If ` f(x) = 2x + cot^(-1) x + log ( sqrt(1+x^2 )-x)` then f(x)

Options

A. increases in ` [0,oo)`
B. decreases in `[0,-oo)`
C. neither increases nor decreases in `(0,oo)`
D. increases in `(-oo,oo)`

Answer

To solve the problem, we need to analyze the function \( f(x) = 2x + \cot^{-1}(x) + \log(\sqrt{1+x^2} - x) \) and determine its behavior (increasing or decreasing) by finding its derivative \( f'(x) \).

### Step 1: Differentiate \( f(x) \)

We start by differentiating each term in \( f(x) \):

1. **Differentiate \( 2x \)**:
\[ \frac{d}{dx}(2x) = 2 \]

2. **Differentiate \( \cot^{-1}(x) \)**:
The derivative of \( \cot^{-1}(x) \) is:
\[ \frac{d}{dx}(\cot^{-1}(x)) = -\frac{1}{1+x^2} \]

3. **Differentiate \( \log(\sqrt{1+x^2} - x) \)**:
Using the chain rule, we have:
\[ \frac{d}{dx}(\log(u)) = \frac{1}{u} \cdot \frac{du}{dx} \]
where \( u = \sqrt{1+x^2} - x \).

First, we find \( \frac{du}{dx} \):
\[ u = \sqrt{1+x^2} - x \]
\[ \frac{du}{dx} = \frac{1}{2\sqrt{1+x^2}} \cdot 2x - 1 = \frac{x}{\sqrt{1+x^2}} - 1 \]

Now, substituting \( u \) back into the derivative:
\[ \frac{d}{dx}(\log(\sqrt{1+x^2} - x)) = \frac{1}{\sqrt{1+x^2} - x} \left(\frac{x}{\sqrt{1+x^2}} - 1\right) \]

### Step 2: Combine the derivatives

Now we can combine all the derivatives to find \( f'(x) \):
\[ f'(x) = 2 - \frac{1}{1+x^2} + \frac{1}{\sqrt{1+x^2} - x} \left(\frac{x}{\sqrt{1+x^2}} - 1\right) \]

### Step 3: Analyze \( f'(x) \)

To determine when \( f'(x) \) is greater than or equal to zero, we need to analyze the expression:
\[ f'(x) = 2 - \frac{1}{1+x^2} + \frac{\frac{x}{\sqrt{1+x^2}} - 1}{\sqrt{1+x^2} - x} \]

1. **Behavior of \( 1+x^2 \)**:
- \( 1 + x^2 \) is always positive for all real \( x \).
- Thus, \( \frac{1}{1+x^2} \) is always positive and less than or equal to 1.

2. **Behavior of \( \sqrt{1+x^2} \)**:
- \( \sqrt{1+x^2} \geq 1 \) for all \( x \).

### Step 4: Conclusion about \( f'(x) \)

Since \( f'(x) \) consists of a constant (2) minus a positive term \( \frac{1}{1+x^2} \) and additional terms that are also non-negative, we can conclude that:
- \( f'(x) \geq 0 \) for all \( x \).

### Final Result

Thus, the function \( f(x) \) is increasing for all real numbers \( x \).