2018 H2 Maths Paper 1 Q3 — Worked Solution

Part (i)

3 marks

Since $y = ux^{2}$ , where $u$ is a function of $x$ , we differentiate using the product rule:

\frac{dy}{dx} = x^{2}\frac{du}{dx} + 2xu.

Substituting $y = ux^{2}$ and this expression for $\frac{dy}{dx}$ into $x\frac{dy}{dx} = 2y - 6$ :

x\left( x^{2}\frac{du}{dx} + 2xu \right) = 2\left( ux^{2} \right) - 6

x^{3}\frac{du}{dx} + 2x^{2}u = 2x^{2}u - 6

x^{3}\frac{du}{dx} = - 6

\frac{du}{dx} = - \frac{6}{x^{3}}\ \ \left( \text{shown} \right)

$\therefore f(x) = - \frac{6}{x^{3}}$ .

Part (ii)

4 marks

Approach

Having reduced the equation to $\frac{du}{dx} = f(x)$ in (i), we integrate directly to find $u$ , then convert back to $y$ using $y = ux^{2}$ and apply the given condition.

Integrating $\frac{du}{dx} = - 6x^{- 3}$ with respect to $x$ :

u = \int - 6x^{- 3}\, dx

= 3x^{- 2} + C

= \frac{3}{x^{2}} + C,

where $C$ is an arbitrary constant. Since $y = ux^{2}$ ,

y = \left( \frac{3}{x^{2}} + C \right)x^{2}

= 3 + Cx^{2}.

Using the given condition $y = 2$ when $x = 1$ :

2 = 3 + C

C = - 1

$\therefore y = 3 - x^{2}$ .

To check your answer: with

y = 3 - x^{2}

\frac{dy}{dx} = - 2x

, so

x\frac{dy}{dx} = - 2x^{2}

, and

2y - 6 = 2\left( 3 - x^{2} \right) - 6 = - 2x^{2}

. The two sides agree, and

y(1) = 3 - 1 = 2

matches the given condition. ✓

2018 H2 Maths Paper 1, Question 3

Part (i)

Part (ii)

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