Problem 27 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

\(f\) is defined in \([-5,5]\) as \(f(x)=x\) if \(x\) is rational \(=-x\) if \(x\) is irrational. Then (a) \(f(x)\) is continuous at every \(x\), except \(x=0\) (b) \(f(x)\) is discontinuous at every \(x\), except \(x=0\) (c) \(f(x)\) is continuous everywhere (d) \(f(x)\) is discontinuous everywhere

Step-by-Step Solution

Answer

(b) \(f(x)\) is discontinuous at every \(x\), except \(x=0\).

Step 1: Understanding Continuity at a Point

A function is continuous at a point if the limit of the function as it approaches the point from both sides is equal to the function's value at that point.

Step 2: Determine Continuity at x = 0

Consider the limit of \(f(x)\) as \(x\) approaches 0 from both rational and irrational numbers. For rational \(x\), \(f(x) = x\) approaches 0. For irrational \(x\), \(f(x) = -x\) also approaches 0. So, \(f(x)\) is continuous at \(x = 0\).

Step 3: Determine Continuity at x ≠ 0

Consider any \(x_0 eq 0\). As \(x\) approaches \(x_0\), if \(x\) is rational, \(f(x) = x\) approaches \(x_0\). If \(x\) is irrational, \(f(x) = -x\) approaches \(-x_0\). Since these two limits are not equal, \(f\) is discontinuous at every \(x_0 eq 0\).

Step 4: Conclusion

Based on the analysis, \(f(x)\) is continuous only at \(x = 0\) and discontinuous everywhere else.

Key Concepts

Piecewise Functions

Piecewise functions are a special type of function that are defined by different expressions based on the input value. Think of them as a collection of several "smaller" functions, each applicable on specific parts of the domain. In this exercise, we have a piecewise function where:

• For rational numbers, the function is defined as \(f(x) = x\).
• For irrational numbers, the function is defined as \(f(x) = -x\).

This setup allows the function to behave differently based on whether the input \(x\) is rational or irrational. It's a flexible way to construct functions with different "rules" for different scenarios and enhances our ability to model complex systems where continuity and behavior changes across a domain.

Continuity

When we talk about continuity in functions, we're referring to the idea that small changes in the input cause small changes in the output. A function is continuous at a point if the limit as we approach that point from the left and from the right matches the actual function value at that point. For the function in our exercise:

• At \(x = 0\), the function is continuous. This means as \(x\) approaches 0 from either rational numbers \(f(x) = x\) or irrational numbers \(f(x) = -x\), the output approaches 0.
• For any \(x eq 0\), the function is discontinuous. This is because, for rational numbers \(x\), the value is \(x\), but for irrational numbers \(x\), the value is \(-x\). These won't match unless \(x\) is 0.

Understanding these shifts in continuity is crucial for grasping how functions can change behavior drastically, especially in the context of piecewise defined scenarios.

Rational and Irrational Numbers

To fully appreciate piecewise functions like the one in the exercise, it's essential to understand rational and irrational numbers. Rational numbers can be expressed as a fraction \(\frac{a}{b}\) where \(a\) and \(b\) are integers, and \(b eq 0\). Examples include integers, fractions like \(\frac{1}{2}\), and decimals that terminate or repeat like \(0.333\ldots\).Irrational numbers cannot be written as such fractions and have non-repeating, non-terminating decimals. Classic examples are \(\pi\) and \(\sqrt{2}\). In mathematics, they fill the "gaps" between rational numbers on the number line.In our function, \(f(x)\) tackles both types with different rules, highlighting the distinction between these two categories and how this discontinuity arises due to changes in the underlying number type when moving from rational to irrational inputs.