If f(x) =If `f(x)={:{(x^2+3x+a"," " " x le 1),( " "bx +2"," " " x gt 1):}` is differentiable for all values of x, than find the values of 'a' and 'b'.

Answer

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

\[ f(x) = \begin{cases} x^2 + 3x + a & \text{if } x \leq 1 \\ bx + 2 & \text{if } x > 1 \end{cases} \]

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

For \( f(x) \) to be continuous at \( x = 1 \), the left-hand limit as \( x \) approaches 1 must equal the right-hand limit at \( x = 1 \):

1. **Left-hand limit**:
\[ \lim_{x \to 1^-} f(x) = 1^2 + 3(1) + a = 1 + 3 + a = 4 + a \]

2. **Right-hand limit**:
\[ \lim_{x \to 1^+} f(x) = b(1) + 2 = b + 2 \]

Setting the left-hand limit equal to the right-hand limit for continuity:
\[ 4 + a = b + 2 \]
This simplifies to:
\[ a - b = -2 \quad \text{(Equation 1)} \]

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

For \( f(x) \) to be differentiable at \( x = 1 \), the derivative from the left must equal the derivative from the right:

1. **Derivative for \( x \leq 1 \)**:
\[ f'(x) = 2x + 3 \quad \text{(for } x \leq 1\text{)} \]
Evaluating at \( x = 1 \):
\[ f'(1) = 2(1) + 3 = 2 + 3 = 5 \]

2. **Derivative for \( x > 1 \)**:
\[ f'(x) = b \quad \text{(for } x > 1\text{)} \]
Thus, at \( x = 1 \):
\[ f'(1) = b \]

Setting the left-hand derivative equal to the right-hand derivative:
\[ 5 = b \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations

Now we have two equations:
1. \( a - b = -2 \) (Equation 1)
2. \( b = 5 \) (Equation 2)

Substituting the value of \( b \) from Equation 2 into Equation 1:
\[ a - 5 = -2 \]
Solving for \( a \):
\[ a = -2 + 5 = 3 \]

### Final Values

Thus, the values of \( a \) and \( b \) are:
\[ a = 3, \quad b = 5 \]

### Summary

The final answer is:
\[ \boxed{a = 3 \text{ and } b = 5} \]