2022 H2 Maths Paper 1 Q4 — Worked Solution

Part (a)

2 marks

We write $\cot x$ in terms of sine and cosine, then differentiate with the quotient rule.

\cot x = \frac{\cos x}{\sin x}

\frac{d}{dx}\left( \cot x \right) = \frac{\left( - \sin x \right)\left( \sin x \right) - \left( \cos x \right)\left( \cos x \right)}{\sin^{2}x}

= \frac{- \sin^{2}x - \cos^{2}x}{\sin^{2}x}

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

Since $\sin^{2}x + \cos^{2}x = 1$ ,

\frac{d}{dx}\left( \cot x \right) = \frac{- 1}{\sin^{2}x} = - {cosec}^{2}x\ \ \left( \text{shown} \right)

To check your answer: at

x = \frac{\pi}{4}

,

\cot x = 1

, and numerically differentiating

\cot x

there gives a gradient close to

- 2

, matching

- {cosec}^{2}\left( \frac{\pi}{4} \right) = - \left( \sqrt{2} \right)^{2} = - 2

✓

Part (b)

1 mark

Using $\sin 2x = 2\sin x\cos x$ and $\tan x = \frac{\sin x}{\cos x}$ ,

\sin 2x\tan x = \left( 2\sin x\cos x \right)\left( \frac{\sin x}{\cos x} \right)

= 2\sin^{2}x\ \ \left( \text{shown} \right)

To check your answer: at

x = \frac{\pi}{6}

, LHS

\text{=}\sin\frac{\pi}{3}\tan\frac{\pi}{6} = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}} = \frac{1}{2}

; RHS

\text{=}2\sin^{2}\frac{\pi}{6} = 2\left( \frac{1}{2} \right)^{2} = \frac{1}{2}

✓

Part (c)

4 marks

Approach

the integrand is $cosec\, 6x\cot 3x$ , which is not a standard form. Part (b) supplies an identity relating $\sin 2x$ , $\tan x$ and $\sin^{2}x$ ; replacing $x$ by $3x$ turns it into an identity for $\sin 6x$ , $\tan 3x$ and $\sin^{2}3x$ , from which $cosec\, 6x\cot 3x$ can be rewritten as a multiple of ${cosec}^{2}3x$ , a form we can integrate directly.

Replacing $x$ by $3x$ in (b), $\sin 6x\tan 3x = 2\sin^{2}3x$ . Rearranging for $\cot 3x$ ,

\tan 3x = \frac{2\sin^{2}3x}{\sin 6x}

\Rightarrow \cot 3x = \frac{\sin 6x}{2\sin^{2}3x}

Multiplying both sides by $cosec\, 6x = \frac{1}{\sin 6x}$ ,

cosec\, 6x\cot 3x = \frac{1}{\sin 6x} \cdot \frac{\sin 6x}{2\sin^{2}3x}

= \frac{1}{2}{cosec}^{2}3x

To check your answer: at

x = \frac{\pi}{12}

,

cosec\left( \frac{\pi}{2} \right)\cot\left( \frac{\pi}{4} \right) = (1)(1) = 1

;

\frac{1}{2}\text{cosec}^{2}\left( \frac{\pi}{4} \right) = \frac{1}{2}\left( \sqrt{2} \right)^{2} = 1

✓

By (a) generalized with the chain rule, $\frac{d}{dx}\left\lbrack \cot(3x) \right\rbrack = - 3\,{cosec}^{2}(3x)$ , so

\int cosec\, 6x\cot 3x\ dx = \int\frac{1}{2}{cosec}^{2}3x\ dx

= \frac{1}{2}\left( - \frac{1}{3}\cot 3x \right) + c

= - \frac{1}{6}\cot 3x + c

Evaluating from $x = \frac{\pi}{18}$ to $x = \frac{\pi}{9}$ : at $x = \frac{\pi}{9}$ , $3x = \frac{\pi}{3}$ , so $\cot\left( \frac{\pi}{3} \right) = \frac{1}{\sqrt{3}}$ ; at $x = \frac{\pi}{18}$ , $3x = \frac{\pi}{6}$ , so $\cot\left( \frac{\pi}{6} \right) = \sqrt{3}$ .

\int_{\pi/18}^{\pi/9}{cosec}\, 6x\cot 3x\ dx = \left\lbrack - \frac{1}{6}\cot 3x \right\rbrack_{\pi/18}^{\pi/9}

= - \frac{1}{6}\left( \frac{1}{\sqrt{3}} - \sqrt{3} \right)

= - \frac{1}{6}\left( \frac{1 - 3}{\sqrt{3}} \right)

= \frac{2}{6\sqrt{3}}

= \frac{1}{3\sqrt{3}}

Rationalizing,

\frac{1}{3\sqrt{3}} = \frac{\sqrt{3}}{9}

To check your answer:

\frac{\sqrt{3}}{9} \approx 0.1925

, matching a numerical estimate of

\int_{\pi/18}^{\pi/9}{cosec}\, 6x\cot 3x\ dx \approx 0.1925

✓

$\therefore$ $\int_{\pi/18}^{\pi/9}{cosec}\, 6x\cot 3x\ dx = \frac{\sqrt{3}}{9}$ .

2022 H2 Maths Paper 1, Question 4

Part (a)

Part (b)

Part (c)

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