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

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{2x+k}{x-2},x\in \mathbb{R},\mathrm{ }x\ne 2$ where $k$ is a positive constant. Sketch the graph of $y=\mathrm{f}\left( x \right)$, stating the equations of any asymptotes and the coordinates of any points where the curve crosses the $x$ and $y$ axes.[3] Describe fully a sequence of transformations which would transform the curve $y=\frac{1}{x}$ onto $y=\mathrm{f}\left( x \right)$.[4] Find ${{\mathrm{f}}^{-1}}$ in a similar form and write down the range of ${{\mathrm{f}}^{-1}}$.[3] Hence or otherwise, find ${{\mathrm{f}}^{2}}$. Find the value of ${{\mathrm{f}}^{2017}}\left( \frac{1}{2} \right)$, leaving your answer in terms of $k$.[4]

Accepted Answer

(i) Asymptotes $x=2$ and $y=2$; intercepts $\left(-\dfrac k2,0\right)$ and $\left(0,-\dfrac k2\right)$, with branches upper-right and lower-left about $(2,2)$ (ii) Translate $2$ units right, stretch parallel to the $y$-axis by scale factor $k+4$, then translate $2$ units up (iii) $\mathrm f^{-1}(x)=\dfrac{2x+k}{x-2}$, $x\ne2$, and $R_{\mathrm f^{-1}}=\mathbb R\setminus\{2\}$ (iv) $\mathrm f^2(x)=x$ for $x\ne2$, and $\mathrm f^{2017}\left(\dfrac12\right)=-\dfrac23(k+1)$

2016 CJC P1 Q11

2016 CJC P1 Q11

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