Question

Let $f\left(x\right)=\sqrt{x-1}+\sqrt{x+24-10\sqrt{x-1}};1<x<26$ be a real valued function. Then $f^{\prime }\left(x\right)$ for $1<x<26$ is

• A. $0$
• B. $\frac{1}{\sqrt{x-1}}$
• C. $2\sqrt{x-1}-5$
• D. none of these

Answer 1

The correct option is A $0$

Let, $f\left(x\right)=\sqrt{x-1}+\sqrt{x+24-10\sqrt{x-1}}$
$1<x<26$ be a real valued function ,
We have,
$f\left(x\right)=\sqrt{x-1}+\sqrt{x+24-10\sqrt{x-1}}$
This can be written as
$f\left(x\right)=\sqrt{x-1}+\sqrt{{(5-\sqrt{x-1})}^2}$
$\because 1<x<26$
$f\left(x\right)=\sqrt{x-1}+5-\sqrt{x-1}$
$f\left(x\right)=5$
On differentiating wrt x , we get
$f{}^{\prime }\left(x\right)=0$
Hence ,Option A