JEE MAIN - Mathematics (2006 - No. 5)
The function $$f\left( x \right) = {x \over 2} + {2 \over x}$$ has a local minimum at
$$x=2$$
$$x=-2$$
$$x=0$$
$$x=1$$
Explanation
$$f\left( x \right) = {x \over 2} + {2 \over x} \Rightarrow f'\left( x \right) = {1 \over 2} - {2 \over {{x^2}}} = 0$$
$$ \Rightarrow {x^2} = 4$$ or $$x=2,-2;$$ $$\,\,\,\,\,f''\left( x \right) = {4 \over {{x^3}}}$$
$$f''{\left. {\left( x \right)} \right]_{x = 2}} = + ve \Rightarrow f\left( x \right)$$
has local min at $$x=2.$$
$$ \Rightarrow {x^2} = 4$$ or $$x=2,-2;$$ $$\,\,\,\,\,f''\left( x \right) = {4 \over {{x^3}}}$$
$$f''{\left. {\left( x \right)} \right]_{x = 2}} = + ve \Rightarrow f\left( x \right)$$
has local min at $$x=2.$$
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