`f(x) = (4x^3-3x^2)/6 -2Sinx +(2x-1)Cosx` , then `f(x)` is

Options

A. Increases in `(1/2 , oo)`
B. Decreases in `(1/2 , oo)`
C. Increases in `(-oo ,1/2)`
D. Decreases in `(-oo ,1/2)`

Answer

To determine the behavior of the function \( f(x) = \frac{4x^3 - 3x^2}{6} - 2\sin x + (2x - 1)\cos x \), we need to find its derivative \( f'(x) \) and analyze the intervals of increase and decrease.

### Step 1: Differentiate the function
We start by differentiating \( f(x) \):

\[ f'(x) = \frac{d}{dx}\left(\frac{4x^3 - 3x^2}{6}\right) - \frac{d}{dx}(2\sin x) + \frac{d}{dx}((2x - 1)\cos x) \]

Calculating each term separately:
- The derivative of \( \frac{4x^3 - 3x^2}{6} \) is \( \frac{1}{6}(12x^2 - 6x) = 2x^2 - x \).
- The derivative of \( -2\sin x \) is \( -2\cos x \).
- For \( (2x - 1)\cos x \), we use the product rule:
\[ \frac{d}{dx}((2x - 1)\cos x) = (2x - 1)(-\sin x) + 2\cos x = -2x\sin x + \sin x + 2\cos x \]

Combining these results, we get:
\[ f'(x) = (2x^2 - x) - 2\cos x - 2x\sin x + \sin x + 2\cos x \]
Simplifying this gives:
\[ f'(x) = 2x^2 - x + \sin x - 2x\sin x \]
Rearranging terms:
\[ f'(x) = 2x^2 - x + \sin x(1 - 2x) \]

### Step 2: Set the derivative to zero
To find the critical points, we set \( f'(x) = 0 \):
\[ 2x^2 - x + \sin x(1 - 2x) = 0 \]

This equation is not straightforward to solve analytically, but we can identify two critical points:
1. From \( 2x - 1 = 0 \) which gives \( x = \frac{1}{2} \).
2. From \( x = \sin x \).

### Step 3: Analyze the critical points
To analyze the intervals of increase and decrease, we need to evaluate the sign of \( f'(x) \) around the critical points.

1. For \( x < \frac{1}{2} \):
- Choose a test point, say \( x = 0 \):
\[ f'(0) = 2(0)^2 - 0 + \sin(0)(1 - 2(0)) = 0 \]
This does not help, so we check another point, say \( x = 0.1 \):
\[ f'(0.1) > 0 \quad (\text{since } \sin(0.1) \text{ is positive and } 2(0.1)^2 - 0.1 \text{ is also positive}) \]

2. For \( x = \frac{1}{2} \):
- Check \( f'(\frac{1}{2}) \):
\[ f'(\frac{1}{2}) = 2\left(\frac{1}{2}\right)^2 - \frac{1}{2} + \sin\left(\frac{1}{2}\right)(1 - 2\left(\frac{1}{2}\right)) = 0 \]

3. For \( x > \frac{1}{2} \):
- Choose a test point, say \( x = 1 \):
\[ f'(1) = 2(1)^2 - 1 + \sin(1)(1 - 2(1)) < 0 \quad (\text{since } \sin(1) \text{ is positive and } 1 - 2 < 0) \]

### Conclusion
From the analysis:
- \( f(x) \) is increasing on \( (-\infty, \frac{1}{2}) \)
- \( f(x) \) is decreasing on \( (\frac{1}{2}, \infty) \)

Thus, the final answer is that \( f(x) \) increases on \( (-\infty, \frac{1}{2}) \) and decreases on \( (\frac{1}{2}, \infty) \).