First slide

Functions (XII)

Question

The domain of the function
$f (x) = \sin^{- 1} \{\log_{2} (\frac{1}{2} x^{2})\} is$

Moderate

A

$[- 2, - 1) \cup [1, 2]$
B

$(- 2, - 1] \cup [1, 2]$
C

$[- 2, - 1] \cup [1, 2]$
D

$(- 2, - 1) \cup (1, 2)$

Solution

For f (x) to be defined, we must have
$- 1 \leq \log_{2} (\frac{1}{2} x^{2}) \leq 1 \Rightarrow 2^{- 1} \leq \frac{1}{2} x^{2} \leq 2^{1} [∵ the base = 2 > 1]$
$\Rightarrow 1 \leq x^{2} \leq 4 . . . . . . . . (1) Now, 1 \leq x^{2} \Rightarrow x^{2} - 1 \geq 0 i.e. (x - 1) (x + 1) \geq 0 \Rightarrow x \leq - 1 or x \geq 1 . . . . . . . . (2) Also, x^{2} \leq 4 \Rightarrow x^{2} - 4 \leq 0 i.e. (x - 2) (x + 2) \leq 0 \Rightarrow - 2 \leq x \leq 2 . . . . . . . . (3)$
From Eq. (2) and (3), we get the domain of f
$= ((- \infty, - 1] \cup [1, \infty)) \cap [- 2, 2] = [- 2, - 1] \cup [1, 2]$

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