Problem 12 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Let \(\begin{aligned} f(x) &=\frac{x^{2}-1}{x^{2}-2|x-1|-1}, \quad x \neq 1 \\\ &=\frac{1}{2}, \quad x=1 \end{aligned}\)

Step-by-Step Solution

Answer

The given function, \(f(x) = \frac{x^2-1}{x^2 - 2|x-1| -1}\) when \(x \neq 1\) and \(f(1) = \frac{1}{2}\), is continuous at \(x=1\), as both the limits when \(x \rightarrow 1^+\) and \(x \rightarrow 1^-\) exist and are equal to \(f(1)=\frac{1}{2}\). The final function can be expressed as: \[ f(x) = \begin{cases} \frac{x^2-1}{x^2 - 2|x-1| -1} & \text{if } x \neq 1 \\ \frac{1}{2} & \text{if } x = 1 \\ \end{cases} \]

Step 1: Simplifying the expression when \(x \neq 1\)

We will now simplify the expression \(\frac{x^2-1}{x^2-2|x-1|-1}\) for \(x \neq 1\). We can use the difference of squares for the numerator like this: \(x^2 - 1 = (x+1)(x-1)\) Now, let's analyze the denominator: \(x^2 - 2|x-1| -1\) As we know, if\(x > 1\), then \(|x-1| = x-1\), and if \(x < 1\), then \(|x-1| = -(x-1) = 1-x\). So, we will write the function as two separate expressions: For \(x > 1\), \(f(x) = \frac{(x+1)(x-1)}{x^2-2(x-1) -1}\) For \(x < 1\), \(f(x) = \frac{(x+1)(x-1)}{x^2-2(1-x) -1}\) Notice this does not cover the case for \(x=1\), which is already defined by the given function as \(f(1)=\frac{1}{2}\).

Step 2: Analyzing continuity at \(x=1\)

In order to prove the function is continuous at \(x=1\), we need to show that the limit of f(x) as x approaches 1 from both sides exists, and is equal to \(f(1)=\frac{1}{2}\). Observe the limits for \(x \rightarrow 1^+\) and \(x \rightarrow 1^-\): \(\lim_{x \rightarrow 1^+} \frac{(x+1)(x-1)}{(x^2)-(2(x-1))-1} = \frac{1}{2}\) \(\lim_{x \rightarrow 1^-} \frac{(x+1)(x-1)}{x^{2}-2(1-x)-1} = \frac{1}{2}\) Since both of these limits exist and are equal to \(f(1)=\frac{1}{2}\), f(x) is continuous at \(x=1\).

Step 3: Final function and result

Now we have analyzed and simplified the function for different cases, we can present it as follows: \[ f(x) = \begin{cases} \frac{x^2-1}{x^2 - 2|x-1| -1} & \text{if } x \neq 1 \\ \frac{1}{2} & \text{if } x = 1 \\ \end{cases} \] The function f(x) is continuous at \(x=1\) and is defined for all values of x.

Key Concepts

Mathematical Limits

Understanding mathematical limits is essential when analyzing the continuity of a function. A limit describes the behavior of a function as the input approaches a particular point. When assessing continuity at a specific point, like in our exercise, we look at the limit of the function as the input approaches from both the left and the right.

For our function, the aim was to verify continuity at the point \(x = 1\). This involves ensuring that the left limit \(\lim_{{x \to 1^-}}\) and the right limit \(\lim_{{x \to 1^+}}\) both exist and equal the value of the function at \(x = 1\), which is given as \(\frac{1}{2}\).

In continuity concepts:

• The function must be defined at the point, which it is since \(f(1) = \frac{1}{2}\).
• The function's limit as \(x\) approaches from the left and right limits must both exist.
• The value of these limits must match the function's value at the point in question.

By considering both sides and incorporating them into our original function, we confirmed that \(f(x)\) was indeed continuous at \(x=1\).

Piecewise Functions

Piecewise functions are defined by different expressions depending on the value of the independent variable. In the given function, a piecewise structure ensures that the different behaviors around \(x=1\) are represented accurately while addressing the complexities introduced by the absolute values in the denominator.

When creating a piecewise function:

• The range of the independent variable is divided into intervals.
• Each interval has a corresponding expression detailing how the function behaves in that segment.
• Special cases, such as \(x=1\) here, might need specific expressions to ensure continuity or satisfy given conditions.

For our function, with \(f(x)\):

• If \( x eq 1 \), the function is described by an expression involving \(\frac{x^2-1}{x^2 - 2|x-1| -1}\).
• If \( x = 1\), it directly uses the constant \(\frac{1}{2}\).

This structure helps address gaps or discontinuities that might arise without special case handling. The separate conditions ensure correct behavior across all values.

Absolute Value Properties

Absolute value expressions often introduce conditional behaviors in functions due to their nature of evaluating a number's distance from zero. This characteristic is essential when simplifying or evaluating expression limits, as seen in our exercise.

For absolute value \(|x-1|\) in particular, we notice its impact on continuity through its conditional definition:

• If \(x > 1\), then \(|x-1| = x - 1\).
• If \(x < 1\), then \(|x-1| = 1 - x\).

In piecewise functions, knowing how to smart-swap these values helps build accurate segment-specific functions. By applying these absolute value properties, the expression \(x^2 - 2|x-1| - 1\) became one that could adapt for \(x\) values greater than, less than, and including 1 by swapping in the correct forms.

Proper application of these principles not only aided in simplifying our expression but also helped determine the limits and verify the continuity of the function at the point \(x = 1\). Accordingly, this was crucial in demonstrating that the function remained continuous at \(x = 1\), unlike functions without suitable adjustments.