Domain of the function sin−1⁡[2x−3] where [ · ] denotes the greatest integer function is

We can have [2x-3]=-1 or 0 or 1⇒x=1,32 or 2 - (1)[2x−3]=−1⇒−1<2x−3<02<2x<31<x<32 -(2)[2x−3]=0⇒0<2x−3<13<2x<432<x<2 -(3)[2x−3]=+1⇒1<2x−3<24<2x<52<x<52 -(4)combining (1), (2), (3) and (4), domain of the function is1,52

Domain of the function sin−1⁡[2x−3] where [ · ] denotes the greatest integer function is

We can have [2x-3]=-1 or 0 or 1⇒x=1,32 or 2 - (1)[2x−3]=−1⇒−1<2x−3<02<2x<31<x<32 -(2)[2x−3]=0⇒0<2x−3<13<2x<432<x<2 -(3)[2x−3]=+1⇒1<2x−3<24<2x<52<x<52 -(4)combining (1), (2), (3) and (4), domain of the function is1,52

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Domain of the function $\sin^{- 1} [2 x - 3]$ where [ $\cdot$ ] denotes the greatest integer function is

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[-1,1]

$[1, \frac{3}{2})$

$[1, \frac{5}{2})$

$[\frac{- 1}{2}, 1)$

answer is C.

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Detailed Solution

We can have [2x-3]=-1 or 0 or 1

$\Rightarrow x = 1, \frac{3}{2}$ or 2 - (1)

$\begin{array}{l} [2 x - 3] = - 1 \Rightarrow - 1 < 2 x - 3 < 0 \\ 2 < 2 x < 3 \\ 1 < x < \frac{3}{2} - (2) \end{array}$

$\begin{array}{l} [2 x - 3] = 0 \Rightarrow 0 < 2 x - 3 < 1 \\ 3 < 2 x < 4 \\ \frac{3}{2} < x < 2 - (3) \\ [2 x - 3] = + 1 \Rightarrow 1 < 2 x - 3 < 2 \\ 4 < 2 x < 5 \\ 2 < x < \frac{5}{2} - (4) \end{array}$