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

Option 1)
2 $\sqrt{2}$

Option 2)
3

Option 3)
-3

Option 4)
-2 $\sqrt{2}$

Answers (2)

best_answer

$h(x)= \frac{x^{2} + \frac{1}{x^{2}}}{x-\frac{1}{x}}= \frac{(x-\frac{1}{x})^{2}+2}{x-\frac{1}{x}}$

$=\left ( x-\frac{1}{x} \right )+ \frac{2}{(x-\frac{1}{x})}$

If $\left ( x-\frac{1}{x} \right )= y$

$h(x)= y+\frac{2}{y}$

AM $\geq$ GM

thus $y+ \frac{y}{2}\geq 2\sqrt{y\times \frac{2}{y}}$

$y+ \frac{y}{2}\geq 2\sqrt2$

or $y+ \frac{y}{2}\leq -2\sqrt2$

Local minimum value $= 2\sqrt2$

Local maximum and Local minimum -

Let y = f(x) be the given function then $x=x_{\circ }$ is a point of local maximum if there exists an open interval containing $x_{\circ }$ such that $f(x_{\circ })>f(x)$

for all values of x lying in that interval and for $f(x_{\circ })<f(x)$ then it is local minimum in fig and at

$x=a,x=c\:\:\:Local\:\: maxima$

$and\:at\:\:x=b,x=d\:\:\:Local\:\: minima$

- wherein

Option 1)

2 $\sqrt{2}$

Option 2)

3

Option 3)

-3

Option 4)

-2 $\sqrt{2}$

Posted by

Himanshu

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