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.