MCQ
The function $f:R \to \left[ { - \frac{1}{2},\frac{1}{2}} \right],$ defined as $f\left( x \right) = \frac{x}{{1 + {x^2}}}$ is
- Aneither injective nor surjective
- Binvertible
- Cinjective but not surjective
- ✓surjective not injective
$f\left( x \right) = \frac{x}{{1 + {x^2}}}\forall x \in R$
$ \Rightarrow f'\left( x \right) = \frac{{\left( {1 + {x^2}} \right).1 - x.2x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \frac{{ - \left( {x + 1} \right)\left( {x - 1} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}$
$ \Rightarrow f'\left( x \right)$ changes sign in different intervals.
$\therefore $ Not injective
Now $y = \frac{x}{{1 + {x^2}}}$
$ \Rightarrow y + y{x^2} = x$
$ \Rightarrow y{x^2} - x + y = o$
For $y \ne 0,D = 1 - 4{y^2} \ge 0$
$ \Rightarrow y \in \left[ {\frac{{ - 1}}{2},\frac{1}{2}} \right] - \left\{ 0 \right\}$
For $y = 0 \Rightarrow x = 0$
$\therefore $ Range is $\left[ {\frac{{ - 1}}{2},\frac{1}{2}} \right]$
$ \Rightarrow $ surjective but not injective
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$f(x) = sin^{-1} \left( {\frac{{\,\,1 - \,\,\left| x \right|}}{3}} \right) + cos^{-1}\left( {\frac{{\left| x \right|\,\, - \,\,3}}{5}} \right)$ .
Then domain of $f(x)$ is given by :