Let `f(x) =cos (cos x)`. Then which one is not correct?

Options

A. f (x) is increasing in `(-pi/2, pi/2)`
B. f (x) is decreasing in `(-pi/2,0)`
C. f (x) is increasing in `(0,pi/2)`
D. f (x) is decreasing in `((pi)/(2),pi)`

Answer

To solve the problem, we need to analyze the function \( f(x) = \cos(\cos x) \) and determine which statement about its behavior is not correct. We will differentiate the function and examine its intervals of increase and decrease.

### Step 1: Differentiate the function
We start by differentiating \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\cos(\cos x)) \]
Using the chain rule, we have:
\[ f'(x) = -\sin(\cos x) \cdot \frac{d}{dx}(\cos x) = -\sin(\cos x)(-\sin x) = \sin(\cos x) \sin x \]

### Step 2: Analyze the sign of \( f'(x) \)
Next, we need to analyze the sign of \( f'(x) = \sin(\cos x) \sin x \).

1. **Interval \( (0, \frac{\pi}{2}) \)**:
- In this interval, \( \cos x \) ranges from \( 1 \) to \( 0 \), so \( \sin(\cos x) \) is positive.
- \( \sin x \) is also positive.
- Therefore, \( f'(x) > 0 \) implies that \( f(x) \) is increasing.

2. **Interval \( (\frac{\pi}{2}, \pi) \)**:
- In this interval, \( \cos x \) ranges from \( 0 \) to \( -1 \), so \( \sin(\cos x) \) is still positive (since \( \sin(0) = 0 \) and \( \sin(-1) \) is negative).
- However, \( \sin x \) is negative.
- Therefore, \( f'(x) < 0 \) implies that \( f(x) \) is decreasing.

3. **Interval \( (-\frac{\pi}{2}, 0) \)**:
- In this interval, \( \cos x \) ranges from \( 0 \) to \( 1 \), so \( \sin(\cos x) \) is positive.
- \( \sin x \) is negative.
- Therefore, \( f'(x) < 0 \) implies that \( f(x) \) is decreasing.

### Step 3: Conclusion
From our analysis, we conclude:
- \( f(x) \) is increasing on \( (0, \frac{\pi}{2}) \).
- \( f(x) \) is decreasing on \( (\frac{\pi}{2}, \pi) \).
- \( f(x) \) is also decreasing on \( (-\frac{\pi}{2}, 0) \).

Now we can check the options provided in the question to identify which one is incorrect.