Question 3 - A-Level Maths

Edexcel FP2 2014 June — Question 3 8 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2014
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Taylor series
Type	Maclaurin series for composite exponential/root functions
Difficulty	Standard +0.8 This requires finding a Maclaurin series for a composite function involving both a square root and exponential. Students must apply the chain rule multiple times to find derivatives, substitute x=0, and simplify coefficients carefully. While the technique is standard for FP2, the nested composition and algebraic manipulation of terms like 1/(2√9) make it moderately challenging beyond routine Taylor series questions.
Spec	4.08a Maclaurin series: find series for function 4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n

3. $$y = \sqrt { 8 + \mathrm { e } ^ { x } } , \quad x \in \mathbb { R }$$ Find the series expansion for $y$ in ascending powers of $x$, up to and including the term in $x ^ { 2 }$, giving each coefficient in its simplest form.

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Question 3:

Answer	Marks	Guidance
Answer	Mark	Guidance
$y = (8+e^x)^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = \frac{1}{2}(8+e^x)^{-\frac{1}{2}} \times e^x$	M1A1	M1: $\frac{dy}{dx} = k(8+e^x)^{-\frac{1}{2}} \times e^x$. A1: Correct differentiation
$\frac{d^2y}{dx^2} = \frac{1}{2}(8+e^x)^{-\frac{1}{2}} \times e^x - \frac{1}{4}(8+e^x)^{-\frac{3}{2}} \times e^{2x}$	M1A1	M1: Correct use of product rule: $\pm K(8+e^x)^{-\frac{3}{2}} \times e^{2x}$. A1: Correct second derivative with $e^x \times e^x$ or $e^{2x}$
$f(0) = 3$	B1	May only appear in the expansion
$f'(0) = \frac{1}{6},\quad f''(0) = \frac{17}{108}$	M1	Attempt both $f'(0)$ and $f''(0)$ with their derivatives found above
$(y=)\ 3 + \frac{1}{6}x + \frac{17}{216}x^2$	M1 A1cso (8)	M1: Uses correct Maclaurin series with their values. Accept 2 or 2! in $x^2$ term. A1: Correct expression

# Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| $y = (8+e^x)^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = \frac{1}{2}(8+e^x)^{-\frac{1}{2}} \times e^x$ | M1A1 | M1: $\frac{dy}{dx} = k(8+e^x)^{-\frac{1}{2}} \times e^x$. A1: Correct differentiation |
| $\frac{d^2y}{dx^2} = \frac{1}{2}(8+e^x)^{-\frac{1}{2}} \times e^x - \frac{1}{4}(8+e^x)^{-\frac{3}{2}} \times e^{2x}$ | M1A1 | M1: Correct use of product rule: $\pm K(8+e^x)^{-\frac{3}{2}} \times e^{2x}$. A1: Correct second derivative with $e^x \times e^x$ or $e^{2x}$ |
| $f(0) = 3$ | B1 | May only appear in the expansion |
| $f'(0) = \frac{1}{6},\quad f''(0) = \frac{17}{108}$ | M1 | Attempt both $f'(0)$ and $f''(0)$ with their derivatives found above |
| $(y=)\ 3 + \frac{1}{6}x + \frac{17}{216}x^2$ | M1 A1cso (8) | M1: Uses correct Maclaurin series with their values. Accept 2 or 2! in $x^2$ term. A1: Correct expression |

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3.

$$y = \sqrt { 8 + \mathrm { e } ^ { x } } , \quad x \in \mathbb { R }$$

Find the series expansion for $y$ in ascending powers of $x$, up to and including the term in $x ^ { 2 }$, giving each coefficient in its simplest form.\\

\hfill \mbox{\textit{Edexcel FP2 2014 Q3 [8]}}

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