Question

If \[f\left( x \right) = \left\{ \begin{array}{l}x,x \in {\bf{Q}}\\ - x,x \notin {\bf{Q}}\end{array} \right.\] , then \[f\] is continuous at-
A.Only at zero
B.Only at 0, 1
C.All real numbers
D.All rational numbers

Answer 1

Hint: Here, we will find the point where the given function is continuous. We will consider a rational number to find the function, whether it is continuous at the point. A continuous function is a function where the function does not have any abrupt changes in the interval.

Complete step-by-step answer:
We are given that \[f\left( x \right) = \left\{ \begin{array}{l}x,x \in {\bf{Q}}\\ - x,x \notin {\bf{Q}}\end{array} \right.\]
Here \[{\bf{Q}}\] represents the rational numbers.
Let us consider P to be any rational number.
Thus from the given function, we get
\[f\left( P \right) = P\]
Taking the left hand limit \[x \to {P^ - }\]
\[L.H.L = \mathop {\lim }\limits_{x \to {P^ - }} f\left( x \right)\]
\[ \Rightarrow L.H.L = \mathop {\lim }\limits_{x \to {P^ - }} \left( { - x} \right)\]
Substituting the limits, we get
\[ \Rightarrow L.H.L = \left( { - P} \right)\]
Taking the right hand limit \[x \to {P^ + }\]
\[R.H.L = \mathop {\lim }\limits_{x \to {P^ + }} \left( { - x} \right)\]
Substituting the limits, we get
\[ \Rightarrow R.H.L = \left( { - P} \right)\]
So, we have at \[x = P \in {\bf{Q}}\] , both the right hand limits and the left hand limits are equal.
\[L.H.L = R.H.L = - P \ne f\left( P \right)\]
So, at \[x = P \in {\bf{Q}}\] , \[f\left( P \right)\] is discontinuous at any point of the rational number.
Now, consider \[L.H.L = R.H.L = f\left( P \right)\]
\[ \Rightarrow L.H.L = R.H.L = f\left( P \right)\]
\[ \Rightarrow - P = P\]
\[ \Rightarrow P = 0\]
So, at \[x = 0\] , \[L.H.L = R.H.L = f\left( P \right)\] is continuous at the point zero only.
Thus the given function is continuous at \[P = 0\].
Therefore, the function \[f\left( x \right) = \left\{ \begin{array}{l}x,x \in {\bf{Q}}\\ - x,x \notin {\bf{Q}}\end{array} \right.\] , then \[f\] is continuous at only at zero point.
Thus option (A) is the correct answer.

Note: A continuous function can also be defined as the function where the graph of the function has no unbroken curve. We are considering the limits, by considering the number line, which has a rational number surrounded by the irrational number. Thus every rational number has an irrational number on all its sides. A function is continuous only when we attain a constant at both limits.