2018 H2 Maths Paper 1 Q2 — Worked Solution

Part (i)

2 marks

Since $A$ and $B$ lie on both the curve and the line, we substitute $y = \frac{3}{x}$ into $y + 2x = 7$ .

\frac{3}{x} + 2x = 7

3 + 2x^{2} = 7x

2x^{2} - 7x + 3 = 0

(2x - 1)(x - 3) = 0

$\therefore$ the $x$ -coordinates of $A$ and $B$ are $x = \frac{1}{2}$ and $x = 3$ .

Part (ii)

4 marks

Approach

The region bounded by the curve and the line lies between $x = \frac{1}{2}$ and $x = 3$ . Since the two boundaries are a curve and a line (not one curve and the $x$ -axis), the volume of revolution is the volume swept by the line minus the volume swept by the curve, not a single $\pi\int y^{2}\, dx$ .

Writing the line as $y = 7 - 2x$ , we check which boundary is on top: at $x = 1$ , the line gives $y = 5$ and the curve gives $y = 3$ , so the line lies above the curve on $\frac{1}{2} < x < 3$ . The required volume is therefore

V = \pi\int_{1/2}^{3}\left\lbrack (7 - 2x)^{2} - \left( \frac{3}{x} \right)^{2} \right\rbrack dx.

Expanding the integrand,

V = \pi\int_{1/2}^{3}\left( 4x^{2} - 28x + 49 - \frac{9}{x^{2}} \right)dx

= \pi\left\lbrack \frac{4x^{3}}{3} - 14x^{2} + 49x + \frac{9}{x} \right\rbrack_{1/2}^{3}

At $x = 3$ :

\frac{4(3)^{3}}{3} - 14(3)^{2} + 49(3) + \frac{9}{3} = 36 - 126 + 147 + 3 = 60

At $x = \frac{1}{2}$ :

\frac{4\left( \frac{1}{2} \right)^{3}}{3} - 14\left( \frac{1}{2} \right)^{2} + 49\left( \frac{1}{2} \right) + \frac{9}{\frac{1}{2}} = \frac{1}{6} - \frac{7}{2} + \frac{49}{2} + 18 = \frac{235}{6}

V = \pi\left( 60 - \frac{235}{6} \right)

= \frac{125\pi}{6}

$\therefore$ the exact volume generated is $\frac{125\pi}{6}\ \text{units}^{3}$ .

2018 H2 Maths Paper 1, Question 2

Part (i)

Part (ii)

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