Question

Let $f:R\to R$ defined as $f(x)=x^5+e^{\frac{x}{2}}+e^{x^3}$ and $g(x)=f^{-1}(x)$, then the value of $g^{\prime }(2)$, is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

$fC(x)=x^5+e^{x/2}+e^{x^3}$
$f^{\prime }(x)=5x^4+\frac{1}{2}{e}^{x/2}+3x^2{e}^{x^3}>0$
$\therefore f(x)$ is one one funciton Clearly given function will be onto function as its range is $R$.
$\because g(f(x))=x\Rightarrow g^{\prime }(f(x))\cdot f^{\prime }(x)=1\Rightarrow g^{\prime }(f(x))=\frac{1}{f^{\prime }(x)}$
$f(x)=2\text{ when }x=0$
$\therefore g^{\prime }(2)=\frac{1}{f^{\prime }(0)}=\frac{1}{\left(\frac{1}{2}\right)}=2$