Functions: MI 2024 Promo Q7

(i) [2]

$$R_{\mathrm g}=(3,\infty). \qquad D_{\mathrm f}=[0,\infty).$$

Since $R_{\mathrm g}=(3,\infty)\subseteq[0,\infty)=D_{\mathrm f}$, fg exists.

7(ii) [2] $$\begin{aligned} \mathrm{fg}(x)&=\mathrm f[\mathrm g(x)]=\mathrm f(\mathrm e^x+3)\\ &=5-(\mathrm e^x+3-2)^2\\ &=5-(\mathrm e^x+1)^2 \end{aligned}$$ $$D_{\mathrm{fg}}=D_{\mathrm g}=(-\infty,\infty)\quad\text{or }\mathbb R$$

7(iii) [2]

Method 1: GC/Graph $y=\mathrm{fg}(x)$

Method 2: Mapping $$D_{\mathrm{fg}}=(-\infty,\infty)\xrightarrow{\ \mathrm g\ }(3,\infty) \xrightarrow[\ R_{\mathrm g}\ ]{\ \mathrm f\ }(-\infty,4)=R_{\mathrm{fg}}$$

$$R_{\mathrm{fg}}=(-\infty,4)$$

7(iv) [1]

Method 1: Horizontal Line Test

Since there exists a horizontal line $y=3$ that intersects the graph of $y=\mathrm f(x)$ more than once, $\mathrm f$ is not one-one and $\mathrm f^{-1}$ does not exist.

Method 2: Counterexamples $$\begin{aligned} \mathrm f(0)&=5-(0-2)^2=1\\ \mathrm f(4)&=5-(4-2)^2=1 \end{aligned}$$ Since $\mathrm f(0)=\mathrm f(4)=1$, $\mathrm f$ is not one-one and $\mathrm f^{-1}$ does not exist.

7(v) [1]

For $\mathrm f^{-1}$ to exists under domain $[k,\infty)$, $\mathrm f$ must be a one-one function.

Hence, the smallest value of $k=2$.

7(vi) [3]

Let $y=\mathrm f(x)$ $$\begin{aligned} y&=5-(x-2)^2, &&\text{for }x\in[2,\infty)\\ y-5&=-(x-2)^2\\ -y+5&=(x-2)^2\\ x-2&=\pm\sqrt{-y+5}\\ x&=2\pm\sqrt{5-y} \end{aligned}$$ Since $x\geq2$, $$x=2+\sqrt{5-y}\quad[\text{reject }2-\sqrt{5-y}\text{ as }x\geq2]$$ $$\therefore\ \mathrm f^{-1}(x)=2+\sqrt{5-x}$$ $$D_{\mathrm f^{-1}}=R_{\mathrm f}=(-\infty,5]$$