Consider the following statements :
1. The function `f(x)=x^(2)+2cosx` is increasing in the interval `(0,pi)`
2. The function `f(x)=1n(sqrt(1+x^(2)-x))` is decreasing in the interval `(-oo,oo)`
Which of the above statements is/ are correct ?

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Explanation

To determine the correctness of the given statements regarding the functions, we will analyze each statement step by step.

### Statement 1: The function \( f(x) = x^2 + 2\cos x \) is increasing in the interval \( (0, \pi) \).

**Step 1: Differentiate the function.**

To check if the function is increasing, we need to find its derivative:

\[ f'(x) = \frac{d}{dx}(x^2 + 2\cos x) = 2x - 2\sin x \]

**Hint:** The derivative of a function indicates whether it is increasing or decreasing. If \( f'(x) > 0 \), the function is increasing.

**Step 2: Analyze the derivative in the interval \( (0, \pi) \).**

We need to check if \( f'(x) > 0 \) for \( x \in (0, \pi) \):

\[ f'(x) = 2(x - \sin x) \]

**Step 3: Check the sign of \( x - \sin x \).**

We need to analyze \( x - \sin x \):

- At \( x = 0 \), \( x - \sin x = 0 - 0 = 0 \).
- At \( x = \pi \), \( x - \sin x = \pi - 0 = \pi > 0 \).

Since \( \sin x \) is less than \( x \) for \( x \in (0, \pi) \) (as \( \sin x \) is increasing and \( x \) is also increasing), we can conclude that \( x - \sin x > 0 \) for \( x \in (0, \pi) \).

Thus, \( f'(x) > 0 \) in the interval \( (0, \pi) \), which means \( f(x) \) is increasing in that interval.

**Conclusion for Statement 1:** True.

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### Statement 2: The function \( f(x) = \ln(\sqrt{1 + x^2 - x}) \) is decreasing in the interval \( (-\infty, \infty) \).

**Step 1: Differentiate the function.**

To check if this function is decreasing, we find its derivative:

\[ f'(x) = \frac{d}{dx} \left( \ln(\sqrt{1 + x^2 - x}) \right) = \frac{1}{\sqrt{1 + x^2 - x}} \cdot \frac{d}{dx}(\sqrt{1 + x^2 - x}) \]

Using the chain rule, we find:

\[ \frac{d}{dx}(\sqrt{1 + x^2 - x}) = \frac{1}{2\sqrt{1 + x^2 - x}} \cdot (2x - 1) \]

Thus,

\[ f'(x) = \frac{1}{\sqrt{1 + x^2 - x}} \cdot \frac{(2x - 1)}{2\sqrt{1 + x^2 - x}} = \frac{2x - 1}{2(1 + x^2 - x)} \]

**Hint:** The sign of the derivative will tell us if the function is increasing or decreasing.

**Step 2: Analyze the sign of \( f'(x) \).**

We need to determine when \( f'(x) < 0 \):

\[ 2x - 1 < 0 \implies x < \frac{1}{2} \]

For \( x < \frac{1}{2} \), \( f'(x) < 0 \), indicating that the function is decreasing in that interval.

**Step 3: Check the behavior for \( x \geq \frac{1}{2} \).**

For \( x \geq \frac{1}{2} \), \( f'(x) \) can be positive or zero, indicating that the function is not necessarily decreasing for all \( x \).

**Conclusion for Statement 2:** False.

### Final Conclusion

- Statement 1 is **True**.
- Statement 2 is **False**.

Thus, the correct answer is that only Statement 1 is correct.