Problem 74 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x-y), f(x) \cdot f(y)\) and \(f(x+y)\) are in A.P. for all \(x, y\) and \(f(0) \neq 0\), then (A) \(f(2)=f(-2)\) (B) \(f(3)+f(-3)=0\) (C) \(f^{\prime}(2)+f^{\prime}(-2)=0\) (D) \(f^{\prime}(3)=f^{\prime}(-3)\)

Step-by-Step Solution

Answer

The correct answer is (A) and (D).

Step 1: Understanding the Problem

Given that \( f(x-y), f(x) \cdot f(y) \), and \( f(x+y) \) are in arithmetic progression (A.P.). In an A.P., the difference of middle term with first term is equal to the difference of last term with the middle term. Hence, the condition becomes \[ 2f(x) \cdot f(y) = f(x-y) + f(x+y). \] We also know \( f(0) eq 0 \).

Step 2: Simplifying the Given Condition

Substitute \( y = 0 \) in the equation: \( 2f(x)f(0) = f(x) + f(x) \). Simplifying this we get \[ 2f(x)f(0) = 2f(x). \] Since \( f(0) eq 0 \), it can be concluded that \( f(x) = f(0) \). Thus, \( f \) is a constant function.

Step 3: Verifying the Options

Since \( f(x) = f(0) \), it implies \( f \) is constant, so whatever the value of \( x \) or \( y \) is, \( f(x) = f(y) \). Examining the options:- \( A) \) \( f(2) = f(-2) \) is true given \( f(x) = f(y) \).- \( B) \) \( f(3) + f(-3) = 0 \) is false since \( f(3) = f(-3) \), hence \( f(3) + f(-3) = 2f(0) eq 0\).- \( C) \) \( f'(2) + f'(-2) = 0 \) does not apply as the derivative of a constant function is zero.- \( D) \) \( f'(3) = f'(-3) \) is true since the derivative of a constant function is zero.

Key Concepts

Constant Function

A constant function is a very simple yet fundamental concept in mathematics. It is defined as a function that always returns the same value, no matter what input you give it. In mathematical terms, if a function \( f \) is constant, then there exists some constant \( c \) such that for every input \( x \), \( f(x) = c \). This means that the graph of a constant function is a horizontal line parallel to the x-axis.

Here are some key points about constant functions:

• They do not change in value and therefore have no slope.
• The derivative of a constant function is always zero, because there’s no rate of change.
• In our exercise, since \( f(x) = f(0) \), this indicates \( f \) is a constant function. Thus, \( f(x) = c\) for some constant \( c \).

Constant functions are crucial when learning about more complex functions as they set a foundationally stable behavior and often serve as bounds in calculus.

Differential Calculus

Differential calculus focuses on the concept of change. It provides tools to determine how a function changes as its input changes. The fundamental instrument in differential calculus is the derivative, which measures the rate of change or the slope of the function at any point.

For a function \( f(x) \), the derivative is denoted as \( f'(x) \) and is calculated as the limit:
\[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h}\]

When a function is constant, as in the case in our exercise, every derivative for any value of \( x \) is zero. This physical interpretation is that there is no change at any point, and hence no slope.

Key takeaways include:

• The derivative of a constant function is always zero.
• It quickly verifies functional status: zero derivative implies constancy.
• In our context, options involving derivatives (as in the choices C and D) leverage the fact that any derivative of our constant function, \( f \), remains zero.

Understanding derivatives is crucial not only in mathematics but also in fields like physics and engineering, where predicting change is essential.

Functional Equations

Functional equations are equations in which the unknowns are functions rather than simple variables. Solving these equations involves finding a function or set of functions that satisfy certain conditions given by the equation.

In our exercise, a functional equation arose from putting conditions on \( f(x-y), f(x) \cdot f(y) \), and \( f(x+y) \) being in arithmetic progression. The requirement introduces an equation:
\[2f(x) \,f(y) = f(x-y) + f(x+y)\]

This equation had to be satisfied for all \( x \) and \( y \). By exploring provided conditions such as \( f(0) eq 0 \), we discovered that the functional equation leads to a constant solution \( f(x) = f(0) \).

Points to solidify understanding include:

• Functional equations may involve operations like addition, multiplication of the function's arguments.
• Recognizing patterns such as common function forms aids in simplifying solutions.
• These are applicable in finding consistent solutions for problems defined across variable domains.

Overall, functional equations help us explore deeper relationships between variables and pave the way for understanding symmetries and consistencies in mathematical problems.