SPM.1.AHL.TZ0.9

pestleMathematicsAAHLPaper 1SPM· ahl-2-14-odd-and-even-functions-self-inverse-inverse-and-domain-restrictionsource ↗

The function $f$ is defined by $f\left( x \right) = {{\text{e}}^{2x}} - 6{{\text{e}}^x} + 5{\text{,}}\,\,x \in \mathbb{R}{\text{,}}\,\,x \leqslant a$ . The graph of $y = f\left( x \right)$ is shown in the following diagram.

Find the largest value of $a$ such that $f$ has an inverse function.

[3]

For this value of $a$ , find an expression for ${f^{ - 1}}\left( x \right)$ , stating its domain.

[5]

Markscheme / solution

attempt to differentiate and set equal to zero M1

$f'\left( x \right) = 2{{\text{e}}^{2x}} - 6{{\text{e}}^x} = 2{{\text{e}}^x}\left( {{{\text{e}}^x} - 3} \right) = 0$ A1

minimum at $x = {\text{ln}}\,3$

$a = {\text{ln}}\,3$ A1

[3 marks]

Note: Interchanging $x$ and $y$ can be done at any stage.

$y = {\left( {{{\text{e}}^x} - 3} \right)^2} - 4$ (M1)

${{\text{e}}^x} - 3 = \pm \sqrt {y + 4}$ A1

as $x \leqslant {\text{ln}}\,3$ , $x = {\text{ln}}\left( {3 - \sqrt {y + 4} } \right)$ R1

so ${f^{ - 1}}\left( x \right) = {\text{ln}}\left( {3 - \sqrt {x + 4} } \right)$ A1

domain of ${f^{ - 1}}$ is $x \in \mathbb{R}$ , $- 4 \leqslant x < 5$ A1

[5 marks]

Examiners’ report

[N/A]