Question
The real number $$x$$ when added to its inverse gives the minimum value of the sum at $$x$$ equal to
A.
$$ - 2$$
B.
$$2$$
C.
$$1$$
D.
$$ - 1$$
Answer :
$$1$$
Solution :
$$y = x + \frac{1}{x}\,\,{\text{or}}\,\,\,\frac{{dy}}{{dx}} = 1 - \frac{1}{{{x^2}}}$$
For max. or min., $$1 - \frac{1}{{{x^2}}} = 0$$
$$\eqalign{
& \Rightarrow x = \pm 1 \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{2}{{{x^3}}} \cr
& \Rightarrow \,\,{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 2}} = 2\left( { + ve\,\,{\text{minima}}} \right) \cr
& \therefore \,\,x = 1 \cr} $$