2020 H2 Maths Paper 1 Q3 — Worked Solution

Part (i)

3 marks

We differentiate $f(x) = \ln\left( 1 + \sin 3x \right)$ using the chain rule.

f'(x) = \frac{3\cos 3x}{1 + \sin 3x}

Differentiating again using the quotient rule, with numerator $u = 3\cos 3x$ (so $u'= - 9\sin 3x$ ) and denominator $v = 1 + \sin 3x$ (so $v'= 3\cos 3x$ ),

f''(x) = \frac{u'v - uv'}{v^{2}}

= \frac{\left( - 9\sin 3x \right)\left( 1 + \sin 3x \right) - \left( 3\cos 3x \right)\left( 3\cos 3x \right)}{\left( 1 + \sin 3x \right)^{2}}

= \frac{- 9\sin 3x - 9\sin^{2}(3x) - 9\cos^{2}(3x)}{\left( 1 + \sin 3x \right)^{2}}

Using $\sin^{2}(3x) + \cos^{2}(3x) = 1$ ,

f''(x) = \frac{- 9\sin 3x - 9}{\left( 1 + \sin 3x \right)^{2}}

= \frac{- 9\left( 1 + \sin 3x \right)}{\left( 1 + \sin 3x \right)^{2}}

= \frac{- 9}{1 + \sin 3x}\ \ \left( \text{shown} \right)

To check your answer: at

x = 0

f''(0) = \frac{- 9}{1 + \sin 0} = - 9

, matching direct numerical differentiation of

f(x)

twice at

x = 0

. ✓

$\therefore$ $k = - 9$ .

Part (ii)

4 marks

Approach

Since $f(0) = 0$ , the constant term of the Maclaurin series vanishes, so the first non-zero term is $f'(0)x$ . We need the third derivative to reach the third non-zero term, found by differentiating the compact form of $f''(x)$ from part (i) rather than differentiating $f'(x)$ directly again.

We evaluate $f(x)$ and its derivatives at $x = 0$ .

f(0) = \ln\left( 1 + \sin 0 \right) = \ln 1 = 0

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

f''(0) = \frac{- 9}{1 + \sin 0} = - 9

Differentiating $f''(x) = - 9\left( 1 + \sin 3x \right)^{- 1}$ using the chain rule,

f'''(x) = - 9 \times ( - 1)\left( 1 + \sin 3x \right)^{- 2} \times 3\cos 3x

= \frac{27\cos 3x}{\left( 1 + \sin 3x \right)^{2}}

f'''(0) = \frac{27\cos 0}{\left( 1 + \sin 0 \right)^{2}} = 27

The Maclaurin series is $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3} + \cdots$ .

f(x) = 0 + 3x + \frac{- 9}{2}x^{2} + \frac{27}{6}x^{3} + \cdots

= 3x - \frac{9}{2}x^{2} + \frac{9}{2}x^{3} + \cdots

To check your answer: a computer-algebra Maclaurin expansion of

\ln\left( 1 + \sin 3x \right)

up to

x^{3}

confirms

3x - \frac{9}{2}x^{2} + \frac{9}{2}x^{3}

. ✓

$\therefore$ the first three non-zero terms are $3x - \frac{9}{2}x^{2} + \frac{9}{2}x^{3}$ .

2020 H2 Maths Paper 1, Question 3

Part (i)

Part (ii)

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