A2 June 2025 Paper 1 Q11

AQACurrent spec7 marksDifferentiation & Maclaurin

11 The function \(\mathrm{f}\) is defined by

\[\mathrm{f}(x) = \frac{1}{1 + \mathrm{e}^x} \qquad (x \in \mathbb{R})\]

(a) Show that\[\mathrm{f}^{\prime\prime}(x) = \frac{-\mathrm{e}^x + \mathrm{e}^{2x}}{(1 + \mathrm{e}^x)^3}\] [3 marks]

(b) Hence, find the Maclaurin expansion of \(\mathrm{f}(x)\) up to and including the term in \(x^3\) [4 marks]

Mark scheme (a)

Scheme	Marks	AO
Obtains \(\dfrac{-\mathrm{e}^x}{(1 + \mathrm{e}^x)^2}\) OE	B1	1.1b
Uses a correct method to obtain the second derivative.	M1	1.1a
Completes a reasoned argument to obtain \(\mathrm{f}^{\prime\prime}(x) = \dfrac{-\mathrm{e}^x + \mathrm{e}^{2x}}{(1 + \mathrm{e}^x)^3}\) AG	R1	2.1
	(3)

Typical solution

\[\mathrm{f}(x) = \frac{1}{1 + \mathrm{e}^x}\]\[\mathrm{f}^{\prime}(x) = \frac{-\mathrm{e}^x}{(1 + \mathrm{e}^x)^2}\]\[\begin{aligned}\mathrm{f}^{\prime\prime}(x) &= \frac{(1 + \mathrm{e}^x)^2(-\mathrm{e}^x) - (-\mathrm{e}^x)(2\mathrm{e}^x)(1 + \mathrm{e}^x)}{(1 + \mathrm{e}^x)^4} \\ &= \frac{-\mathrm{e}^x(1 + \mathrm{e}^x) + 2\mathrm{e}^{2x}}{(1 + \mathrm{e}^x)^3} \\ &= \frac{-\mathrm{e}^x + \mathrm{e}^{2x}}{(1 + \mathrm{e}^x)^3}\end{aligned}\]

Mark scheme (b)

Scheme	Marks	AO
Obtains \(\mathrm{f}(0) = \dfrac{1}{2}\) OE \(\mathrm{f}^{\prime}(0) = -\dfrac{1}{4}\) OE \(\mathrm{f}^{\prime\prime}(0) = 0\)	B1	1.1b
Obtains \(\mathrm{f}^{(3)}(0) = 0.125\) OE	B1	1.1b
Substitutes their four values into Maclaurin’s series.	M1	1.1a
Obtains \(\dfrac{1}{2} - \dfrac{1}{4}x + \dfrac{1}{48}x^3\)	A1	1.1b
	(4)
	(7 marks)

Typical solution

\[\mathrm{f}^{(3)}(x) = \frac{(1 + \mathrm{e}^x)^3(-\mathrm{e}^x + 2\mathrm{e}^{2x}) - (-\mathrm{e}^x + \mathrm{e}^{2x})(3\mathrm{e}^x)(1 + \mathrm{e}^x)^2}{(1 + \mathrm{e}^x)^6}\]\[\mathrm{f}(0) = \frac{1}{2}\]\[\mathrm{f}^{\prime}(0) = -\frac{1}{4}\]\[\mathrm{f}^{\prime\prime}(0) = 0\]\[\mathrm{f}^{(3)}(0) = \frac{2^3(-1 + 2) - 0}{2^6} = \frac{8}{64} = \frac{1}{8}\]\[\mathrm{f}(x) = \frac{1}{2} - \frac{1}{4}x + \frac{1}{48}x^3 + \ldots\]