Let fx-x2-1x2 and gx-x-1x- x- -1- 0- 1 - If hx-fxgx- then the local minimum value of hx is -

We have, f(x)=x^2+1x^2, g(x)=x 1x Now, nbsp;h(x)=f(x)g(x)=x^2+1x^2x 1x ⇒ h(x)= ( x 1x )^2+2 ( x 1x ) ⇒ nbsp;h(x)= ( x 1x )+2x 1x Let x 1x=t. Then, nbsp; n ...

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Byju's Answer

Standard XII

Mathematics

Definition of Function

Let fx=x2+1x2...

Question

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 :

2 \sqrt{2}

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- 2 \sqrt{2}

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Solution

The correct option is A $2 \sqrt{2}$
$f (x) = x^{2} + \frac{1}{x^{2}}$ $g (x) = x - \frac{1}{x}$

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

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

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

$h^{'} (x) = (1 + \frac{1}{x^{2}}) ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1 - \frac{2}{{(x - \frac{1}{x})}^{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

$\Rightarrow h^{'} (x) = 0 at x - \frac{1}{x} = \pm \sqrt{2}$

Hence, local minimum value is $2 \sqrt{2}$

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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 :

The correct option is A 2√2f(x)=x2+1x2 g(x)=x−1x h(x)=f(x)g(x)=x2+1x2x−1x =(x−1x)2+2(x−1x) h(x)=(x−1x)+2x−1x h′(x)=(1+1x2)⎡⎢ ⎢ ⎢ ⎢ ⎢⎣1−2(x−1x)2⎤⎥ ⎥ ⎥ ⎥ ⎥⎦ ⇒h′(x)=0 at x−1x=±√2 Hence, local minimum value is 2√2

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 :