If -x- denotes the integral part of x- then domain of the function fx-3-x-x-1x-2x-3-sin -1-3x-2-2- is

If $[2 s i n x] + [c o s x] = - 3$ then the range of the function $f (x) = s i n x + \sqrt{3} c o s x i n [0, 2 π]$ is (where $[\cdot]$ denotes greatest integer function)

Q. The domain of the function

f (x) = \frac{\sqrt{5 - x}}{(x + 1) (x - 2) (x - 3)} + a {sin}^{- 1} [\frac{2 x - 5}{3}]; a \neq 0

and

[\cdot]

denotes greatest integer function, is:

The domain of the function $f (x) = \sqrt{\frac{(x + 1) (x - 3)}{x - 2}}$ is

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If [x] denotes the integral part of x, then domain of the function f(x)=√3−x(x−1)(x−2)(x−3)+sin−1[3x−22] is

The correct option is B [0,2)−{1}Conditions for f(x) to be existing :-−1≤3x−22<2−2≤3x−2<40≤3x<60≤x<2and x≠1,2,3 and 3−x>0⇒x<3So, xϵ[0,2]−{1}

If $[x]$ denotes the integral part of $x$ , then domain of the function
$f (x) = \frac{\sqrt{3 - x}}{(x - 1) (x - 2) (x - 3)} + {sin}^{- 1} [\frac{3 x - 2}{2}]$ is

The correct option is B $[0, 2) - {1}$
Conditions for $f (x)$ to be existing :-
$- 1 \leq \frac{3 x - 2}{2} < 2$
$- 2 \leq 3 x - 2 < 4$
$0 \leq 3 x < 6$
$0 \leq x < 2$
and $x \neq 1, 2, 3$ and $3 - x > 0 \Rightarrow x < 3$
So, $x ϵ [0, 2] - {1}$