Let `a` and `b` are real numbers such that the function `f(x)={(-3ax^(2)-2, xlt1),(bx+a^(2),xge1):}` is differentiable of all `xepsilonR`, then

Options

A. `a=b=7`
B. `ab=12`
C. `a+b=15`
D. `ab=-6`

Explanation

To solve the problem, we need to ensure that the function \( f(x) \) is both continuous and differentiable at \( x = 1 \).

The function is defined as:
\[ f(x) = \begin{cases} -3ax^2 - 2, & \text{if } x < 1 \\ bx + a^2, & \text{if } x \geq 1 \end{cases} \]

### Step 1: Check Continuity at \( x = 1 \)

For \( f(x) \) to be continuous at \( x = 1 \), we need:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \]

Calculating the left-hand limit:
\[ \lim_{x \to 1^-} f(x) = -3a(1)^2 - 2 = -3a - 2 \]

Calculating the right-hand limit:
\[ \lim_{x \to 1^+} f(x) = b(1) + a^2 = b + a^2 \]

Setting these equal for continuity:
\[ -3a - 2 = b + a^2 \]

### Step 2: Check Differentiability at \( x = 1 \)

For \( f(x) \) to be differentiable at \( x = 1 \), we need:
\[ f'(1^-) = f'(1^+) \]

Calculating the derivatives:
- For \( x < 1 \):
\[ f'(x) = \frac{d}{dx}(-3ax^2 - 2) = -6ax \]
Thus,
\[ f'(1^-) = -6a(1) = -6a \]

- For \( x \geq 1 \):
\[ f'(x) = \frac{d}{dx}(bx + a^2) = b \]
Thus,
\[ f'(1^+) = b \]

Setting these equal for differentiability:
\[ -6a = b \]

### Step 3: Substitute \( b \) into the Continuity Equation

From the differentiability condition, we have \( b = -6a \). Substituting this into the continuity equation:
\[ -3a - 2 = -6a + a^2 \]

Rearranging gives:
\[ a^2 - 3a + 6a + 2 = 0 \implies a^2 + 3a + 2 = 0 \]

### Step 4: Factor the Quadratic Equation

Factoring:
\[ (a + 1)(a + 2) = 0 \]

Thus, the solutions for \( a \) are:
\[ a = -1 \quad \text{or} \quad a = -2 \]

### Step 5: Find Corresponding \( b \) Values

Using \( b = -6a \):
1. If \( a = -1 \):
\[ b = -6(-1) = 6 \]
2. If \( a = -2 \):
\[ b = -6(-2) = 12 \]

### Step 6: Check the Options

Now we have two pairs:
1. \( (a, b) = (-1, 6) \)
2. \( (a, b) = (-2, 12) \)

We can check the options:
- \( a - b = -1 - 6 = -7 \) (not an option)
- \( a - b = -2 - 12 = -14 \) (not an option)
- \( ab = -1 \cdot 6 = -6 \) (correct)
- \( ab = -2 \cdot 12 = -24 \) (not an option)

### Conclusion

The correct options are:
- \( ab = -6 \)