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Question

# Let f(x)=x2+1x2 and g(x)=x−1x,xϵR−{−1,0,1}. If h(x)=f(x)g(x). Then the local minimum value of h(x) is:

A
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Solution

## The correct option is A 2√2Given:f(x)=x2+1x2g(x)=x−1xh(x)=f(x)g(x)To find:Minimum value of h(x)Solution:h(x)=f(x)g(x)⇒h(x)=x2+1x2x−1x⇒h(x)=(x−1x)2+2x−1x⇒h(x)=(x−1x)+2x−1xDifferentiate both the sides⇒h′(x)=(1+1x2)−2(x−1x)2(1+1x2)⇒h′(x)=(1+1x2)⎛⎜ ⎜ ⎜ ⎜ ⎜⎝1−2(x−1x)2⎞⎟ ⎟ ⎟ ⎟ ⎟⎠⇒h′(x)=(1+1x2)⎛⎜ ⎜ ⎜ ⎜ ⎜⎝(x−1x)2−2(x−1x)2⎞⎟ ⎟ ⎟ ⎟ ⎟⎠⇒h′(x)=(1+1x2)(x−1x)2((x−1x)2−2)⇒h′(x)=x2+1(x−1x)2⋅x2((x2−1)2−2x2x2)Equating h'(x)=0, we get⇒h′(x)=x2+1(x−1x)2⋅x4((x2−1)2−2x2)=0⇒((x2−1)2−2x2)=0⇒(x4+1−2x2−2x2)=0⇒(x4+1−4x2)=0⇒(x2−2)2−3=0⇒(x2−2)2=3⇒(x2−2)=±√3⇒x2=2±√3⇒x2=(√3±1)22⇒x=±√3±1√2As x∈R−{−1,0,1}∴x=√3+1√2and⇒1x=√2√3+1⇒1x=√2√3+1⋅√3−1√3−1⇒1x=√3−1√2Now,⇒x−1x=√3+1√2−√3−1√2⇒x−1x=2√2⇒x−1x=√2Also,⇒h(x)=(x−1x)+2x−1xputting the value,⇒h(x)=√2+2√2⇒h(x)=2√2

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