Consider the function `f:(-oo,oo)vec(-oo,oo)` defined by `f(x)=(x^2+a)/(x^2+a),a >0,` which of the following is not true? maximum value of `f` is not attained even though `f` is bounded. `f(x)` is increasing on `(0,oo)` and has minimum at `,=0` `f(x)` is decreasing on `(-oo,0)` and has minimum at `x=0.` `f(x)` is increasing on `(-oo,oo)` and has neither a local maximum nor a local minimum at `x=0.`

Options

A. ` f (x)` is decreasing on ` (- 1, 1)` and has a local minimum at ` x = 1 `
B. `f (x)` is increasing on ` (-1, 1 )` and has a local maximum at ` x = 1 `
C. ` f(x)` is increasing on ` (-1, 1 )` but has neither a local maximum nor a local minimum at ` x = 1 `
D. ` f (x)` is decreasing on ` (-1, 1)` but has neither a local maximum nor a local minimum at ` x = 1`

Answer

a

Explanation

When `x in (-1, 1)`
` x ^(2) lt 1 rArr x ^(2) - 1 lt 0 `
`therefore f (x) lt 0 , f(x)` is decreasing.
Also, at ` x = 1 , f'' (1) = ( 4a )/( (a + 2)^(2)) gt 0 " "[ because 0 lt a lt 2 ]`
So,`f(x)` has local minimum at ` x = 1 `