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

f(x)=x2+1x2 and g(x)=x−1x⇒h(x)=f(x)g(x)∴h(x)=x2+1x2x−1x=(x−1x)2+2x−1x⇒h(x)=(x−1x)+2x−1xx−1x>0,(x−1x)+2x−1x∈[22,∞)x−1x<0,(x−1x)+2x−1x∈(−∞,-22]∴ Local minimum value is 22.

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

f(x)=x2+1x2 and g(x)=x−1x⇒h(x)=f(x)g(x)∴h(x)=x2+1x2x−1x=(x−1x)2+2x−1x⇒h(x)=(x−1x)+2x−1xx−1x>0,(x−1x)+2x−1x∈[22,∞)x−1x<0,(x−1x)+2x−1x∈(−∞,-22]∴ Local minimum value is 22.

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Let $f (x) = x^{2} + \frac{1}{x^{2}} and g (x) = x - \frac{1}{x}, x \in R - {- 1, 0, 1} . If h (x) = \frac{f (x)}{g (x)},$ then the local minimum value of $h (x)$ is

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$2 \sqrt{2}$

$- 2 \sqrt{2}$

$- 3$

answer is B.

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Detailed Solution

$\begin{matrix} f (x) = x^{2} + \frac{1}{x^{2}} and g (x) = x - \frac{1}{x} \Rightarrow h (x) = \frac{f (x)}{g (x)} \\ ∴ h (x) = \frac{x^{2} + \frac{1}{x^{2}}}{x - \frac{1}{x}} = \frac{{(x - \frac{1}{x})}^{2} + 2}{x - \frac{1}{x}} \\ \Rightarrow h (x) = (x - \frac{1}{x}) + \frac{2}{x - \frac{1}{x}} \\ x - \frac{1}{x} > 0, (x - \frac{1}{x}) + \frac{2}{x - \frac{1}{x}} \in [2 \sqrt{2}, \infty) \\ x - \frac{1}{x} < 0, (x - \frac{1}{x}) + \frac{2}{x - \frac{1}{x}} \in (- \infty, - 2 \sqrt{2}] \end{matrix}$