Question

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.

Answer 1

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

\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.