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Let f(x)=x^2+(1/x^2) and g(x)=x-1/x x i...

Let `f(x)=x^2+(1/x^2)` and `g(x)=x-1/x` `x in R-{-1,0,1}`. If `h(x)=(f(x)/g(x))` then the local minimum value of `h(x)` is: (1) 3 (2) `-3` (3) `-2sqrt(2)` (4) `2sqrt(2)`

A

3

B

`-3`

C

`-2sqrt(2)`

D

`2sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
We have ,
`f(x)=x^(2)+(1)/(x^(2))andg(x) =x-(1)/(x) rArr h(x)=(f(x))/(g(x))`
`:. h(x)=(x^(2)+(1)/(x^(2)))/(x-(1)/(x))=((x-(1)/(x))^(2)+2)/(x-(1)/(x))`
`rArr h(x)=(x-(1)/(x))+(2)/(x-(1)/(2))`
`x-(1)/(x)gt0, (x-(1)/(x))+(2)/(x-(1)/(x))in [ 2 sqrt(2),oo]`
`x-(1)/(x)lt0, (x-(1)/(x))+(2)/(x-(1)/(x))in [oo, 2sqrt(2)]`
`:.` Local minimum value is `2sqrt(2)`
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