Question

Let $f\left(x\right)=\sqrt{x-1}+\sqrt{x+24-10\sqrt{x-1}}$ , $1<x<26$ be real-valued function, the $f\text{'}\left(x\right)$ for $1<x<26$ is:

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

Answer 1

The correct option is A

$0$

Explanation for the correct answer

Given that: $f\left(x\right)=\sqrt{x-1}+\sqrt{x+24-10\sqrt{x-1}}$

Simplifying the given expression:

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

Now, as $\sqrt{(x-1)}<5$ for $1<x<26$, as on putting any value of $x$ between $(1,26)$, the square root will always be $<5$.
$\therefore f\left(x\right)=5$

Differentiating the function with respect to $x$.

$\Rightarrow f\text{'}\left(x\right)=0$

Hence, the correct answer is option (A).