Problem Statement
Find the values of $a$ and $b$ such that the function $f$ defined by
$$f(x) = \begin{cases} x^2 + 3x + a, & x \le 1 \\ bx + 2, & x > 1 \end{cases}$$
is differentiable at $x = 1$.
Verified Solution & Marking Scheme
Condition of Continuity at x = 1
Since differentiability implies continuity, $f$ must be continuous at $x = 1$:
$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$$
LHL: $\lim_{x \to 1^-}(x^2 + 3x + a) = 1 + 3 + a = 4 + a$.
RHL: $\lim_{x \to 1^+}(bx + 2) = b + 2$.
Therefore, $4 + a = b + 2 \implies b - a = 2 \quad \text{--- (1)}$
Condition of Differentiability at x = 1
For $f$ to be differentiable at $x = 1$, Left Hand Derivative (LHD) must equal Right Hand Derivative (RHD):
$$L f'(1) = \lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0} \frac{(1 - h)^2 + 3(1 - h) + a - (4 + a)}{-h}$$
$$= \lim_{h \to 0} \frac{1 - 2h + h^2 + 3 - 3h + a - 4 - a}{-h} = \lim_{h \to 0} \frac{-5h + h^2}{-h} = \lim_{h \to 0}(5 - h) = 5$$
Evaluate RHD and equate to LHD
$$R f'(1) = \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0} \frac{b(1 + h) + 2 - (4 + a)}{h}$$
From equation (1), $4 + a = b + 2$, so:
$$= \lim_{h \to 0} \frac{b + bh + 2 - (b + 2)}{h} = \lim_{h \to 0} \frac{bh}{h} = b$$
Equating LHD = RHD gives $b = 5$.
Solve for a
Substitute $b = 5$ into equation (1):
$$5 - a = 2 \implies a = 3$$