1. Tardigrade
  2. Question
  3. Mathematics
  4. Let f(x) = x2 + (1/x2) and g(x) = x - (1/x) , x ∈ R - - 1, 0, 1 . If h(x) = (f(x) /g(x)), then the local minimum value of h(x) is

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Q. 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

JEE MainJEE Main 2018Application of Derivatives

Solution:

$h(x)=\frac{x^{2}+\frac{1}{x^{2}}}{x-1 / x}$
$=(x-1 / x)+\frac{2}{(x-1 / x)}$
$x-\frac{1}{x}>0,(x-1 / x)+\frac{2}{(x-1 / x)} \in(2 \sqrt{2}, \infty]$
$x-\frac{1}{x}<0,(x-1 / x)+\frac{2}{(x-1 / x)} \in(-\infty,-2 \sqrt{2}]$
Local minimum is $2 \sqrt{2}$