Find $f\left( {\frac{\pi }{2}} \right)$ .

Find $(g \circ f)\left( {\frac{\pi }{2}} \right)$ .

Given that $(g \circ f)(x)$ can be written as $\cos (kx)$ , find the value of k, $k \in \mathbb{Z}$ .

$f\left( {\frac{\pi }{2}} \right) = \cos \pi$     (A1)

$= - 1$     A1     N2

[2 marks]

$(g \circ f)\left( {\frac{\pi }{2}} \right) = g( - 1)$ $( = 2{( - 1)^2} - 1)$    (A1)

$= 1$     A1     N2

[2 marks]

$(g \circ f)(x) = 2{(\cos (2x))^2} - 1$ $( = 2{\cos ^2}(2x) - 1)$     A1

evidence of $2{\cos ^2}\theta - 1 = \cos 2\theta$ (seen anywhere)     (M1)

$(g \circ f)(x) = \cos 4x$

$k = 4$     A1     N2

[3 marks]

In part (a), a number of candidates were not able to evaluate $\cos \pi$ , either leaving it or evaluating it incorrectly.

Almost all candidates evaluated the composite function in part (b) in the given order, many earning follow-through marks for incorrect answers from part (a). On both parts (a) and (b), there were candidates who correctly used double-angle formulas to come up with correct answers; while this is a valid method, it required unnecessary additional work.

Candidates were not as successful in part (c). Many tried to use double-angle formulas, but either used the formula incorrectly or used it to write the expression in terms of $\cos x$ and went no further. There were a number of cases in which the candidates "accidentally" came up with the correct answer based on errors or lucky guesses and did not earn credit for their final answer. Only a few candidates recognized the correct method of solution.