Problem 114 - Calculus for JEE ADVANCED

Question

$$ \left(x+\frac{1}{2}\right)+f\left(x-\frac{1}{2}\right)=f(x) $$ Replacing \(x\) by \(\left(x+\frac{1}{2}\right)\), we get $$ f(x+1)+f(x)=f\left(x+\frac{1}{2}\right) $$ of \(f(x+1)+f\left(x-\frac{1}{2}\right)=0\) (from (1) and (2)) or \(f\left(x+\frac{3}{2}\right)=-f(x)\) (Replacing \(x\) by \(\left(x+\frac{1}{2}\right)\) ) of \(f(x+3)=-f\left(x+\frac{3}{2}\right)=f(x) \quad\) (Replacing \(x\) by \(\left.\left(x+\frac{3}{2}\right)\right)\) Therefore, \(f(x)\) is periodic with period 3 .

Step-by-Step Solution

Answer

The function \( f(x) \) is periodic with period 3.

Step 1: Substitute and Simplify Original Equation

Given the equation \( \left(x+\frac{1}{2}\right)+f\left(x-\frac{1}{2}\right)=f(x) \), replace \( x \) with \( x + \frac{1}{2} \). This results in the equation \( (x + 1) + f(x) = f\left(x + \frac{1}{2}\right) \).

Step 2: Extract Key Relationships

From the equations derived, namely \( f(x+1) + f(x) = f\left(x + \frac{1}{2}\right) \) and the original \( f(x) + f\left(x - \frac{1}{2}\right) = f(x) \), deduce two relationships: \( f(x+1) + f\left(x - \frac{1}{2}\right) = 0 \) and \( f\left(x + \frac{3}{2}\right) = -f(x) \).

Step 3: Identify Periodicity

Substitute \( x \) with \( x + \frac{3}{2} \) in the relationship \( f\left(x + \frac{3}{2}\right) = -f(x) \) to obtain \( f(x+3) = f(x) \). This implies the function \( f(x) \) is periodic with a period of 3.

Key Concepts

Function Transformation

Function transformation involves altering the format or position of a function without changing its core properties. One common transformation is translating the function horizontally or vertically. In the exercise, by substituting \( x \) with \( x + \frac{1}{2} \), we effectively change the input of the function. This transformation helps us analyze the function by introducing a shift.

Through substitution, new equations can be derived, such as \( f(x+1) + f(x) = f(x + \frac{1}{2}) \). This indicates how the function responds when the input is shifted. Understanding these changes is critical for solving more complex problems, especially when looking for patterns or properties like periodicity.

Periodicity in Calculus

Periodicity refers to the tendency of a function to repeat its values at regular intervals. In this exercise, periodicity plays a crucial role in solving the problem.

The key was identifying that the function meets the condition \( f(x+3) = f(x) \). This equation shows that the function repeats every 3 units along the x-axis. Recognizing periodicity is essential because it allows us to predict the behavior of the function over different intervals.

• Periodic functions repeat their outputs at consistent intervals.
• This is a valuable tool in calculus for solving and simplifying functions.

Such understanding is especially useful in applying calculus concepts to real-world periodic phenomena.

Equations Solving Techniques

In solving equations, various techniques are utilized to find valid solutions. Sometimes, these involve substitutions to simplify the problem or to discover relationships between terms.

In this exercise, substitution was used to transform the original function and identify new relationships. By strategically replacing \( x \) with shifted versions like \( x + \frac{1}{2} \), the equations were manipulated to reveal crucial patterns. These steps lead to relationships such as \( f(x+1) + f\left(x-\frac{1}{2}\right) = 0 \).

Techniques like these help isolate parts of the equation, making it simpler to determine properties like periodicity or symmetry, ultimately leading to a clearer understanding of the function's behavior.