It is given that \[f\left( x \right) = \left\{ \begin{array}{l}
{x^2} + ax + 1:x \in Q\\
a{x^2} + bx + 1:x \notin Q
\end{array} \right\}\], find a and b if f(x) is continuous at x=1 and x=2.

Hint:
 Here we will be using the given information and apply the rule of continuity of a function at a point.

Complete step by step solution:
It is given that \[f\left( x \right) = \left\{ \begin{array}{l}
{x^2} + ax + 1:x \in Q\\
a{x^2} + bx + 1:x \notin Q
\end{array} \right\}\], we need to find a and b if f(x) is continuous at x=1 and x=2.
Now, we know that the rule of continuity of a function at a point states that if the function f(x) is continuous at the point x=a, then
\[\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\].
Since it is given that f(x) is continuous at x=1, so
\[\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)\\
 \Rightarrow \mathop {\lim }\limits_{x \to 1} \left( {a{x^2} + bx + 1} \right) = {1^2} + a\left( 1 \right) + 1\\
 \Rightarrow a + b + 1 = 1 + a + 1\\
 \Rightarrow b = 1
\end{array}\]
Since it is given that f(x) is continuous at x=2, so
\[\begin{array}{l}
\mathop {\lim }\limits_{x \to 2} f\left( x \right) = f\left( 1 \right)\\
 \Rightarrow \mathop {\lim }\limits_{x \to 2} \left( {a{x^2} + bx + 1} \right) = {\left( 2 \right)^2} + a\left( 2 \right) + 1\\
 \Rightarrow 4a + 2b + 1 = 4 + 2a + 1\\
 \Rightarrow 2a + 2b = 4\\
 \Rightarrow a + b = 2
\end{array}\]
Putting b=1, we get that
\[\begin{array}{l}
a + b = 2\\
 \Rightarrow a + 1 = 2\\
 \Rightarrow a = 1
\end{array}\]
Hence, the required values of a and b are respectively 1 and 1.

Note:
In problems like these, it is to be always remembered that the concept of continuity at a point is entirely different than the continuity in an interval.