2019 H2 Maths Paper 1 Q2 — Worked Solution

Part (i)

1 mark

Since $C$ crosses the $x$ -axis at $(a,0)$ , $a$ satisfies

x^{3} + x - 1 = 0

By GC, this cubic has only one real root.

\Rightarrow a = 0.682\ \left( 3\ \text{d.p.} \right)

$\therefore$ $a = 0.682$ .

Part (ii)

4 marks

Approach

We need to know, on each interval, whether $C$ lies above or below the $x$ -axis, since that decides whether an integral of $y$ gives the area directly or its negative. Since $y'= 3x^{2} + 1 > 0$ for all $x$ , $C$ is strictly increasing throughout, and $a = 0.682$ is its only zero. So $y < 0$ for $x < a$ and $y > 0$ for $x > a$ .

On $\lbrack - 1,0\rbrack$ , since $0 < a$ , $C$ lies entirely below the $x$ -axis, so $P$ is the region between $C$ and the $x$ -axis there and

\text{Area of}\ P = - \int_{- 1}^{0}\left( x^{3} + x - 1 \right)\, dx

= - \left\lbrack \frac{x^{4}}{4} + \frac{x^{2}}{2} - x \right\rbrack_{- 1}^{0}

= - \left( 0 - \left( \frac{1}{4} + \frac{1}{2} + 1 \right) \right)

= \frac{7}{4}

On $\lbrack a,b\rbrack$ , since $b > a$ , $C$ lies entirely above the $x$ -axis, so

\text{Area of}\ Q = \int_{a}^{b}\left( x^{3} + x - 1 \right)\, dx = \left\lbrack \frac{x^{4}}{4} + \frac{x^{2}}{2} - x \right\rbrack_{a}^{b}

Using the given condition that the area of $Q$ is $2$ times the area of $P$ ,

\left\lbrack \frac{x^{4}}{4} + \frac{x^{2}}{2} - x \right\rbrack_{a}^{b} = 2\left( \frac{7}{4} \right) = \frac{7}{2}

This is an equation in $b$ alone, since $a = 0.682327\ldots$ is already known. By GC, solving numerically,

b = 1.892\ \left( 3\ \text{d.p.} \right)

To check your answer: with

a = 0.682327803828

and

b = 1.892432535925

, numerically integrating

x^{3} + x - 1

from

a

b

gives

1.750000\ldots = 2 \times \frac{7}{4}

. ✓

$\therefore$ $b = 1.892$ .

2019 H2 Maths Paper 1, Question 2

Part (i)

Part (ii)

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