Download App

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 $(1)]$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Functions→Functions — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Choose the correct answer from the given four option. Solution of differential equation $x\ dy - y \ dx = 0$ represents :
→
If $ \text{x} \in \left ( \frac{3\pi}{2}, 2\pi \right )$ then the value of the expression $ \sin^{-1}[\cos({\cos^{-1}(\cos \, \text{x})}+\sin^{-1}(\sin \, \text{x}))]$ is:
→
The binary operation $\times $ defind on set $R,$ given by $\text{a}\times\text{b}=\frac{\text{a}+\text{b}}{2}$ for all $a, b \in R$ is:
→
Evaluate: $\int\frac{1-\cos\text{x}}{cos\text{x}(1+cos\text{x})}\text{dx}.$
→
If $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right]$ and $B=\left[\begin{array}{rrr}3 & -1 & 3 \\ -1 & 0 & 2\end{array}\right],$ then find $2 A-B$.
→
If $\vec a = \hat i + \hat j + \hat k,\,\,\vec b = \hat i - \hat j + \hat k,\,\,\vec c = \hat i + 2\hat j + \hat k$ ,  then the value of $\left| {\begin{array}{*{20}{c}}
  {\vec a.\vec a}&{\vec a.\vec b}&{\vec a.\vec c} \\ 
  {\vec b.\vec a}&{\vec b.\vec b}&{\vec b.\vec c} \\ 
  {\vec c.\vec a}&{\vec c.\vec b}&{\vec c.\vec c} 
\end{array}} \right|$ is
→
If $\vec a,\vec b,\vec c$ are non- zero and non-coplanar vectors such that $\left( {\vec a + \lambda \vec b} \right).\left[ {\left( {\vec b + 3\vec c} \right) \times \left( {\vec c - 4\vec a} \right)} \right] = 0$ , then $\lambda $ is equal to
→
If $a,b,c$  are vectors such that  $[abc\,]=4$   , then $[a\times b\,\,b\times c\,\,c\times a]$  =  
→
If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0,$ then the angle between the lines is:
→
If $\text{A} = \begin{bmatrix} 2 &\text{amp; } 3\\ 6 &\text{amp; x} \end{bmatrix}, \text{B} = \begin{bmatrix} 2 &\text{amp; 3}\\ \text{p} &\text{amp; }2 \end{bmatrix}$ and $\text{A} = \text{B}, $ then $\text{p}$ and $ \text{x} $ are:
→