Applications Of Derivatives Question 322
Question: The real number x when added to its inverse gives the minimum value of the sum at x equal to
[RPET 2000; AIEEE 2003]
Options:
A) - 2
B) 2
C) 1
D) - 1
Show Answer
Answer:
Correct Answer: C
Solution:
Let the positive number $ ( x+\frac{1}{x} ) $ will be minimum, when $ \frac{dy}{dx}=0 $ and $ \frac{d^{2}y}{dx^{2}}>0 $ .
Differentiate with respect to x, we have $ 1-\frac{1}{x^{2}}=0 $
therefore $ x=-1,,1 $ and $ \frac{d^{2}y}{dx^{2}}=+\frac{2}{x^{3}} $
therefore $ {{( \frac{d^{2}y}{dx^{2}} )}_{x=1}}>0 $
So, at $ x=1,,( x+\frac{1}{x} ) $ will be minimum.
BETA
