Problem 70 - A Complete Resource Book in Mathematics for JEE Main

Question

The function \(f(x)=\left\\{\begin{array}{cll}\frac{x-1}{\frac{1}{x-1}+1} & , & x \neq 1 \\ 0, & x=1\end{array}\right.\) (A) is continuous (B) has removable discontinuity (C) has jump discontinuity (D) has infinite discontinuity

Step-by-Step Solution

Answer

The function is continuous at all points, so (A) is correct.

Step 1: Analyze the function for x ≠ 1

For the function given, when \(x eq 1\), the expression for \(f(x)\) is \(\frac{x-1}{\frac{1}{x-1}+1}\). Simplifying gives \(\frac{x-1}{\frac{x}{x-1}} = x - 1\). Thus, for \(x eq 1\), this simplifies to \(f(x) = x - 1\).

Step 2: Evaluate the function at x = 1

The function defines \(f(1) = 0\) explicitly. To check continuity at \(x = 1\), let's evaluate the limit of \(f(x) = x - 1\) as \(x\) approaches 1.

Step 3: Calculate the limit as x approaches 1

The limit of \(f(x) = x - 1\) as \(x \to 1\) is \(1 - 1 = 0\). Thus, \(\lim_{{x \to 1}} f(x) = 0\).

Step 4: Check continuity at x = 1

For \(f(x)\) to be continuous at \(x = 1\), \(\lim_{{x \to 1}} f(x) = f(1)\) must hold. We have \(\lim_{{x \to 1}} f(x) = 0\) and \(f(1) = 0\). Both are equal, implying \(f(x)\) is continuous at \(x = 1\).

Step 5: Conclude the type of discontinuity or continuity

Since \(f(x)\) is continuous at \(x = 1\) and there are no other points of discontinuity in the given piecewise function, option (A) is correct, and there is no type of discontinuity present.

Key Concepts

Piecewise Functions

Piecewise functions are a fantastic tool in mathematics, allowing us to define a function with multiple expressions over different intervals of the domain. In essence, they let us glue together several different functions, providing flexibility and precision where a single expression might fall short.

In this exercise, we see a piecewise function defined as follows:

• For values of \(x\) that are not equal to 1, the function is expressed as \(f(x) = \frac{x-1}{\frac{1}{x-1}+1}\). Upon simplifying, this converts to \(f(x) = x-1\).
• For \(x = 1\), the function is directly stated as \(f(1) = 0\).

Piecewise functions like this help in scenarios where a function needs to behave differently across different intervals. In practical applications, think of scenarios like tax brackets, where different income ranges are taxed at various rates. By understanding the concept of piecewise functions, students can appreciate how these can be applied both in theoretical math and everyday problems.

Removable Discontinuity

In calculus, a removable discontinuity is a kind of discontinuity where a function is undefined or does not match the function's limit at a point but can be "made" continuous by filling in a single point. It's like finding a hole in the graph of the function that can be easily patched by redefining the function at one point.

In the original problem, at \(x = 1\), we might suspect removable discontinuity initially because the function changes its form at this point. However, upon calculating the limit of \(f(x) = x - 1\) as \(x\) approaches 1, we find that \[\lim_{{x \to 1}} f(x) = 0\]which matches the given value \(f(1) = 0\). This equation demonstrates that even though the function is initially defined in two parts, it results in continuous behavior at that specific point as the limit seamlessly connects with the defined value. Hence, no patching or redefining is necessary, indicating there's no actual removable discontinuity.

Limits in Calculus

Limits are a fundamental building block in calculus, providing the tools to understand behavior of functions at specific points. They help us to explore the value that a function approaches as the input moves towards some specific value. This becomes particularly crucial when analyzing continuity and discontinuities.

In this exercise, limits are used to determine whether the function \(f(x)\) is continuous at \(x = 1\). The limit of \(f(x) = x - 1\) as \(x\) approaches 1 is calculated, giving us \[\lim_{{x \to 1}} f(x) = 0\]This limit is crucial because it tells us the behavior of \(f(x)\) as it comes infinitely close to \(x = 1\). Since this matches the function's actual value at \(x = 1\), \(f(1) = 0\), we learn that the function continues smoothly through this point. Thus, confirming the concept of limit helps establish continuity, solidifying its importance in not only understanding the function's graph but also in guaranteeing the function's behavior aligns with expectations.