First slide

Functions (XII)

Question

$The domain of f (x) = \sin^{- 1} \{\log_{9} (x^{2} / 4)\} is$

Easy

A

$[- 6, \frac{- 2}{3}] \cup [\frac{2}{3}, 6]$
B

$[- 6, \frac{- 2}{3}]$
C

$[\frac{2}{3}, 6]$
D

$[- 6, \frac{- 2}{3}] \cap [\frac{2}{3}, 6]$

Solution

$We have f (x) = \sin^{- 1} \{\log_{9} (\frac{x^{2}}{4})\}$
$The domain of \sin^{- 1} x is [- 1, 1] . Therefore,$
$f (x) = \sin^{- 1} \{\log_{9} (\frac{x^{2}}{4})\} i s d e f i n e d i f$
$- 1 \leq \log_{9} (\frac{x^{2}}{4}) \leq 1$
$\Rightarrow 9^{- 1} \leq \frac{x^{2}}{4} \leq 9^{1}$
$\Rightarrow \frac{4}{9} \leq x^{2} \leq 36$
$\Rightarrow \frac{2}{3} \leq | x | \leq 6$
$\Rightarrow x \in [- 6, - \frac{2}{3}] \cup [\frac{2}{3}, 6] \begin{matrix} (∵ a \leq | x | \leq b \Leftrightarrow \\ x \in [- b, - a] \cup [a, b]) \end{matrix}$
$Hence, the domain of f (x) is [- 6, - \frac{2}{3}] \cup [\frac{2}{3}, 6] .$

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