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Question: Domain of f(x) = sin<sup>−1</sup>\(\left( \frac{\lbrack x\rbrack}{\{ x\}} \right)\), where [.] and {...

Domain of f(x) = sin−1([x]{x})\left( \frac{\lbrack x\rbrack}{\{ x\}} \right), where [.] and {.} denote the greatest integer function and fractional part respectively, is

A

(0, 1)

B

(-1, 1) ~ {0}

C

(-2, 2) ~ {-1, 0, 1}

D

None of these

Answer

(0, 1)

Explanation

Solution

We must have, −1 ≤ [x]{x}\frac { [ x ] } { \{ x \} } ≤ 1

From −1 ≤ [x]{x}\frac { [ x ] } { \{ x \} }, we get {x}+[x]{x}\frac { \{ x \} + [ x ] } { \{ x \} } ≥ 0

x{x}\frac { x } { \{ x \} } > 0 ⇒ x ∈ (0, ∞) ~ I+ .

From, [x]{x}\frac { [ x ] } { \{ x \} } ≤ 1 we get, [x]{x}{x}\frac { [ x ] - \{ x \} } { \{ x \} }≤ 0

⇒ [x] ≤ {0} ⇒ x ∈ (-∞, 0)∪(0, 1)~I+

Thus domain is (0, 1).