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. `2`
C. `1`
D. `-1`

Answer

C

Explanation

Let `A=x+(1)/(x)implies(dA)/(dx)=1-(1)/(x^(2))`
For maximum or minima, put `(dA)/(dx)=0`
`implies1-(1)/(x^(2))=0 impliesx=-1,1`
Again differentiating, we get
`(d^(2)A)/(dx^(2))=(2)/(x^(3))implies((d^(2)A)/(dx^(2)))=2gt0`
`therefore`A is minimum at x=1.