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1.Relation and Function
normal
વિધેય $f(x) = \frac{{x + 2}}{{|x + 2|}}$ નો વિસ્તાર મેળવો.
A
$\{0, 1\}$
B
$\{-1, 1\}$
C
$R$
D
$R - \{ - 2\} $
Solution
(b) $f(x) = \frac{{x + 2}}{{|x + 2|}}$
$f(x) = \left\{ {\begin{array}{*{20}{c}}{ – 1,}\\{\,\,1,}\end{array}} \right.\,\,\,\,\,\,\begin{array}{*{20}{c}}{x < – 2}\\{x > – 2}\end{array}$
$\therefore$ Range of $f(x)$ is $\{ – 1,\,1\} $.
Standard 12
Mathematics