Problem 79 - A Complete Resource Book in Mathematics for JEE Main

Question

The function \(f(x)=\max .\\{(1-x),(1+x), 2\\}\), \(x \in(-\infty, \infty)\), is (A) continuous at all points (B) differentiable at all points (C) differentiable at all points except at \(x=1\) and \(x\) \(=-1\) (D) continuous at all points except at \(x=1\) and \(x=-1\), where it is discontinuous.

Step-by-Step Solution

Answer

C: Differentiable at all points except at \(x=1\) and \(x=-1\).

Step 1: Analyze the Definition

The function given is defined as \( f(x) = \max\{ (1-x), (1+x), 2 \} \). This means for any \( x \), we take the largest of the three quantities \( 1-x \), \( 1+x \), and \( 2 \). Our task is to determine the continuity and differentiability of this function over the entire real line.

Step 2: Identify Critical Points

The expressions \( 1-x \), \( 1+x \), and \( 2 \) change their order at specific points. Setting the first two expressions equal, \( 1-x = 1+x \), we find \( x = 0 \). For \( 1-x = 2 \), we find \( x = -1 \), and for \( 1+x = 2 \), we find \( x = 1 \). The values \( x = -1, 0, 1 \) are critical and may affect continuity or differentiability.

Step 3: Evaluate Function at Critical Points

To understand continuity, evaluate the function at \( x = -1, 0, 1 \). For \( x < -1 \), \( f(x) = 2 \). At \( x = -1 \), \( 1-x = 2 \) causing \( f(x) = 2 \). Switch values: for \( x \) slightly greater than \(-1\), \( 1-x < 2 \), but \( 2 \) stays the maximum. Repeat: At \( x = 1 \), test limits, finding non-continuity. Hence, \( x = -1 \) and \( x = 1 \) are points of concern for discontinuity.

Step 4: Check Continuity

Since \( f(x) = 2 \) around \( x = -1 \) and \( x = 1 \), but changes occur abruptly at these points instead of gradually, \( f(x) \) is continuous everywhere except at \( x = -1 \) and \( x = 1 \).

Step 5: Check Differentiability

Differentiability requires continuity, so any discontinuities prevent differentiability. Furthermore, even where \( f(x) \) switches quickly at points like \( x = 0 \), evaluate derivatives around them to find points of non-differentiability, showing \( x = 1 \) and \( x = -1 \) break this rule more.

Key Concepts

Piecewise functions

A piecewise function is a mathematical expression that uses different formulas for different intervals of the domain. In the case of the function given by the exercise, it is defined as \( f(x) = \max\{ (1-x), (1+x), 2 \} \). This indicates that for any value of \( x \), \( f(x) \) will take the largest value among \( 1-x \), \( 1+x \), and \( 2 \).
Piecewise functions are often used to model scenarios where a rule or behavior changes at specific points. They allow us to express complex behavior concisely.

• At \( x = -1 \), \( 1-x \) and \( 2 \) both can be maximum, potentially changing the output swiftly as \( x \) changes around this point.
• Similarly, the expressions can change their ordinal ranking at \( x = 1 \) and \( x = 0 \), causing piecewise function transitions.

Piecewise functions are very useful in modeling real-world systems where rules need to be adjusted based on varying conditions.

Critical points

Critical points are values in the domain of a function where certain behavior changes. For piecewise functions, these points are where the function definition might change its formula.
In this exercise, critical points are found by setting the expressions equal to each other to determine where the maximum value might jump from one expression to another. We have:

• \( 1-x = 1+x \) when \( x = 0 \)
• \( 1-x = 2 \) when \( x = -1 \)
• \( 1+x = 2 \) when \( x = 1 \)

At each of these points, the function can have a sudden change, such as becoming less smooth or potentially discontinuous. These critical points are important to identify as they are likely where discontinuities or non-differentiabilities occur.

Limits and continuity

The concept of limits and continuity expresses how smoothly a function's output changes as its input varies. A function is continuous if it doesn't have any abrupt jumps or breaks.
To check continuity at a point, we need to see if the limit of the function as \( x \) approaches the point is the same from both sides and equals the function's value at that point.
In this exercise, \( f(x) \) was found to be continuous everywhere except at \( x = -1 \) and \( x = 1 \).

• At \( x = -1 \), there's a sudden change in maximum from \( 1-x \) to \( 2 \).
• Similarly, at \( x = 1 \), \( 1+x \) matches \( 2 \), resulting in another abrupt shift.

Since the function cannot be smoothly transitioned over these points, it becomes discontinuous there. Understanding and calculating limits help in confirming and explaining these changes clearly.