ACJC 2025 H2 Math Prelim Paper 2 — Functions (worked solution)

Functions: Inverse & composite — ACJC 2025 H2 Math Prelim Paper 2

Anglo-Chinese Junior College2025 PrelimPaper 2●●○ Standard9 marks

What this question tests

Inverse & composite.

Question

The function f is defined by \(\mathrm{f} : x \mapsto \dfrac{1}{x^2 + 6x + 5}\), for \(x \geq -3\), \(x \neq -1\).

Find \(\mathrm{f}^{-1}(x)\).
Find algebraically the range of f.

The function g is defined by \(\mathrm{g} : x \mapsto \mathrm{e}^x\), for \(x \in \mathbb{R}\).

Find the exact range of \(\mathrm{gf}\).

Show full worked solution▾

The function is \(f(x) = \dfrac{1}{x^2+6x+5}\), where \(x \geq -3\).

Part (a)

Let \(y = \dfrac{1}{x^2+6x+5}\). Then: \[\begin{aligned} x^2 + 6x + 5 &= \frac{1}{y}\\ (x+3)^2 - 4 &= \frac{1}{y}\\ (x+3)^2 &= \frac{1}{y} + 4\\ x+3 &= \sqrt{\frac{1}{y}+4} \qquad (\text{since } x \geq -3)\\ x &= -3 + \sqrt{\frac{1}{y}+4} \end{aligned}\]

Therefore \(f^{-1}(x) = -3 + \sqrt{\dfrac{1}{x}+4}\).

When taking the square root, both \(\pm\) must be considered. The positive root is chosen because \(x \geq -3 \Rightarrow x+3 \geq 0\). Common incorrect factorisations: \((x+1)(x+5) = \frac{1}{y} \Rightarrow x+1 = \frac{1}{y}\) (wrong — cannot split a product this way).

Part (b)

Let \(y = \dfrac{1}{x^2+6x+5}\). For \(y \in R_f\), the equation \(yx^2 + 6yx + 5y - 1 = 0\) must have real solutions in \(x\). The discriminant must be non-negative: \[\begin{aligned} 36y^2 - 4y(5y-1) &\geq 0\\ 16y^2 + 4y &\geq 0\\ 4y(4y+1) &\geq 0 \end{aligned}\] So \(y \leq -\dfrac{1}{4}\) or \(y > 0\) (since \(y \neq 0\)).

Many students attempted partial fractions and differentiation to find the stationary point, with no test to confirm it is a maximum. The discriminant method is the cleanest algebraic approach. Note: \((-\infty, 0)\) is incorrect notation — use \((-\infty, 0)\) only for the open interval; the closed bracket at \(-\frac{1}{4}\) is essential.

Part (c)

\(\displaystyle R_f = \left(-\infty, -\frac{1}{4}\right] \cup (0, \infty) \xrightarrow{g} \left(0, e^{-0.25}\right] \cup (1, \infty)\) \(\displaystyle R_{gf} = \left(0, e^{-0.25}\right] \cup (1, \infty)\)

Answer: (a) \(f^{-1}(x) = -3 + \sqrt{\dfrac{1}{x}+4}\). (b) \(R_f = \left(-\infty,\,-\dfrac{1}{4}\right] \cup (0,\,\infty)\). (c) \(R_{\mathrm{gf}} = \left(0,\,e^{-1/4}\right] \cup (1,\,\infty)\).

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