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Consider the function f:(-oo,oo)vec(-oo,...

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.`

A

`g ' (x)` is positive on ` (-oo, 0) ` and negative on ` (0, oo)`

B

` g' (x) ` is negative on ` (-oo, 0) ` and positive on ` (0, oo)`

C

` g' (x)` changes sign on both ` (-oo, 0) and ( 0, oo)`

D

` g' (x) ` does not changes sign ` (-oo, oo)`

Text Solution

Verified by Experts

The correct Answer is:
a
`g' (x) = (f ' (e ^(x)))/( 1+ ( e^(x)) ^(2)) * e ^(x)`
`" " = 2a [ ( e ^(2x ) - 1) /( ( e ^( 2x ) + ae ^(x) + 1)^(2)) ] (( e^(x))/( 1 + e ^( 2x)))`
` " " g ' (x) = 0, if e ^(2x) - 1 = 0`, i.e . `x = 0`
If `" " x lt 0, e ^( 2x) lt 1 rArr g ' (x) lt 0`
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