2017 H2 Maths Paper 1 Q3 — Worked Solution

Part (i)

4 marks

Approach

The equation of $C$ is not given as $y$ in terms of $x$ , so we differentiate implicitly to obtain $\frac{dy}{dx}$ , then set it to zero.

Differentiating $y^{2} - 2xy + 5x^{2} - 10 = 0$ implicitly with respect to $x$ :

2y\frac{dy}{dx} - 2y - 2x\frac{dy}{dx} + 10x = 0

\Rightarrow \frac{dy}{dx}(2y - 2x) = 2y - 10x

\Rightarrow \frac{dy}{dx} = \frac{2y - 10x}{2y - 2x} = \frac{y - 5x}{y - x}\ \ (y \neq x)

At a stationary point, $\frac{dy}{dx} = 0$ , so

y - 5x = 0

\Rightarrow y = 5x

Substituting into the equation of $C$ :

(5x)^{2} - 2x(5x) + 5x^{2} - 10 = 0

\Rightarrow 25x^{2} - 10x^{2} + 5x^{2} - 10 = 0

\Rightarrow 20x^{2} = 10

\Rightarrow x^{2} = \frac{1}{2}

\Rightarrow x = \pm \frac{\sqrt{2}}{2}

Note: at both points $y = 5x \neq x$ since $x \neq 0$ , so $\frac{dy}{dx}$ is genuinely defined and equal to $0$ there, confirming both are valid stationary points.

$\therefore$ $x = \frac{\sqrt{2}}{2}$ or $x = - \frac{\sqrt{2}}{2}$ .

Part (ii)

3 marks

Approach

We differentiate $\frac{dy}{dx} = \frac{y - 5x}{y - x}$ implicitly a second time to get $\frac{d^{2}y}{dx^{2}}$ . Since the numerator $y - 5x$ is zero exactly at the stationary point, the term involving it drops out, leaving a short expression.

Differentiating $\frac{dy}{dx}(y - x) = y - 5x$ implicitly with respect to $x$ :

\frac{d^{2}y}{dx^{2}}(y - x) + \frac{dy}{dx}\left( \frac{dy}{dx} - 1 \right) = \frac{dy}{dx} - 5

At the stationary point, $\frac{dy}{dx} = 0$ , so this reduces to

\frac{d^{2}y}{dx^{2}}(y - x) = - 5

\Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{- 5}{y - x}

For the stationary point with $x > 0$ : $x = \frac{\sqrt{2}}{2}$ , $y = 5x = \frac{5\sqrt{2}}{2}$ , so

y - x = 4x = 2\sqrt{2} > 0

giving

\frac{d^{2}y}{dx^{2}} = \frac{- 5}{2\sqrt{2}} < 0

Since $\frac{d^{2}y}{dx^{2}} < 0$ , the stationary point with $x > 0$ is a maximum.

$\therefore$ the stationary point with $x > 0$ is a maximum point.

2017 H2 Maths Paper 1, Question 3

Part (i)

Part (ii)

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