The function 'f' defined as :
`f(x)={{:(x^(2)+3x+a", if "xle1),(bx+c", if "xgt1):}`
is derivable for every x. Find the value of 'a' and 'b'.

Answer

To determine the values of 'a' and 'b' for the function

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

such that it is differentiable for every \( x \), we need to ensure that the function is both continuous and differentiable at the point where the definition of the function changes, which is at \( x = 1 \).

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

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

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

Calculating the left-hand limit:

\[ f(1) = 1^2 + 3(1) + a = 1 + 3 + a = 4 + a \]

Calculating the right-hand limit:

\[ \lim_{x \to 1^+} f(x) = b(1) + c = b + c \]

Setting the two limits equal for continuity:

\[ 4 + a = b + c \quad (1) \]

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

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

Calculating the derivative for \( x \leq 1 \):

\[ f'(x) = 2x + 3 \quad \text{for } x \leq 1 \]

Evaluating at \( x = 1 \):

\[ f'(1) = 2(1) + 3 = 2 + 3 = 5 \]

Calculating the derivative for \( x > 1 \):

\[ f'(x) = b \quad \text{for } x > 1 \]

Setting the derivatives equal for differentiability:

\[ 5 = b \quad (2) \]

### Step 3: Substitute \( b \) back into the continuity equation

From equation (2), we have:

\[ b = 5 \]

Substituting \( b \) into equation (1):

\[ 4 + a = 5 + c \]

Rearranging gives:

\[ a - c = 1 \quad (3) \]

### Step 4: Determine values of \( a \) and \( c \)

Since we have one equation (3) with two unknowns (a and c), we can express \( a \) in terms of \( c \):

\[ a = c + 1 \]

### Conclusion

We have found:

- \( b = 5 \)
- \( a = c + 1 \)

To find specific values for \( a \) and \( c \), we can choose any value for \( c \). For example, if we let \( c = 0 \):

\[ a = 0 + 1 = 1 \]

Thus, one possible solution is:

\[ a = 1, \quad b = 5 \]

### Final Answer

The values are:

\[ \boxed{a = 1, \quad b = 5} \]