Functions: RI 2024 Promo Q10

(a)[1]

By taking $x=0$ in $\mathrm f(x)=\mathrm f(x+3)$, $$\mathrm f(0)=\mathrm f(3)=\frac14[-3(3)+14]=\frac54.$$

(b)[3]

(c)[4]

Finding the $x$-coordinates of the points of intersection $A$ and $B$,

Point A: $$\begin{aligned} \frac43(2-x)&=\frac12x^2,\\ 3x^2+8x-16&=0,\\ (3x-4)(x+4)&=0,\\ x&=\frac43\quad\text{or}\quad x=-4\quad\text{(reject since $1<x<2$ at A)}. \end{aligned}$$ Point B: $$\begin{aligned} \frac43(x-2)&=-\frac34x+\frac72,\\ 16(x-2)&=-9x+42,\\ x&=\frac{74}{25}\quad\text{or }2.96. \end{aligned}$$ Hence the solution to the inequality is $\left(\dfrac43,\dfrac{74}{25}\right)$.

(d)[4]

From the graph above, $R_{\mathrm f}=\left[\dfrac12,2\right]$ and $D_{\mathrm h}=[0,2]$.

Since $R_{\mathrm f}\subseteq D_{\mathrm h}$, $\mathrm{hf}$ exists.

$\therefore R_{\mathrm{hf}}=[-2,\sqrt3]$.

$$R_{\mathrm f}=\left[\frac12,2\right]\xrightarrow{\ h\ }[-1,\sqrt3+1].$$ $$\therefore R_{\mathrm{hf}}=[-1,\sqrt3+1].$$