2025 NJC P1 Q11
2025 NJC P1 Q11
10 marks
The function \(\mathrm{f}\) is defined by
\[\mathrm{f}(x) = \begin{cases} \frac{a}{2} + \frac{4}{3}\left(x - \frac{1}{2}\right)^2 & \text{for}\hspace{0.5em} x \in \mathbb{R}, \frac{1}{2} < x < 2, \\ \frac{a}{x} & \text{for}\hspace{0.5em} x \in \mathbb{R}, x \ge 2, \end{cases}\]
where \(a\) is a positive constant.
- Find the range of \(\mathrm{f}\).[3]
- Find \(\mathrm{f}^{-1}(x)\) and state its domain.[3]
The function \(\mathrm{g}\) is defined by \(\mathrm{g}(x) = 3 + \mathrm{e}^x\), for \(x \in \mathbb{R}\).
- Show that \(\mathrm{f}\mathrm{g}\) exists.[1]
- Find the exact value of \(k\) for which \(\mathrm{f}\mathrm{g}(k) = \frac{a}{7}\).[3]
Solution:
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Answer:(i) \(R_f=\left(0,\frac a2+3\right)\). (ii) \(f^{-1}(x)=\frac ax\) for \(0<x\le\frac a2\); \(f^{-1}(x)=\frac12+\sqrt{\frac34\left(x-\frac a2\right)}\) for \(\frac a2<x<\frac a2+3\). Domain \(\left(0,\frac a2+3\right)\). (iii) \(R_g\subset D_f\), so \(fg\) exists. (iv) \(k=\ln4\).
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