Functions: EJC 2024 Promo Q10

Eunoia Junior College2024 PromoPromo Q10●●● Challenging9 marks

What this question tests

Functions.

Question

The figure shows a sketch of the graphs $y = \dfrac{1}{4}x(12 - x)$ and $y = \dfrac{1}{2}x + 4$ , for $0 \leq x \leq 8$ .

$Eunoia Junior College 2024 Promo Q10: question diagram$

The function $\mathrm{f}$ is defined as $\mathrm{f}: x \mapsto \begin{cases} \dfrac{1}{4}x(12 - x) & \text{for } 0 \leq x \leq c,\\[8pt] \dfrac{1}{2}x + 4 & \text{for } c < x \leq 8, \end{cases}$ where the value of $c$ satisfies $0 < c < 8$ .

State the domain of $\mathrm{f}$ . [1]
1. Explain why the inverse function $\mathrm{f}^{-1}$ does not exist when $c = 6$ . [2]
2. Find the set of values of $c$ for which the function $\mathrm{f}^{-1}$ exists. [1]
1. State the range of $\mathrm{f}$ when $c = 6$ . [1]
2. Find the range of $\mathrm{f}$ when $c = 1$ . [2]
3. Find the set of values of $c$ for which the composite function $\mathrm{f}^2$ exists. [2]

Show full worked solution▾

(a)

$[0,8]$

(b)

(i) $\because\ f(4)=f(8)=8$

$f$ is a many-to-one function and $f^{-1}$ does not exist.

(b)

(ii) $0<c\leq2$

(c)

(i) $[0,9]$

(c)

(ii) When $c=1$ : From graph,

For $0\leq x\leq1$ , $0\leq f(x)\leq2.75$ ;

For $1<x\leq8$ , $4.5<f(x)\leq8$ .

Hence, the range is $[0,2.75]\cup(4.5,8]$ .

(c)

(iii) For $f^2$ to exist, we need the range of $f$ to be a subset of $[0,8]$ the domain of $f$ , i.e. $R_f\subseteq D_f=[0,8]$ .

From graph, we need $0<c\leq4$ .

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