First slide

Functions (XII)

Question

The domain of f(x)=sin^-1( $|x - 1| - 2$ ) is

Moderate

A

[-2,0] $\cup$ [2,4]
B

(-2,0) $\cup$ (2,4)
C

[-2,0] $\cup$ [1,3]
D

none of these

Solution

given f(x)=sin^-1( $|x - 1| - 2$ )
Since,the domain of sin^-1 x is [-1,1],f(x) is defined,if
-1 $\leq$ $|x - 1| - 2 \leq 1$
$\Rightarrow 1 \leq |x - 1| \leq 3$
$\Rightarrow - 3 \leq x - 1 \leq - 1 o r 1 \leq x - 1 \leq 3 \Rightarrow - 2 \leq x \leq 0 o r 2 \leq x \leq 4 ∴ D o m a i n = [- 2, 0] \cup [2, 4]$

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