- Home
- Standard 12
- Mathematics
1.Relation and Function
hard
વિધેય $f(x) = \frac{x}{{1 + \left| x \right|}},\,x \in R,$ નો વિસ્તાર મેળવો.
A
$R$
B
$(-1,1)$
C
$R-\{0\}$
D
$[-1,1]$
(AIEEE-2012)
Solution
$f\left( x \right) = \frac{x}{{1 + \left| x \right|}},x \in R$
If $x > 0,\left| x \right| = x \Rightarrow f\left( x \right) = \frac{x}{{1 + x}}$
which is not defined for $x=-1$
If $x < 0,\left| x \right| = – x \Rightarrow f\left( x \right) = \frac{x}{{1 – x}}$
which is not defined for $x=1$
Thus $f\left( x \right)$ defined for all value of $R$ except $1$ and $-1$
Hence, range $=(-1,1)$.
Standard 12
Mathematics