Let `f(x)=2x-tan^-1 x- ln(x+sqrt(1+x^2)); x in R,` then

Options

A. strictly increases `AA x in R`
B. strictly increases only in `(0,oo)`
C. strictly decreases `AA x in R`
D. strictly decreases in `(o,oo)` and strictly increases I `(-oo, 0)`

Answer

To solve the problem, we need to analyze the function \( f(x) = 2x - \tan^{-1}(x) - \ln(x + \sqrt{1 + x^2}) \) and determine whether it is an increasing or decreasing function. We will do this by finding the derivative \( f'(x) \) and analyzing its sign.

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

We start by differentiating \( f(x) \):

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

Calculating each derivative:

1. The derivative of \( 2x \) is \( 2 \).
2. The derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1 + x^2} \).
3. For \( \ln(x + \sqrt{1 + x^2}) \), we use the chain rule. Let \( u = x + \sqrt{1 + x^2} \), then:

\[ \frac{d}{dx}(\ln(u)) = \frac{1}{u} \cdot \frac{du}{dx} \]

Now, we find \( \frac{du}{dx} \):

\[ \frac{du}{dx} = 1 + \frac{x}{\sqrt{1 + x^2}} \]

Therefore,

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

### Step 2: Combine the derivatives

Now, we can write \( f'(x) \):

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

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

We can simplify \( f'(x) \):

1. The term \( \frac{1 + \frac{x}{\sqrt{1 + x^2}}}{x + \sqrt{1 + x^2}} \) can be simplified further.
2. We can find a common denominator and combine the terms.

After simplification, we have:

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

### Step 4: Analyze the sign of \( f'(x) \)

To determine if \( f'(x) \) is always positive, we need to analyze the terms:

- The term \( 1 + x^2 \) is always positive for all \( x \in \mathbb{R} \).
- The term \( \sqrt{1 + x^2} \) is also always positive.
- Therefore, both \( \frac{1}{1 + x^2} \) and \( \frac{1 + x/\sqrt{1 + x^2}}{x + \sqrt{1 + x^2}} \) are positive.

Since \( 2 \) is greater than both of these terms, we conclude:

\[ f'(x) > 0 \quad \text{for all } x \in \mathbb{R} \]

### Conclusion

Since \( f'(x) > 0 \) for all \( x \), the function \( f(x) \) is an increasing function for all \( x \in \mathbb{R} \).

### Final Answer

The function \( f(x) \) is strictly increasing for \( x \in \mathbb{R} \).

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