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

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

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

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

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

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

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

(ii)
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.

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

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

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

(iii)

$\int_0^1 y dx$

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

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

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

(iv)
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, $\mathrm{tan} \theta = \sqrt{2}$

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

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

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.

• […] Question 11 Question 11 […]

Those who got A for 2013 paper around how much did they get for Paper 2 and paper 1 separately? 🙂 And also for the last part if i leave answer as tan tether = 3/a how many marks I get?

• KS Teng

From what I understand, mostly >75. I had one student who left 30 marks blanks in paper 1 due to time constraint, she was very confident of paper 2 getting > 95 though. And she still got an A. But she was really confident, like she new how she till manage. In any case, the past year distinctions should not be much of an indication, since it largely still depends on your cohort and how you stack against the rest. Like I mentioned here, Bell curve distributes the score of students. It will help the best student for sure, But the weaker students are subjected to how well the best students did. I probably just lose the 1 mark.

• aaa

ooh so >75 for P1 and >75 for P2?

• aaa

• KS Teng

Its quite hard to suggest a sufficient mark. Necessary mark, 70/100? Safe and guaranteed mark 170/100. The definition need to be really precise here. :/

• aaa

thanks for answering 🙂 hmmm what was the lowest mark you know that got A in 2014?

for 11ii) if I did everything including proving second derivative except find the coordinates how many marks do I get? Thank u!!

• KS Teng

I think if you left it in trigonometry form, probably lose 2 marks maximum. :/