Problem 33 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2} \forall x, y\) and \(f(x)\) is continuous at \(x=0\), then check the continuity of \(f(x)\).

Step-by-Step Solution

Answer

The function \(f(x)\) is continuous for all \(x\) in its domain.

Step 1: Using the mid-point theorem property

The given property of the function \(f(x)\), i.e., \(f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2} \forall x, y\), tells us that the function produces the average of \(f(x)\) and \(f(y)\) when it operates on the average of \(x\) and \(y\). This is a hint towards continuity of the function in general as continuity is related to the idea that small changes in input yield small changes in output.

Step 2: Employing the definition of continuity

By definition, a function is continuous at a point if the limit of the function as x approaches that point from both sides is equal to the function's value at that point. Since it's given that the function is continuous at \(x=0\), we need to prove this for every other x-value. We will hence use the equation and the given condition of continuity at \(x = 0\).

Step 3: Proving continuity for all x

Now we choose \(y = 0\) and \(x = 2h\) in the given equation. We get \(f(h) = \frac{f(2h)}{2}+ \frac{f(0)}{2}\). Since function is continuous at zero, as \(h\) approaches to zero, \(f(h)\) becomes equal to \(f(0)\). Hence, the function \(f(x)\) is continuous at all \(h\). Which indicates that the function \(f(x)\) is continuous everywhere in its domain.

Key Concepts

Mid-point Theorem

The Mid-point Theorem is a fascinating and useful concept in mathematics. It provides important insights about the structure of functions. In our specific problem, the expression \( f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2} \) illustrates the mid-point property. What this means is that the function value at the average of two points \( x \) and \( y \) is actually the average of the function values at these points.
This is a special kind of symmetry property which suggests smoothness in the behavior of the function.

• It tells us that the function does not make sudden jumps around the mid-point.
• In general, a function with such a mid-point property hints at uniform behavior across its domain.
• This aligns with characteristics of continuous functions that do not show abrupt changes.

Thus, the mid-point theorem is a strong indicator that the function might be continuous not only at the investigated point but possibly throughout its domain.

Definition of Continuity

Understanding the definition of continuity is key to determining if a function is continuous. According to mathematics, a function is continuous at a point \(x = a\) if:
- The limit of the function as \(x\) approaches \(a\) exists.
- The limit is equal to the function value at that point, \(f(a)\).
- The function is defined at that point, meaning \(f(a)\) exists.

For the function \(f\), it was given that it is continuous at \(x=0\). Leveraging this information means that we know:

• \(\lim_{x \to 0} f(x) = f(0)\).
• This equality holds from both directions (from the left and the right).

Continuity is a critical concept, as it ensures the function's behavior is predictable and smooth at that point. In our solution, we extend this definition to validate the continuity of the function for every other value by showing how the function behaves uniformly across its range.

Function Continuity

The overall continuity of a function is an important aspect in calculus and analysis. For our function \(f(x)\), proving it is continuous everywhere requires using given conditions effectively. We know it is continuous at \(x=0\) and the mid-point theorem property holds.

To establish continuity for all \(x\), we use:

• Setting \(y = 0\) and \(x = 2h\), we derive that \(f(h) = \frac{f(2h)}{2} + \frac{f(0)}{2}\).
• As \(h\) approaches zero, since \(f(x)\) is continuous at zero, \(f(h)\) tends to \(f(0)\).

This argument extends to all \(h\), showing that the function is continuous throughout its domain.
This strategy demonstrates the function's invariant continuity and shows how to leverage specific conditions and theorem properties to generalize results. Thus, confirming that \(f(x)\) is continuous for every value within the function's domain efficiently combines these pieces of information.