Let ` ƒ(x) = sinx + (x^3 – 3x^2 + 4x – 2)cosx` for `x` in `(0, 1)`. Consider the following statements
I. ƒ has a zero in (0, 1)
II. ƒ is monotone in (0, 1)
Then

Options

A. I and II are true
B. I is true and II are false
C. I is false and II are true
D. I and II are false

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin x + (x^3 - 3x^2 + 4x - 2) \cos x \) for \( x \) in the interval \( (0, 1) \) and determine the validity of the two statements provided.

### Step 1: Evaluate \( f(0) \) and \( f(1) \)

First, we will calculate \( f(0) \) and \( f(1) \):

1. **Calculate \( f(0) \)**:
\[ f(0) = \sin(0) + (0^3 - 3 \cdot 0^2 + 4 \cdot 0 - 2) \cos(0) \]
\[ = 0 + (-2) \cdot 1 = -2 \]

2. **Calculate \( f(1) \)**:
\[ f(1) = \sin(1) + (1^3 - 3 \cdot 1^2 + 4 \cdot 1 - 2) \cos(1) \]
\[ = \sin(1) + (1 - 3 + 4 - 2) \cos(1) \]
\[ = \sin(1) + 0 \cdot \cos(1) = \sin(1) \]
Since \( \sin(1) > 0 \), we conclude that \( f(1) > 0 \).

### Step 2: Apply the Intermediate Value Theorem

Since \( f(0) < 0 \) and \( f(1) > 0 \), by the Intermediate Value Theorem, there exists at least one \( c \) in \( (0, 1) \) such that \( f(c) = 0 \).

Thus, **Statement I**: \( f \) has a zero in \( (0, 1) \) is **true**.

### Step 3: Check if \( f(x) \) is Monotonic

To determine if \( f(x) \) is monotonic in \( (0, 1) \), we need to find the derivative \( f'(x) \) and check its sign.

1. **Differentiate \( f(x) \)**:
\[ f'(x) = \cos x + \frac{d}{dx}[(x^3 - 3x^2 + 4x - 2) \cos x] \]
Using the product rule:
\[ f'(x) = \cos x + (3x^2 - 6x + 4) \cos x + (x^3 - 3x^2 + 4x - 2)(-\sin x) \]
\[ = \cos x (3x^2 - 6x + 4 + 1) - (x^3 - 3x^2 + 4x - 2) \sin x \]
\[ = \cos x (3x^2 - 6x + 5) - (x^3 - 3x^2 + 4x - 2) \sin x \]

2. **Analyze the sign of \( f'(x) \)**:
- For \( x \in (0, 1) \), \( \cos x > 0 \).
- The term \( 3x^2 - 6x + 5 \) is a quadratic function that opens upwards. Evaluating it at the endpoints:
- At \( x = 0 \): \( 3(0)^2 - 6(0) + 5 = 5 > 0 \)
- At \( x = 1 \): \( 3(1)^2 - 6(1) + 5 = 2 > 0 \)
- Therefore, \( 3x^2 - 6x + 5 > 0 \) for all \( x \in (0, 1) \).

Since \( \sin x > 0 \) for \( x \in (0, 1) \) and \( \cos x > 0 \), we can conclude that \( f'(x) > 0 \) for all \( x \in (0, 1) \).

Thus, **Statement II**: \( f \) is monotone in \( (0, 1) \) is also **true**.

### Final Conclusion

Both statements are true. Therefore, the answer is:

**Both statements are true.**