Let a and b be real numbers such that the function
`g(x)={{:(,-3ax^(2)-2,x lt 1),(,bx+a^(2),x ge1):}` is differentiable for all `x in R`
Then the possible value(s) of a is (are)

Options

A. 1,2
B. 3,4
C. 5,6
D. 8,9

Explanation

To solve the problem, we need to ensure that the function \( g(x) \) is both continuous and differentiable at \( x = 1 \). The function is defined as follows:

\[ g(x) = \begin{cases} -3ax^2 - 2 & \text{if } x < 1 \\ bx + a^2 & \text{if } x \geq 1 \end{cases} \]

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

For \( g(x) \) to be continuous at \( x = 1 \), we need:

\[ \lim_{x \to 1^-} g(x) = \lim_{x \to 1^+} g(x) \]

Calculating the left-hand limit:

\[ \lim_{x \to 1^-} g(x) = -3a(1)^2 - 2 = -3a - 2 \]

Calculating the right-hand limit:

\[ \lim_{x \to 1^+} g(x) = b(1) + a^2 = b + a^2 \]

Setting the two limits equal for continuity:

\[ -3a - 2 = b + a^2 \tag{1} \]

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

For \( g(x) \) to be differentiable at \( x = 1 \), the left-hand derivative must equal the right-hand derivative.

Calculating the left-hand derivative:

\[ g'(x) = \frac{d}{dx}(-3ax^2 - 2) = -6ax \]
Evaluating at \( x = 1 \):

\[ g'(1^-) = -6a(1) = -6a \]

Calculating the right-hand derivative:

\[ g'(x) = \frac{d}{dx}(bx + a^2) = b \]
Evaluating at \( x = 1 \):

\[ g'(1^+) = b \]

Setting the two derivatives equal for differentiability:

\[ -6a = b \tag{2} \]

### Step 3: Solve the System of Equations

Now we have two equations (1) and (2):

1. \( -3a - 2 = b + a^2 \)
2. \( -6a = b \)

Substituting equation (2) into equation (1):

\[ -3a - 2 = -6a + a^2 \]

Rearranging gives:

\[ a^2 - 3a + 6a - 2 + 2 = 0 \]
\[ a^2 + 3a = 0 \]
Factoring out \( a \):

\[ a(a + 3) = 0 \]

Thus, \( a = 0 \) or \( a = -3 \).

### Step 4: Find Corresponding Values of \( b \)

Using \( b = -6a \):

1. If \( a = 0 \):
\[ b = -6(0) = 0 \]

2. If \( a = -3 \):
\[ b = -6(-3) = 18 \]

### Conclusion

The possible values of \( a \) are:

\[ \boxed{0 \text{ and } -3} \]