Problem 116 - A Complete Resource Book in Mathematics for JEE Main

Question

In the following questions an Assertion \((A)\) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is a correct explanation for Assertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion: If \(f(x)=\operatorname{sgn}(x)\) and \(g(x)=x\left(1-x^{2}\right)\), then \(f o g(x)\) and \(g o f(x)\) are continuous everywhere Reason: \(f o g(x)=\left\\{\begin{aligned}-1, & x \in(-1,0) \cup(1, \infty) \\\ 0, & x \in\\{-1,0,1\\} \\ 1, & x \in(-\infty,-1) \cup(0,1) \end{aligned}\right.\) and, \(g o f(x)=0, \forall x \in R\)

Step-by-Step Solution

Answer

Assertion is False; Reason is True. Option (D) is correct.

Step 1: Understanding Functions

First, we understand the given functions: \( f(x) = \operatorname{sgn}(x) \) is the signum function which outputs -1 if \( x < 0 \), 0 if \( x = 0 \), and 1 if \( x > 0 \). The function \( g(x) = x(1-x^2) \) is a cubic function with roots at \( x = 0 \), \( x = 1 \), and \( x = -1 \).

Step 2: Composition of Functions

The composition \( f \circ g(x) \) means applying \( g(x) \) first and then \( f(x) \). Similarly, \( g \circ f(x) \) means applying \( f(x) \) first and then \( g(x) \).

Step 3: Analyzing \( f \circ g(x) \)

Evaluate \( g(x) = x(1-x^2) \): - If \( x \in (-1,0) \cup(1, \infty) \), then \( g(x) < 0 \) so \( f(g(x)) = -1 \).- If \( x \in \{-1, 0, 1\} \), then \( g(x) = 0 \) so \( f(g(x)) = 0 \).- If \( x \in (-\infty, -1) \cup (0, 1) \), then \( g(x) > 0 \) so \( f(g(x)) = 1 \).This matches the reason given.

Step 4: Analyzing \( g \circ f(x) \)

For all real \( x \), \( f(x) \) is either -1, 0, or 1. Hence, - If \( f(x) = 0 \), \( g(f(x)) = g(0) = 0. \)- If \( f(x) = \pm 1 \), \( g(f(x)) = g(1) = g(-1) = 0. \)Hence, \( g(f(x)) = 0 \) for all \( x \in \mathbb{R} \), which matches the reason.

Step 5: Continuity Check

\( g \circ f(x) = 0 \) is clearly continuous everywhere because it is a constant function. However, \( f \circ g(x) \) is not continuous at \( x = -1, 0, 1 \) because of the discontinuity in the signum function. It changes values abruptly at these points.

Step 6: Conclusion

Assertion (A) is false because \( f \circ g(x) \) is not continuous at \( x = -1, 0, 1 \). Reason (R) is true because it correctly describes the behavior of both compositions. Therefore, option (D) is correct: Assertion(A) is False, Reason(R) is True.

Key Concepts

Function Composition

Function composition is a valuable concept in calculus. It involves combining two functions to form a new function. When we have two functions, say \( f(x) \) and \( g(x) \), the composition \( f \circ g(x) \) means you first apply \( g(x) \), followed by applying \( f \) to the result of \( g \). Similarly, \( g \circ f(x) \) means to apply \( f \) first and then \( g \) to the result.
This sequence is essential when determining how functions transform inputs and interact with each other.

• The order matters: \( f \circ g(x) \) is generally not the same as \( g \circ f(x) \).
• By understanding the individual behavior of \( f \) and \( g \), one can predict the behavior of the composed function.

Applying these ideas to the given problem: \( f(x) = \operatorname{sgn}(x) \), which is the signum function, and \( g(x) = x(1-x^2) \), a cubic function, we need to understand the distinct roles each plays in the composition. This helps in exploring deeper mathematical questions such as limits, continuity, and differentiability when dealing with composite functions.

Continuity

In calculus, continuity of a function means having no interruptions or gaps. A function is continuous if you can draw its graph without lifting your pencil. This means for every point in the function's domain, the limit of the function as it approaches that point is equal to the function's value at that point.
For composed functions, evaluating continuity involves considering the continuity of each individual function.

• For \( f \circ g(x) \): Even if \( g(x) \) is continuous, \( f \) might introduce discontinuities if it changes values suddenly, like the signum function.
• For \( g \circ f(x) \): If \( f \) results in constant values and \( g \) is continuous over those values, the composition can be continuous.

In our example, \( g \circ f(x) \) results in a constant zero function, which is trivially continuous. However, \( f \circ g(x) \) faces discontinuities due to how the signum function behaves around zero and at the points where it switches values.

Signum Function

The signum function, often denoted \( \operatorname{sgn}(x) \), is a simple yet intriguing function in calculus used to extract or identify the sign of a real number. Its definition is:

• \( \operatorname{sgn}(x) = -1 \) if \( x < 0 \)
• \( \operatorname{sgn}(x) = 0 \) if \( x = 0 \)
• \( \operatorname{sgn}(x) = 1 \) if \( x > 0 \)

This function is important due to its simplicity and its role in various mathematical transformations and analyses.
However, it introduces discontinuities where it changes its output abruptly:

• The abrupt changes at integers like -1, 0, and 1 cause the function to be non-continuous at these points.
• When used in compositions like \( f \circ g(x) \), these discontinuities can result in the composed function also being non-continuous at certain key points, unless the inner function handles these transitions smoothly.

Understanding the signum function's behavior allows us to foresee potential difficulties in operations involving limits, derivatives, or integrations where continuity can be essential.