Let f((x+y)/2) = (f(x)+f(y))/2 for all real x and y. If f'(0) exists and equals -1 and f(0) = 1, fin

Q: Why does this equation force the function to be linear? Couldn't it be a parabola?

Great question! A parabola has changing curvature. If you take the midpoint of two points on a parabola, the average of the function values will not equal the function value at the midpoint. Only a straight line, where the rate of change is constant, satisfies this property perfectly.

Q: Why is $f'(0) = a$?

Since we established that $f(x) = ax + b$, the derivative $f'(x)$ is simply the derivative of $ax + b$ with respect to $x$, which is $a$. Because $a$ is a constant, $f'(x)$ is $a$ for any value of $x$, including $x=0$.

Q: What if the function wasn't differentiable? Would Jensen's equation still imply linearity?

Actually, for continuous functions, Jensen's equation is sufficient to prove linearity. Even without assuming differentiability initially, the midpoint property is strong enough to force the function to be linear. The fact that $f'(0)$ exists just makes our job easier!

Question

Let $f\left(\frac{x+y}{2}\right) = \frac{f(x)+f(y)}{2}$ for all real $x$ and $y$. If $f'(0)$ exists and equals $-1$ and $f(0) = 1$, find $f(2)$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey. Today, we are going to unravel a problem that might look like a cryptic puzzle at first glance, but beneath the surface, it holds a beautiful, elegant truth.

We are dealing with a functional equation:

$$f\left(\frac{x+y}{2}\right) = \frac{f(x)+f(y)}{2}$$

This isn't just a random collection of symbols; it is a profound statement about the nature of a function. Imagine you are standing on a graph. This equation tells you that if you pick any two points on the $x$-axis, the function's value at their midpoint is exactly the average of the function's values at those two points.

In the world of mathematics, this is a signature property. It is the hallmark of a straight line. This is known as **Jensen's Functional Equation**, and it is a classic trap-catcher in competitive exams because students often try to overcomplicate it with complex calculus when the answer lies in the simple, rigid geometry of a line.

### The Linear Assumption

Once we recognize the Jensen property, the path forward becomes clear. If a function satisfies this midpoint property, it must be linear.

So, we can confidently assume the general form of a linear function:

$$f(x) = ax + b$$

Here, $a$ is the slope, and $b$ is the $y$-intercept. This is our toolkit. We have reduced a terrifying-looking functional equation into a simple algebraic model.

Now, our mission is to find the specific values of $a$ and $b$ that define our unique function. We have two clues, or boundary conditions, provided in the problem statement. These are our keys to unlocking the mystery.

### Decoding the Clues

First, we are given $f(0) = 1$. Geometrically, this is a gift. It tells us that when $x=0$, the function value is $1$.

If we plug $x=0$ into our linear form $f(x) = ax + b$, we get:

$$f(0) = a(0) + b \Rightarrow b = 1$$

Just like that, we have found our $y$-intercept! The line crosses the $y$-axis at $1$.

Now, we need the slope $a$. The problem gives us $f'(0) = -1$. Since our function is $f(x) = ax + 1$, its derivative is simply $f'(x) = a$.

Because the slope of a straight line is constant, the derivative is the same everywhere, including at $x=0$. Therefore:

$$f'(0) = a = -1$$

We have found both our constants: $a = -1$ and $b = 1$.

### The Final Reveal

With $a = -1$ and $b = 1$, our mystery function is fully revealed:

$$f(x) = -x + 1$$

This is the line that satisfies all the conditions. Now, the final step is to find $f(2)$.

We simply substitute $x=2$ into our function:

$$f(2) = 1 - 2 = -1$$

The final answer is **$-1$**. It is elegant, it is precise, and it is satisfying. Remember, in JEE, the most complex-looking problems often have the most beautiful, simple solutions if you can just see the underlying structure.

Animated Solution for Mathematics - Differentiation: Let $f (\frac{x + y}{2}) = \frac{f ( x ) + f ( y )}{2}$ for all real $x$ and $y$ . If $f^{'} (0)$ exists and equals $- 1$ and $f (0) = 1$ , find $f (2)$ .

Visualized Solution

Analyzing the Functional Equation

Jensen's Functional Equation

Assuming the Linear Form

Applying the First Boundary Condition

Calculating the Intercept $b$

Differentiating the Function

Applying the Derivative Condition

Formulating the Exact Function

Calculating $f (2)$

Final Answer and Geometric Verification

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

The Linear Assumption

Decoding the Clues

The Final Reveal

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(2x+y​)=2f(x)+f(y)​ for all real x and y. If f′(0) exists and equals −1 and f(0)=1, find f(2).