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. Maximum value of f is not attained even though f is bounded
B. f(x) is increasing on `(0,oo)` and has minimum at x=0
C. f(x) is decreasing on `(-oo,0)` and has minimum at x=0
D. f(x) is increasing on `(-oo,oo)` and has neither a local maximum nor a local minimum at x=0

Answer

4

Explanation

we have `f(X)=(x^(2)-a)/(x^(2)+a)=1-(2a)/(x^(2)+a)`
clearly range of f is [-1,1) now
`f(X)=(4ax)/(x^(2)+a^(3))`
and `f(X)=(4a)/(x^(2)+a)^(3)(a-3x^(2))`
Sign scheme of f(X) is as follows

Thus f(X) is decreasing on `(-oo,0)` and increasing on `(0,oo)`
Therefore f(X) has a local minimum at x=0