Given function is f(x) \(=[sin^{-1}x]\) + [sin x]
Since domain of function \(sin^{-1}x\) is [-1, 1]
∴ Domain of function f(x) is [-1, 1]
The range of \(sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)
i.e. \(-\frac{\pi}{2}\) ≤ sin-1x ≤ \(\frac{\pi}{2}\)
⇒ \([-\frac{\pi}{2}]\) ≤ [sin-1x] ≤ \([\frac{\pi}{2}]\)
⇒ - 2 ≤ [sin-1x] ≤ 1
⇒ [sin-1x] \(\in\) {-2, -1, 0, 1} -------(1)
(∵ \([\frac{\pi}{2}]\) = [1.57] = 1 and \([-\frac{\pi}{2}]\) = [-1.57] = -2)
∵ The range of sin x is [-1, 1]
But \(\frac{\pi}{2}\) \(\notin\) Domain of f(x)
Similarly \(-\frac{\pi}{2}\) \(\notin\)Domain of f(x).
And -1 < sin x < 1 ∀ x \(\in\) [-1, 1].
⇒ -1 ≤ [sin x] ≤ 0
⇒ [sin x] \(\in\) {-1, 0} -----(2)
(∵ [x] < a, a \(\in\) I then [x] = a - 1)
From equations (1) and (2), we get
[sin-1 x] + [sin x] = -2 -1 = -3, (At x = -1, [sin-1 x] = [-sin-1 1] = [- \(\frac{\pi}{2}\)] = -2 and [sin (-1)] = [-sin 1] = -1)
[sin-1 x] + [sin x] = -1 -1 = -2, (At x = -1/2, [sin-1 x] = [- sin-1 1/2] = [- \(\frac{\pi}{6}\)] = -1 and [sin (-1/2)] = [-sin 1/2] = -1)
But [sin-1 x] + [sin x] never be -1 (Because both [sin-1 x] and [sin x] are either positive or negative together)
[sin-1 x] + [sin x] = 0 + 0 = 0 (At x = 0, [sin-1 x] = [sin-1 0] = 0 and [sin 0] = [0] = 0)
And [sin-1 x] + [sin x] = 1 + 0 = 1 (At x = 1, [sin-1 1] = [\(\frac{\pi}{2}\)] = 1 and [sin 1] = 0)
Therefore, [sin-1 x] + [sin x] \(\in\) {-3, -2, 0, 1}
⇒ f(x) \(\in\) {-3, -2, 0, 1}
Hence, the range of f(x) is {-3, -2, 0, 1}.