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AP EAPCET 07-Jul-2022 Shift 1 Paper

Section: Mathematics

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Question : 11 of 160

Marks: +1, -0

Let

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

g(x)=x-\frac{1}{x}

x \in \mathbb{R}\setminus\{-1,0,+1\}

then, the local minimum of

\;\frac{f(x)}{g(x)}

is

$-3$
$2\sqrt{2}$
$2-2\sqrt{2}$
$3$

Solution: 👈: Video Solution

f(x)=x^2+\;\frac{1}{x^2}= \left(x-\;\frac{1}{x}\right) +2, g(x)=x-\;\frac{1}{x}

Let

x-\;\frac{1}{x}=t

f(x)=t^2+2, g(x)=t

\frac{f(x)}{g(x)}=t+\frac{2}{t}=h(t)

h'(t)=1-\frac{2}{t^2}

On putting

h'(t)=0

\Rightarrow 1-\frac{2}{t^2}=0

\Rightarrow t=\pm \sqrt{2}.

\Rightarrow h'(t)=\frac{4}{t^3}

h''(-\sqrt{2})<0

and

h''(\sqrt{2})>0.

\therefore t=\sqrt{2}

is a point of minima. Local minimum value of

h(t)=\sqrt{2}+\frac{2}{\sqrt{2}}=2\sqrt{2}

\therefore

Required local minimum value

=2\sqrt{2}

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