First slide

Functions (XII)

Question

The domain of the function $f (x) = \sqrt{|\sin^{- 1} (\sin x)| - \cos^{- 1} (\cos x)}$ in $[0, 2 π]$ is

Moderate

A

$[0, \frac{π}{2}] \cup [\frac{3 π}{2}, 2 π]$
B

$[π, 2 π]$
C

$[0, π] - \{\frac{π}{2}\}$
D

$[0, 2 π] - \{\frac{π}{2}, \frac{3 π}{2}\}$

Solution

As, $|\sin^{- 1} (\sin x)|$ could be sketched, as
and $\cos^{- 1} (\cos x)$ could be sketched as

$∴ |\sin^{- 1} (\sin x)| > \cos^{- 1} (\cos x)$ is not possible.
Only equality holds as
Thus, $|\sin^{- 1} (\sin x)| = \cos^{- 1} (\cos x)$
When $x \in [0, \frac{π}{2}] \cup [\frac{3 π}{2}, 2 π]$
So, domain for $f (x) is [0, \frac{π}{2}] \cup [\frac{3 π}{2}, 2 π]$
Hence, (a) is the correct answer.

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