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

(i)
$\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}$

$\frac{dy}{dx} = 6t^{2} \times \frac{1}{6t}$

$\frac{dy}{dx} = t$

Equation of tangent: $y - 2t^{3} = t(x-3t^{2})$

$\therefore, y = tx - t^{3}$

(ii)
Tangent at P: $y = px - p^{3}$

Tangent at Q: $y = qx - q^{3}$

$px-p^{3} = qx - q^{3}$

$x(p-q) = p^{3} - q^{3}$

$x(p-q) = (p-q)(p^{2}+pq+q^{2})$

$x= p^{2}+pq+q^{2}$

$y = p(p^{2}+pq+q^{2}) - p^{3}$

$y = p^{2}q+pq^{2}$

Since $pq=-1$, $y = -p - q = - (p + q)$

$x = p^{2} + pq + q^{2}$

$x = p^{2} + 2pq + q^{2} - pq$

$x = (p+q)^{2} -pq$

$x = (-y)^{2} + 1$

$x= y^{2} + 1$

$\therefore, R ~\mathrm{lies~on~the~curve~} x=y^{2} + 1$

(iii)
Equating the equations of both C and L,

$x = y^{2} + 1$

$3t^{2} = (2t^{3})^{2} +1$

$4t^{6} - 3t^{2} +1 = 0$

let $t^{2} = w$,

$4w^{3} - 3w + 1 =0$

Using GC, $w = \frac{1}{2}$

$t^{2} = \frac{1}{2}$

$t = \frac{1}{\sqrt{2}} or - \frac{1}{\sqrt{2}} (\mathrm{rej~ since} y>0)$

$\therefore, x= \frac{3}{2}, y=\frac{1}{\sqrt{2}}~\Rightarrow M(\frac{3}{2}, \frac{1}{\sqrt{2}})$

(iv)
$\int_{0}^{1.5} 2t^{3} dx - \int_{1}^{1.5} \sqrt{x-1} dx$

$= \int_{0}^{\frac{1}{\sqrt{2}}} 2t^{3} (6t) dt - \int_{1}^{1.5} \sqrt{x-1} dx$

$= \int_{0}^{\frac{1}{\sqrt{2}}} 12t^{4} dt - \int_{1}^{1.5} \sqrt{x-1} dx$

$= \frac{12t^{5}}{5}\biggl|_0^{\frac{1}{\sqrt{2}}} - \frac{(x-1)^{1.5}}{1.5} \biggl|_1^{1.5}$

$= \frac{3}{5 \sqrt{2}} - \frac{1}{3 \sqrt{2}}$

$= \frac{4}{15 \sqrt{2}}$

This question is not difficult. The only problem I noticed is that students were panicking by the time they reached this question, partly due to the heavy weightage of this question. But panicking caused them to make further mistakes.

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