A2 October 2021 Paper 1 Q2

2 You are given that \(\mathrm{f}(x) = \tan^{-1}(1 + x)\).

(a)

(i) Find the value of \(\mathrm{f}(0)\). [1]

(ii) Determine the value of \(\mathrm{f}'(0)\). [2]

(iii) Show that \(\mathrm{f}''(0) = -\dfrac{1}{2}\). [3]

(b) Hence find the Maclaurin series for \(\mathrm{f}(x)\) up to and including the term in \(x^2\). [2]

Mark scheme (a)

Scheme	Marks	AO
(i) \(\mathrm{f}(0) = \dfrac{\pi}{4}\)	B1	1.1
	[1]
(ii) \(\mathrm{f}'(x) = \dfrac{1}{1 + (1 + x)^2} \Rightarrow \mathrm{f}'(0) = \dfrac{1}{2}\)	M1 A1	2.1 1.1
	[2]
(iii) \(\mathrm{f}'(x) = \dfrac{1}{1 + (1 + x)^2} = \dfrac{1}{2 + 2x + x^2}\)
\(\Rightarrow \mathrm{f}''(x) = \dfrac{1}{\left(2 + 2x + x^2\right)^2} \times (-1) \times (2 + 2x)\)	M1	2.1
\(= \dfrac{-(2 + 2x)}{\left(2 + 2x + x^2\right)^2}\)	A1	2.1
\(\Rightarrow \mathrm{f}''(0) = \left(\dfrac{-2}{4}\right) = -\dfrac{1}{2}\)	A1	2.1
	[3]

(a)(i)

B1: Not for \(45^\circ\)

(a)(ii)

M1: Diffn – Must be seen
\(\mathrm{f}'(x) = \dfrac{1}{1 + x^2}\) is M0

(a)(iii)

M1: Diffn their \(\mathrm{f}'(x)\)

A1: oe, e.g. \(\mathrm{f}''(x) = -\dfrac{2(1 + x)}{\left(1 + (1 + x)^2\right)^2}\)

A1: \(\mathrm{f}''(0)\) must be seen. The substitution must be seen (implied by \(-\frac{2}{4}\))
AG

Scheme	Marks
\(y = \tan^{-1}(1 + x) \Rightarrow 1 + x = \tan y\) \(\Rightarrow 1 = \sec^2 y\,.\,\dfrac{\mathrm{d}y}{\mathrm{d}x}\) \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sec^2 y} = \dfrac{1}{1 + \tan^2 y} = \dfrac{1}{1 + (1 + x)^2}\)

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{f}(x) = \mathrm{f}(0) + \mathrm{f}'(0)x + \mathrm{f}''(0)\dfrac{x^2}{2}\) \(= \dfrac{\pi}{4} + \dfrac{1}{2}x - \dfrac{1}{2} \times \dfrac{x^2}{2}\)	M1	1.1
\(= \dfrac{\pi}{4} + \dfrac{x}{2} - \dfrac{x^2}{4}\)	A1	2.2a
	[2]

M1: Using the formula and substituting their value for f;(0)

A1: ft their values from (a)