(i) Write down the range of the function f .

(ii) Consider $f(x) = 1$ , $0 \le x \le 2\pi$ . Write down the number of solutions to this equation. Justify your answer.

Find $f'(x)$ , giving your answer in the form $a{\sin ^p}x{\cos ^q}x$ where $a{\text{, }}p{\text{, }}q \in \mathbb{Z}$ .

Let $g(x) = \sqrt 3 \sin x{(\cos x)^{\frac{1}{2}}}$ for $0 \le x \le \frac{\pi }{2}$ . Find the volume generated when the curve of g is revolved through $2\pi$ about the x-axis.

(i) range of f is $[ - 1{\text{, }}1]$ , $( - 1 \le f(x) \le 1)$     A2     N2

(ii) ${\sin ^3}x \Rightarrow 1 \Rightarrow \sin x = 1$     A1

justification for one solution on $[0{\text{, }}2\pi ]$    R1

e.g. $x = \frac{\pi }{2}$ , unit circle, sketch of $\sin x$

1 solution (seen anywhere)     A1     N1

[5 marks]

$f'(x) = 3{\sin ^2}x\cos x$     A2     N2

[2 marks]

using $V = \int_a^b {\pi {y^2}{\rm{d}}x}$     (M1)

$V = \int_0^{\frac{\pi }{2}} {\pi (\sqrt 3 } \sin x{\cos ^{\frac{1}{2}}}x{)^2}{\rm{d}}x$     (A1)

$= \pi \int_0^{\frac{\pi }{2}} {3{{\sin }^2}x\cos x{\rm{d}}x}$     A1

$V = \pi \left[ {{{\sin }^3}x} \right]_0^{\frac{\pi }{2}}$ $\left( { = \pi \left( {{{\sin }^3}\left( {\frac{\pi }{2}} \right) - {{\sin }^3}0} \right)} \right)$     A2

evidence of using $\sin \frac{\pi }{2} = 1$ and $\sin 0 = 0$     (A1)

e.g. $\pi \left( {1 - 0} \right)$

$V = \pi$     A1     N1

[7 marks]

This question was not done well by most candidates. No more than one-third of them could correctly give the range of $f(x) = {\sin ^3}x$ and few could provide adequate justification for there being exactly one solution to $f(x) = 1$ in the interval $[0{\text{, }}2\pi ]$ .

This question was not done well by most candidates.

This question was not done well by most candidates. No more than one-third of them could correctly give the range of $f(x) = {\sin ^3}x$ and few could provide adequate justification for there being exactly one solution to $f(x) = 1$ in the interval $[0{\text{, }}2\pi ]$ . Finding the derivative of this function also presented major problems, thus making part (c) of the question much more difficult. In spite of the formula for volume of revolution being given in the Information Booklet, fewer than half of the candidates could correctly put the necessary function and limits into $\pi \int_a^b {{y^2}{\rm{d}}x}$ and fewer still could square $\sqrt 3 \sin x{\cos ^{\frac{1}{2}}}x$ correctly. From those who did square correctly, the correct antiderivative was not often recognized. All manner of antiderivatives were suggested instead.