(a)     If A, B and C have x-coordinates $a\frac{\pi }{2}$, $b\frac{\pi }{6}$ and $c\frac{\pi }{2}$, where $a$ , $b$, $c \in \mathbb{N}$ , find the values of $a$ , $b$ and $c$ .

(b)     Find the range of $f$ .

(c)     Find the equation of the normal to the graph of f at the point C, giving your answer in the form $y = px + q$ .

(d)     The region $S$ is rotated through ${2\pi }$ about the x-axis to generate a solid.

  (i)     Write down an integral that represents the volume $V$ of this solid.

  (ii)     Show that $V = 6{\pi ^2}$ .

(a)     METHOD 1

using GDC

$a = 1$, $b = 5$, $c = 3$     A1A2A1

METHOD 2

$x = x + 2\cos x \Rightarrow \cos x = 0$

$\Rightarrow x = \frac{\pi }{2}$, $\frac{{3\pi }}{2}$ ...     M1

$a = 1$, $c = 3$     A1

$1 - 2\sin x = 0$     M1

$\Rightarrow \sin x = \frac{1}{2} \Rightarrow x = \frac{\pi }{6}$ or $\frac{{5\pi }}{6}$

$b = 5$     A1

Note: Final M1A1 is independent of previous work.

[4 marks]

(b)     $f\left( {\frac{{5\pi }}{6}} \right) = \frac{{5\pi }}{6} - \sqrt 3$   (or $0.886$)     (M1)

$f(2\pi ) = 2\pi  + 2$ (or $8.28$)     (M1)

the range is $\left[ {\frac{{5\pi }}{6} - \sqrt 3 ,{\text{ }}2\pi  + 2} \right]$ (or [$0.886$, $8.28$])     A1

[3 marks]

(c)     $f'(x) = 1 - 2\sin x$     (M1)

$f'\left( {\frac{{3\pi }}{6}} \right) = 3$     A1

gradient of normal $= - \frac{1}{3}$     (M1)

equation of the normal is $y - \frac{{3\pi }}{2} = - \frac{1}{3}\left( {x - \frac{{3\pi }}{2}} \right)$     (M1)

$y = - \frac{1}{3}x + 2\pi$   (or equivalent decimal values)     A1     N4

[5 marks]

(d)     (i)     $V = \pi \int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left( {{x^2} - {{\left( {x + 2\cos x} \right)}^2}} \right)} {\text{d}}x$   (or equivalent)     A1A1

Note: Award A1 for limits and A1 for $\pi$ and integrand.

(ii)     $V = \pi \int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left( {{x^2} - {{\left( {x + 2\cos x} \right)}^2}} \right)} {\text{d}}x$

$= - \pi \int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left( {4x\cos x + 4{{\cos }^2}x} \right)} {\text{d}}x$

using integration by parts     M1

and the identity $4{\cos ^2}x = 2\cos 2x + 2$ ,     M1

$V = - \pi \left[ {\left( {4x\sin x + 4\cos x} \right) + \left( {\sin 2x + 2x} \right)} \right]_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}}$     A1A1

Note: Award A1 for ${4x\sin x + 4\cos x}$ and A1 for sin ${2x + 2x}$ .

$= - \pi \left[ {\left( {6\pi \sin \frac{{3\pi }}{2} + 4\cos \frac{{3\pi }}{2} + \sin 3\pi  + 3\pi } \right) - \left( {2\pi sin\frac{\pi }{2} + 4\cos \frac{\pi }{2} + \sin \pi  + \pi } \right)} \right]$     A1

$= - \pi \left( { - 6\pi  + 3\pi  - \pi } \right)$

$= 6{\pi ^2}$     AG     N0

Note: Do not accept numerical answers.

[7 marks]

Total [19 marks]

Generally there were many good attempts to this, more difficult, question. A number of students found $b$ to be equal to 1, rather than 5. In the final part few students could successfully work through the entire integral successfully.