Functions Ques 14
- Let $f: R \rightarrow R$ be defined by $f(x)=\frac{x}{1+x^2}$, $x \in R$. Then the range of $f$ is
(2019 Main, 11 Jan I)
(a) $\left[-\frac{1}{2}, \frac{1}{2}\right]$
(b) $(-1,1)-\{0\}$
(c) $R-\left[-\frac{1}{2}, \frac{1}{2}\right]$
(d) $R-[-1,1]$
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Answer:
Correct Answer: 14.(a)
Solution: (a) We have, $f(x)=\frac{x}{1+x^2}, x \in R$
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
$ \left[-\frac{1}{2}, \frac{1}{2}\right] $
IInd Method $f(x)=\frac{1}{x+\frac{1}{x}}$
If $x>0$, then by $\mathrm{AM} \geq \mathrm{GM}$, we get $x+\frac{1}{x} \geq 2$
$\Rightarrow \quad \frac{1}{x+\frac{1}{x}} \leq \frac{1}{2} \quad \Rightarrow 0<f(x) \leq \frac{1}{2}$
If $x<0$, then by AM $\geq$ GM, we get $x+\frac{1}{x} \leq-2$
$\Rightarrow \quad \frac{1}{x+\frac{1}{x}} \geq-\frac{1}{2} \quad \Rightarrow-\frac{1}{2} \leq f(x)<0$
If $x=0$, then $f(x)=\frac{0}{1+0}=0$
Thus, $\quad-\frac{1}{2} \leq f(x) \leq \frac{1}{2}$
Hence, $f(x) \in\left[-\frac{1}{2}, \frac{1}{2}\right]$
IIIrd Method Let $y=\frac{x}{1+x^2} \Rightarrow y x^2-x+y=0$
$\because x \in R$, so $D \geq 0$
$\Rightarrow \quad 1-4 y^2 \geq 0$
$\Rightarrow(1-2 y)(1+2 y) \geq 0 \Rightarrow y \in\left[-\frac{1}{2}, \frac{1}{2}\right]$
So, range is $\left[-\frac{1}{2}, \frac{1}{2}\right]$.
BETA
