If f(x-y) = f(x) · g(y) - f(y) · g(x) and g(x-y) = g(x) · g(y) + f(x) · f(y) for all x, y ∈ R. If ri

Q: Wait, why do we assume $g(0) = 1$? Couldn't $g(0)$ be $0$?

That is a fantastic question! If we look at the equation $g(0) = g(0)^2 + f(0)^2$, and we know $f(0) = 0$, then $g(0) = g(0)^2$. This implies $g(0)(g(0) - 1) = 0$. So, $g(0)$ could be $0$ or $1$. However, if $g(0) = 0$, then $g(x) = 0$ for all $x$ (the trivial solution). In JEE problems, we are almost always looking for the non-trivial solution unless specified otherwise. If $g(x)$ were identically zero, the derivative would be zero anyway, but assuming non-triviality is standard practice.

Q: Can I just assume $f(x) = \sin(x)$ and $g(x) = \cos(x)$ and differentiate them?

While your intuition is spot on—these equations are indeed the defining properties of sine and cosine—in a rigorous JEE Advanced exam, simply assuming the functions is risky. You must prove the properties (like symmetry and the derivative value) using the functional equations provided. Using the functions as a 'mental model' is excellent, but the derivation must be algebraic to earn full marks.

Q: Why does $g'(0) = -g'(0)$ imply $g'(0) = 0$? Is there any other way?

Algebraically, if $x = -x$, then $2x = 0$, which forces $x = 0$. There is no other real number that satisfies this condition. Conceptually, this result tells us that the slope of the tangent at the peak of an even function must be horizontal. If the slope were positive or negative, the function would be increasing or decreasing at the origin, which would violate the symmetry of the even function.

Question

If $f(x-y) = f(x) \cdot g(y) - f(y) \cdot g(x)$ and $g(x-y) = g(x) \cdot g(y) + f(x) \cdot f(y)$ for all $x, y \in R$. If right hand derivative at $x=0$ exists for $f(x)$. Find derivative of $g(x)$ at $x=0$.

Sigma Percentile · Accepted Answer

### The Beauty of Functional Equations

Imagine you are standing at the edge of a vast, uncharted mathematical landscape. You are given two mysterious equations:

$$f(x-y) = f(x)g(y) - f(y)g(x)$$

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

To the untrained eye, these are just abstract symbols. But to a JEE aspirant, they whisper secrets. They are the structural DNA of the sine and cosine addition formulas. While we won't blindly assume they are trigonometric functions, we will use that intuition to guide our journey. Our mission is to find the derivative of $g(x)$ at $x=0$.

### Phase 1: The Origin

In any functional equation, the origin is our best friend. It is the anchor point. Let us substitute $x=0$ and $y=0$ into our first equation:

$$f(0-0) = f(0)g(0) - f(0)g(0)$$

The right side collapses into nothingness, leaving us with $f(0) = 0$. This is a profound realization! It tells us that the function $f(x)$ is pinned to the origin.

Now, let us apply the same logic to the second equation:

$$g(0-0) = g(0)g(0) + f(0)f(0)$$

Since $f(0)=0$, this simplifies to $g(0) = g(0)^2$. For a non-trivial solution, $g(0)$ must be $1$. We have our coordinates: $f(0)=0$ and $g(0)=1$.

### Phase 2: The Hidden Geometry

Now, let us uncover the deeper relationship between these two functions. Substitute $y=x$ into the second equation:

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

The left side becomes $g(0)$, which we know is $1$. The right side becomes $g(x)^2 + f(x)^2$.

Thus, we arrive at the beautiful identity:

$$f(x)^2 + g(x)^2 = 1$$

This is the fundamental trigonometric identity! It tells us that the values of $f(x)$ and $g(x)$ are bounded, living on a unit circle in the phase plane.

### Phase 3: The Symmetry

Next, we must understand the symmetry of $g(x)$. Let us substitute $x=0$ and $y=x$ into the second equation:

$$g(0-x) = g(0)g(x) + f(0)f(x)$$

Using our known values, this becomes $g(-x) = (1)g(x) + (0)f(x)$, which simplifies to $g(-x) = g(x)$.

This is the definition of an **even function**. Just like the cosine function, $g(x)$ is perfectly symmetric across the y-axis.

### Phase 4: The Calculus

We are now ready for the final act. We need to find $g'(0)$. By the first principle of derivatives:

$$g'(0) = \lim_{h 	o 0} \frac{g(h) - g(0)}{h}$$

This looks daunting, but remember our symmetry property! Since $g(x)$ is even, $g(h) = g(-h)$. Let us substitute this into our limit:

$$g'(0) = \lim_{h 	o 0} \frac{g(-h) - g(0)}{h}$$

Now, let us perform a change of variable. Let $k = -h$. As $h 	o 0$, $k$ also approaches $0$. The expression becomes:

$$g'(0) = \lim_{k 	o 0} \frac{g(k) - g(0)}{-k}$$

We can pull that negative sign out:

$$g'(0) = - \lim_{k 	o 0} \frac{g(k) - g(0)}{k}$$

Look closely—the limit on the right is exactly the definition of $g'(0)$! We have arrived at the elegant conclusion:

$$g'(0) = -g'(0)$$

### The Final Conclusion

What real number is equal to its own negative? Only zero. Thus, $2g'(0) = 0$, which forces **$g'(0) = 0$**.

Visually, this makes perfect sense. The tangent line at the peak of an even, differentiable function must be horizontal. We have solved the mystery, not by guessing, but by peeling back the layers of the functional equations one by one.

Enter Numerical Value:

Visualized Solution

Introduction to Functional Equations

Finding the value of $f (0)$

Finding the value of $g (0)$

Establishing a Powerful Identity

Proving $g (x)$ is an Even Function

Setting up the Derivative $g^{'} (0)$

Applying the Even Function Property

Evaluating the Limit

Final Conclusion

The Sigma Insight: Differentiability of a Function

The Beauty of Functional Equations

Phase 1

Phase 2

Phase 3

Phase 4

The Final Conclusion

Similar Questions

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