For ` x , t in R ` let
` p_(t) (x)= ( sin t) x^(2) - (2 cost ) x + sin t` be a family of quadratic polynomials in x with variable coefficients. Let `A(t) = int_(0)^(1) p_(t) ( x)dx ` , Which of the following statements are true?
(I) `A(t) lt 0 `for all t.
(II) A(t) has infinitely many critical points.
(III) A(t) = 0 for infinitely many t.
(IV) `A'(t) lt 0` for all t.

Options

A. ) (I) and (II) only
B. (II) and (III) only
C. (III) and (IV) only
D. (IV) and (I) only

Answer

To solve the problem, we need to analyze the function \( A(t) \) defined by the integral of the quadratic polynomial \( p_t(x) \) over the interval from 0 to 1. The polynomial is given as:

\[ p_t(x) = \sin(t) x^2 - 2 \cos(t) x + \sin(t) \]

We need to compute \( A(t) = \int_0^1 p_t(x) \, dx \).

### Step 1: Compute the integral

First, we will compute the integral \( A(t) \):

\[ A(t) = \int_0^1 \left( \sin(t) x^2 - 2 \cos(t) x + \sin(t) \right) \, dx \]

We can split this integral into three parts:

\[ A(t) = \int_0^1 \sin(t) x^2 \, dx - 2 \cos(t) \int_0^1 x \, dx + \int_0^1 \sin(t) \, dx \]

### Step 2: Evaluate each integral

1. **First integral**:

\[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \]

Thus,

\[ \int_0^1 \sin(t) x^2 \, dx = \sin(t) \cdot \frac{1}{3} = \frac{\sin(t)}{3} \]

2. **Second integral**:

\[ \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \]

Thus,

\[ -2 \cos(t) \int_0^1 x \, dx = -2 \cos(t) \cdot \frac{1}{2} = -\cos(t) \]

3. **Third integral**:

\[ \int_0^1 \sin(t) \, dx = \sin(t) \cdot 1 = \sin(t) \]

### Step 3: Combine the results

Now, we can combine all parts to find \( A(t) \):

\[ A(t) = \frac{\sin(t)}{3} - \cos(t) + \sin(t) = \frac{4\sin(t)}{3} - \cos(t) \]

### Step 4: Analyze the statements

Now we analyze the statements based on \( A(t) = \frac{4\sin(t)}{3} - \cos(t) \):

1. **Statement I**: \( A(t) < 0 \) for all \( t \).

This is not necessarily true since \( \sin(t) \) and \( \cos(t) \) can take values that make \( A(t) \) positive.

2. **Statement II**: \( A(t) \) has infinitely many critical points.

To find critical points, we take the derivative \( A'(t) \):

\[ A'(t) = \frac{4}{3} \cos(t) + \sin(t) \]

Setting \( A'(t) = 0 \) gives us a trigonometric equation which has infinitely many solutions.

3. **Statement III**: \( A(t) = 0 \) for infinitely many \( t \).

The equation \( \frac{4\sin(t)}{3} - \cos(t) = 0 \) can be solved for \( t \) and will yield infinitely many solutions.

4. **Statement IV**: \( A'(t) < 0 \) for all \( t \).

This is not true as \( A'(t) \) can be positive or negative depending on the values of \( \sin(t) \) and \( \cos(t) \).

### Conclusion

The true statements are:

- **Statement II**: \( A(t) \) has infinitely many critical points.
- **Statement III**: \( A(t) = 0 \) for infinitely many \( t \).