Question 4 - A-Level Maths

Edexcel FP1 2020 June — Question 4 8 marks

Exam Board	Edexcel
Module	FP1 (Further Pure Mathematics 1)
Year	2020
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Taylor series
Type	Direct multiplication of series
Difficulty	Challenging +1.8 This question requires applying Leibniz's theorem for the 8th derivative of a product, then evaluating at x=π and simplifying. While the technique is standard for FP1, it involves substantial algebraic manipulation across 8 derivatives, careful tracking of binomial coefficients, and evaluation of trigonometric functions at π. The multi-step nature and potential for algebraic errors place it above average difficulty.
Spec	1.07q Product and quotient rules: differentiation 4.08a Maclaurin series: find series for function 4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n

4. $$f ( x ) = x ^ { 4 } \sin ( 2 x )$$ Use Leibnitz's theorem to show that the coefficient of $( x - \pi ) ^ { 8 }$ in the Taylor series expansion of $\mathrm { f } ( x )$ about $\pi$ is $$\frac { a \pi + b \pi ^ { 3 } } { 315 }$$ where $a$ and $b$ are integers to be determined. The Taylor series expansion of $\mathrm { f } ( \mathrm { x } )$ about $\mathrm { x } = \mathrm { k }$ is given by $$f ( x ) = f ( k ) + ( x - k ) f ^ { \prime } ( k ) + \frac { ( x - k ) ^ { 2 } } { 2 ! } f ^ { \prime \prime } ( k ) + \ldots + \frac { ( x - k ) ^ { r } } { r ! } f ^ { ( r ) } ( k ) + \ldots$$

Show mark scheme Show mark scheme source

Question 4:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$u' = 4x^3,\ u'' = 12x^2,\ u''' = 24x,\ u^{(4)} = 24$ (and $u^{(n)}=0$ for $n>4$)	M1	Establishes non-disappearing derivatives of $x^4$; allow slips in coefficients but powers must decrease
$v' = 2\cos(2x),\ v'' = -4\sin(2x),\ v''' = -8\cos(2x),\ v^{(4)} = 16\sin(2x)$, $v^{(5)} = 32\cos(2x),\ v^{(6)} = -64\sin(2x),\ v^{(7)} = -128\cos(2x)$, $v^{(8)} = 256\sin(2x)$	M1 A1 A1	Identifies relevant derivatives for $\sin(2x)$ up to 8th; correct sizes for coefficients (allow sign errors); all derivatives correctly established
$f^{(8)}(x) = x^4\times 256\sin(2x) + 8\times 4x^3\times -128\cos(2x)$ $+\frac{8\times7}{2}\times 12x^2\times -64\sin(2x)+\frac{8\times7\times6}{6}\times 24x\times 32\cos(2x)$ $+\frac{8\times7\times6\times5}{24}\times 24\times 16\sin(2x)$	M1	Applies Leibniz's theorem to get 8th derivative; binomial coefficients must be present
$f^{(8)}(\pi) = 0 - 4096\pi^3 - 0 + 1344\times 2^5\pi + 0\ \left(= -4096\pi^3 + 43008\pi\right)$	M1	Evaluates their 8th derivative at $\pi$
$\text{Coefficient} = \frac{f^{(8)}(\pi)}{8!} = \frac{1344\times 2^5\pi - 4096\pi^3}{8!\ \text{or}\ \{40320\}}$	M1	Uses Taylor series — divides value for $f^{(8)}(\pi)$ by $8!$
$= \frac{336\pi-32\pi^3}{315}$ (So $a=336$ and $b=-32$)	A1	Simplifies to correct answer

Note: If Leibniz's theorem not used, maximum M0 M0 A0 A0 M0 M1 M1 A0

# Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $u' = 4x^3,\ u'' = 12x^2,\ u''' = 24x,\ u^{(4)} = 24$ (and $u^{(n)}=0$ for $n>4$) | **M1** | Establishes non-disappearing derivatives of $x^4$; allow slips in coefficients but powers must decrease |
| $v' = 2\cos(2x),\ v'' = -4\sin(2x),\ v''' = -8\cos(2x),\ v^{(4)} = 16\sin(2x)$, $v^{(5)} = 32\cos(2x),\ v^{(6)} = -64\sin(2x),\ v^{(7)} = -128\cos(2x)$, $v^{(8)} = 256\sin(2x)$ | **M1** **A1** **A1** | Identifies relevant derivatives for $\sin(2x)$ up to 8th; correct sizes for coefficients (allow sign errors); all derivatives correctly established |
| $f^{(8)}(x) = x^4\times 256\sin(2x) + 8\times 4x^3\times -128\cos(2x)$ $+\frac{8\times7}{2}\times 12x^2\times -64\sin(2x)+\frac{8\times7\times6}{6}\times 24x\times 32\cos(2x)$ $+\frac{8\times7\times6\times5}{24}\times 24\times 16\sin(2x)$ | **M1** | Applies Leibniz's theorem to get 8th derivative; binomial coefficients must be present |
| $f^{(8)}(\pi) = 0 - 4096\pi^3 - 0 + 1344\times 2^5\pi + 0\ \left(= -4096\pi^3 + 43008\pi\right)$ | **M1** | Evaluates their 8th derivative at $\pi$ |
| $\text{Coefficient} = \frac{f^{(8)}(\pi)}{8!} = \frac{1344\times 2^5\pi - 4096\pi^3}{8!\ \text{or}\ \{40320\}}$ | **M1** | Uses Taylor series — divides value for $f^{(8)}(\pi)$ by $8!$ |
| $= \frac{336\pi-32\pi^3}{315}$ (So $a=336$ and $b=-32$) | **A1** | Simplifies to correct answer |

**Note:** If Leibniz's theorem not used, maximum **M0 M0 A0 A0 M0 M1 M1 A0**

---

Show LaTeX source

4.

$$f ( x ) = x ^ { 4 } \sin ( 2 x )$$

Use Leibnitz's theorem to show that the coefficient of $( x - \pi ) ^ { 8 }$ in the Taylor series expansion of $\mathrm { f } ( x )$ about $\pi$ is

$$\frac { a \pi + b \pi ^ { 3 } } { 315 }$$

where $a$ and $b$ are integers to be determined.

The Taylor series expansion of $\mathrm { f } ( \mathrm { x } )$ about $\mathrm { x } = \mathrm { k }$ is given by

$$f ( x ) = f ( k ) + ( x - k ) f ^ { \prime } ( k ) + \frac { ( x - k ) ^ { 2 } } { 2 ! } f ^ { \prime \prime } ( k ) + \ldots + \frac { ( x - k ) ^ { r } } { r ! } f ^ { ( r ) } ( k ) + \ldots$$

\hfill \mbox{\textit{Edexcel FP1 2020 Q4 [8]}}

This paper (8 questions)

View full paper

Q1 5 Q2 6 Q3 9 Q4 8 Q5 7 Q6 10 Q7 14 Q8 16