Applications Of Derivatives Question 400
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
[MP PET 2000]
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:
From Rolle-s theorem in (1, 26), $ f(1)=f(26)=5 $ . In given interval, function satisfy all the conditions of Rolle’s theorem, therefore in [1, 26], at least, there is a point for which $ {f}’(x)=0 $ .
BETA
