The diagram below shows a curve with equation $y = 1 + k\sin x$ , defined for $0 \leqslant x \leqslant 3\pi$ .

The point ${\text{A}}\left( {\frac{\pi }{6}, - 2} \right)$ lies on the curve and ${\text{B}}(a,{\text{ }}b)$ is the maximum point.

(a)     Show that k = – 6 .

(b)     Hence, find the values of a and b .

(a)     $- 2 = 1 + k\sin \left( {\frac{\pi }{6}} \right)$     M1

$- 3 = \frac{1}{2}k$     A1

$k = - 6$     AG     N0

(b)     METHOD 1

maximum $\Rightarrow \sin x = - 1$     M1

$a = \frac{{3\pi }}{2}$     A1

$b = 1 - 6( - 1)$

$= 7$     A1     N2

METHOD 2

$y' = 0$     M1

$k\cos x = 0 \Rightarrow x = \frac{\pi }{2},{\text{ }}\frac{{3\pi }}{2},{\text{ }} \ldots$

$a = \frac{{3\pi }}{2}$     A1

$b = 1 - 6( - 1)$

$= 7$     A1     N2

Note: Award A1A1 for $\left( {\frac{{3\pi }}{2},{\text{ }}7} \right)$ .

[5 marks]

This was the most successfully answered question in the paper. Part (a) was done well by most candidates. In part (b), a small number of candidates used knowledge about transformations of functions to identify the coordinates of B. Most candidates used differentiation.