Question

The function $f(\theta)=\sin \theta+\cos \theta,\ 0\leq \theta \leq 2\pi$ is decreasing in the interval:

• A. \(0\leq \theta < \pi\)
• B. \(0\leq \theta \leq \frac{\pi}{4}\)
• C. \(\frac{\pi}{4}\leq \theta \leq \frac{5\pi}{4}\)
• D. \(\frac{5\pi}{4}\leq \theta \leq 2\pi\)
• E. \(\frac{\pi}{2}\leq \theta \leq \frac{3\pi}{2}\)

Hint:

To find where a trigonometric function is increasing or decreasing, always differentiate first and then solve the sign of the derivative carefully over the given interval.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A function \( f(x) \) is decreasing in an interval where its first derivative, \( f'(x) \), is less than or equal to zero (\( f'(x) \le 0 \)). We need to find the derivative of the given function and then determine the interval(s) in \( [0, 2\pi] \) where it is non-positive. Note: The question likely has a typo and should be \( f(x) = \sin x + \cos x \) instead of using \( \theta \), as the function is defined in terms of \( x \).
Step 2: Key Formula or Approach:
1. Find the first derivative, \( f'(x) \).
2. Solve the inequality \( f'(x) \le 0 \) for \( x \) in the domain \( [0, 2\pi] \).
Step 3: Detailed Explanation:
Given the function \( f(x) = \sin x + \cos x \).
First, we compute the derivative with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x \] For the function to be decreasing, we must have \( f'(x) \le 0 \).
\[ \cos x - \sin x \le 0 \] \[ \cos x \le \sin x \] To solve this inequality in the interval \( [0, 2\pi] \), we first find where \( \cos x = \sin x \). This occurs when \( \tan x = 1 \). In the interval \( [0, 2\pi] \), the solutions are:
\[ x = \frac{\pi}{4} \quad \text{and} \quad x = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \] These points divide the interval \( [0, 2\pi] \) into three sub-intervals: \( [0, \frac{\pi}{4}] \), \( [\frac{\pi}{4}, \frac{5\pi}{4}] \), and \( [\frac{5\pi}{4}, 2\pi] \). We test a point from each interval to check the sign of \( f'(x) = \cos x - \sin x \).

Interval \( [0, \frac{\pi}{4}] \): Let's test \( x=0 \). \( f'(0) = \cos 0 - \sin 0 = 1 - 0 = 1>0 \). The function is increasing.
Interval \( [\frac{\pi}{4}, \frac{5\pi}{4}] \): Let's test \( x = \frac{\pi}{2} \). \( f'(\frac{\pi}{2}) = \cos \frac{\pi}{2} - \sin \frac{\pi}{2} = 0 - 1 = -1<0 \). The function is decreasing.
Interval \( [\frac{5\pi}{4}, 2\pi] \): Let's test \( x = \frac{3\pi}{2} \). \( f'(\frac{3\pi}{2}) = \cos \frac{3\pi}{2} - \sin \frac{3\pi}{2} = 0 - (-1) = 1>0 \). The function is increasing.
The function is decreasing in the interval where \( f'(x) \le 0 \), which is \( [\frac{\pi}{4}, \frac{5\pi}{4}] \).
Step 4: Final Answer:
The function is decreasing in the interval \( \frac{\pi}{4} \le x \le \frac{5\pi}{4} \). This corresponds to option (C).