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

$0$

Explanation for the correct option:

Find the value of $f\prime \left(x\right)$:

Given,

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

Simplifying the $f\left(x\right)$.

$\begin{array}{rcl}f\left(x\right)& =& \sqrt{x-1}+\sqrt{x+24-10\sqrt{x-1}}\\ & =& \sqrt{x-1}+\sqrt{25+\left(x-1\right)-10\sqrt{x-1}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[\because a^2+b^2-2·a·b={\left(a-b\right)}^2\right]\\ & =& \sqrt{x-1}+\sqrt{{\left(5-\sqrt{x-1}\right)}^2}\\ & =& \sqrt{x-1}+\left(5-\sqrt{x-1}\right)\\ & & \end{array}$

As, $\sqrt{x-1}<5$ for $1<x<26$

Therefore, $f\left(x\right)=5$ and

$f\text{'}\left(x\right)=0$ for $x\in (1,26)$

Hence, the correct option is A.