The function $f$ is defined by

$$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\left| {x - 2} \right| + 1}&{x < 2} \\ {a{x^2} + bx}&{x \geqslant 2} \end{array}} \right.$$

where $a$ and $b$ are real constants

Given that both $f$ and its derivative are continuous at $x = 2$, find the value of $a$ and the value of $b$.

considering continuity at $x = 2$

$\mathop {{\text{lim}}}\limits_{x \to {2^ - }} f\left( x \right) = 1$ and $\mathop {{\text{lim}}}\limits_{x \to {2^ + }} f\left( x \right) = 4a + 2b$    (M1)

$4a + 2b = 1$     A1

considering differentiability at $x = 2$

$f'\left( x \right) = \left\{ {\begin{array}{*{20}{c}} { - 1}&{x < 2} \\ {2ax + b}&{x \geqslant 2} \end{array}} \right.$    (M1)

$\mathop {{\text{lim}}}\limits_{x \to {2^ - }} f'\left( x \right) =  - 1$ and $\mathop {{\text{lim}}}\limits_{x \to {2^ + }} f'\left( x \right) = 4a + b$     (M1)

Note: The above M1 is for attempting to find the left and right limit of their derived piecewise function at $x = 2$.

$4a + b =  - 1$     A1

$a =  - \frac{3}{4}$ and $b = 2$     A1

[6 marks]