I think function has to be modified a little bit,
$f(x) =\left\{\begin{matrix} \large x^2, & x\leq 1\\ \large{2ax^2 + bx + c},&1 < x \leq 2\\ \large{x+2d},&x>2 \end{matrix}\right.$
Continuity of $\large{f(x)}$:-
$1)$ At $\large{x=1}$:
$f(1) = 1^2 = 1$
Left hand limit:- $\large{\lim_{x \to 1^-} x^2 = 1^2 =1 = f(1)}$
Right Hand limit:- $\large{\lim_{x \to 1^+}2ax^2 + bx + c = 2a + b + c}$
In order to be continuous at $\large{x=1}$, $\large{2a + b + c = 1}$ ------$(1)$
$1)$ At $\large{x=2}$:
$f(2) = \large{2a(2)^2 + b(2) + c = 8a + 2b + c}$
Left hand limit:- $\large{\lim_{x \to 2^-} 2ax^2 + bx + c = 8a + 2b + c = f(2)}$
Right Hand limit:- $\large{\lim_{x \to 2^+}x + 2d = 2 + 2d}$
In order to be continuous at $\large{x=2}$, $\large{8a + 2b + c = 2 + 2d}$ ------$(2)$
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Differentiability of $\large{f(x)}$:-
$f'(x) =\left\{\begin{matrix} \large 2x, & x < 1\\ \large{4ax + b },&1 < x < 2\\ \large{1},&x>2 \end{matrix}\right.$
$1)$ At $\large{x=1}$
Left Hand Derivative:- $\large{f'(1^-) = 2(1) = 2}$
Right Hand Derivative:- $\large{f'(1^+) = 4a(1) + b = 4a + b}$
In order to be differentiable at $\large{x =1}$, $\large{4a + b = 2}$ ------$(3)$
$2)$ At $\large{x=2}$
Left Hand Derivative:- $\large{f'(2^-) = 4a(2) + b = 8a +b}$
Right Hand Derivative:- $\large{f'(2^+) = 1}$
In order to be differentiable at $\large{x =2}$, $\large{8a + b = 2}$ ------$(4)$
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Subtract $3$ from $4$ it results in $\large{a=0}$, $\large{b=2}$, substitute these values in $1 ,2$, it results in $\large{c = -1, d=\dfrac{1}{2}}$