Let f x = x 2 + 1 x 2 and g x = x - 1 x ,   x ∈ R - - 1 ,   0 ,   1 . If h x = f x g x …

Let fx=x2+1x2 and gx=x-1x, x∈R--1, 0, 1. If hx=fxgx , then the local minimum value of hx is:
 
  1. 22
  2. 3
  3. -3
  4. -22

Solution

fx=x2+1x2=x-1x2+2

gx=x-1x

x∈R--1,0,1

Let x-1x=t

hx=fxgx=t2+2t=t+2t

t∈R-0

hx=t+2t

AM≥GM

Minimum of hx⇒ t+2t2 ≥ t×2t

t+2t≥22

Local minimum =22

Asked in: JEE Main 2018 (08 Apr)

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