JEE MAIN - Mathematics (2003 - No. 39)
The real number $$x$$ when added to its inverse gives the minimum sum at $$x$$ equal :
-2
2
1
-1
Explanation
$$f(x) = y = x + {1 \over x}$$ or $${{dy} \over {dx}} = 1 - {1 \over {{x^2}}}$$
For max. or min, $$1 - {1 \over {{x^2}}} = 0 \Rightarrow x = \pm 1$$
$${{{d^2}y} \over {d{x^2}}} = {2 \over {{x^3}}} $$
$$ {\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{x = 1}} = 2$$ ($$+ve$$)
$$\therefore$$ f(x) will be minima at $$x=1$$.
For max. or min, $$1 - {1 \over {{x^2}}} = 0 \Rightarrow x = \pm 1$$
$${{{d^2}y} \over {d{x^2}}} = {2 \over {{x^3}}} $$
$$ {\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{x = 1}} = 2$$ ($$+ve$$)
$$\therefore$$ f(x) will be minima at $$x=1$$.
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