Question

The range of the function $y = \dfrac{x}{{1 + {x^2}}}$ is
1. $\left( { - \infty , - \infty } \right)$
2. $\left[ { - 1,1} \right]$
3. $\left[ {\left( {\dfrac{{ - 1}}{2}} \right),\left( {\dfrac{1}{2}} \right)} \right]$
4. $\left[ { - \sqrt 2 ,\sqrt 2 } \right]$

Answer 1

Hint: Solve the given function till the required equation will be a quadratic equation. The range of the function is basically the value of its real roots. Solve the required quadratic equation and find the real roots using the quadratic formula.

Formula Used:
General quadratic equation –
$a{x^2} + bx + c$
Quadratic formula to find the roots –
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step Solution:
Given that,
$y = \dfrac{x}{{1 + {x^2}}}$
$y\left( {1 + {x^2}} \right) = x$
$y + y{x^2} = x$
$y{x^2} - x + y = 0$
Compare above quadratic equation with general quadratic equation $a{x^2} + bx + c$
$ \Rightarrow a = y,b = - 1,c = y$
Roots of the quadratic equation are
Using quadratic formula,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
For real roots, ${b^2} - 4ac \geqslant 0$ this condition should be true
${( - 1)^2} - 4(y)(y) \geqslant 0$
$1 - 4{y^2} \geqslant 0$
${(1)^2} - {(2y)^2} \geqslant 0$
$(1 + 2y)(1 - 2y) \geqslant 0$
$(2y + 1) \geqslant 0,(2y - 1) \leqslant 0$
$y \geqslant \dfrac{{ - 1}}{2},y \leqslant \dfrac{1}{2}$
$ \Rightarrow \dfrac{{ - 1}}{2} \leqslant y \leqslant \dfrac{1}{2}$
$y \in \left[ {\dfrac{{ - 1}}{2},\dfrac{1}{2}} \right]$
Hence, the range of given function $y = \dfrac{x}{{1 + {x^2}}}$ is $y \in \left[ {\dfrac{{ - 1}}{2},\dfrac{1}{2}} \right]$.
$ \Rightarrow $Option (3) is the correct.

Hence, the correct option is 3

Note: In such type of question, one should know that for real roots in quadratic formula ${b^2} - 4ac$ should be greater than and equal to zero. Solve the inequality the same as equality but multiplying and dividing any number in inequality change the sign of inequality. If the values are equal also use a square(closed) bracket and if the values are only less or greater then apply a circular(open) bracket.