Given that:\[f(x) = \sqrt 1 - 2x + 2{\sin ^{ - 1}}\left( {\dfrac{{3x - 1}}{2}} \right)\]is a real function. Find its domain values.
(a) $ \left( { - \dfrac{1}{3},\dfrac{1}{2}} \right) $
(b) $ \left( {\dfrac{1}{2}, - \dfrac{1}{3}} \right) $
(c) Cannot be determine
(d) None of these

Hint: The given problem revolves around the concepts of trigonometry. Any trigonometric functions exist within the boundary limits as (especially for $ \sin $ term) $ - \dfrac{\pi }{2} \leqslant {\sin ^{ - 1}}\theta \leqslant \dfrac{\pi }{2} $ . Then simplifying and then multiplying $ 3 $ further in desired calculations to obtain the solution.

Complete step-by-step answer:
Since, we have given the function ,
\[f(x) = \sqrt 1 - 2x + 2{\sin ^{ - 1}}\left( {\dfrac{{3x - 1}}{2}} \right)\]
The equation can also revolved as mathematically, we get
\[f(x) = \left( {\sqrt 1 - 2x} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{{3x - 1}}{2}} \right)\]
Since, we have given that, the condition exists the real values, we can predict that
 $ \Rightarrow \sqrt 1 - 2x \geqslant 0 $
The equation becomes,
 $ \Rightarrow 1 \geqslant 2x $
Transferring two on left hand side of the above equation, we get
 $ \Rightarrow \dfrac{1}{2} \geqslant x $ … (i)
As a result, of the function ‘ $ {\sin ^{ - 1}}x $ ’ we know that the values exists between $ - \dfrac{\pi }{6} $ and $ \dfrac{\pi }{6} $ respectively
(Where,\[\sin {30^ \circ } = \sin \dfrac{\pi }{6} = \dfrac{1}{2}\])
Hence, the inverse function exists that is $ {\sin ^{ - 1}} $ , so the equation becomes
 $ - 1 \leqslant \left( {\dfrac{{3x - 1}}{2}} \right) \leqslant 1 $
[Where, $ x = \dfrac{1}{2} $ satisfies the above equation]
Multiplying the equation $ - 1 \leqslant \left( {\dfrac{{3x - 1}}{2}} \right) \leqslant 1 $ by $ \sin $ , the equation becomes
\[ - 2 \leqslant \left( {3x - 1} \right) \leqslant 2\]
Adding $ 1 $ predominantly, we get
\[ - 1 \leqslant 3x \leqslant 3\]
Dividing the equation by $ 3 $ , we get
 \[ - \dfrac{1}{3} \leqslant x \leqslant 1\]
But, from (i) that is $ \dfrac{1}{2} \geqslant x $
\[ - \dfrac{1}{3} \leqslant \dfrac{x}{2} \leqslant 1\]
Hence, $ x $ tends to,
 $ x \in \left( { - \infty ,\dfrac{1}{2}} \right) $
Where, infinity $ \infty = \dfrac{1}{3} $
 $ \therefore x \equiv \left( { - \dfrac{1}{3},\dfrac{1}{2}} \right) $
 $ \Rightarrow \therefore $ Hence, the option (a) is correct!
So, the correct answer is “Option a”.

Note: One should know the condition of trigonometric terms such as ‘sine’, ‘cosine’, etc. while solving a question (here, we used the condition of ‘sine’ term). We should also know the basic concepts of simplification of the problem studied in earlier classes. As a result, to get the accurate answer we must take care of the calculations so as to be sure of our final answer.