Question

If $f\left(x\right)=x+\tan x\text{and}f$ Is inverse of g, then g $\left(x\right)$ is equal to

Is inverse of g, then g

• A. $\frac{1}{1+{(g\left(x\right)-x)}^2}$
• B. $\frac{1}{1-{(g\left(x\right)-x)}^2}$
• C. $\frac{1}{2+{(g\left(x\right)-x)}^2}$
• D. $\frac{1}{2-{(g\left(x\right)-x)}^2}$

Answer 1

The correct option is C $\frac{1}{2+{(g\left(x\right)-x)}^2}$

Given :

$f\left(x\right)=x+tanx$

$f\left(f^{-1}\right(y\left)\right)=f^{-1}\left(y\right)+tan\left[f^{-1}\right(y\left)\right]$

$y=g\left(y\right)+tan\left[g\right(y\left)\right]$

$x=g\left(x\right)+tan\left[g\right(x\left)\right]$

Differentiating above w.r.t. to $x$,

$1=g^{\prime }\left(x\right)+se{c}^2\left[g\right(x\left)\right].g^{\prime }\left(x\right)$

$\Rightarrow g^{\prime }\left(x\right)\{1+se{c}^2[g\left(x\right)\left]\right\}=1$

$\Rightarrow g^{\prime }\left(x\right)=\frac{1}{1+se{c}^2\left[g\right(x\left)\right]}=\frac{1}{2+\left[g\right(x)-x^2]}$