Which of the following function is decreasing of `(0,(pi)/(2))`

Options

A. sin 2 x
B. tan x
C. cos x
D. cos 3 x

Answer

To determine which of the given functions is decreasing on the interval \( (0, \frac{\pi}{2}) \), we will analyze each function by finding its derivative and checking where the derivative is negative.

### Step-by-step Solution:

1. **Function 1: \( f(x) = \sin(2x) \)**

- Find the derivative:
\[ f'(x) = \frac{d}{dx}(\sin(2x)) = 2\cos(2x) \]
- Set the derivative less than zero to find where it is decreasing:
\[ 2\cos(2x) < 0 \implies \cos(2x) < 0 \]
- The cosine function is negative in the intervals:
\[ \frac{\pi}{2} < 2x < \frac{3\pi}{2} \implies \frac{\pi}{4} < x < \frac{3\pi}{4} \]
- Since \( \frac{\pi}{4} \) to \( \frac{3\pi}{4} \) is not within \( (0, \frac{\pi}{2}) \), this function is **not decreasing** in the interval.

2. **Function 2: \( f(x) = \tan(x) \)**

- Find the derivative:
\[ f'(x) = \sec^2(x) \]
- Since \( \sec^2(x) \) is always positive for \( x \) in \( (0, \frac{\pi}{2}) \), this function is **not decreasing** in the interval.

3. **Function 3: \( f(x) = \cos(x) \)**

- Find the derivative:
\[ f'(x) = -\sin(x) \]
- Set the derivative less than zero:
\[ -\sin(x) < 0 \implies \sin(x) > 0 \]
- The sine function is positive in the interval:
\[ 0 < x < \pi \]
- Since \( (0, \frac{\pi}{2}) \) is a subset of \( (0, \pi) \), this function is **decreasing** in the interval.

4. **Function 4: \( f(x) = \cos(3x) \)**

- Find the derivative:
\[ f'(x) = -3\sin(3x) \]
- Set the derivative less than zero:
\[ -3\sin(3x) < 0 \implies \sin(3x) > 0 \]
- The sine function is positive in the intervals:
\[ 0 < 3x < \pi \implies 0 < x < \frac{\pi}{3} \]
- Since \( \frac{\pi}{3} < \frac{\pi}{2} \), this function is **not decreasing** in the entire interval \( (0, \frac{\pi}{2}) \).

### Conclusion:
The only function that is decreasing on the interval \( (0, \frac{\pi}{2}) \) is \( f(x) = \cos(x) \).