The inverse of the function f(x) = e^x - e^ - x e^x + e^ - x + 2 … - Vidyadip

MCQ

The inverse of the function $f(x) = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} + 2$ is given by

A
${\log _e}{\left( {\frac{{x - 2}}{{x - 1}}} \right)^{1/2}}$
✓
${\log _e}{\left( {\frac{{x - 1}}{{3 - x}}} \right)^{1/2}}$
C
${\log _e}{\left( {\frac{x}{{2 - x}}} \right)^{1/2}}$
D
${\log _e}{\left( {\frac{{x - 1}}{{x + 1}}} \right)^{ - 2}}$

✓

Answer

Correct option: B.

${\log _e}{\left( {\frac{{x - 1}}{{3 - x}}} \right)^{1/2}}$

b
(b) $y = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} + 2\,\, $

$\Rightarrow \,\,y = \frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}} + 2\,\,$

$ \Rightarrow \,\,{e^{2x}} = \frac{{1 - y}}{{y - 3}} = \frac{{y - 1}}{{3 - y}}\,\, $

$\Rightarrow \,\,x = \frac{1}{2}{\log _e}\,\left( {\frac{{y - 1}}{{3 - y}}} \right)$

$ \Rightarrow {f^{ - 1}}(y) = {\log _e}\,{\left( {\frac{{y - 1}}{{3 - y}}} \right)^{1/2}} $

$\Rightarrow {f^{ - 1}}(x) = {\log _e}{\left( {\frac{{x - 1}}{{3 - x}}} \right)^{1/2}}$.

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 STD 12 - 1. relation and function→STD 12 - 1. relation and function — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

If $\frac{{dy}}{{dx}} + y\tan x = \sin 2x$ and $y(0)\,=1$ , then $y(\pi)$ is equal to

The set of points where the function $f(x) = x |x|$ is differentiable is :

Let $f : R \rightarrow R$ be a differentiable function such that $f^{\prime}(x)+f(x)=\int \limits_0^2 f(t) d t$. If $f(0)=e^{-2}$, then $2 f (0)- f (2)$ is equal to $.........$.

If the vectors $a $ and $b$ are mutually perpendicular, then $a \times \{ a \times \{ a \times (a \times b)\} \} $ is equal to

Choose the correct answer from the given four options. Two dice are thrown. If it is known that the sum of numbers on the dice was less than $6,$ the probability of getting a sum $3,$ is:

Let $f:(-1,1) \rightarrow$ IR be such that $f(\cos 4 \theta)=\frac{2}{2-\sec ^2 \theta}$ for $\theta \in\left(0, \frac{\pi}{4}\right) \cup\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$. Then the value(s) of $f\left(\frac{1}{3}\right)$ is (are)

$(A)$ $1-\sqrt{\frac{3}{2}}$ $(B)$ $1+\sqrt{\frac{3}{2}}$ $(C)$ $1-\sqrt{\frac{2}{3}}$ $(D)$ $1+\sqrt{\frac{2}{3}}$

If $f(x) = \sin x - \cos x,$ the function decreasing in $0 \le x \le 2\pi $ is

Let $A$ be a $3 \times 3$ matrix such that $\operatorname{adj} A=\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1\end{array}\right]$ and $B = adj ($ adj $A )$ If $| A |=\lambda$ and $\left|\left( B ^{-1}\right)^{ T }\right|=\mu,$ then the ordered pair $(|\lambda|, \mu)$ is equal to

Let the position vectors of two points $P$ and $Q$ be $3 \hat{ i }-\hat{ j }+2 \hat{ k }$ and $\hat{ i }+2 \hat{ j }-4 \hat{ k },$ respectively. Let $R$ and $S$ be two points such that the direction ratios of lines $PR$ and $QS$ are $(4,-1,2)$ and $(-2,1,-2),$ respectively. Let lines $PR$ and $QS$ intersect at $T$. If the vector $\overline{ TA }$ is perpendicular to both $\overline{ PR }$ and $\overline{ QS }$ and the length of vector $\overline{ TA }$ is $\sqrt{5}$ units, then the modulus of a position vector of $A$ is

If $f: \mathrm{R} \rightarrow \mathrm{R}$ is a twice differentiable function such that $f^{\prime \prime}(x)>0$ for all $x \in \mathrm{R}$, and $f\left(\frac{1}{2}\right)=\frac{1}{2}, f(1)=1$, then