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}}

KS Comments:

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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