A curve is given by the equation \(y = \sin (\pi \cos x)\).

Find the coordinates of all the points on the curve for which \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0,{\text{ }}0 \leqslant x \leqslant \pi \).

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - \cos (\pi \cos x) \times \pi \sin x\)    M1A1

Note:     Award follow through marks below if their answer is a multiple of the correct answer.

considering either \(\sin x = 0\) or \(\cos (\pi \cos x) = 0\)     (M1)

\(x = 0,{\text{ }}\pi \)    A1

\(\pi \cos x = \frac{\pi }{2},{\text{ }} - \frac{\pi }{2}{\text{ }}\left( { \Rightarrow \cos x = \frac{1}{2}, - \frac{1}{2}} \right)\)    M1

Note:     Condone absence of \( - \frac{\pi }{2}\).

\( \Rightarrow x = \frac{\pi }{3},{\text{ }}\frac{{2\pi }}{3}\)

\((0,{\text{ }}0),{\text{ }}\left( {\frac{\pi }{3},{\text{ 1}}} \right),{\text{ (}}\pi {\text{, 0)}}\)    A1

\(\left( {\frac{{2\pi }}{3},{\text{ }} - 1} \right)\)    A1

[7 marks]

This was not a straight-forward differentiation and it was pleasing to see how many candidates managed to do it correctly. Having done this they found two of the solutions, often three of the solutions, successfully. The final solution was found by only a few candidates. Again candidates lost marks unnecessarily by not close reading the question and realising that they needed both coordinates of the points, not just the \(x\)-coordinates.