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.