Problem 65 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f\left(\frac{x+y}{2}\right)=\frac{1}{2}[f(x)+f(y)]\) for real \(x\) and \(y .\) If \(f^{\prime}(0)\) exists and equals \(-1\) and \(f(0)=1\) then the value of \(f(2)\) is (A) 1 (B) \(-1\) (C) 0 (D) None of these

Step-by-Step Solution

Answer

The value of \( f(2) \) is -1.

Step 1: Identify Function Type

The functional equation given, \( f\left(\frac{x+y}{2}\right)=\frac{1}{2}[f(x)+f(y)] \), suggests that \( f \) is a linear function. This is because such equations often characterize linear functions.

Step 2: Assume a Form for the Function

Assume a linear form for \( f(x) \), such as \( f(x) = ax + b \). This is based on the linear nature suggested by the functional equation.

Step 3: Use Given Derivative Information

The information \( f'(0) = -1 \) implies that \( a = -1 \) because the derivative of \( ax + b \) is \( a \).

Step 4: Use Initial Condition to Solve for Constant

The condition \( f(0) = 1 \) implies \( b = 1 \). Substituting in our linear form \( a = -1 \) and \( b = 1 \), we get \( f(x) = -x + 1 \).

Step 5: Calculate \(f(2)\)

Substitute \( x = 2 \) into the function \( f(x) = -x + 1 \): \( f(2) = -(2) + 1 = -2 + 1 = -1 \).

Key Concepts

Understanding Linear Functions

Linear functions are among the simplest types of functions in mathematics. They are characterized by straight-line graphs and are defined by equations of the form \( f(x) = ax + b \), where \( a \) and \( b \) are constants. This form of equation is called "linear" because it describes a straight line when graphed on the coordinate plane.
Some attributes of linear functions:

• Slope \( a \): This determines the steepness of the line. A larger absolute value of \( a \) makes the line steeper.
• Y-intercept \( b \): This is where the line crosses the y-axis. It represents the value of \( f(x) \) when \( x = 0 \).

The reason linear functions fit the original problem is due to their additivity and scaling properties, which align well with the given functional equation. Identifying the function as linear simplifies solving for \( f(x) \) and determining specific values.

Exploring Derivative at a Point

The derivative of a function at a specific point gives us the slope of the tangent line to the function at that point. In simple terms, it tells us how the function is changing at that exact spot. For a linear function \( f(x) = ax + b \), the derivative \( f'(x) \) is constant and equals the slope \( a \).
In the context of the exercise, we are given \( f'(0) = -1 \), indicating the slope of the tangent line to \( f(x) \) at \( x = 0 \) is \(-1\).
This derivative information is crucial because:

• It helps us identify the coefficient \( a \) in the assumed linear function \( f(x) = ax + b \).
• It confirms the linear characteristic of the function since for non-linear functions, the derivative would typically vary with \( x \).

Overall, understanding derivatives as slopes helps simplify the process of identifying function parameters and confirming the function's nature.

Initial Conditions in Function Analysis

Initial conditions are specific values of a function at certain points which help in fully determining the function. They are essential when trying to identify the constants in a function's equation.
In the problem, the initial condition \( f(0) = 1 \) is given. This means when \( x = 0 \), the output of the function \( f(x) \) is 1.
Utilizing initial conditions effectively:

• Plug \( x = 0 \) into the linear equation \( f(x) = ax + b \). With \( f(0) = 1 \), it leads to \( b = 1 \).
• The initial condition allows us to solve for any unknown constant terms, in this case, \( b \), completing the definition of \( f(x) \).

Applying initial conditions is often the key to unlocking other pieces of a problem by clearly defining the specific behavior of a function at known points. This lays the foundation for confidently predicting the function's behavior at unknown points.