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

Option 1)

$R-\begin{bmatrix} -\frac{1}{2}\: ,\frac{1}{2} \end{bmatrix}$
Option 2)

$R-\begin{bmatrix} -{1}\: ,\1\end{bmatrix}$
Option 3)

$(-1,1)- {\left { 0 \right }}$
Option 4)

$\begin{bmatrix} -\frac{1}{2}\: ,\frac{1}{2} \end{bmatrix}$

Answers (1)

best_answer

Range of function -

All possible values of $f(x)\: \: \forall x\in$ domain $(f)$ is known as Range

-

$f(x)=\frac{x}{\left ( \frac{1}{x}+x \right )}$

$AM\geq GM$

$\Rightarrow \frac{1}{AM}\leq \frac{1}{GM}$ $if x> 0$

$f(x)\leq \frac{1}{2\left ( \frac{1}{x}\times x \right )}$

$\Rightarrow f(x)\leq \frac{1}{2}$

$if \: \: \: x< 0,$

$-AM\geq GM$

$\Rightarrow f(x)\geq -\frac{1}{2}$

Range of $f(x)\equiv \left [ -\frac{1}{2},\frac{1}{2} \right ]$

Option 1)

$R-\begin{bmatrix} -\frac{1}{2}\: ,\frac{1}{2} \end{bmatrix}$

Option 2)

$R-\begin{bmatrix} -{1}\: ,\1\end{bmatrix}$

Option 3)

$(-1,1)- {\left { 0 \right }}$

Option 4)

$\begin{bmatrix} -\frac{1}{2}\: ,\frac{1}{2} \end{bmatrix}$

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