A function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{2x+3}{x-2}$, \(x\in \mathbb{R}\

Question

A function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{2x+3}{x-2}$, $x\in \mathbb{R}$, $x\ne 2$. Find the domain of ${{\mathrm{f}}^{-1}}$.[2] A function $\mathrm{h}$ is said to be self-inverse if $\mathrm{h}$ and ${{\mathrm{h}}^{-1}}$ are the same function. Show that $\mathrm{f}$ is self-inverse.[2] State the value of ${{\mathrm{f}}^{2023}}(6)$.[1] Another function $\mathrm{g}$ has range $\left( k,k+10 \right]$, where $k$ is a constant. Find the range of values of $k$ such that the composite function $\mathrm{fg}$ exists.[3] Assume for the remainder of this question that $\mathrm{fg}$ exists and $k\ne 2$. Find the range of $\mathrm{fg}$ in terms of $k$.[3] Given that $\mathrm{fg}\left( x \right)=\frac{2x+2}{1-6x}$, find $\mathrm{g}\left( x \right)$ in terms of $x$.[3]

Accepted Answer

(i) $D_{\mathrm f^{-1}}=(-\infty,2)\cup(2,\infty)$ (ii) $\mathrm f^{-1}=\mathrm f$, so $\mathrm f$ is self-inverse (iii) $\dfrac{15}{4}$ (iv) $k<-8$ or $k\geq2$ (v) $R_{\mathrm{fg}}=\left[\dfrac{2k+23}{k+8},\dfrac{2k+3}{k-2}\right)$ (vi) $\mathrm g(x)=\dfrac1{2x}-1$

2023 NJC Promo Q7

2023 NJC Promo Q7

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