<p>The figure shows a sketch of the graphs \(y = \frac{1}{4}x(12 - x)\) and \(y = \frac{1}{2}x + 4\), for&nbsp;\(0 \le x \le 8\).</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/pasted/1779450996916-hz7xiq2lv0e.png" style="max-width: min(250px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>The function \(\mathrm{f}\) is defined as</p>
<p style="text-align: center">\[\mathrm{f} : x \mapsto \begin{cases} \frac{1}{4}x(12 - x) & \text{for}\hspace{0.5em} 0 \le x \le c, \\[2ex] \frac{1}{2}x + 4 & \text{for}\hspace{0.5em} c &lt; x \le 8, \end{cases}\]</p>
<p>where the value of \(c\) satisfies \(0 &lt; c &lt; 8\).</p>
<ol class="abc-list"><li><div class="list-item-paragraph">State the domain of \(\mathrm{f}\).<span class="inline-marks">[1]</span></div></li><li><ol class="i-list"><li><div class="list-item-paragraph">Explain why the inverse function \(\mathrm{f}^{-1}\) does not exist when&nbsp;\(c = 6\).<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Find the set of values of \(c\) for which the function \(\mathrm{f}^{-1}\) exists.<span class="inline-marks">[1]</span></div></li></ol></li><li><ol class="i-list"><li><div class="list-item-paragraph">State the range of \(\mathrm{f}\) when&nbsp;\(c = 6\).<span class="inline-marks">[1]</span></div></li><li><div class="list-item-paragraph">Find the range of \(\mathrm{f}\) when&nbsp;\(c = 1\).<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Find the set of values of \(c\) for which the composite function \(\mathrm{f}^2\) exists.<span class="inline-marks">[2]</span></div></li></ol></li></ol>

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

\[\mathrm{f} : x \mapsto \begin{cases} \frac{1}{4}x(12 - x) & \text{for } 0 \le x \le c, \\[2ex] \frac{1}{2}x + 4 & \text{for } c < x \le 8, \end{cases}\]

where the value of \(c\) satisfies \(0 < c < 8\).

2024 EJC Promo Q10

2024 EJC Promo Q10