Applications Of Derivatives Question 386
Question: From mean value theorem $ f(b)-f(a)= $ $ (b-a)f’(x_1); $ $ a<x_1<b $ if $ f(x)=\frac{1}{x} $ , then $ x_1= $
[MP PET 1996]
Options:
A) $ \sqrt{ab} $
B) $ \frac{a+b}{2} $
C) $ \frac{2ab}{a+b} $
D) $ \frac{b-a}{b+a} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ f’(x_1)=\frac{-1}{x_1^{2}} $
$ \therefore \frac{-1}{x_1^{2}}=\frac{\frac{1}{b}-\frac{1}{a}}{b-a}=-\frac{1}{ab}\Rightarrow x_1=\sqrt{ab} $ .
BETA
