2017 H2 Maths Paper 1 Q4 — Worked Solution

Part (i)

3 marks

By the quotient rule, with $u = 4x + 9$ , $v = x + 2$ :

\frac{dy}{dx}

= \frac{4(x + 2) - (4x + 9)(1)}{(x + 2)^{2}}

= \frac{4x + 8 - 4x - 9}{(x + 2)^{2}}

= \frac{- 1}{(x + 2)^{2}}

Since $x \neq - 2$ (excluded from the domain of $C$ ), $(x + 2)^{2} > 0$ for every point on $C$ , so

\frac{dy}{dx} = \frac{- 1}{(x + 2)^{2}} < 0\ \ \text{for all points on }C.

$\therefore$ the gradient of $C$ is negative for all points on $C$ . (shown)

Part (ii)

3 marks

Writing $4x + 9 = a(x + 2) + b$ :

4x + 9 = ax + 2a + b

Comparing coefficients of $x$ : $a = 4$ . Comparing constants: $2a + b = 9 \Rightarrow b = 9 - 8 = 1$ .

\therefore y = 4 + \frac{1}{x + 2}

To check your answer: at

x = 0

,

y = 4 + \frac{1}{2} = \frac{9}{2}

, which matches

\frac{4(0) + 9}{0 + 2} = \frac{9}{2}

. ✓

As $x \rightarrow \infty$ , $\frac{1}{x + 2} \rightarrow 0$ , so $y \rightarrow 4$ ; as $x \rightarrow - 2$ , $\frac{1}{x + 2} \rightarrow \pm \infty$ .

$\therefore$ the asymptotes of $C$ are $y = 4$ and $x = - 2$ .

Part (iii)

2 marks

Approach

From (ii), $y = 4 + \frac{1}{x + 2}$ . Starting from $y = \frac{1}{x}$ , replacing $x$ by $x + 2$ shifts the graph horizontally, and adding $4$ shifts it vertically; these two translations are independent of each other and can be applied in either order.

$\therefore$ a translation of $2$ units in the negative $x$ -direction, followed by a translation of $4$ units in the positive $y$ -direction, transforms the graph of $y = \frac{1}{x}$ onto the graph of $C$ .

2017 H2 Maths Paper 1, Question 4

Part (i)

Part (ii)

Part (iii)

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