The function $\mathrm{f}$ is defined by\[\mathrm{f}(x) = \begin{cases} a^2 - x^2, & 0

Question

The function $\mathrm{f}$ is defined by$$\mathrm{f}(x) = \begin{cases} a^2 - x^2, & 0 < x \le a, \ a(x - a), & a < x \le 2a, \end{cases}$$and $\mathrm{f}(x) = \mathrm{f}(x + 2a)$ for all real values of $x$, where $a$ is a positive constant. Sketch the graph of $y = \mathrm{f}(x)$ for $-2a \le x \le 3a$.[3] Find $\mathrm{f}\left( \frac{101a}{2} \right)$ in terms of $a$.[2] The function $\mathrm{g}$ is defined by$$\mathrm{g} : x \mapsto \ln x, \text{ where } x \in \mathbb{R}, 0 < x \le \mathrm{e}.$$The diagram below shows the graph of the quadratic function $\mathrm{h}$ with domain defined where $0 < x < 4$. The graph has a maximum point at $x = 1$ and has end-points $\left( 0, \frac{\mathrm{e}}{2} \right)$ and $(4, 0)$. Determine whether $\mathrm{hg}$ exist. Justify your answer.[2] State the other value of $x$ for which $\mathrm{h}(x) = \frac{\mathrm{e}}{2}$. Given that $\mathrm{h}^{-1}$ exists, deduce the restricted domain of $\mathrm{h}$ such that domain of $\mathrm{h}^{-1}$ is $\left( 0, \frac{\mathrm{e}}{2} \right]$.[2] Using the restricted domain of $\mathrm{h}$ found in (ii), find the range of $\mathrm{gh}$.[2]

Accepted Answer

(a)(i) Periodic sketch shown. (a)(ii) $\frac{3a^2}{4}$. (b)(i) $hg$ does not exist on $D_g$. (b)(ii) $x=2$, restricted $D_h=[2,4)$. (b)(iii) $R_{gh}=(-\infty,1-\ln2]$.

2025 SAJC P1 Q4

2025 SAJC P1 Q4

Solution:

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