2018 H2 Maths Paper 2 Q1 — Worked Solution

Part (i)

4 marks

Since $\frac{dy}{dx} = \left( \frac{1}{3}y - 15 \right)^{\frac{1}{3}}$ is a function of $y$ only, we separate variables.

\int\frac{dy}{\left( \frac{1}{3}y - 15 \right)^{\frac{1}{3}}} = \int dx

Let $u = \frac{1}{3}y - 15$ , so that $y = 3u + 45$ and $dy = 3\, du$ .

3\int u^{- \frac{1}{3}}\, du = \int dx

Integrating both sides,

\frac{9}{2}u^{\frac{2}{3}} = x + C

Substituting the initial condition $(0,69)$ : when $x = 0$ , $y = 69$ , so $u = \frac{1}{3}(69) - 15 = 8$ .

\frac{9}{2}(8)^{\frac{2}{3}} = C

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

= 18

So $C = 18$ .

So $\frac{9}{2}u^{\frac{2}{3}} = x + 18$ , giving $u^{\frac{2}{3}} = \frac{2}{9}(x + 18)$ , so $u = \left\lbrack \frac{2}{9}(x + 18) \right\rbrack^{\frac{3}{2}}$ .

Since $y = 3u + 45$ ,

f(x) = 45 + \frac{2\sqrt{2}}{9}(x + 18)^{\frac{3}{2}}

To check your answer:

f(0) = 45 + \frac{2\sqrt{2}}{9}(18)^{\frac{3}{2}} = 45 + 24 = 69

, which matches the given point. ✓

$\therefore$ $f(x) = 45 + \frac{2\sqrt{2}}{9}(x + 18)^{\frac{3}{2}}$ .

Part (ii)

2 marks

Approach

The gradient is given directly in terms of $y$ , so it is quicker to find $y$ first from the gradient condition, then use the relation $\frac{9}{2}u^{\frac{2}{3}} = x + 18$ from part (i) to recover $x$ , rather than differentiating $f(x)$ .

Using the given gradient,

\left( \frac{1}{3}y - 15 \right)^{\frac{1}{3}} = 4

\Rightarrow \frac{1}{3}y - 15 = 64

\Rightarrow y = 237

So $u = 64$ at this point, and $u^{\frac{2}{3}} = 16$ . Substituting into $\frac{9}{2}u^{\frac{2}{3}} = x + 18$ ,

\frac{9}{2}(16) = x + 18

\Rightarrow 72 = x + 18

\Rightarrow x = 54

$\therefore$ the required point is $(54,\ 237)$ .

2018 H2 Maths Paper 2, Question 1

Part (i)

Part (ii)

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