2015 H2 Maths Paper 1 Q10 — Worked Solution

Part (i)

4 marks

Approach

From the diagram, $A_{1}$ is bounded below by the $x$ -axis, on the left by $y = \sin x$ (from $O$ to $P$ ) and on the right by $y = \cos x$ (from $P$ to $\left( \frac{1}{2}\pi,\ 0 \right)$ ); $A_{2}$ is bounded on the left by the $y$ -axis, above by $y = \cos x$ (from $(0,\ 1)$ to $P$ ) and below by $y = \sin x$ (from $O$ to $P$ ). Both areas split naturally at $x = \frac{1}{4}\pi$ , the $x$ -coordinate of $P$ .

A_{1} = \int_{0}^{\frac{1}{4}\pi}\sin x\, dx + \int_{\frac{1}{4}\pi}^{\frac{1}{2}\pi}\cos x\, dx

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

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

= 2 - \sqrt{2}.

A_{2} = \int_{0}^{\frac{1}{4}\pi}\left( \cos x - \sin x \right)\, dx

= \left\lbrack \sin x + \cos x \right\rbrack_{0}^{\frac{1}{4}\pi}

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

= \sqrt{2} - 1.

\frac{A_{1}}{A_{2}} = \frac{2 - \sqrt{2}}{\sqrt{2} - 1}

= \frac{\sqrt{2}\left( \sqrt{2} - 1 \right)}{\sqrt{2} - 1}

= \sqrt{2}.\ \ \left( \text{shown} \right)

Part (ii)

8 marks

Approach

The region described has $x = 0$ as its left boundary throughout, so rotating about the $y$ -axis gives a solid of revolution by the disc method, with radius equal to $x$ expressed as a function of $y$ .

For $0 \leq x \leq \frac{1}{4}\pi$ , $y = \sin x$ is increasing, so it has an inverse there; writing $x$ in terms of $y$ ,

x = \sin^{- 1}y.

The region bounded by $y = \sin x$ between $O$ and $P$ , the line $y = \frac{1}{2}\sqrt{2}$ , and the $y$ -axis is the set of points with $0 \leq y \leq \frac{1}{2}\sqrt{2}$ and $0 \leq x \leq \sin^{- 1}y$ . Rotating this about the $y$ -axis through $360^{\circ}$ , each cross-section at height $y$ is a disc of radius $x = \sin^{- 1}y$ , so the volume is

V = \pi\int_{0}^{\frac{1}{2}\sqrt{2}}x^{2}\, dy

= \pi\int_{0}^{\frac{1}{2}\sqrt{2}}\left( \sin^{- 1}y \right)^{2}dy.\ \ \left( \text{shown} \right)

Part (iii)

Approach

The substitution $y = \sin u$ removes the inverse sine, since $\sin^{- 1}y = u$ once $u$ is restricted to the range where $y = \sin u$ has an inverse; the limits transform along with the variable.

Let $y = \sin u$ , so $dy = \cos u\, du$ . For $u \in \left\lbrack 0,\ \frac{1}{4}\pi \right\rbrack \subset \left\lbrack - \frac{1}{2}\pi,\ \frac{1}{2}\pi \right\rbrack$ , the range of the principal branch of $\sin^{- 1}$ , we have $\sin^{- 1}y = \sin^{- 1}\left( \sin u \right) = u$ .

When $y = 0$ , $u = 0$ ; when $y = \frac{1}{2}\sqrt{2}$ , $\sin u = \frac{1}{2}\sqrt{2}$ , so $u = \frac{1}{4}\pi$ .

\pi\int_{0}^{\frac{1}{2}\sqrt{2}}\left( \sin^{- 1}y \right)^{2}dy = \pi\int_{0}^{\frac{1}{4}\pi}u^{2}\cos u\, du.

$\therefore a = 0$ , $b = \frac{1}{4}\pi$ .

Approach

$\int u^{2}\cos u\, du$ needs integration by parts twice, differentiating $u^{2}$ down to a constant while integrating $\cos u$ each time.

\int u^{2}\cos u\, du = u^{2}\sin u - \int 2u\sin u\, du

= u^{2}\sin u - 2\left( - u\cos u + \int\cos u\, du \right)

= u^{2}\sin u + 2u\cos u - 2\sin u\ ( + C).

\pi\int_{0}^{\frac{1}{4}\pi}u^{2}\cos u\, du = \pi\left\lbrack u^{2}\sin u + 2u\cos u - 2\sin u \right\rbrack_{0}^{\frac{1}{4}\pi}

= \pi\left( \frac{\pi^{2}}{16} \cdot \frac{1}{2}\sqrt{2} + 2 \cdot \frac{1}{4}\pi \cdot \frac{1}{2}\sqrt{2} - 2 \cdot \frac{1}{2}\sqrt{2} \right)

= \pi\left( \frac{\sqrt{2}\pi^{2}}{32} + \frac{\sqrt{2}\pi}{4} - \sqrt{2} \right)

= \frac{\sqrt{2}\pi^{3}}{32} + \frac{\sqrt{2}\pi^{2}}{4} - \sqrt{2}\pi.

$\therefore$ the exact volume is $\frac{\sqrt{2}\pi^{3}}{32} + \frac{\sqrt{2}\pi^{2}}{4} - \sqrt{2}\pi$ , i.e. $\sqrt{2}\pi\left( \frac{\pi^{2}}{32} + \frac{\pi}{4} - 1 \right)$ .

2015 H2 Maths Paper 1, Question 10

Part (i)

Part (ii)

Part (iii)

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