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STD 12 - 1. relation and function — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 1. relation and function4 Marks
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
  • A
    neither  injective nor surjective
  • B
    invertible
  • C
     injective but not surjective
  • ✓
    surjective not injective
✓

Answer

Correct option: D.
surjective not injective
d
$(4)$  we have $f:R \to \left[ { - \frac{1}{2},\frac{1}{2}} \right],$

         $f\left( x \right) = \frac{x}{{1 + {x^2}}}\forall x \in R$

$ \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 

 

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