If $f(x)=\left\{\begin{array}{cl}a x^2+b, & 0 \leq x<1 \\ 4 \quad, & x=1 \\ x+3, & 1<x \leq 2\end{array}\right.$ then the value of (a, b) for which f(x) cannot be continuous at x = 1, is

(5, 12)

We observe that $\lim\limits_{x \rightarrow 1^{+}} f(x)=f(1)$. So, f(x) is right continuous at x = 1. Thus, f(x) will not be continuous at x = 1 only, if

$\lim\limits_{x \rightarrow 1^{-}} f(x) \neq f(1)$

i.e. $\lim\limits_{x \rightarrow 1}\left(a x^2+b\right) \neq 4$ or, $a+b \neq 4$

Clearly, option (4) satisfies this condition.