1.Relation and Function
hard

જો $f :R \to R$ ; $f(x)\,\, = \,\,\frac{x}{{1 + {x^2}}},\,x\, \in \,R$ હોય તો $f$ નો વિસ્તાર મેળવો.

A

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

B

$R\, - [ - 1,1]$

C

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

D

$( - 1,1) - \{ 0\} $

(JEE MAIN-2019)

Solution

$y = \frac{x}{{{x^2} + 1}}$

$y{x^2} – x + y = 0$

$D \ge 0$

$ \Rightarrow 1 – 4{y^2} \ge 0$

$ \Rightarrow {y^2} \le \frac{1}{4}$

$y \in \left[ { – \frac{1}{2}.\frac{1}{2}} \right]$

 

Standard 12
Mathematics

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