The function f, defined by ` f(x)=(x^(2))/(2) +In x - 2 cos x ` increases for ` x in `

Options

A. `R^(-)`
B. `R^(+)`
C. `R-{0}`
D. `[1,oo)`

Answer

To determine the intervals where the function \( f(x) = \frac{x^2}{2} + \ln x - 2 \cos x \) is increasing, we need to find the derivative of the function and analyze where this derivative is positive.

### Step 1: Find the derivative of the function.
The first step is to differentiate the function \( f(x) \).

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

Calculating each term:
- The derivative of \( \frac{x^2}{2} \) is \( x \).
- The derivative of \( \ln x \) is \( \frac{1}{x} \).
- The derivative of \( -2 \cos x \) is \( 2 \sin x \).

Thus, we have:

\[ f'(x) = x + \frac{1}{x} + 2 \sin x \]

### Step 2: Set the derivative greater than zero.
To find where the function is increasing, we set the derivative greater than zero:

\[ f'(x) > 0 \implies x + \frac{1}{x} + 2 \sin x > 0 \]

### Step 3: Analyze the inequality.
We can rearrange the inequality:

\[ 2 \sin x > -\left(x + \frac{1}{x}\right) \]

### Step 4: Understand the range of \( \sin x \).
The sine function \( \sin x \) has a range of \([-1, 1]\). Therefore, we can analyze the inequality:

\[ -1 < -\left(x + \frac{1}{x}\right) \]

This leads to:

\[ x + \frac{1}{x} < 1 \]

### Step 5: Solve the inequality.
We can multiply both sides by \( x \) (noting that \( x > 0 \)):

\[ x^2 - x + 1 < 0 \]

The quadratic \( x^2 - x + 1 \) does not have real roots (its discriminant \( (-1)^2 - 4(1)(1) = -3 < 0 \)), which means it is always positive. Therefore, we cannot find any \( x > 0 \) that satisfies this inequality.

### Step 6: Conclusion about the intervals of increase.
Since \( x + \frac{1}{x} + 2 \sin x \) is always greater than zero for \( x > 0 \), we conclude that the function \( f(x) \) is increasing for all \( x > 0 \).

Thus, the function \( f(x) \) increases for:

\[ x \in (0, \infty) \]