Question

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

Answer 1

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.