Question

Let $g$ is the inverse function of $f_{\text{and }}{f}^{\prime }(x)=\frac{x^{10}}{\left(1+x^2\right)}$. If $g(2)=a$ then $g^{\prime }(2)$ is equal to

• A. $\frac{5}{2^{10}}$
• B. $\frac{1+a^2}{a^{10}}$
• C. $\frac{a^{10}}{1+a^2}$
• D. $\frac{1+a^{10}}{a^2}$

Answer 1

Answer: (B)

$f[g(x)]=x\Rightarrow f^{\prime }[g(x)]\cdot \left[g^{\prime }(x)\right]=1\Rightarrow f^{\prime }[g(2)]g^{\prime }(2)=12\Rightarrow f^{\prime }(a)g^{\prime }(2)[\text{ putting }x=2]$
$\text{ given, }{f}^{\prime }(a)=\frac{a^{10}}{1+a^2}\Rightarrow g^{\prime }(2)=\frac{1+a^2}{a^{10}}$
$\text{ Altemative: }g[f(x)]=x$

$g^{\prime }[f(x)]\cdot f^{\prime }(x)=1$
$\text{ now }g(2)=a\Rightarrow f(a)=2$
$\therefore g\text{ and }f\text{ are inverse of each other }$
$\text{ now }f(x)=2\Rightarrow g(2)=x=a$
$\therefore g^{\prime }(2)\cdot f^{\prime }(a)=1$
$g^{\prime }(2)=\frac{1}{f^{\prime }(a)}=\frac{1+a^2}{a^{10}}$