Problem 10 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

A function \(f(x)\) is defined as \(f(x)=\frac{x^{2}-4 x+3}{x^{2}-1}, x \neq 1\) \(=2, \quad x=1 .\)

Step-by-Step Solution

Answer

The function \(f(x) = \frac{x^2 - 4x + 3}{x^2 - 1}\) for \(x \neq 1\) has a limit of -1 as \(x\) approaches 1. However, the function is defined to be 2 when \(x = 1\), indicating that the function is not continuous at \(x = 1\).

Step 1: Analyze the function for \(x \neq 1\)

For \(x \neq 1\), we have the function as: \(f(x) = \frac{x^2 - 4x + 3}{x^2 - 1}\)

Step 2: Find the limit of the function as \(x\) approaches 1

Now, let's find the limit of \(f(x)\) as \(x\) approaches 1; that is \(\lim_{x \to 1} f(x)\). First, we will factor the numerator and denominator: \(\lim_{x \to 1} \frac{(x - 1)(x - 3)}{(x - 1)(x + 1)}\) Next, we cancel out the common factor \((x-1)\) from the numerator and the denominator, noting that \(x\) cannot equal 1 in this case: \(\lim_{x \to 1} \frac{x - 3}{x + 1}\) Now, we substitute \(x = 1\) in the limit function and evaluate: \(\lim_{x \to 1} \frac{1 - 3}{1 + 1} = \frac{-2}{2} = -1\)

Step 3: Verify the value of the function when \(x = 1\)

According to the given problem, the value of \(f(x)\) when \(x = 1\) is 2. \(f(1) = 2\) This means that the function is not continuous at \(x = 1\), as the limit when \(x\) approaches 1 is -1, but the function is defined to be 2 when \(x = 1\).

Key Concepts

Function Analysis

When we are dealing with functions, it is important to understand the behavior of the function at various points, especially points where the function might change its nature. The function provided in this exercise is a rational function, as it is expressed as the ratio of two polynomials.

• The function is defined differently for the point \(x=1\) than it is for all other points, showcasing an example of a piecewise function.
• For \(x eq 1\), the function is \(f(x) = \frac{x^2 - 4x + 3}{x^2 - 1}\).
• For \(x = 1\), the function is given as \(f(x) = 2\).

Understanding when functions change their definition helps in analyzing their behavior and determining continuity and limits.

Limit Evaluation

Evaluating the limit of a function helps us understand the behavior of the function as the input approaches a specific value. In this case, we are interested in the limit of \(f(x)\) as \(x\) approaches 1.

• The limit expression for \(f(x)\) as \(x\) tends towards 1 is: \[ \lim_{x \to 1} \frac{x^2 - 4x + 3}{x^2 - 1} \]
• Factor both the numerator and the denominator to simplify the limit expression. This ends up as: \[ \lim_{x \to 1} \frac{(x - 1)(x - 3)}{(x - 1)(x + 1)} \]
• Since \(x eq 1\), we can cancel out the \((x-1)\) factor, leaving us with: \[ \lim_{x \to 1} \frac{x - 3}{x + 1} \]
• Substitute \(x = 1\) into the simplified expression to find: \[ \frac{-2}{2} = -1 \]

Thus, the limit of the function as \(x\) approaches 1 is \(-1\).

Piecewise Functions

Piecewise functions are defined in segments based on different conditions or intervals. In this case, the function is modified such that it has a specific value at one point, different from its general formula.

• For most values of \(x\) (except \(x = 1\)), the function follows a standard form: \(f(x) = \frac{x^2 - 4x + 3}{x^2 - 1}\).
• However, for \(x = 1\), the value of \(f(x)\) is specifically defined as \(2\).
• This definition causes a discontinuity at \(x = 1\) because the limit of the function at this point \(-1\) does not match the actual function value \(2\).

Understanding piecewise functions and their discontinuities is essential in calculus as it involves evaluating the continuity and behavior of functions at specified intervals.