Functions Ques 14

  1. 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 $\left[-\frac{1}{2}, \frac{1}{2}\right]$.



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