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Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsFunctions1 Mark
MCQ
The inverse of the function $\text{f}:\text{R}\rightarrow\{\text{x}\in\text{R}:\text{x}<1\}$ given by $\text{f(x)}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^\text{x}+\text{e}^{-\text{x}}}$ is:
  • ✓
    $\frac{1}{2}\log\frac{1+\text{x}}{1-\text{x}}$
  • B
    $\frac{1}{2}\log\frac{2+\text{x}}{2-\text{x}}$
  • C
    $\frac{1}{2}\log\frac{1-\text{x}}{1+\text{x}}$
  • D
    None of these
✓

Answer

Correct option: A.
$\frac{1}{2}\log\frac{1+\text{x}}{1-\text{x}}$
Let $f^{-1}(x) = y .....(1)$
$\Rightarrow\ \text{f(y)}=\text{x}$
$\Rightarrow\ \frac{\text{e}^{\text{y}}-\text{e}^{-\text{y}}}{\text{e}^{\text{y}}+\text{e}^{-\text{y}}}=\text{x}$
$\Rightarrow\ \frac{\text{e}^{-\text{y}}(\text{e}^{2\text{y}}-1)}{\text{e}^{-\text{y}}(\text{e}^{2\text{y}}+1)}=\text{x}$
$\Rightarrow\ (\text{e}^{2\text{y}}-1)=\text{x}(\text{e}^{2\text{y}}+1)$
$\Rightarrow\ \text{e}^{2\text{y}}-1=\text{xe}^{2\text{y}}+\text{x}$
$\Rightarrow\ \text{e}^{2\text{y}}=\frac{1+\text{x}}{1-\text{x}}$
$\Rightarrow\ 2\text{y}=\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
$\Rightarrow\ \text{y}=\frac{1}{2}\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{1}{2}\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big) [$From $(1)]$

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