A real valued function f(x) satisfies the functional equation f(x - y) = f(x)f(y) - f(a - x)f(a + y)

Q: Why did we substitute $y = x$ at the very beginning? Is that just a random guess?

It is not a guess, it is a strategy! We saw $f(0) = 1$ in the problem statement. By setting $y = x$, the left side becomes $f(x - x) = f(0)$. This is the 'key' that unlocks the door to the rest of the problem, allowing us to simplify the equation significantly.

Q: Wait, how did we know $f(a) = 0$? Is it possible for $f(a)$ to be something else?

Great question! We did not 'know' it; we derived it. By substituting $x = 0$ into the equation $1 = f(x)^2 - f(a - x)f(a + x)$, we forced the equation to reveal the value of $f(a)$. It is a logical necessity, not an assumption. Keep trusting the algebra!

Q: Can I just assume $f(x)$ is a specific function like $\cos(x)$ or $\sin(x)$ and solve it that way?

While guessing a function might work in a multiple-choice setting, it is dangerous in JEE Advanced. If you guess wrong, you lose marks. The algebraic method we used is robust and guaranteed to work regardless of the function's form. Always prioritize derivation over intuition.

Question

A real valued function $f(x)$ satisfies the functional equation $f(x - y) = f(x)f(y) - f(a - x)f(a + y)$ where $a$ is a given constant and $f(0) = 1, f(2a - x)$ is equal to

Options:

(A) $-f(x)$

(B) $f(x)$

(C) $f(a) + f(a - x)$

(D) $f(-x)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future IITians! Today, we are going to embark on a journey into the heart of functional equations. These problems are the ultimate test of your mathematical intuition and algebraic discipline.

They are not just about plugging in numbers; they are about uncovering the hidden symmetry of a function. Imagine you are a detective, and the functional equation

$$f(x - y) = f(x)f(y) - f(a - x)f(a + y)$$

is your crime scene. We have a few clues: a real-valued function $f(x)$, a constant $a$, and the golden key, $f(0) = 1$. Our mission is to find the value of $f(2a - x)$.

### Phase 1: The First Breakthrough

Functional equations are notoriously intimidating because they define a function by its relationship with itself. The most powerful tool in your arsenal is strategic substitution.

We need to find intermediate values to simplify the landscape. Given $f(0) = 1$, we look at the left-hand side of our equation: $f(x - y)$. If we set $y = x$, the left side becomes $f(x - x) = f(0)$.

Let us perform this substitution:

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

$$f(0) = f(x)^2 - f(a - x)f(a + x)$$

Since we know $f(0) = 1$, we have:

$$1 = f(x)^2 - f(a - x)f(a + x)$$

This is a massive simplification. We have reduced a complex functional equation into a relationship between $f(x)$ and the product of two other terms.

### Phase 2: Isolating the Constant

Now, we need to find the value of $f(a)$. How do we isolate it? Look at our new equation: $1 = f(x)^2 - f(a - x)f(a + x)$.

If we set $x = 0$, we get:

$$1 = f(0)^2 - f(a - 0)f(a + 0)$$

Since $f(0) = 1$, this becomes:

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

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

Solving this simple algebraic equation, we find:

$$f(a)^2 = 0 \Rightarrow f(a) = 0$$

This is a beautiful moment! We have discovered that at the point $a$, the function vanishes. This will be the key to our final step.

### Phase 3: The Final Transformation

Now, let us refocus on our ultimate target: finding $f(2a - x)$. We need to choose $x$ and $y$ in the original equation such that $x - y = 2a - x$.

Let us substitute $x = a$ and $y = x - a$. The left side becomes:

$$f(a - (x - a)) = f(a - x + a) = f(2a - x)$$

This is exactly what we wanted! Now, let us apply this to the right side of the original equation:

$$f(a)f(x - a) - f(a - a)f(a + (x - a))$$

Substituting $f(a) = 0$ and $f(0) = 1$:

$$0 \cdot f(x - a) - f(0)f(x)$$

$$0 - 1 \cdot f(x) = -f(x)$$

And there it is! The entire expression collapses into $-f(x)$.

The elegance of this cancellation is why we love mathematics. We started with a terrifying functional equation and, through careful, logical steps, arrived at the final result:

**$f(2a - x) = -f(x)$**

Animated Solution for Mathematics - Functions: A real valued function $f (x)$ satisfies the functional equation $f (x - y) = f (x) f (y) - f (a - x) f (a + y)$ where $a$ is a given constant and $f (0) = 1, f (2 a - x)$ is equal to

Select Answer:

Visualized Solution

The Functional Equation

Our Goal

Substituting $y = x$

Simplifying with $f (0)$

Substituting $x = 0$

Calculating $f (a)$

Setting up for $f (2 a - x)$

Applying $x = a$ and $y = x - a$

Simplifying the Expression

Substituting Known Values

The Final Result

The Sigma Insight: Classification of Functions

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

The Functional Equation

Our Goal

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=x

Simplifying with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0)

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Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(a)

Setting up for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(2a−x)

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