The numbers A, B and C such that a function of the form `f(x) = Ax^(2) + Bx + C` satisfies the conditions `f'(1)= 8, f(2) + f''(2) = 33 and int_(0)^(1) f(x) dx=7//3`, are

Options

A. `A=1, B=-4, C=2`
B. `A=7, B=-6, C=3`
C. `A=8, B=-6, C=3`
D. none of these

Answer

To solve the problem, we need to find the values of A, B, and C for the function \( f(x) = Ax^2 + Bx + C \) given the conditions:

1. \( f'(1) = 8 \)
2. \( f(2) + f''(2) = 33 \)
3. \( \int_0^1 f(x) \, dx = \frac{7}{3} \)

### Step 1: Find \( f'(x) \) and use the first condition

The derivative of \( f(x) \) is:
\[ f'(x) = 2Ax + B \]

Using the first condition \( f'(1) = 8 \):
\[ f'(1) = 2A(1) + B = 2A + B = 8 \quad \text{(Equation 1)} \]

### Step 2: Find \( f''(x) \) and use the second condition

The second derivative of \( f(x) \) is:
\[ f''(x) = 2A \]

Now, we need to calculate \( f(2) \):
\[ f(2) = A(2^2) + B(2) + C = 4A + 2B + C \]

Using the second condition \( f(2) + f''(2) = 33 \):
\[ f(2) + f''(2) = (4A + 2B + C) + 2A = 6A + 2B + C = 33 \quad \text{(Equation 2)} \]

### Step 3: Evaluate the integral and use the third condition

Now we calculate the integral:
\[ \int_0^1 f(x) \, dx = \int_0^1 (Ax^2 + Bx + C) \, dx \]

Calculating the integral:
\[ \int_0^1 Ax^2 \, dx = \left[ \frac{A}{3} x^3 \right]_0^1 = \frac{A}{3} \]
\[ \int_0^1 Bx \, dx = \left[ \frac{B}{2} x^2 \right]_0^1 = \frac{B}{2} \]
\[ \int_0^1 C \, dx = [Cx]_0^1 = C \]

Thus,
\[ \int_0^1 f(x) \, dx = \frac{A}{3} + \frac{B}{2} + C \]

Setting this equal to \( \frac{7}{3} \):
\[ \frac{A}{3} + \frac{B}{2} + C = \frac{7}{3} \quad \text{(Equation 3)} \]

### Step 4: Solve the system of equations

Now we have three equations:
1. \( 2A + B = 8 \) (Equation 1)
2. \( 6A + 2B + C = 33 \) (Equation 2)
3. \( \frac{A}{3} + \frac{B}{2} + C = \frac{7}{3} \) (Equation 3)

#### From Equation 1:
We can express \( B \) in terms of \( A \):
\[ B = 8 - 2A \quad \text{(Substituting into other equations)} \]

#### Substitute \( B \) into Equation 2:
\[ 6A + 2(8 - 2A) + C = 33 \]
\[ 6A + 16 - 4A + C = 33 \]
\[ 2A + C = 17 \quad \text{(Equation 4)} \]

#### Substitute \( B \) into Equation 3:
\[ \frac{A}{3} + \frac{8 - 2A}{2} + C = \frac{7}{3} \]
\[ \frac{A}{3} + 4 - A + C = \frac{7}{3} \]
\[ \frac{A}{3} - A + C = \frac{7}{3} - 4 \]
\[ \frac{A}{3} - A + C = -\frac{5}{3} \]
\[ -\frac{2A}{3} + C = -\frac{5}{3} \quad \text{(Equation 5)} \]

### Step 5: Solve Equations 4 and 5

From Equation 4:
\[ C = 17 - 2A \]

Substituting \( C \) into Equation 5:
\[ -\frac{2A}{3} + (17 - 2A) = -\frac{5}{3} \]
\[ -\frac{2A}{3} + 17 - 2A = -\frac{5}{3} \]
Multiplying through by 3 to eliminate fractions:
\[ -2A + 51 - 6A = -5 \]
\[ -8A + 51 = -5 \]
\[ -8A = -56 \]
\[ A = 7 \]

### Step 6: Find \( B \) and \( C \)

Using \( A = 7 \) in Equation 1:
\[ B = 8 - 2(7) = 8 - 14 = -6 \]

Using \( A = 7 \) in Equation 4:
\[ C = 17 - 2(7) = 17 - 14 = 3 \]

### Final Values
Thus, the values are:
\[ A = 7, \quad B = -6, \quad C = 3 \]