2018 H2 Maths Paper 1 Q5 — Worked Solution

Solution

5 marks

Approach

The condition $ff = g$ means $f\left( f(x) \right) = x$ for every $x$ in the domain, i.e. $f$ is a self-inverse function. Rather than guessing, we compute $f\left( f(x) \right)$ explicitly and force it to equal $x$ identically; matching coefficients then pins down $b$ . Since $f$ turns out to be self-inverse, $f^{- 1}$ falls out for free once $b$ is known.

Since $f:x \mapsto \frac{x + a}{x + b}$ ,

f\left( f(x) \right)

= \frac{\frac{x + a}{x + b} + a}{\frac{x + a}{x + b} + b}

= \frac{(x + a) + a(x + b)}{(x + a) + b(x + b)}

= \frac{(1 + a)x + a(1 + b)}{(1 + b)x + a + b^{2}}

For $ff = g$ , this must equal $x$ for every $x$ in the domain, so

(1 + a)x + a(1 + b) = x\left\lbrack (1 + b)x + a + b^{2} \right\rbrack.

Comparing coefficients of $x^{2}$ on both sides forces $1 + b = 0$ , i.e. $b = - 1$ .

(Checking the remaining coefficients with $b = - 1$ : the $x^{1}$ terms give $1 + a = a + b^{2} = a + 1$ , true for all $a$ ; the constant term gives $a(1 + b) = 0 = 0$ , also automatic. So $b = - 1$ is the only requirement, consistent with $a \neq - 1$ being given separately to keep $f$ well-defined and non-constant.)

$\therefore b = - 1$ .

With $b = - 1$ , $f:x \mapsto \frac{x + a}{x - 1}$ for $x \in \mathbb{R}$ , $x \neq 1$ .

Note: Since $ff = g$ means $f$ is its own inverse, we expect $f^{- 1}(x) = f(x)$ ; we find $f^{- 1}$ directly to confirm this.

Let $y = \frac{x + a}{x - 1}$ . Then

y(x - 1) = x + a

yx - y = x + a

x(y - 1) = a + y

x = \frac{a + y}{y - 1}.

Interchanging $x$ and $y$ ,

f^{- 1}(x) = \frac{x + a}{x - 1}.

To check your answer:

f^{- 1}(x) = \frac{x + a}{x - 1} = f(x)

, matching the self-inverse property forced by

ff = g

. ✓

$\therefore f^{- 1}(x) = \frac{x + a}{x - 1}$ .

2018 H2 Maths Paper 1, Question 5

Solution

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