All the values of `a` for which `int_1^2 [a^2+(4-4a)x+4x^3]dxle12` are given by (A) `a=3` (B) `ale4` (C) `0lealt3` (D) none of these

Options

A. `a=3`
B. `a le 4`
C. `0 le a lt 3`
D. None of these

Answer

To solve the inequality \( \int_1^2 [a^2 + (4 - 4a)x + 4x^3] \, dx \leq 12 \), we will follow these steps:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_1^2 [a^2 + (4 - 4a)x + 4x^3] \, dx \]

### Step 2: Split the integral
We can split the integral into three parts:
\[ I = \int_1^2 a^2 \, dx + \int_1^2 (4 - 4a)x \, dx + \int_1^2 4x^3 \, dx \]

### Step 3: Evaluate each integral
1. **First integral**:
\[ \int_1^2 a^2 \, dx = a^2 \int_1^2 1 \, dx = a^2 [x]_1^2 = a^2 (2 - 1) = a^2 \]

2. **Second integral**:
\[ \int_1^2 (4 - 4a)x \, dx = (4 - 4a) \int_1^2 x \, dx = (4 - 4a) \left[\frac{x^2}{2}\right]_1^2 = (4 - 4a) \left(\frac{4}{2} - \frac{1}{2}\right) = (4 - 4a) \cdot \frac{3}{2} = 6 - 6a \]

3. **Third integral**:
\[ \int_1^2 4x^3 \, dx = 4 \int_1^2 x^3 \, dx = 4 \left[\frac{x^4}{4}\right]_1^2 = [x^4]_1^2 = 2^4 - 1^4 = 16 - 1 = 15 \]

### Step 4: Combine the results
Now we can combine the results of the integrals:
\[ I = a^2 + (6 - 6a) + 15 = a^2 - 6a + 21 \]

### Step 5: Set up the inequality
We need to solve the inequality:
\[ a^2 - 6a + 21 \leq 12 \]

### Step 6: Rearrange the inequality
Subtract 12 from both sides:
\[ a^2 - 6a + 9 \leq 0 \]

### Step 7: Factor the quadratic
The left-hand side can be factored:
\[ (a - 3)^2 \leq 0 \]

### Step 8: Solve the inequality
Since a square is always non-negative, the only solution occurs when:
\[ (a - 3)^2 = 0 \implies a - 3 = 0 \implies a = 3 \]

### Conclusion
The only value of \( a \) that satisfies the inequality is:
\[ \boxed{3} \]