QJEEM18Ma8046: JEE Main 2018 Mathematics - Class 12 JEEM18Ma8046 Solution: Let and , . If , then the local minimum value of h(x) is

Given $f\left( x \right) = {x^2} + {1 \over {{x^2}}}$ and $g\left( x \right) = x - {1 \over x}$

As $h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$

= ${{{x^2} + {1 \over {{x^2}}}} \over {x - {1 \over x}}}$

= ${{{{\left( {x - {1 \over x}} \right)}^2} + 2.x.{1 \over x}} \over {x - {1 \over x}}}$

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

Let ${x - {1 \over x} = t}$

So $f\left( t \right) = {{{t^2} + 2} \over t}$ = $t + {2 \over t}$

then $f'\left( t \right) = 1 - {2 \over {{t^2}}}$

At maximum or minimum $f'\left( t \right) = 0$ .

$\therefore$ $1 - {2 \over {{t^2}}} = 0$

$\Rightarrow t = \pm \sqrt 2$

Now we need to find $f''\left( t \right)$ which will tell at which point among $+ \sqrt 2$ and $- \sqrt 2$ will be maximum and minimum.

$f''\left( t \right) = + {4 \over {{t^3}}}$

So when $t = + \sqrt 2$ , then $f''\left( t \right) = + {4 \over {{{\left( {\sqrt 2 } \right)}^3}}}$ = $+ \sqrt 2$

As f''\left( t \right) > 0 when $t = + \sqrt 2$ then at $t = + \sqrt 2$ the function $f\left( t \right)$ will be minimum.

Min of $f\left( t \right) = {{{{\left( {\sqrt 2 } \right)}^2} + 2} \over {\sqrt 2 }}$ = ${4 \over {\sqrt 2 }}$ = $2\sqrt 2$

So local minimum of $f\left( t \right)$ = $2\sqrt 2$