Functions: ASRJC 2024 Promo Q7

Anderson Serangoon Junior College2024 PromoPromo Q7●●● Challenging8 marks

What this question tests

Functions.

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}: x \mapsto \sqrt{5 + 2x}$ , for $x \in \mathbb{R}$ , $x > -\dfrac{5}{2}$ .

Find the value of $a$ such that $\mathrm{f}^2(a) = 5$ . [2]

Another function $\mathrm{g}$ is defined by $\mathrm{g}: x \mapsto \left|\dfrac{x}{1 - 2x}\right|$ , for $x \in \mathbb{R}$ , $x \leq k$ . It is now given that $k = -1$ .

State the largest value of $k$ for $\mathrm{g}^{-1}$ to exist. [1]
Find $\mathrm{g}^{-1}(x)$ and state its domain. [3]
Explain why the composite function $\mathrm{fg}$ exists. State the exact range of $\mathrm{fg}$ . [2]

Show full worked solution▾

(a)

$\begin{aligned} \mathrm f^2(a)&=5\\ \mathrm f\!\left(\sqrt{5+2a}\right)&=5\\ \sqrt{5+2\sqrt{5+2a}}&=5\\ \sqrt{5+2a}&=\frac{25-5}{2}\\ a&=47.5 \end{aligned}$

(b)

largest $k=0$

(c)

$\text{Let }y=\left|\frac{x}{1-2x}\right|$ $\begin{aligned} y&=-\frac{x}{1-2x}\qquad\because x\leq0\\ y-2xy&=-x\\ x&=\frac y{2y-1} \end{aligned}$ $\mathrm g^{-1}(x)=\frac{x}{2x-1}$ Domain of $\mathrm g^{-1}=\left[\dfrac13,\dfrac12\right)$

(d)

$R_{\mathrm g}=\left[\dfrac13,\dfrac12\right)$ $D_{\mathrm f}=\left(-\frac52,\infty\right)$ Since $R_{\mathrm g}\subseteq D_{\mathrm f}$ , so $\mathrm{gf}$ exists. $(-\infty,-1]\xrightarrow{\ \mathrm g\ }\left[\frac13,\frac12\right) \xrightarrow{\ \mathrm f\ }\left[\sqrt{\frac{17}{3}},\sqrt6\right)$ So $R_{\mathrm{fg}}=\left[\sqrt{\dfrac{17}{3}},\sqrt6\right)$

Answers

(a) $a = 47.5$ (b)largest $k = 0$ (c) $\mathrm{g}^{-1}(x) = \dfrac{x}{2x - 1}$ , domain $\left[\dfrac{1}{3}, \dfrac{1}{2}\right)$
(d)range of $\mathrm{fg}$ is $\left[\sqrt{\dfrac{17}{3}}, \sqrt{6}\right)$

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