JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

The range of

f(x)=\frac{x}{1+x^2}

is given by

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because by setting

f(x) = y

and analyzing the discriminant of the resulting quadratic equation, we find the range of

y

to be between -1/2 and 1/2.

More Questions from FUNCTIONS

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f:(-\infty, 2] \longrightarrow(-\infty, 4]

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f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

f\left( x \right) = x - \left[ x \right]

{f^{ - 1}}\left( x \right)

f: R \rightarrow R

f(x)=x^4

f(x)=(x+1)^2-1(x \geq-1)

S=\left\{x: f(x)=f^{-1}(x)\right\}