Question

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

• 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

$0$

Explanation for the correct option:

Finding the value of $f\text{'}\left(x\right)$

The given function.

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

Rearranging it

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

Now differentiating it with respect to. $x$

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

Hence, the correct option is (A)