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RELATIONS AND FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsRELATIONS AND FUNCTIONS3 Marks
Question
Let $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{R}$ be a function defined as $\text{f(x)}=\frac{4\text{x}}{3\text{x}+4}.$ Show that $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{Range (f)}$ is one-one and onto. Hence find $f^{-1}$.
✓

Answer

We have given that
$f : R → (0, 2)$ defined by
$\text{f(x)}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+1$ is invertible.
let f(x) = y
$\Rightarrow\ \frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+1=\text{y}$
$\Rightarrow\ \frac{2\text{e}^{\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}=\text{y}$
$\Rightarrow\ \frac{2\text{e}^{2\text{x}}}{\text{e}^{2\text{x}}+1}=\text{y}$
$\Rightarrow\ 2\text{e}^{2\text{x}}=\text{y}(\text{e}^{2\text{x}}+1)$
$\Rightarrow\ \text{e}^{2\text{x}}(2-\text{y})=\text{y}$
$\Rightarrow\ \text{e}^{2\text{x}}=\frac{\text{y}}{2-\text{y}}\Rightarrow\ \text{x}=\frac{1}{2}\log_\text{e}\Big(\frac{\text{y}}{2-\text{y}}\Big)$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{1}{2}\log_\text{e}\Big(\frac{\text{x}}{2-\text{x}}\Big)$

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