Functions $\mathrm{f}$ and $\mathrm{g}$ are defined by \(\mathrm{f}:x\mapsto 2\left| x-p \right|

Question

Functions $\mathrm{f}$ and $\mathrm{g}$ are defined by $\mathrm{f}:x\mapsto 2\left| x-p \right|+1$ for $x\in \mathbb{R}$, $p>1$, $\mathrm{g}:x\mapsto x\left( x-q \right)$ for $x\in \mathbb{R}$, $q<0$. Explain why $\mathrm{f}$ does not have an inverse.[2] The domain of $\mathrm{f}$ is now restricted to $x\le k$. Write down the largest value of $k$ for which the function ${{\mathrm{f}}^{-1}}$ exists. Hence find ${{\mathrm{f}}^{-1}}\left( x \right)$ and state the domain of ${{\mathrm{f}}^{-1}}$.[4] Sketch on the same diagram the graphs of $y=\mathrm{f}\left( x \right)$ and $y={{\mathrm{f}}^{-1}}\left( x \right)$, giving the coordinates of the axial intercepts. Hence solve $\mathrm{f}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)$.[4] Find the range of $\mathrm{g}{{\mathrm{f}}^{-1}}$.[2]

Accepted Answer

(i) $f$ is not one-to-one, so $f^{-1}$ does not exist on $\mathbb R$. (ii) $k=p$, $f^{-1}(x)=p+\frac{1-x}{2}$, $\mathrm D_{f^{-1}}=[1,\infty)$ (iii) $x=\frac{2p+1}{3}$ (iv) $\mathrm R_{gf^{-1}}=\left[-\frac{q^2}{4},\infty\right)$

2019 NJC P1 Q9

2019 NJC P1 Q9

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