Given that f and its derivative, $f'$ , are continuous for all values in the domain of f , find the values of a and b .

Show that f is a one-to-one function.

Obtain expressions for the inverse function ${f^{ - 1}}$ and state their domains.

f continuous $\Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ \div }} f(x)$     M1

$4a + 2b = 8$     A1

$f'(x) = \left\{ {\begin{array}{*{20}{c}}   {2,}&{x < 2} \\   {2ax + b,}&{2 < x < 3} \end{array}} \right.$     A1

$f'{\text{ continuous}} \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} f'(x) = \mathop {\lim }\limits_{x \to {2^ \div }} f'(x)$

$4a + b = 2$     A1

solve simultaneously     M1

to obtain a = –1 and b = 6     A1

[6 marks]

for $x \leqslant 2,{\text{ }}f'(x) = 2 > 0$     A1

for $2 < x < 3,{\text{ }}f'(x) = - 2x + 6 > 0$     A1

since $f'(x) > 0$ for all values in the domain of f , f is increasing     R1

therefore one-to-one     AG

[3 marks]

$x = 2y - 1 \Rightarrow y = \frac{{x + 1}}{2}$     M1

$x = - {y^2} + 6y - 5 \Rightarrow {y^2} - 6y + x + 5 = 0$     M1

$y = 3 \pm \sqrt {4 - x}$

therefore

${f^{ - 1}}(x) = \left\{ {\begin{array}{*{20}{c}}   {\frac{{x + 1}}{2},}&{x \leqslant 3} \\   {3 - \sqrt {4 - x} ,}&{3 < x < 4} \end{array}} \right.$     A1A1A1

Note: Award A1 for the first line and A1A1 for the second line.

[5 marks]