Question

The range of the function \[y = \dfrac{x}{{1 + {x^2}}}\] is

Answer 1

Hint: We are given with a function y. We have to find the range of the function. We will first rewrite the function in simplified form such that it appears to be a quadratic equation. Then we will find the root of the function and that is only the range of the function.

Complete step-by-step answer:
Given that,
\[y = f\left( x \right)\]
We know the given function,
\[y = \dfrac{x}{{1 + {x^2}}}\]
On cross multiplying the denominator,
\[y\left( {1 + {x^2}} \right) = x\]
Opening the bracket,
\[y + y{x^2} = x\]
Rearranging the terms,
\[y{x^2} - x + y = 0\]
This equation is similar to the standard quadratic equation \[a{x^2} + bx + c = 0\]
On comparing we will get the roots of the equation as,
\[x = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4 \times y \times y} }}{{2y}}\]
\[x = \dfrac{{1 \pm \sqrt {1 - 4{y^2}} }}{{2y}}\]
Now since the roots are real we should equate the radical to zero.
\[1 - 4{y^2} = 0\]
\[1 = 4{y^2}\]
\[{y^2} = \dfrac{1}{4}\]
Taking the roots on both the sides,
\[y = \pm \dfrac{1}{2}\]
Thus this is the range of the given function.\[ - \dfrac{1}{2} \leqslant y \leqslant \dfrac{1}{2}\]
But do remember that \[y \ne 0\].
So the range is defined as, \[y = \left[ { - \dfrac{1}{2},\dfrac{1}{2}} \right]\]
So, the correct answer is “Option B”.

Note: Note that, range of a function can be defined as a set of all possible values as output. We know that if y equals to zero then the function will be meaningless. So we have cleared that y will not equal zero.
Generally we write the given function of the form \[y = f\left( x \right)\] in the form of something like \[x = g\left( y \right)\]. Then finding the domain of \[g\left( y \right)\] will be the range of the function so given.