# Function (Composite/ Inverse Function)

(CIE 0606/2021/m/22/Q9)
The function $\mathrm{f}$ is defined by $\mathrm{f}(x)=\dfrac{\sqrt{4 x^{2}-1}}{2 x}$ for $0.5 \leqslant x \leqslant 1.5$.

The diagram shows a sketch of $y=\mathrm{f}(x)$.

(a) (i) It is given that $\mathrm{f}^{-1}$ exists. Find the domain and range of $\mathrm{f}^{-1}$.

(ii) Find an expression for $\mathrm{f}^{-1}(x)$.

(b) The function $g$ is defined by $g(x)=\mathrm{e}^{x^{2}}$ for all real $x$. Show that $\operatorname{gf}(x)=\mathrm{e}^{\left(1-\dfrac{a}{b x^{2}}\right)}$, where $a$ and $b$ are integers.

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*********math solution*************

$\begin{array}{ll}&f(x)=\dfrac{\sqrt{4 x^{2}-1}}{2 x}, \cdot 5 \leqslant x \leqslant 1.5 \\&\text { When } x=.5,y=\dfrac{\sqrt{4(.5)^{2}-1}}{2(.5)}=0 \\&x=1.5, y=\dfrac{\sqrt{4(1.5)^{2}-1}}{2(1.5)}=\dfrac{2 \sqrt{2}}{3} \\&\text { (a) Doman of } f^{-1}: 0 \leqslant x \leqslant \dfrac{2 \sqrt{2}}{3} \\&\text { Range of } \bar{f}^{-1}: 0.5 \leqslant f^{-1} \leqslant 1.5 \\&\text { (ii) Let } \begin{array}[t]{rl}f^{-1}(x) &=y \cdot \\\text { Then } x &=f(y)=\dfrac{\sqrt{4 y^{2}-1}}{2 y} \\2 x y &=\sqrt{4 y^{2}-1} \\4 x^{2} y^{2} &=4 y^{2}-1 \\4 y^{2}-4 x^{2} y^{2} &=1 \\y^{2}\left(4-4 x^{2}\right) &=1\\y &=\dfrac{1}{\sqrt{4-4 x^{2}}} \\\therefore f^{-1}(x)&=\dfrac{1}{\sqrt{4-4 x^{2}}} \end{array}\end{array}$

$\begin{array}{ll}\text { (b) } g(x) &=e^{x^{2}} \\\operatorname{gf}(x) &=g(f(x)) \\&=g\left(\dfrac{\sqrt{4 x^{2}-1}}{2 x}\right) \\&=e^{\left(\dfrac{\sqrt{4 x^{2}-1}}{2 x}\right)^2} \\&=e^{\left(\dfrac{4 x^{2}-1}{4 x^{2}}\right)} \\&=e^{\left(1-\dfrac{1}{4 x^{2}}\right)}\end{array}$
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