2019 H2 Maths Paper 1 Q5 — Worked Solution

Part (i)

3 marks

Let $y = f(x) = e^{2x} - 4$ . To find $f^{- 1}$ , we write $x$ in terms of $y$ and then swap the letters.

y = e^{2x} - 4

e^{2x} = y + 4

2x = \ln(y + 4)

x = \frac{1}{2}\ln(y + 4)

Interchanging $x$ and $y$ ,

f^{- 1}(x) = \frac{1}{2}\ln(x + 4).

Note: the domain of $f^{- 1}$ is the range of $f$ , not all $x$ for which $\ln(x + 4)$ is defined by itself.

Since $x \in \mathbb{R}$ , we have $e^{2x} > 0$ for every $x$ , so

f(x) = e^{2x} - 4 > - 4,

and as $x \rightarrow \infty$ , $f(x) \rightarrow \infty$ , while as $x \rightarrow - \infty$ , $f(x) \rightarrow - 4^{\text{+}}$ . So the range of $f$ is $( - 4,\ \infty)$ .

$\therefore$ $f^{- 1}(x) = \frac{1}{2}\ln(x + 4)$ , with domain $x > - 4$ .

To check your answer:

f(0) = e^{0} - 4 = - 3

, and

f^{- 1}( - 3) = \frac{1}{2}\ln(1) = 0

. ✓

Part (ii)

3 marks

Approach

$fg(x)$ means $f\left( g(x) \right)$ , so we substitute $g(x) = x + 2$ into $f$ before solving.

Since $g(x) = x + 2$ ,

fg(x) = f(x + 2)

= e^{2(x + 2)} - 4

= e^{2x + 4} - 4.

Using the given equation $fg(x) = 5$ ,

e^{2x + 4} - 4 = 5

e^{2x + 4} = 9

Taking $\ln$ of both sides,

Since $\ln 9 = \ln 3^{2} = 2\ln 3$ ,

2x + 4 = 2\ln 3.

Solving:

x = \ln 3 - 2

$\therefore$ $x = \ln 3 - 2$ .

2019 H2 Maths Paper 1, Question 5

Part (i)

Part (ii)

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