Question

Let $f\left(x\right)=\sqrt{x-1}+\sqrt{x+24-10\sqrt{x-1}},1\le x\le 26$ 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. $2\sqrt{x-1}$
• D. $1$

Answer 1

The correct option is A $0$

We have,
$f\left(x\right)=\sqrt{x-1}+\sqrt{x+24-10\sqrt{x-1}},1<x<26$

Rearrange the terms so as to get ${(a-b)}^2=a^2-2ab+b^2$

$f\left(x\right)=\sqrt{x-1}+\sqrt{(x-1)+25-10\sqrt{x-1}}$
$\therefore f\left(x\right)=\sqrt{x-1}+\sqrt{(\sqrt{x-1}-5{)}^2}$
$f\left(x\right)=\sqrt{x-1}+|\sqrt{x-1}-5|$
$\therefore f\left(x\right)=\sqrt{x-1}-(\sqrt{x-1}-5)[\because \sqrt{x-1}-5<0\text{for}1<x<26]$
Now, differentiating w.r. to $x$, we get
$\therefore f^{\prime }\left(x\right)=0$ for all $x\in (1,26)$