Let $f (x) = x^{2} + \frac{1}{x^{2}} a n d g (x) = x - \frac{1}{x}, x Ε R - {- 1, 0, 1} .$
if $h (x) = \frac{f (x)}{g (x)}$ , then the local minimum value of h(x) is:

$2 \sqrt{2}$

3

-3

$- 2 \sqrt{2}$

Solution

$2 \sqrt{2}$

$h (x) = \frac{x^{2} + \frac{1}{x^{2}}}{x - \frac{1}{x}} = (x - \frac{1}{x}) + \frac{2}{(x - \frac{1}{x})} x - \frac{1}{x} > 0, (x - \frac{1}{x}) + \frac{2}{(x - \frac{1}{x})} \in (2 \sqrt{2}, \infty) x - \frac{1}{x} < o, (x - \frac{1}{x}) + \frac{2}{(x - \frac{1}{x})} \in (- \infty, - 2 \sqrt{2)} L o c a l m i n i m u m (2 \sqrt{2)}$

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Inverse Trigonometric Functions

Let $f (x) = x^{2} + \frac{1}{x^{2}} a n d g (x) = x - \frac{1}{x}, x Ε R - {- 1, 0, 1} .$
if $h (x) = \frac{f (x)}{g (x)}$ , then the local minimum value of h(x) is:

$2 \sqrt{2}$

3

-3

$- 2 \sqrt{2}$

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Inverse Trigonometric Functions

Let f(x) = x2 + 1x2 and g (x) = x - 1x, x Ε R - {-1,0,1}.if h(x) = f(x)g(x), then the local minimum value of h(x) is: 22 3 -3 -22

Some More Questions From Inverse Trigonometric Functions Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Let $f (x) = x^{2} + \frac{1}{x^{2}} a n d g (x) = x - \frac{1}{x}, x Ε R - {- 1, 0, 1} .$
if $h (x) = \frac{f (x)}{g (x)}$ , then the local minimum value of h(x) is:

$2 \sqrt{2}$

3

-3

$- 2 \sqrt{2}$