Problem 41 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(f(x+y)+f(x-y)=2 f(x) f(y) \forall x, y \in R\) and \(f(0) \neq 0\), then determine that \(f(x)\) is an even function or odd function or neither.

Step-by-Step Solution

Answer

The function \(f(x)\) is neither even nor odd.

Step 1: Check for even function property

To check for the even function property, choose \(y = x\). Substitute this into the given equation to get: \(f(2x) + f(0) = 2f(x)f(x)\)

Step 2: Check for odd function property

To check for the odd function property, choose \(y = -x\). Substitute this into the given equation to get: \(f(0) + f(2x) = 2f(x)f(-x)\)

Step 3: Determine if the function is even, odd, or neither

Now, let's analyze the equations obtained in step 1 and step 2 to see if the function is even, odd, or neither. From Step 1, we get: \(f(2x) + f(0) = 2f(x)f(x)\), from which we cannot determine that \(f(x) = f(-x)\) because there is no term with a negative sign. Therefore, we cannot determine if the function is even from this equation. From Step 2, we get: \(f(0) + f(2x) = 2f(x)f(-x)\) However, we still cannot determine if \(f(x) = -f(-x)\) from this equation because there is no term with a negative sign associated with \(f\). Therefore, we cannot determine if the function is odd from this equation. Since we cannot determine if the function is even or odd, we conclude that the function is neither even nor odd.

Key Concepts

Even Function

An even function is a type of function that shows a specific symmetry. For a function to be classified as even, it must satisfy the condition: \( f(x) = f(-x) \) for all valid values of \( x \). This equation implies that the graph of the function is symmetric about the y-axis.
Understanding this concept through an example, consider the function \( f(x) = x^2 \). By substituting \( -x \) into the function, we get \( f(-x) = (-x)^2 = x^2 \). Therefore, \( f(x) = f(-x) \) holds true, confirming that \( f(x) = x^2 \) is an even function.
In the original exercise, when checking the even property, the given equation \( f(x+y) + f(x-y) = 2f(x)f(y) \) was transformed by letting \( y = x \), resulting in \( f(2x) + f(0) = 2f(x)f(x) \). However, solving this equation did not definitively lead to \( f(x) = f(-x) \). Thus, we could not conclude that \( f(x) \) was an even function.

Odd Function

An odd function is defined by its distinctive symmetry about the origin. For a function to be odd, it must meet the criterion: \( f(-x) = -f(x) \) for all \( x \). This implies that when you reflect the function's graph over the origin, it remains unchanged.
Take for example the function \( f(x) = x^3 \). By substituting \( -x \) for \( x \), you find \( f(-x) = (-x)^3 = -x^3 \). Hence, \( f(-x) = -f(x) \), confirming the odd nature of the function.
In the exercise, testing for an odd function involved substituting \( y = -x \) into the functional equation, resulting in \( f(0) + f(2x) = 2f(x)f(-x) \). However, analyzing this expression did not yield \( f(-x) = -f(x) \). Therefore, the function was not determined to be odd.

Symmetry in Functions

Symmetry in functions is a foundational aspect of understanding mathematical equations and graphs. It allows us to classify functions into even, odd, or neither, based on their graphical symmetry properties.

• Even functions display symmetry about the y-axis, such that the left and right sides of the graph are mirror images.
• Odd functions show symmetry about the origin, meaning if you rotate the graph 180 degrees about the origin, it appears unchanged.
• Functions that don't match the criteria for even or odd functions may not show any simple symmetry, making them neither even nor odd.

In the context of the original exercise, analysis of the given functional equation did not reveal any clear symmetry in the function \( f(x) \). Despite attempts to apply definitions of even and odd functions through substitution and evaluation, conclusions for symmetry about the y-axis or origin could not be drawn. Consequently, the function was classified as neither even nor odd, illuminating the complexity of symmetry in functional equations.