If the function

$f(x) = \begin{cases} ax + 2, & x \leq 1 \\ x^2 + 3x + b, & x > 1 \end{cases}$

is differentiable at $x = 1$, then the value of $(2a + b)$ is

For a function to be differentiable at a point, it must be continuous at that point and the derivatives from both sides must match.

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For continuity at $x = 1$, the value from the left side must equal the value from the right side.

From the left side (using $ax + 2$ since $x \leq 1$):

$f(1) = a(1) + 2$

$f(1) = a + 2$

From the right side (using $x^2 + 3x + b$ since $x > 1$):

$\lim_{x \to 1^+} f(x) = (1)^2 + 3(1) + b$

$\lim_{x \to 1^+} f(x) = 4 + b$

Setting them equal:

$a + 2 = 4 + b$

$a - b = 2$ ...(Equation 1)

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For differentiability at $x = 1$, the left-hand derivative must equal the right-hand derivative.

Left-hand derivative (derivative of $ax + 2$):

$f'(x) = a$

$f'(1^-) = a$

Right-hand derivative (derivative of $x^2 + 3x + b$):

$f'(x) = 2x + 3$

$f'(1^+) = 2(1) + 3$

$f'(1^+) = 5$

Setting them equal:

$a = 5$ ...(Equation 2)

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From Equation 2: $a = 5$

Substituting into Equation 1:

$5 - b = 2$

$b = 3$

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$2a + b = 2(5) + 3$

$2a + b = 10 + 3$

$2a + b = 13$

Therefore, the value of $(2a + b) = 13$.