Problem 90 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f: R \rightarrow R\) be a function satisfying the condition \(f\left(\frac{x+y}{k}\right)=\frac{f(x)+f(y)}{k}\), where \(k \neq 0,2\). The function \(f(x)\) is differentiable on \(R\) and \(f^{\prime}(0)=m\). The function \(f(x)\) is equal is (A) \(m x\) (B) \(m x+1\) (C) \(-2 m x\) (D) None of these

Step-by-Step Solution

Answer

The function is \(f(x) = mx\), option (A).

Step 1: Function Property Setup

We are given that the function \(f\) satisfies the condition \(f\left(\frac{x+y}{k}\right)=\frac{f(x)+f(y)}{k}\). Begin by examining this functional equation. Let \(y = 0\), then the equation becomes \(f\left(\frac{x}{k}\right) = \frac{f(x) + f(0)}{k}\). This gives us the relation: \(k \cdot f\left(\frac{x}{k}\right) = f(x) + f(0)\).

Step 2: Differentiation at Zero

Since \(f(x)\) is differentiable and \(f'(0) = m\), consider the derivative by substituting \(x = 0\) in the functional equation. Differentiating both sides with respect to \(x\) at \(x = 0\), \(\frac{d}{dx}\left[f\left(\frac{x+y}{k}\right)\right] = \frac{1}{k} f'(0)\). So \(f'(0) = m = k \cdot f'(\frac{0}{k})\). This relation confirms that the derivative at zero yields \(f'(0) = m\).

Step 3: Analysing Linear Function

The form of \(f\) is likely to be linear because differentiability at zero straightforwardly suggests a polynomial form of degree one or less for \(f(x)\). Set \(f(x) = m \cdot x + c\), and find \(c\) by using derived properties. Notice \(f(0)=0\) because the function is homogeneous as seen from the symmetry in the functional equation (\(f(0)/k = f(0)\)).

Step 4: Validation

Evaluate \(f(x) = m \cdot x\) given \(f(0) = 0\). Verify: \(f\left(\frac{x+y}{k}\right) = \frac{m\cdot(x+y)/k}{1} = \frac{m\cdot x}{k} + \frac{m\cdot y}{k} = \frac{f(x)+f(y)}{k}\). Thus, \(f(x) = m \cdot x\) satisfies the functional equation.

Key Concepts

Differentiability

Differentiability is an essential concept in calculus. It tells us about the smoothness of a function. When a function is differentiable at a point, it means we can find its derivative at that point. This derivative gives us a precise rate of change or slope at any given location on the function's curve.
To say a function is differentiable everywhere, like in our exercise, means we can do this process at every point in its domain. This ensures there are no sharp turns or corners. If a function isn't differentiable at a certain point, it might still be continuous there, but with a corner or cusp, it won't have a clear slope.
In our specific exercise, differentiability at zero is crucial because it sets the foundation for testing if the function could be linear, which would have a constant slope throughout its span.

Linear Function

A linear function is one of the simplest types of functions. It is represented by the equation:
\[ f(x) = mx + c \]
where \( m \) is the slope and \( c \) is the y-intercept. Linear functions have a constant rate of change. This means that the graph of a linear function forms a straight line.
The concept of linearity comes in handy when studying functional equations, as in the problem given. The initial analysis may lead you to consider that the solution is a linear function due to how differentiability interacts with constant rates or slopes.

• Constant slope: The function increases or decreases at a consistent rate.
• Form of the line: Clearly, a straight line indicates no change in the slope.
• Simple graphing: These functions are predictable and easy to work with.

In the exercise's context, if \( f(x) = mx \), the properties of linear functions ensure it satisfies the given conditions.

Polynomial Function

Polynomial functions are more complex than linear functions. They consist of terms that are powers of \( x \), like \( f(x) = ax^n + bx^{n-1} + \, \ldots \, + d \). The degree of the highest power determines the polynomial's degree.
These functions can be linear (degree 1), quadratic (degree 2), cubic (degree 3), and so on. Each degree has varied curves and forms, giving unique behavior regarding roots and symmetry.
In the current problem, you might initially consider both linear and polynomial forms, given the differentiability condition, but ultimately, a linear function becomes most fitting due to constraints posed by differentiability and the functional equation's symmetry.

• Varying degrees: More complex behaviors as degrees increase
• Potential turning points: Depending on the degree, could have multiple peaks and valleys
• Symmetry: Many polynomial functions display symmetrical properties, especially evenly powered ones

The task simplifies considerably when realizing only first-degree (linear) forms satisfy the exercise specifics when evaluated under the conditions provided.

Derivative

The derivative is a fundamental tool in calculus, capturing the idea of change. It's defined as the limit of the function's average rate of change as the interval approaches zero. In simpler terms, it gives the slope of the tangent at any point on a curve.
Mathematically, it's expressed as: \[f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}\] Derivatives help us understand how one quantity changes with another. In our exercise, the derivative at zero, \(f'(0) = m\), signifies how the function is initially changing right at the origin.
This initial slope is key to determining the function's linear nature. Because linear functions have constant derivatives, knowing the derivative at zero helps form the conclusion that the function takes on the simple form \(f(x) = mx\).
By recognizing that any deviation in the form would alter this derivative, it aligns with our purpose."