Let `f: R to R ` be defined by `f(x)=(x)/(1+x^(2)), x in R.` Then, the range of f is (A) `[-(1)/(2),(1)/(2)]` (B) `(-1,1)-{0}` (C) `R-[-(1)/(2),(1)/(2)]` (D) `R-[-1,1]`

Options

A. `[-(1)/(2),(1)/(2)]`
B. `(-1,1)-{0}`
C. `R-[-(1)/(2),(1)/(2)]`
D. `R-[-1,1]`

Answer

A

Explanation

We have,` f(x)=(x)/(1+x^(2)), x in R`
Ist Method f(x) is an odd function and maximum occur at x = 1

From the graph it is clear that range of f(x) is
`[-(1)/(2),(1)/(2)]`
IInd Method `f(x)=(1)/(x+(1)/(x))`
If `x gt 0`, then by `AM ge GM`, we get `x+(1)/(x) ge 2`
`rArr (1)/(x+(1)/(x)) le (1)/(2) rArr 0 lt f(x) le (1)/(2)`
If `x lt 0`, then by `AM ge GM`, we get `x+(1)/(x) le -2`
`rArr (1)/(x+(1)/(x)) ge -(1)/(2) rArr -(1)/(2) le f(x) le 0`
If x = 0, then `f(x)=(0)/(1+0)=0`
Thus, `-(1)/(2) le f(x) le (1)/(2)`
Hence, `f(x) in [-(1)/(2),(1)/(2)]`
IIIrd Method
Let `y=(x)/(1+x^(2)) rArr yx^(2)-x+y=0`
` because x in R, " so "D ge 0`
`rArr 1-4y^(2) ge 0`
`rArr (1-2y)(1+2y) ge 0 rArr y in [-(1)/(2),(1)/(2)]`

So, range is `[-(1)/(2),(1)/(2)]`.