If `f(x)=abs(2x^2+3x-2) +sinx cosx , x in [0,1]`, then the sum of absolute maxima and absolute minima is

Options

A. `3+1/2(sin 2+sin1)`
B. `3-1/2(sin 2+sin1)`
C. `3-1/2(sin 2-sin1)`
D. `3+1/2(sin 2-sin1)`

Answer

To solve the problem, we need to analyze the function \( f(x) = |2x^2 + 3x - 2| + \sin x \cos x \) over the interval \( [0, 1] \).

### Step 1: Determine the critical points of the polynomial inside the absolute value.
First, we need to find the roots of the quadratic equation \( 2x^2 + 3x - 2 = 0 \).

Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = 3, c = -2 \):
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4} \]
Calculating the roots:
\[ x_1 = \frac{2}{4} = \frac{1}{2}, \quad x_2 = \frac{-8}{4} = -2 \]
Only \( x_1 = \frac{1}{2} \) lies within the interval \( [0, 1] \).

### Step 2: Evaluate \( f(x) \) at the critical point and endpoints.
Now we need to evaluate \( f(x) \) at the endpoints \( x = 0 \) and \( x = 1 \), and at the critical point \( x = \frac{1}{2} \).

1. **At \( x = 0 \)**:
\[ f(0) = |2(0)^2 + 3(0) - 2| + \sin(0) \cos(0) = | -2 | + 0 = 2 \]

2. **At \( x = 1 \)**:
\[ f(1) = |2(1)^2 + 3(1) - 2| + \sin(1) \cos(1) = |2 + 3 - 2| + \sin(1) \cos(1) = |3| + \frac{1}{2} \sin(2) = 3 + \frac{1}{2} \sin(2) \]

3. **At \( x = \frac{1}{2} \)**:
\[ f\left(\frac{1}{2}\right) = |2\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2}\right) - 2| + \sin\left(\frac{1}{2}\right) \cos\left(\frac{1}{2}\right) \]
\[ = |2 \cdot \frac{1}{4} + \frac{3}{2} - 2| + \sin\left(\frac{1}{2}\right) \cos\left(\frac{1}{2}\right) \]
\[ = | \frac{1}{2} + \frac{3}{2} - 2 | + \frac{1}{2} \sin(1) = |0| + \frac{1}{2} \sin(1) = \frac{1}{2} \sin(1) \]

### Step 3: Compare the values.
Now we have:
- \( f(0) = 2 \)
- \( f(1) = 3 + \frac{1}{2} \sin(2) \)
- \( f\left(\frac{1}{2}\right) = \frac{1}{2} \sin(1) \)

To find the absolute maximum and minimum, we need to compare these values.

### Step 4: Determine the maximum and minimum values.
1. **Minimum**: Compare \( 2 \) and \( \frac{1}{2} \sin(1) \). Since \( \sin(1) \) is approximately \( 0.8415 \), we have:
\[ \frac{1}{2} \sin(1) \approx 0.42075 < 2 \]
Thus, the minimum value is \( \frac{1}{2} \sin(1) \).

2. **Maximum**: Compare \( 3 + \frac{1}{2} \sin(2) \) with \( 2 \). Since \( \sin(2) \) is approximately \( 0.9093 \):
\[ 3 + \frac{1}{2} \sin(2) \approx 3 + 0.45465 = 3.45465 > 2 \]
Thus, the maximum value is \( 3 + \frac{1}{2} \sin(2) \).

### Step 5: Calculate the sum of absolute maxima and minima.
Now we can calculate the sum:
\[ \text{Sum} = \text{Maximum} + \text{Minimum} = \left(3 + \frac{1}{2} \sin(2)\right) + \left(\frac{1}{2} \sin(1)\right) \]

### Final Answer:
The sum of the absolute maxima and absolute minima is:
\[ 3 + \frac{1}{2} \sin(2) + \frac{1}{2} \sin(1) \]