Question

What are the absolute maximum value and the absolute minimum value of a function \( f(x) = \sin x + \cos x \) in the interval \( [0, \pi] \)?

• A. \( \sqrt{2} \) and 1
• B. \( \sqrt{2} \) and -1
• C. 2 and 1
• D. \( \sqrt{2} \) and 0

Hint:

To find the absolute maximum and minimum of trigonometric functions, express the function in a form that involves a single trigonometric term and use the known maximum and minimum values of sine or cosine.

Answer 1

The Correct Option is B

Step 1: Analyze the function.
We are given the function \( f(x) = \sin x + \cos x \) in the interval \( [0, \pi] \). First, let’s express \( f(x) \) in a more convenient form. We know that: \[ f(x) = \sin x + \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right) \] Using the angle addition identity: \[ \sin x + \cos x = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) \] Step 2: Find the maximum and minimum values.
The maximum and minimum values of \( \sin \left( x + \frac{\pi}{4} \right) \) occur at \( \pm 1 \), so: \[ \text{Maximum value of } f(x) = \sqrt{2} \times 1 = \sqrt{2} \] \[ \text{Minimum value of } f(x) = \sqrt{2} \times (-1) = -\sqrt{2} \] Step 3: Determine the values within the interval.
In the interval \( [0, \pi] \), the function \( \sin \left( x + \frac{\pi}{4} \right) \) reaches its maximum at \( x = \frac{\pi}{4} \) and its minimum at \( x = \frac{5\pi}{4} \). However, since the interval is only \( [0, \pi] \), the minimum is \( -1 \) (since \( \sin \left( \frac{5\pi}{4} \right) = -1 \)). Thus, the maximum value is \( \sqrt{2} \) and the minimum value is \( -1 \). Final Answer: \[ \boxed{\sqrt{2} \text{ and } -1} \]