Differentiation Question 464
Question: Let $ f(x)=\sqrt{x-1}+\sqrt{x+24-10\sqrt{x-1}}; $ $ 1<x<26 $ Be real valued function. Then $ f’(x) $ for $ 1<x<26 $ is
Options:
A) 0
B) $ \frac{1}{\sqrt{x-1}} $
C) $ 2\sqrt{x-1}-5 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ f(x)=\sqrt{x-1}+\sqrt{25+(x-1)-10\sqrt{x-1}} $ $ =\sqrt{x-1}+\sqrt{{{(5-\sqrt{x-1})}^{2}}} $ $ =\sqrt{x-1}+| 5-\sqrt{x-1} |=5 $ $ [\because \sqrt{x-1}<5for1<x<26] $
$ \therefore f’(x)=0 $
BETA
