JEE Main On 08 April 2018 Question 10
Question: Let $ f(x)=x^{2}+\frac{1}{x^{2}} $ and $ g(x)=x-\frac{1}{x}, $ $ x\in R-{-1,0,1}. $ If $ h(x)=\frac{f(x)}{g(x)} $ , then the local minimum value of $ h(x) $ is: [JEE Main Online 08-04-2018]
Options:
A) $ -2\sqrt{2} $
B) $ 2\sqrt{2} $
C) 3
D) -3
Show Answer
Answer:
Correct Answer: B
Solution:
$ F(x)=x^{2}+\frac{1}{x^{2}},g(x)=x-\frac{1}{x} $
$ h(x)=\frac{F(x)}{g(x)} $ $ h(x)=\frac{x^{2}+\frac{1}{x^{2}}}{x-\frac{1}{x}} $
$ f(\alpha )=\frac{{{\alpha }^{2}}+2}{\alpha } $ $ f(\alpha )=\alpha +\frac{2}{\alpha } $
$ F(\alpha )={{( \sqrt{\alpha }-\sqrt{\frac{2}{\alpha }} )}^{2}}+2\sqrt{2} $
Minimum value will be $ 2\sqrt{2} $ At $ \sqrt{\alpha }-\sqrt{\frac{2}{\alpha }}=0 $
$ \alpha =\sqrt{2} $ which is possible as $ x-\frac{1}{x}\in R $
