The domain of sin-1[x] is given by

Question

The domain of sin-1[x] is given by (A) [-1, 1] (B) [-1, 2) (C)  -1, 0, 1  (D) None of these

Tardigrade Admin · Accepted Answer

Domain of sin-1 x is [-1, 1] and Domain of sin-1[x] is x: -1 le [x] le 1 = x: [x] = -1, 0, 1 But [x] = begincases -1 ∀ -1 le x < 0 0 ∀ 0 le x < 1 1 ∀ 1 le x < 2 endcases ∴ x ∈ [-1, 2) i.e., -1 le x < 2

Q. The domain of $s i n^{- 1} [x]$ is given by

$[- 1, 1]$

$[- 1, 2)$

${- 1, 0, 1}$

None of these

Domain of $s i n^{- 1} x$ is $[- 1, 1]$
and Domain of $s i n^{- 1} [x]$ is ${x : - 1 \leq [x] \leq 1}$
$= {x : [x] = - 1, 0, 1}$
But $[x] = ⎩ ⎨ ⎧ - 1 01 \forall - 1 \leq x < 0 \forall 0 \leq x < 1 \forall 1 \leq x < 2$
$∴ x \in [- 1, 2)$ i.e., $- 1 \leq x < 2$

Q. The domain of sin−1[x] is given by

[−1,1]

[−1,2)

{−1,0,1}