Solveeit Logo
Login

Question

Question: What is the domain of \({{\sin }^{-1}}2x\) ? (a) \(\left[ \dfrac{-1}{2},\dfrac{1}{2} \right]\) ...

What is the domain of sin⁡−12x{{\sin }^{-1}}2x ?
(a) [−12,12]\left[ \dfrac{-1}{2},\dfrac{1}{2} \right]
(b) [−2,1]\left[ -2,1 \right]
(c) R
(d) [−1,1]\left[ -1,1 \right]

Explanation

Solution

Hint:To calculate the domain of inverse function of sin we have to find the range of sin because the range of any function is equal to the domain of its inverse function. And with that much information we can solve this question.

Complete step-by-step answer:
As we have discussed in the hint we will first write the function from its inverse.
Let sin⁡−12x{{\sin }^{-1}}2x= y.
Therefore, 2x=sin⁡y2x=\sin y
⇒x=sin⁡y2\Rightarrow x=\dfrac{\sin y}{2}
Now we have to use this fact that the range of sin y is [−1,1]\left[ -1,1 \right].
With this fact we can find the range of x by putting the minimum and maximum values of sin y.
Now putting sin⁡y=−1\sin y=-1 we get,
x=−12x=\dfrac{-1}{2}
And by putting sin⁡y=1\sin y=1 we get,
x=12x=\dfrac{1}{2}
So, now we can conclude that the maximum and minimum values of x are 12 and −12\dfrac{1}{2}\text{ and }\dfrac{-1}{2} .
Now the domain of sin⁡−12x{{\sin }^{-1}}2x is [−12,12]\left[ \dfrac{-1}{2},\dfrac{1}{2} \right].
Hence, option (a) is the correct answer.

Note: There is an alternate method to solve this question but it requires a good understanding of graphs and how to draw graph of inverse functions. The alternate method is as follows, first we have to draw the graph of sin⁡−12x{{\sin }^{-1}}2x. And as we know that the range in which it is invertible is [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right] .
And now using this much information we can put those two maximum and minimum values in place of y in the function y = sin⁡−12x{{\sin }^{-1}}2x and then we can find the values of x or the domain of function.