If $f(x) = \begin{cases} ax+b, & 0 < x \le 1 \\ 2x^2-x, & 1 < x < 2 \end{cases}$ is a differentiable function in $(0, 2)$, then find the values of $a$ and $b$.

$a = 3, b = -2$

The correct answer is Option (2) → $a = 3, b = -2$ ##

$f(x) = \begin{cases} ax + b; & 0 < x \le 1 \\ 2x^2 - x; & 1 < x < 2 \end{cases}$

$∴f(1) = a + b \quad \dots(i)$

$∴f'(1^+) = f'(1^-)$

$\lim\limits_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim\limits_{h \to 0^-} \frac{f(1-h) - f(1)}{h}$

$\lim\limits_{h \to 0^+} \frac{2(1+h)^2 - (1+h) - (a+b)}{h} = \lim\limits_{h \to 0^-} \frac{a(1-h) + b - (a+b)}{h}$

$\lim\limits_{h \to 0} \frac{2h^2 + 3h + 1 - a - b}{h} = \lim\limits_{h \to 0} \frac{-ah}{h} = a$

$\lim\limits_{h \to 0^+} \frac{2h^2 + 3h + 1 - a - b}{h} = a$

$f(x)$ is also continuous at $x = 1$.

$∴\lim\limits_{x \to 1^+} f(x) = f(1) = \lim\limits_{x \to 1^-} f(1)$

$∴a + b = 1 \quad \dots(ii)$

$\lim\limits_{h \to 0^+} \frac{2h^2 + 3h + 1 - 1}{h} = a$

$\lim\limits_{h \to 0} \frac{h(2h + 3)}{h} = a$

$\lim\limits_{h \to 0} (2h + 3) = a$

$a = 3$

From $(ii)$,

$3 + b = 1 ⇒b = -2$

$∴a = 3 \text{ and } b = -2$