The function $\mathrm{f}$ is given by $\mathrm{f}:x\mapsto 1+\sqrt{x}$, for \(x\in \mathbb{

Question

The function $\mathrm{f}$ is given by $\mathrm{f}:x\mapsto 1+\sqrt{x}$, for $x\in \mathbb{R}$, $x\ge 0$. Find ${{\mathrm{f}}^{-1}}\left( x \right)$ and state the domain of ${{\mathrm{f}}^{-1}}$.[3] Show that if $\mathrm{ff}\left( x \right)=x$ then ${{x}^{3}}-4{{x}^{2}}+4x-1=0$. Hence find the value of $x$ for which $\mathrm{ff}\left( x \right)=x$. Explain why this value of $x$ satisfies the equation $\mathrm{f}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)$.[5] The function $\mathrm{g}$, with domain the set of non-negative integers, is given by ${\mathrm{g}(n) = \begin{cases} 1 & \mathrm{for } \, n = 0, \ 2 + \mathrm{g}\left(\frac{1}{2} n\right) & \mathrm{for } \, n \, \mathrm{ even}, \ 1 + \mathrm{g}(n - 1) & \mathrm{for } \, n \, \mathrm{ odd}. \end{cases}}$ Find $\mathrm{g}\left( 4 \right)$, $\mathrm{g}\left( 7 \right)$and $\mathrm{g}\left( 12 \right)$.[3] Does $\mathrm{g}$ have an inverse? Justify your answer.[2]

Accepted Answer

(a)(i) ${\mathrm{f}^{-1}(x)=(x-1)^2\,,\; x \in \mathbb{R},\; x \geq 1}$ (ii) ${2.62}$ (b)(i) $\mathrm{g}\left( 4 \right)=6$, $\mathrm{g}\left( 7 \right)=8$ and $\mathrm{g}\left( 12 \right)=9$. (ii) $\mathrm{g}$ is not $1-1$. Therefore, it doesn’t have an inverse.

N2016 P1 Q10

N2016 P1 Q10

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