The curve $y=\mathrm{f}(x)$ cuts the axes at $\left( 1,0 \right)$ and $\left( 0,-2 \right)$. I

Question

The curve $y=\mathrm{f}(x)$ cuts the axes at $\left( 1,0 \right)$ and $\left( 0,-2 \right)$. It is given that ${{\mathrm{f}}^{-1}}$ exists. State, if it is possible to do so, the coordinates of the points where the following curves cut the axes. $y={{\mathrm{f}}^{-1}}(x)$[1] $y=\mathrm{f}\left( \frac{1}{2}x \right)$[1] $y=\mathrm{f}\left( \frac{1}{2}x+1 \right)$[1] Two functions $\mathrm{g}$ and $\mathrm{h}$ are defined as follows: $\mathrm{g}:x\mapsto \sqrt{x+2}$,$x\in \mathbb{R}$, $x\ge -2$ , $\mathrm{h}:x\mapsto {{x}^{2}}-4x+5$, $x\in \mathbb{R}$, $x\ge 0$. Show that $\mathrm{gh}$ exists and find the exact range of $\mathrm{gh}$.[4] Use a non-calculator method to solve $\mathrm{gh}\left( x \right)=2x$.[3]

Accepted Answer

(a)(i) $(0,1)$ and $(-2,0)$ (a)(ii) $(2,0)$ and $(0,-2)$ (a)(iii) $(0,0)$ only (b)(i) $gh$ exists and $R_{gh}=[\sqrt3,\infty)$ (b)(ii) $x=1$

2023 CJC Promo Q7

2023 CJC Promo Q7

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