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Question: Let \[f\left( x \right) = {x^2} + \dfrac{1}{{{x^2}}}\] and \[g\left( x \right) = x - \dfrac{1}{x}\] ...

Let f(x)=x2+1x2f\left( x \right) = {x^2} + \dfrac{1}{{{x^2}}} and g(x)=x−1xg\left( x \right) = x - \dfrac{1}{x} , x \in R - \left\\{ { - 1,0,1} \right\\} . If h(x)=f(x)g(x)h\left( x \right) = \dfrac{{f\left( x \right)}}{{g\left( x \right)}} , then the local minimum value of h(x)h\left( x \right) is
A −22- 2\sqrt 2
B 222\sqrt 2
C 33
D −3 - 3

Explanation

Solution

Hint : In the given question, as it consists of two functions f(x)f\left( x \right) and g(x)g\left( x \right) , in which it is mentioned that h(x)=f(x)g(x)h\left( x \right) = \dfrac{{f\left( x \right)}}{{g\left( x \right)}} , hence by substituting the given functions we can solve for h(x)h\left( x \right) and then applying AM ⩾\geqslant GM we can get the local minimum value of it.

Complete step-by-step answer :
Let us write the given data:
f(x)=x2+1x2f\left( x \right) = {x^2} + \dfrac{1}{{{x^2}}}
g(x)=x−1xg\left( x \right) = x - \dfrac{1}{x}
As mentioned,
h(x)=f(x)g(x)h\left( x \right) = \dfrac{{f\left( x \right)}}{{g\left( x \right)}}
Let us substitute the functions of f(x)f\left( x \right) and g(x)g\left( x \right) by the given data as:
h(x)=f(x)g(x)h\left( x \right) = \dfrac{{f\left( x \right)}}{{g\left( x \right)}}
h(x)=x2+1x2x−1xh\left( x \right) = \dfrac{{{x^2} + \dfrac{1}{{{x^2}}}}}{{x - \dfrac{1}{x}}}
Simplifying the terms, we get
h(x)=(x−1x)2+2x−1xh\left( x \right) = \dfrac{{{{\left( {x - \dfrac{1}{x}} \right)}^2} + 2}}{{x - \dfrac{1}{x}}}
h(x)=(x−1x)+2x−1xh\left( x \right) = \left( {x - \dfrac{1}{x}} \right) + \dfrac{2}{{x - \dfrac{1}{x}}}
Applying AM ⩾\geqslant GM we get
h(x−1x)+2x−1x2⩾(x−1x)⋅2(x−1x)\dfrac{{h\left( {x - \dfrac{1}{x}} \right) + \dfrac{2}{{x - \dfrac{1}{x}}}}}{2} \geqslant \sqrt {\left( {x - \dfrac{1}{x}} \right) \cdot \dfrac{2}{{\left( {x - \dfrac{1}{x}} \right)}}}
Solving the above inequality, we get
(x−1x)+2(x−1x)⩾22\left( {x - \dfrac{1}{x}} \right) + \dfrac{2}{{\left( {x - \dfrac{1}{x}} \right)}} \geqslant 2\sqrt 2
Thus, the local minimum value of h(x)h\left( x \right) is 222\sqrt 2 .
Hence, option B is the right answer.
So, the correct answer is “Option B”.

Note : AM or Arithmetic Mean is the mean or average of the set of numbers which is computed by adding all the terms in the set of numbers and dividing the sum by total number of terms. GM or Geometric Mean is the mean value or the central term in the set of numbers in geometric progression.
The arithmetic mean, or less precisely the average, of a list of n numbers x1,x2,.....xn{x_1},{x_2},.....{x_n} is the sum of the numbers divided by n and is given as:
x1+x2+.....+xnn\dfrac{{{x_1} + {x_2} + ..... + {x_n}}}{n}
The geometric mean is similar, except that it is only defined for a list of nonnegative real numbers, and uses multiplication and a root in place of addition and division, given as:
x1⋅x2⋅.....xnn\sqrt[n] {{{x_1} \cdot {x_2} \cdot .....{x_n}}}