Differentiation Question 119
Question: The values of x, at which the first derivative of the function $ {{( \sqrt{x}+\frac{1}{\sqrt{x}} )}^{2}} $ w.r.t. x is $ \frac{3}{4} $ , are
[MP PET 1998]
Options:
A) $ \pm 2 $
B) $ \pm \frac{1}{2} $
C) $ \pm \frac{\sqrt{3}}{2} $
D) $ \pm \frac{2}{\sqrt{3}} $
Show Answer
Answer:
Correct Answer: A
Solution:
Given $ f(x)={{( \sqrt{x}+\frac{1}{\sqrt{x}} )}^{2}} $
$ f(x)=x+\frac{1}{x}+2 $ and $ f’(x)=3/4 $
So, $ 1-\frac{1}{x^{2}}=\frac{3}{4} $
Therefore $ \frac{1}{x^{2}}=\frac{1}{4} $
Therefore $ x=+2,-2 $ .
BETA
