The real number `x` when added to its inverse given the minimum value of the sum at `x` equal to 1 (b) `-1` (c) `-2` (d) 2

Options

A. 2
B. 1
C. -1
D. -2

Answer

B

Explanation

Let `f(x)=x+1/x`
`impliesf'(x)=1-1/(x^(2))`
For maxima and minima put `f'(x)=0`
`implies 1-1/(x^(2))=0impliesx=+-1`
Now `f''(x)=2/(x^(3))`
At `x=1, f''(x)=+ve`, minima
and at `x=-1, f''(x)=-ve` maxima
Thus `f(x)` attains minimum value at `x=1`.