) = 4 - 2 = 2\) | B1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x) = x - x^2 + \frac{x^3}{3}\) | M1, A1 | |
| [2] | ||
| Alternative: Write down correct series expansion for \(e^{-x}\) and \(\sin x\) and multiply | M1, A1 |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f'(x) = -\sin x \cdot e^{-x} + \cos x \cdot e^{-x}$ | M1 | Differentiation using product correctly |
| $\Rightarrow f'(0) = 1$ | A1 | |
| $f(0) = 0$ | A1 | For **both** values |
| **[3]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f'(x) = \cos x \cdot e^{-x} - \sin x \cdot e^{-x} = \cos x \cdot e^{-x} - f(x)$ | M1 | Differentiate |
| $f''(x) = -f'(x) - \cos x \cdot e^{-x} - f(x)$ | | |
| $= -f'(x) - f'(x) - f(x) - f(x)$ | | |
| $f''(x) = -2f'(x) - 2f(x)$ OR $-2\cos x \cdot e^{-x}$ | A1 | |
| Showing the two equal | A1 | |
| $f''(0) = -2$ | A1 | |
| **[4]** | | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f''(x) = -2f'(x) - 2f(x)$ | | |
| $\Rightarrow f'''(x) = -2f''(x) - 2f'(x)$ | B1 | Not involving trig or exp functions; $= -f''+2f$ or $2f'+4f$ |
| $\Rightarrow f'''(0) = 4 - 2 = 2$ | B1 | |
| **[2]** | | |
### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) = x - x^2 + \frac{x^3}{3}$ | M1, A1 | |
| **[2]** | | |
| **Alternative:** Write down correct series expansion for $e^{-x}$ and $\sin x$ and multiply | M1, A1 | |5 You are given that $\mathrm { f } ( x ) = \mathrm { e } ^ { - x } \sin x$.\\
(i) Find $f ( 0 )$ and $f ^ { \prime } ( 0 )$.\\
(ii) Show that $\mathrm { f } ^ { \prime \prime } ( x ) = - 2 \mathrm { f } ^ { \prime } ( x ) - 2 \mathrm { f } ( x )$ and hence, or otherwise, find $\mathrm { f } ^ { \prime \prime } ( 0 )$.\\
(iii) Find a similar expression for $\mathrm { f } ^ { \prime \prime \prime } ( x )$ and hence, or otherwise, find $\mathrm { f } ^ { \prime \prime \prime } ( 0 )$.\\
(iv) Find the Maclaurin series for $\mathrm { f } ( x )$ up to and including the term in $x ^ { 3 }$.
\hfill \mbox{\textit{OCR FP2 2013 Q5 [11]}}