The real number x when added to its inverse gives the minimum value of the sum at x equal to
(a) 1 

(b) –1 

(c) –2 

(d) 2.

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1 Answers

(a) : f(x) = x + 1/x
f '(x) = 1 – 1/x2 and f "(x) = 2/x3
now f '(x) = 0
= x = ± 1
Therefore,  f "(1) > 0
=> x = 1 is point of minima.

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