Question

Let $f:[0,\infty )\to [2,\infty )$ be a differentiable increasing and onto function satisfying $f^2\left(x\right)-2f\left(x\right)-\sqrt{x}=f^2\left(y\right)-2f\left(y\right)-\sqrt{y}$
If $g\left(x\right)$ be the inverse of $f\left(x\right)$, then $g^{\prime }\left(4\right)$ is

• A. $16$
• B. $\frac{1}{16}$
• C. $96$
• D. $\frac{1}{96}$

Answer 1

The correct option is C $96$

$2f\left(x\right)f^{\prime }\left(x\right)-2f^{\prime }\left(x\right)-\frac{1}{2\sqrt{x}}=0$
$2\left(f\right(x)-1)f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}$
Integrating both the sides
$\left(f\right(x)-1{)}^2-\sqrt{x}-C=0$
$f\left(0\right)=2\Rightarrow C=1$
$f\left(x\right)=1+\sqrt{\sqrt{x}+1}$
$g\left(x\right)$ is inverse of $f\left(x\right)$
Let $f\left(x\right)=y$
$\Rightarrow (y-1{)}^2=\sqrt{x}+1\Rightarrow ((y-1{)}^2-1{)}^2=x\Rightarrow g(x)=((x-1{)}^2-1{)}^2\Rightarrow g^{\prime }(x)=2((x-1{)}^2-1)\times 2(x-1)\Rightarrow g^{\prime }\left(4\right)=96$