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.