Problem 57 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(x+|y|=2 y\), then \(y\) as a function of \(x\), at \(x=0\) is (a) differentiable but not continuous (b) continuous but not differentiable (c) continuous as well as differentiable (d) neither continuous nor differentiable

Step-by-Step Solution

Answer

(b) continuous but not differentiable.

Step 1: Understand the given equation

Given the equation \(x + |y| = 2y\), we need to understand it in terms of conditions that might affect continuity and differentiability. By examining the expression \(|y|\), recognize that splitting it based on the value of \(y\) could simplify the problem.

Step 2: Split equation into cases

Since \(|y|\) can be \(y\) if \(y \geq 0\) and \(-y\) if \(y < 0\), handle these separately: 1. If \(y \geq 0\), the equation becomes \(x + y = 2y\). Rearrange to get \(y = x\).2. If \(y < 0\), the equation becomes \(x - y = 2y\). Rearrange to get \(y = \frac{x}{3}\).

Step 3: Analyze conditions at x = 0

We need to examine the function behavior for both cases when \(x=0\): 1. For \(y \geq 0\), we get \(y = x = 0\).2. For \(y < 0\), we get \(y = \frac{x}{3} = 0\).Since both cases give \(y = 0\) at \(x = 0\), the function is indeed continuous at this point.

Step 4: Check differentiability at x = 0

Next, check if the function is differentiable at \(x = 0\). For the positive side, \(y = x\), the derivative \(\frac{dy}{dx} = 1\). For the negative side, \(y = \frac{x}{3}\), the derivative is \(\frac{dy}{dx} = \frac{1}{3}\). Since these derivatives do not match at \(x = 0\), the function is not differentiable at this point.

Key Concepts

Continuity

Continuity is a fundamental concept that ensures a function does not suddenly jump or break at a certain point. Simply put, if you can draw the graph of a function without lifting your pencil from the paper, it's continuous.
In mathematical terms, a function is continuous at a point if the following three conditions are met:

• The function is defined at that point.
• The limit of the function as it approaches that point exists.
• The limit of the function as it approaches that point is equal to the function's value at that point.

In our problem, we examined the behavior of the function given by the equation \(x + |y| = 2y\) at \(x = 0\). After splitting the function into two cases, based on whether \(y\) is positive or negative, we determined that for both situations, \(y = 0\) at \(x = 0\). Since both cases agree on the value of \(y\), this confirms the function is continuous at \(x = 0\). It doesn't exhibit any jumps or breaks at this point.

Piecewise Functions

Piecewise functions are those that have different rules or expressions based on the input value. These functions can switch between different formulas across their domain.
In the context of the problem, the function is defined in two pieces based on conditions applied to \(y\):

• If \(y \geq 0\), we have the equation \(y = x\).
• If \(y < 0\), we have \(y = \frac{x}{3}\).

This piecewise setup is useful in handling cases like absolute value functions, where the function behaves differently depending on the sign of the input.
When analyzing piecewise functions for continuity and differentiability, focus on the points where the different pieces meet—in this problem at \(x = 0\). Here, we found that \(y\)'s value remains consistent across the pieces, ensuring continuity, but the differing slopes (derivatives) on each side mean that the function isn't differentiable at that point.

Absolute Value Function

The absolute value function, denoted as \(|y|\), is a special type of piecewise function often encountered in mathematics. It measures the distance of a number from zero on the number line, always resulting in a non-negative value.
In its simplest form, the absolute value function can be described as:

• \(|y| = y\) if \(y \geq 0\)
• \(|y| = -y\) if \(y < 0\)

The problem makes use of this property by transforming the original equation \(x + |y| = 2y\) into two separate cases based on the value of \(y\).
This transformation allowed us to simplify the function's behavior and examine it in terms of continuity and differentiability. The absolute value function's non-linearity introduces a sharp corner at points where changes occur, such as where \(y = 0\). This is because the slopes from either side of the corner don't match, leading to non-differentiability at those points, as seen in our solution.