Problem 18 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(f\) be a composite function of \(x\) defined by \(f(u)=\frac{1}{u^{2}+u-2}, u(x)=\frac{1}{x-1}\). Then the number of points \(x\) where \(f\) is discontinuous is : (a) 4 (b) 3 (c) 2 (d) 1

Step-by-Step Solution

Answer

(b) 3

Step 1: Understand the Components of the Composite Function

We have a composite function \( f(x) = f(u(x)) \). \( f(u) = \frac{1}{u^2 + u - 2} \) and \( u(x) = \frac{1}{x-1} \). Our goal is to identify where \( f(x) \) is discontinuous.

Step 2: Determine Discontinuities of Inner Function

Evaluate \( u(x) = \frac{1}{x-1} \). The function is undefined when \( x-1 = 0 \), i.e., at \( x = 1 \). This is a point of discontinuity for \( u(x) \).

Step 3: Simplify Outer Function

Express \( f(u) \) in factored form to determine discontinuities. \( u^2 + u - 2 \) factors as \( (u-1)(u+2) \). Thus, \( f(u) = \frac{1}{(u-1)(u+2)} \), and \( f(u) \) is undefined at \( u = 1 \) and \( u = -2 \).

Step 4: Determine Discontinuities in Composite Function

Substitute \( u(x) = \frac{1}{x-1} \) into the conditions \( u = 1 \) and \( u = -2 \). For \( u = 1 \), solve \( \frac{1}{x-1} = 1 \), giving \( x = 2 \). For \( u = -2 \), solve \( \frac{1}{x-1} = -2 \), giving \( x = \frac{1}{3} \).

Step 5: Count Total Points of Discontinuity

The points \( x = 1, 2, \) and \( \frac{1}{3} \) are where \( f(x) \) is discontinuous. Therefore, there are three discontinuities.

Key Concepts

Composite functions

A composite function is formed when one function is applied to the results of another function. Imagine it like a conveyor belt; the output of the first process becomes the input for the second. In mathematical terms, if we have two functions, say, \( u(x) \) and \( f(u) \), the composite function \( f(u(x)) \) represents this chain of operations.

In the exercise, we have \( f(u) = \frac{1}{u^2 + u - 2} \) and \( u(x) = \frac{1}{x-1} \). Thus, the overall function \( f(x) \) is created by substituting \( u(x) \) into \( f(u) \). This substitution helps us understand how changes in \( x \) affect the final output of \( f(x) \).

Grasping the concept of composite functions is essential for analyzing more complex behaviors of functions like points of discontinuity. The interplay between \( f \) and \( u \) must be understood to identify where a function might "break" or become undefined. This understanding starts with breaking down each function individually.

Factoring polynomials

Factoring polynomials is like searching for the prime components of a number, but in a polynomial context. It allows us to understand how a polynomial can be decomposed into smaller, more manageable pieces.

In the exercise, \( f(u) = \frac{1}{u^2 + u - 2} \) requires factoring the polynomial \( u^2 + u - 2 \). This expression can be decomposed into factors: \( (u-1)(u+2) \).

This process of factoring is crucial because it reveals potential points where the function becomes undefined. Each factor contributes a potential point of discontinuity. For instance, if \( u^2 + u - 2 = 0 \), then \( u = 1 \) and \( u = -2 \) are the solutions, which are critical when evaluating where the composite function breaks.

Points of discontinuity

Points of discontinuity occur where a function "jumps," "breaks," or is simply not defined. For composite functions, these are crucial spots to find as they significantly affect the behavior of the function.

In the given exercise, the function \( f(x) = f(u(x)) \) has points of discontinuity at places where either \( f(u) \) or \( u(x) \) become undefined. This leads us to explore:

• Discontinuity due to \( u(x) \), which becomes undefined at \( x = 1 \).
• Discontinuity due to \( f(u) \), which becomes undefined at \( u = 1 \) and \( u = -2 \). Substituting \( u(x) \) into these conditions gives us new points of discontinuity: \( x = 2 \) (from \( u = 1 \)) and \( x = \frac{1}{3} \) (from \( u = -2 \)).

Overall, understanding these points helps us better handle the behaviors of complex functions, making us aware of where they could potentially "misbehave."

Rational functions

Rational functions are like fractions but with polynomials. They take the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials, and they introduce special characteristics, like discontinuities, depending on their denominators.

For \( f(u) = \frac{1}{u^2 + u - 2} \), knowing it's a rational function means there can be points where it becomes undefined—exactly where the denominator equals zero. This nature requires examining and factoring \( Q(u) = u^2 + u - 2 \) to identify these points. The factors \((u-1)(u+2)\) tell us \( f(u) \) is undefined at \( u = 1 \) and \( u = -2 \), indicating potential trouble spots.

Rational functions are not always smooth and continuous across their domain. Recognizing and finding where these "gaps" occur helps us both understand their limitations and represent them accurately.