To make a function like invertible, check if it is one-to-one. For trigonometric functions, like sine, you can use the property that the function is one-to-one in specific intervals where it does not repeat. In this case, the interval ensures that the sine function is one-to-one and thus invertible.
The correct answer is: (D) .
We are given the functions:
First, we compute the composite function :
To simplify this, let's first expand :
Using the Pythagorean identity , we get:
Thus, becomes:
Using the double angle identity for sine, , we get:
To make invertible, the function must be one-to-one. A sine function is one-to-one in the interval where it doesn't repeat. The sine function is one-to-one in the interval , since will cover only one period of the sine wave in this interval. Therefore, is invertible in the domain . Thus, the correct answer is (D) .
List - I | List - II | ||
(P) | If a = 0, b = 1, c = 0 and d = 0, then | (1) | h is one-one. |
(Q) | If a = 1, b = 0, c = 0 and d = 0, then | (2) | h is onto. |
(R) | If a = 0, b = 0, c = 1 and d = 0, then | (3) | h is differentiable on |
(S) | If a = 0, b = 0, c = 0 and d = 1, then | (4) | the range of h is [0, 1]. |
(5) | the range of h is {0, 1}. |
Let , . Then the domain of the function is equal to: