Let f: R → R be defined by f(x) = (x/1+ x2) , x ∈ R. Then the range of f is :

Question

Let f: R → R be defined by f(x) = (x/1+ x2) , x ∈ R. Then the range of f is : (A) (-1, 1) - 0 (B) [- (1/2) , (1/2)] (C) - [- (1/2) , (1/2)] (D) R - [-1, 1]

Tardigrade Admin · Accepted Answer

f(0) = 0 f(x) is odd. Further, if x > 0 then f(x) = (1/x+ (1)x) ∈ (0, (1/2)] f(x)∈ [-(1/2) , (1/2)]

Q. Let $f : R \to R$ be defined by $f (x) = \frac{x}{1 + x ^{2}}, x \in R$ . Then the range of $f$ is :

(-1, 1) - {0}

$[- \frac{1}{2}, \frac{1}{2}]$

$- [- \frac{1}{2}, \frac{1}{2}]$

$R - [- 1, 1]$

$f (0) = 0$ & $f (x)$ is odd.
Further, if $x > 0$ then
$f (x) = \frac{1}{x + \frac{1}{x}} \in (0, \frac{1}{2}]$
$f (x) \in [- \frac{1}{2}, \frac{1}{2}]$

Q. Let f:R→R be defined by f(x)=1+x2x​,x∈R. Then the range of f is :

(-1, 1) - {0}

[−21​,21​]

−[−21​,21​]

R−[−1,1]

f(0)=0 & f(x) is odd. Further, if x>0 then f(x)=x+x1​1​∈(0,21​] f(x)∈[−21​,21​]

Q. Let $f : R \to R$ be defined by $f (x) = \frac{x}{1 + x ^{2}}, x \in R$ . Then the range of $f$ is :

$[- \frac{1}{2}, \frac{1}{2}]$

$- [- \frac{1}{2}, \frac{1}{2}]$

$R - [- 1, 1]$

$f (0) = 0$ & $f (x)$ is odd.
Further, if $x > 0$ then
$f (x) = \frac{1}{x + \frac{1}{x}} \in (0, \frac{1}{2}]$
$f (x) \in [- \frac{1}{2}, \frac{1}{2}]$