Which of the following functions are decreasing on `(0,\ pi//2)` ? (i) `cosx` (ii) `cos2x` (iii) `tanx` (iv) `cos3x`

Options

A. sin2x
B. tanx
C. cosx
D. cos3x

Explanation

To determine which of the given functions are decreasing on the interval \( (0, \frac{\pi}{2}) \), we need to analyze the derivative of each function. A function is decreasing on an interval if its derivative is less than zero throughout that interval.

### Step 1: Analyze \( f(x) = \cos x \)

1. **Differentiate**:
\[ f'(x) = -\sin x \]
2. **Evaluate on \( (0, \frac{\pi}{2}) \)**:
- In this interval, \( \sin x > 0 \).
- Therefore, \( f'(x) = -\sin x < 0 \).
3. **Conclusion**: \( \cos x \) is decreasing on \( (0, \frac{\pi}{2}) \).

### Step 2: Analyze \( f(x) = \cos 2x \)

1. **Differentiate**:
\[ f'(x) = -2\sin 2x \]
2. **Evaluate on \( (0, \frac{\pi}{2}) \)**:
- In this interval, \( 2x \) ranges from \( 0 \) to \( \pi \), where \( \sin 2x > 0 \).
- Therefore, \( f'(x) = -2\sin 2x < 0 \).
3. **Conclusion**: \( \cos 2x \) is decreasing on \( (0, \frac{\pi}{2}) \).

### Step 3: Analyze \( f(x) = \tan x \)

1. **Differentiate**:
\[ f'(x) = \sec^2 x \]
2. **Evaluate on \( (0, \frac{\pi}{2}) \)**:
- In this interval, \( \sec^2 x > 0 \).
- Therefore, \( f'(x) > 0 \).
3. **Conclusion**: \( \tan x \) is increasing on \( (0, \frac{\pi}{2}) \).

### Step 4: Analyze \( f(x) = \cos 3x \)

1. **Differentiate**:
\[ f'(x) = -3\sin 3x \]
2. **Evaluate on \( (0, \frac{\pi}{2}) \)**:
- In this interval, \( 3x \) ranges from \( 0 \) to \( \frac{3\pi}{2} \), where \( \sin 3x \) is positive for \( 0 < 3x < \pi \) and negative for \( \pi < 3x < \frac{3\pi}{2} \).
- Therefore, \( f'(x) \) may be positive or negative in this interval.
3. **Conclusion**: We cannot conclude that \( \cos 3x \) is strictly increasing or decreasing on \( (0, \frac{\pi}{2}) \).

### Final Conclusion

The functions that are decreasing on the interval \( (0, \frac{\pi}{2}) \) are:
- \( \cos x \)
- \( \cos 2x \)