JEE Main | 2018Maxima and MinimaEasy

Question

Let f(x) = x2 + 1x2 and g(x) = x - 1x x∈R - {-1, 0, 1}. If h(x) = f(x)g(x) then the local minimum value of h(x) is : 

Options

A.

-3

B.

-22

C.

22

D.

3

Solution

h(x) = x2+1x2x-1x=x-1x+2x-1x
when x - 1x>0⇒x-1x+2x-1x⩾22
so 22 will be local minimum value

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