Question

Let the functions \( f \) and \( g \) be \[ f : [0, \frac{\pi}{2}] \to \mathbb{R} \text{ given by } f(x) = \sin x \text{ and } g(x) = \cos x, \text{ where } R \text{ is the set of real numbers}. \] Consider the following statements: Statement (I): \( f \) and \( g \) are one-to-one. Statement (II): \( f + g \) is one-to-one. Which of the following is correct?

• A. Statement (I) is false, statement (II) is true.
• B. Both statements (I) and (II) are true.
• C. Both statements (I) and (II) are false.
• D. Statement (I) is true, statement (II) is false.

Hint:

When checking if a function is one-to-one, verify if it is either strictly increasing or decreasing on the given interval.

Answer 1

The Correct Option is A

Let's analyze the statements: - \( f(x) = \sin x \) is one-to-one on \( [0, \frac{\pi}{2}] \), because it is strictly increasing. - \( g(x) = \cos x \) is not one-to-one on \( [0, \frac{\pi}{2}] \), because it is strictly decreasing. - Therefore, statement (I) is false because \( g \) is not one-to-one. Now for \( f+g \): \[ f+g = \sin x + \cos x \] This is a one-to-one function because it is strictly increasing in the interval \( [0, \frac{\pi}{2}] \), and its derivative \( \frac{d}{dx} (\sin x + \cos x) = \cos x - \sin x \) is positive in this interval. Thus, statement (II) is true.