2022 H2 Maths Paper 1 Q2 — Worked Solution

Part (a)

3 marks

f(x) = \tan^{- 1}\left( \sqrt{2} + x \right)

Using $\frac{d}{du}\tan^{- 1}u = \frac{1}{1 + u^{2}}$ with $u = \sqrt{2} + x$ , where $\frac{du}{dx} = 1$ ,

f'(x) = \frac{1}{1 + \left( \sqrt{2} + x \right)^{2}}

Writing $f'(x) = \left\lbrack 1 + \left( \sqrt{2} + x \right)^{2} \right\rbrack^{- 1}$ and differentiating by the chain rule,

f''(x) = - \left\lbrack 1 + \left( \sqrt{2} + x \right)^{2} \right\rbrack^{- 2} \times 2\left( \sqrt{2} + x \right)

= - \frac{2\left( \sqrt{2} + x \right)}{\left\lbrack 1 + \left( \sqrt{2} + x \right)^{2} \right\rbrack^{2}}

To check your answer: at

x = 0

f'(0) = \frac{1}{1 + 2} = \frac{1}{3}

, finite and consistent with

f

being smooth at

x = 0

✓

$\therefore$ $f'(x) = \frac{1}{1 + \left( \sqrt{2} + x \right)^{2}}$ and $f''(x) = - \frac{2\left( \sqrt{2} + x \right)}{\left\lbrack 1 + \left( \sqrt{2} + x \right)^{2} \right\rbrack^{2}}$ .

Part (b)

3 marks

Evaluating the derivatives from (a) at $x = 0$ ,

f(0) = \tan^{- 1}\left( \sqrt{2} \right) \approx 0.9553

f'(0) = \frac{1}{1 + 2} = \frac{1}{3}

f''(0) = - \frac{2\sqrt{2}}{3^{2}} = - \frac{2\sqrt{2}}{9} \approx - 0.3143

Using the Maclaurin series $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^{2} + \cdots$ ,

f(x) \approx 0.9553 + \frac{1}{3}x + \frac{- 2\sqrt{2}/9}{2}x^{2}

= 0.9553 + \frac{1}{3}x - \frac{\sqrt{2}}{9}x^{2}

To check your answer: at

x = 0.1

, the series gives

0.9553 + 0.0333 - 0.00157 \approx 0.9871

; direct computation gives

\tan^{- 1}\left( \sqrt{2} + 0.1 \right) \approx 0.9871

, matching closely for this small

x

✓

Rounding each coefficient to $3$ significant figures ( $\tan^{- 1}\sqrt{2} \approx 0.955$ , $\frac{1}{3} \approx 0.333$ , $- \frac{\sqrt{2}}{9} \approx - 0.157$ ),

f(x) \approx 0.955 + 0.333x - 0.157x^{2}

$\therefore$ the first three terms are $f(x) \approx 0.955 + 0.333x - 0.157x^{2}$ .

2022 H2 Maths Paper 1, Question 2

Part (a)

Part (b)

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