Functions: HCI 2024 Promo Q8

The function $\mathrm f$ is defined by $$\mathrm f(x)=\begin{cases} 2-\ln(x+\mathrm{e})&\text{for }-\mathrm{e}<x<0,\\ \cos\left(\dfrac\pi2x\right)&\text{for }0\leq x\leq1. \end{cases}$$

(a)

Given that the inverse function $\mathrm f^{-1}$ exists, find the exact value of $m$ if $\mathrm f^{-1}(m)=-1$. [2]

Suggested Solutions

(a)

$$\begin{aligned} \mathrm f^{-1}(m)&=-1\\ m &= \mathrm{f}(-1)\\ &= 2 - \ln(-1 + \mathrm{e})\\ &= 2 - \ln(\mathrm{e} - 1). \end{aligned}$$

Another function $\mathrm g$ is defined by $$\mathrm g(x)=\sin x,\qquad\text{for }0\leq x\leq\frac\pi4.$$

(b)

Explain why the composite function $\mathrm{fg}$ exists. [2]

Suggested Solutions

(b)

$$\mathrm R_{\mathrm g}=\left[0,\frac1{\sqrt2}\right]$$ $$\mathrm D_{\mathrm f}=(-\mathrm{e},1]$$ Since $\mathrm R_{\mathrm g}\subseteq\mathrm D_{\mathrm f}$. Hence, $\mathrm{fg}$ exists.

(c)

Find the exact value of $\mathrm{fg}\left(\dfrac\pi6\right)$. [2]

Suggested Solutions

(c)

$$\begin{aligned} \mathrm{fg}\!\left(\frac{\pi}{6}\right) &= \mathrm{f}\!\left(\frac{1}{2}\right)\\ &= \cos\!\left(\frac{\pi}{4}\right)\\ &= \frac{1}{\sqrt{2}}. \end{aligned}$$

The domain of $\mathrm f$ is further restricted to $0\leq x\leq1$.

(d)

By considering $\mathrm f^2(1)$ and $\mathrm f^3(1)$, deduce the value(s) of $\mathrm f^n(1)$ where $n$ is a positive integer. [2]

Suggested Solutions

(d)

$$\begin{aligned} \mathrm f^2(1)&=\mathrm f[\mathrm f(1)]=\mathrm f(0)=1\\ \mathrm f^3(1)&=\mathrm f[\mathrm f^2(1)]=\mathrm f(1)=0 \end{aligned}$$ $$\mathrm f^n(1)=\begin{cases} 0&\text{if $n$ is odd},\\ 1&\text{if $n$ is even}. \end{cases}$$