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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
