Question

The domain of Inverse trigonometric function \[{{\sin }^{-1}}\dfrac{2x+1}{3}\] is:
A. (-2, 1)
B. [-2, 1]
C. R
D. [-1. 1]

Answer 1

Hint:For the above question we will have to know about the domain of any function. The domain of a function is the complete set of possible values of the independent variables. We know that the domain of \[{{\sin }^{-1}}x\] is \[-1\le x\le 1\] so we will solve the inequality and thus find the condition on x so that it satisfies the condition of domain of \[{{\sin }^{-1}}x\].

Complete step-by-step answer:
We have been given \[{{\sin }^{-1}}\dfrac{2x+1}{3}\] as a function.
We know that the domain of \[{{\sin }^{-1}}x\] is \[-1\le x\le 1\].
\[\Rightarrow -1\le \dfrac{2x+1}{3}\le 1\]
On multiplying 3 to the above equation, we get as following:
\[-1\times 3\le 3\times \left( \dfrac{2x+1}{3} \right)\le 1\times 3\]
On simplification, we get,
\[-3\le 2x+1\le 3\]
On subtracting 1 from the above equation, we get as following:
\[-3-1\le 2x+1-1\le 3-1\]
On simplification, we get the above equation as follows:
\[-4\le 2x\le 2\]
On dividing the equation by 2, we get the equations as follows:
\[\dfrac{-4}{2}\le \dfrac{2x}{2}\le \dfrac{2}{2}\]
On simplification, we get the above equation as follows:
\[-2\le x\le 1\]
Hence, \[x\in \left[ -2,1 \right]\].
Therefore, the correct option of the above question is option B.

Note: Just be careful while solving the inequality as there is a chance that you might make silly mistakes and you will get an incorrect answer.
Also, we can simplify the inequality separately as \[-1\le \dfrac{2x+1}{3}\] and \[\dfrac{2x+1}{3}\le 1\] and then after simplification we will merge the result to get the answer.
Also, be careful while choosing the options as option A and option B seem to be similar but the equality holds at the extremes so option B is the correct answer.