2024 EJC Promo Q10
2024 EJC Promo Q10
The figure shows a sketch of the graphs \(y = \frac{1}{4}x(12 - x)\) and \(y = \frac{1}{2}x + 4\), for \(0 \le x \le 8\).
The function \(\mathrm{f}\) is defined as
\[\mathrm{f} : x \mapsto \begin{cases} \frac{1}{4}x(12 - x) & \text{for}\hspace{0.5em} 0 \le x \le c, \[2ex] \frac{1}{2}x + 4 & \text{for}\hspace{0.5em} c < x \le 8, \end{cases}\]
where the value of \(c\) satisfies \(0 < c < 8\).
- State the domain of \(\mathrm{f}\).[1]
- Explain why the inverse function \(\mathrm{f}^{-1}\) does not exist when \(c = 6\).[2]
- Find the set of values of \(c\) for which the function \(\mathrm{f}^{-1}\) exists.[1]
- State the range of \(\mathrm{f}\) when \(c = 6\).[1]
- Find the range of \(\mathrm{f}\) when \(c = 1\).[2]
- Find the set of values of \(c\) for which the composite function \(\mathrm{f}^2\) exists.[2]
Solution:
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(a) \([0,8]\).
(b)(i) \(\mathrm f(4)=\mathrm f(8)=8\), so no inverse. (ii) \(0<c\le2\).
(c)(i) \([0,9]\). (ii) \(\left[0,\frac{11}{4}\right]\cup\left(\frac92,8\right]\). (iii) \(0<c\le4\).
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