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1.Relation and Function
medium
The domain of the function $f(x) = \frac{{{{\sin }^{ - 1}}(3 - x)}}{{\ln (|x|\; - 2)}}$ is
A
$[2, 4]$
B
$(2, 3) \cup (3, 4]$
C
$[2,\infty)$
D
$( - \infty ,\; - 3) \cup [2,\;\infty )$
Solution
(b) $f(x) = \frac{{{{\sin }^{ – 1}}(3 – x)}}{{\log \left[ {|x| – 2} \right]}}$
Let $g(x) = {\sin ^{ – 1}}(3 – x)$ ==> $ – 1 \le 3 – x \le 1$
Domain of $g(x)$ is $[2, 4]$
and let $h(x) = \log \left[ {|x| – 2} \right]$ ==> $|x| – 2 > 0$
==> $|x|\, > 2$ ==> $x < – 2$ or $x > 2$
==> $( – \infty ,\, – 2) \cup (2,\,\infty )$
we know that
$(f/g)(x) = $ $\frac{{f(x)}}{{g(x)}}\forall x \in {D_1} \cap {D_2} – \left\{ {x \in R:g(x) = 0} \right\}$
$\therefore$ Domain of $f(x) = (2,\,4] – \{ 3\} $$ = (2,\,3) \cup (3,\,4]$.
Standard 12
Mathematics
