Problem 58 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The function \(f(x)=\frac{1}{u^{2}+u-2}\), where \(u=\frac{1}{x-1}\), is discontinuous at the points (A) \(x=-2,1, \frac{1}{2}\) (B) \(x=\frac{1}{2}, 1,2\) (C) \(x=1,0\) (D) None of these

Step-by-Step Solution

Answer

The function is discontinuous at points in option (B): \(x=\frac{1}{2}, 1, 2\).

Step 1: Substituting for u

Given the function \( f(x) = \frac{1}{u^2 + u - 2} \) with \( u = \frac{1}{x - 1} \), first substitute \( u \) in terms of \( x \): \( f(x) = \frac{1}{(\frac{1}{x-1})^2 + \frac{1}{x-1} - 2} \).

Step 2: Simplifying the Equation

Rewrite \( f(x) \) using the substitution: \[ f(x) = \frac{(x-1)^2}{1 + (x-1) - 2(x-1)^2} = \frac{(x-1)^2}{(x-1)^2 + (x-1) - 2} \].

Step 3: Finding Discontinuities in Terms of u

The function is discontinuous where the denominator equals zero, thus solve: \( u^2 + u - 2 = 0 \). Factor this quadratic expression as \( (u-1)(u+2) = 0 \), giving roots \( u = 1 \) or \( u = -2 \).

Step 4: Converting Discontinuities to x

Convert these discontinuities from \( u \) back to \( x \) using \( u = \frac{1}{x-1} \). For \( u = 1 \), we have \( \frac{1}{x-1} = 1 \rightarrow x = 2 \). For \( u = -2 \), \( \frac{1}{x-1} = -2 \rightarrow x = \frac{1}{2} \).

Step 5: Identifying Additional Discontinuities

Note that the function also has a discontinuity at \( x = 1 \) because \( u = \frac{1}{x-1} \) \( x - 1 \) must not be zero. Thus, the function is undefined at \( x = 1 \).

Step 6: Compiling All Discontinuities

The function is discontinuous at \( x = \frac{1}{2}, 1, 2 \) based on the analyses of the numerator and denominator.

Step 7: Answering the Question

Compare these points with the choices provided. The points \( x = \frac{1}{2}, 1, 2 \) correspond to choice (B).

Key Concepts

Rational Functions

To understand rational functions, picture them as the fraction form of two polynomials. A rational function is any function that can be represented by the equation \( f(x) = \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomial functions, and \( Q(x) eq 0 \). In our problem, the function \( f(x) = \frac{1}{u^2 + u - 2} \) is a rational function because its structure reflects that of a ratio between two polynomial expressions.

Here are some important notes about rational functions:

• The denominator Q(x) must not equal zero since division by zero is undefined.
• This type of function may have vertical asymptotes and points of discontinuity where the denominator equals zero.
• An understanding of the zeros of both the denominator and numerator can help in graphing rational functions.

Recognizing the type of function helps in understanding how to deal with it mathematically, including simplifying, finding discontinuities, and figuring its limits.

Quadratic Equations

Quadratic equations are polynomial equations of degree two generally formatted as \( ax^2 + bx + c = 0 \). These equations are central when dealing with rational functions, particularly in identifying discontinuity points.

Key characteristics of quadratic equations include:

• They always form a parabola when graphed.
• The solutions or roots can be found using factoring, completing the square, or the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \).
• Understanding the roots is crucial as they help determine where the function equals zero or is undefined.

In our case, we solved \( u^2 + u - 2 = 0 \) by using factoring as \( (u-1)(u+2) = 0 \), giving the roots \( u = 1 \) and \( u = -2 \). Understanding these roots helps translate the solutions to the variable \( x \) which plays a critical role in the broader context of solving for discontinuities.

Discontinuity Points

Discontinuity points on a graph are where the function is not continuous. These points are particularly crucial in rational functions because they can tell us where the function does not exist or behaves differently than usual.

For the function we've been studying, we found discontinuity points by determining where the denominator equals zero. The original function was \( f(x) = \frac{1}{u^2 + u - 2} \), and our task was to solve \( u^2 + u - 2 = 0 \). Upon solving, it revealed \( u = 1 \) and \( u = -2 \) as the keys to further determination of x-values.

It's important to:

• Translate these u-values back into x-values correctly.
• Consider any intrinsic discontinuity points introduced by the structure of the function, such as where \( x = 1 \) in our function where \( u = \frac{1}{x-1} \).
• Recognize that a point of discontinuity may also be a point of undefined behavior.

In conclusion, the discontinuity points found in this rational function analysis were \( x = \frac{1}{2}, 1, \) and \( 2 \). Identifying these points can guide in sketching the function and explaining its behavior across its domain.