1) 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 A) 3 B) -3 C) −2√2 D) 2√2 Answer: Option DExplanation:We have f(x)=x2+1x2 and g(x)=x−1x ⇒ h(x)=g(x)h(x) ∴ h(x)=x2+1x2x−1x=(x−1x)2+2x−1x ∴ h(x)=(x−1x)+2x−1x x−1x>0, (x−1x)+2x−1x∈[2√2,∞] x−1x<0 (x−1x)+2x−1x∈(−∞,2√2) ∴ Local minimum value is 2√2