Consider the function \(f:x \to \sqrt {\frac{\pi }{4} - \arccos x} \).

(a)     Find the largest possible domain of f.

(b)     Determine an expression for the inverse function, \({f^{ - 1}}\), and write down its domain.

(a)     \(\frac{\pi }{4} - \arccos x \geqslant 0\)

\(\arccos x \leqslant \frac{\pi }{4}\)     (M1)

\(x \geqslant \frac{{\sqrt 2 }}{2}\,\,\,\,\,\left( {{\text{accept }}x \geqslant \frac{1}{{\sqrt 2 }}} \right)\)     (A1)

since \( - 1 \leqslant x \leqslant 1\)     (M1)

\( \Rightarrow \frac{{\sqrt 2 }}{2} \leqslant x \leqslant 1\,\,\,\,\,\left( {{\text{accept }}\frac{1}{{\sqrt 2 }} \leqslant x \leqslant 1} \right)\)     A1

Note: Penalize the use of \( < \) instead of \( \leqslant \) only once.

(b)     \(y = \sqrt {\frac{\pi }{4} - \arccos x}  \Rightarrow x = \cos \left( {\frac{\pi }{4} - {y^2}} \right)\)     M1A1

\({f^{ - 1}}:x \to \cos \left( {\frac{\pi }{4} - {x^2}} \right)\)     A1

\(0 \leqslant x \leqslant \sqrt {\frac{\pi }{4}} \)     A1

[8 marks]

Very few correct solutions were seen to (a). Many candidates realised that \(\arccos x \leqslant \frac{\pi }{4}\) but then concluded incorrectly, not realising that cos is a decreasing function, that \(x \leqslant \cos \left( {\frac{\pi }{4}} \right)\). In (b) candidates often gave an incorrect domain.