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If `x gt 0 and xy =1`, the minimum value of `(x + y)` isA. `-2`B. 1C. 2D. none of these |
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Answer» Correct Answer - C `xy = 1 rArr y = (1)/(x)` Let `S = x + y = x + (1)/(x)`. Then, `(ds)/(dx) = (1 - (1)/(x^(2))) = ((x^(2) -1))/(x^(2)) and (d^(2)s)/(dx^(2)) = (2)/(x^(3))` `(ds)/(dx) = 0 rArr x^(2) - 1 = 0 rArr x = +- 1` `(d^(2)s)/(dx^(2))]_((x = -1)) = -2 lt 0 and (d^(2)s)/(dx^(2))]_((x =1)) = 2 gt 0` `:.` S is minimum at x = 1 and minimum value of `S = (1 + 1) = 2` |
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