All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

\frac{dy}{d\theta} = 6\mathrm{sin}\theta \mathrm{cos^2}\theta - 3 \mathrm{sin^3} \theta

\frac{dx}{d\theta} = 3 \mathrm{sin^2}\theta \mathrm{cos}\theta

\frac{dy}{dx} = \frac{6\mathrm{sin}\theta \mathrm{cos^2}\theta3 - \mathrm{sin^3} \theta}{3 \mathrm{sin^2}\theta \mathrm{cos}\theta}

\frac{dy}{dx} = \frac{3\mathrm{cos^2}\theta - \mathrm{sin^2} \theta}{ \mathrm{sin^2}\theta \mathrm{cos}\theta}

\frac{dy}{dx} = 2 \frac{\mathrm{cos}\theta}{\mathrm{sin}\theta} - \frac{\mathrm{sin}\theta}{\mathrm{cos}\theta}

\frac{dy}{dx} = 2 \mathrm{cot} \theta - \mathrm{tan} \theta

Let \frac{dy}{dx} = 0,  \frac{2}{\mathrm{tan}\theta} - \mathrm{tan}\theta = 0

\mathrm{tan^2} \theta = 2

\mathrm{tan} \theta = \sqrt{2} or - \sqrt{2} (reject since \theta \in [0, \frac{\pi}{2}]

\therefore, k = 2

When \mathrm{tan} \theta = \sqrt{2}, using trigonometry identity, we find

\mathrm{cos} \theta = \frac{\sqrt{3}}{3} ~and~ \mathrm{sin} \theta = \frac{\sqrt{6}}{3}

From \frac{dy}{dx}, \frac{d^2y}{dx^2} = -2\mathrm{cosec^2} \theta \frac{d\theta}{dx} - \mathrm{sec^2} \theta \frac{d\theta}{dx}

For \theta \in [0, \frac{\pi}{2}], \frac{d^2y}{dx^2} \textless 0 since \mathrm{sin} \theta > 0 and \mathrm{cos} \theta >0 for principal values of \theta

We have that when \mathrm{tan} \theta = \sqrt{2}, its is a maximum point.

x = \mathrm{sin^3} \theta = \frac{2\sqrt{6}}{9}

y = 3 \mathrm{sin^2}\theta \mathrm{cos} \theta = \frac{2 \sqrt{3}}{3}

\therefore, \text{Required}~\text{coordinates}~ = (\frac{2\sqrt{6}}{9}, \frac{2\sqrt{3}}{3})


\int_0^1 y dx

= \int_0^{\frac{\pi}{2}} 3 \mathrm{sin^2}\theta \mathrm{cos} \theta 3 \mathrm{sin^2}\theta \mathrm{cos}\theta d\theta

= \int_0^{\frac{\pi}{2}} 9 \mathrm{sin^4}\theta \mathrm{cos^2} \theta d\theta

Using GC, Area = 0.88357 \approx 0.884 \mathrm{units^2}

At P, y = ax
\Rightarrow 3 \mathrm{sin^2}\theta \mathrm{cos} \theta  = a \mathrm{sin^3} \theta

\mathrm{tan} \theta = \frac{3}{a}

At maximum point, \mathrm{tan} \theta = \sqrt{2}

\Rightarrow \sqrt{2} = \frac{3}{a}

\therefore, a = \frac{3\sqrt{2}}{2}

Back to 2015 A-level H2 Mathematics (9740) Paper 1 Suggested Solutions

KS Comments:

To find the exact coordinates, students need to be able to either identify the correct trigonometry identify, or draw the right angled triangle to find \mathrm{sin} \theta and \mathrm{cos} \theta carefully. As for the show part for the parametric integration, students need to start from the definition of area then proceed carefully, not forgetting to change the limits.


Leave a Reply