The function $\mathrm{f}$ is defined by ${f} : x \to 2x - \frac{1}{2x}, \, 0

Question

The function $\mathrm{f}$ is defined by ${f} : x \to 2x - \frac{1}{2x}, \, 0 < x < 2.$ It is given that $\mathrm{f}^{-1}$ exists. Define $\mathrm{f}^{-1}$ in a similar form.[3] Sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}^{-1}(x)$ on the same diagram.[2] The region $R$ is bounded by the curve $y = \mathrm{f}^{-1}(x)$, $y = x$ and the $y$-axis. Find the exact area of $R$.[3] Another function $\mathrm{g}$ is defined by ${g} : x \to \begin{cases} \frac{3}{2} + \frac{3}{3x - 11} & \text{for } x \le 3, \ \left| x - \frac{1}{3}x^2 \right| & \text{for } 3 < x \le 5. \end{cases}$ Show that the composite function $\mathrm{gf}$ exists and find the range of $\mathrm{gf}$.[2]

Accepted Answer

(a) $f^{-1}:x\mapsto\frac{x+\sqrt{x^2+4}}4$, $x<\frac{15}{4}$. (b) Restricted graphs shown. (c) $\frac14\ln2$ square units. (d) $gf$ exists; $R_{gf}=[0,\frac32)$.

2025 ASRJC P1 Q10

2025 ASRJC P1 Q10

Solution:

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