Problem 116 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Assertion: If \(f(x)=\operatorname{sgn}(x)\) and \(g(x)=x\left(1-x^{2}\right)\), then \(\operatorname{fog}(x)\) and gof \((x)\) are continuous everywhere Reason: \(\operatorname{fog}(x)= \begin{cases}-1, & x \in(-1,0) \cup(1, \infty) \\\ 0, & x \in\\{-1,0,1\\} \\ 1, & x \in(-\infty,-1) \cup(0,1)\end{cases}\) and, \(g o f(x)=0, \forall x \in R\)

Step-by-Step Solution

Answer

The assertion is false; only gof(x) is continuous everywhere, not fog(x).

Step 1: Understanding the Function Notations

First, we need to understand the function notations. The function notation \( f(x) = \operatorname{sgn}(x) \) indicates the sign function. The function \( g(x) = x(1-x^2) \) represents a cubic polynomial. The goal is to determine whether the composed functions \( \operatorname{f \circ g}(x) \) and \( \operatorname{g \circ f}(x) \) are continuous everywhere.

Step 2: Exploring the Function f(x)

The sign function \( f(x) = \operatorname{sgn}(x) \) outputs -1 for negative values of \( x \), 0 for \( x = 0 \), and +1 for positive values of \( x \). This function is not continuous at \( x = 0 \) due to a jump discontinuity.

Step 3: Analyzing the Function g(x)

The function \( g(x) = x(1-x^2) \) is a polynomial and, as such, is continuous everywhere on its domain. Polynomials are inherently continuous functions.

Step 4: Composing fog(x) = f(g(x))

To determine \( f \circ g (x) = f(g(x)) \), substitute \( g(x) = x(1-x^2) \) into \( f(x) \). Since \( f(x) = \operatorname{sgn}(x) \), we get: \( \operatorname{sgn}(x(1-x^2)) \). This composite function will be discontinuous where \( x(1-x^2) \leq 0 \) since \( \operatorname{sgn}(x) \) itself is discontinuous at 0.

Step 5: Composing gof(x) = g(f(x))

Now consider \( g \circ f (x) = g(f(x)) \). Since \( f(x) = \operatorname{sgn}(x) \), \( g(f(x)) \) becomes \( g(-1)=g(0)=g(1)=0 \). Therefore, \( g \circ f(x) = 0 \) for all \( x \) in the domain. This is constant and therefore continuous everywhere.

Step 6: Evaluating the Continuity

From Step 4, \( f \circ g(x) \) is discontinuous at points where \( g(x) = 0 \) (i.e., \( x = 0, \pm 1 \)), making it discontinuous overall. However, from Step 5, since \( g \circ f(x) = 0 \) for all \( x \), it is continuous everywhere.

Step 7: Conclusion on the Assertion

The assertion that both \( \operatorname{f \circ g}(x) \) and \( \operatorname{g \circ f}(x) \) are continuous everywhere is false, because \( \operatorname{f \circ g}(x) \) is not continuous everywhere as it is discontinuous at some points.

Key Concepts

Continuity

Continuity is a fundamental concept in mathematics, especially in calculus and analysis. A function is said to be continuous at a point if there is no interruption in its values at that point. Simply put, if you were to draw the function on a graph, you could do so without lifting your pencil off the paper.
A function is considered continuous everywhere on its domain if it does not have any jumps, breaks, or holes. For a function to be continuous over an interval, it must be continuous at each point within that interval.Let's break down continuity with these simple properties:

• A function is continuous at a point if the limit of the function as it approaches that point from both directions is equal to the function’s value at that point.
• The sum, difference, and product of continuous functions are continuous.
• Polynomials, trigonometric functions, exponential functions, and logarithmic functions are examples of continuous functions.

When determining continuity for composite functions like in \(\operatorname{fog}(x)\) and \(\operatorname{gof}(x)\), we must carefully analyze each function's behavior to determine the overall continuity.

Sign Function

The sign function, often denoted as \( \operatorname{sgn}(x) \), is used to determine the sign of a real number. It has three simple steps:

• -1, if \( x < 0 \), indicating a negative number.
• 0, if \( x = 0 \), representing zero itself.
• 1, if \( x > 0 \), denoting a positive number.

The sign function is primarily used to identify whether numbers are positive, negative, or zero. However, it's important to note that the sign function is not continuous at \( x = 0 \). There is a jump from -1 to 1 around this point, which disrupts continuity.In mathematical problems, particularly those involving composite functions like \( f(g(x)) \), the discontinuity at zero in \( \operatorname{sgn}(x) \) can influence the continuity of the entire composition. Therefore, understanding the behavior of \( \operatorname{sgn}(x) \) is crucial in analyzing such functions.

Polynomials

Polynomials are among the most straightforward functions in mathematics and are an essential component of algebra. They are expressions involving variables raised to whole number powers and coefficients. A polynomial can be written in the general form: \( a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \).
Each term consists of a coefficient and a variable raised to a non-negative integer power. Here are some characteristics of polynomial functions:

• They are continuous everywhere on their domain, which means you can draw a polynomial's graph without picking up your pencil.
• They are smooth, having no corners or cusps.
• The degree of the polynomial (the highest power of the variable) often determines the shape and number of turns in the graph.

Polynomials are used in many areas of math due to their simplicity and versatility. Straight lines (linear functions), parabolas (quadratic functions), and cubic curves are all examples of polynomials.
In our specific problem, the function \( g(x) = x(1-x^2) \) is a cubic polynomial, which ensures that \( g(x) \) itself is continuous. This characteristic was crucial in determining the continuity of \( \operatorname{gof}(x) \), as the polynomial's continuity lead to a constant result (and thus continuity) in that particular composition.