Let f:[0,1] → R be such that f(xy) = f(x).f(y) for all x,y ∈ [0,1], and f(0) ≠ 0. If y = y(x) satisf

Q: Wait, why do we reject $f(0)=0$? The equation $f(0) = f(0)^2$ allows for $f(0)=0$ too, right?

Excellent observation! Mathematically, $f(0)=0$ is a solution to $f(0)=f(0)^2$. However, the problem statement explicitly gives us the constraint $f(0) \neq 0$. In JEE Advanced, always respect the constraints—they are the guardrails that keep us on the path to the unique solution.

Q: How can we be sure $f(x)$ is constant? Could it be $x^n$ or something else?

Great question. If you test $f(x) = x^n$, you get $(xy)^n = x^n y^n$, which works! But wait—does $f(x)=x^n$ satisfy $f(0)=1$? If $n > 0$, $f(0)=0$. If $n=0$, $f(x)=1$. Our derivation $f(x) = f(x)f(0)$ with $f(0)=1$ forces $f(x)=1$ for all $x$. There is no room for other functions here.

Q: Why do we need the initial condition $y(0)=1$?

Integration is a process of finding a family of curves. Without the initial condition, we would have $y = x + C$, which is an infinite set of parallel lines. The condition $y(0)=1$ acts like a laser, picking out the one specific line that passes through the point $(0, 1)$.

Question

Let $f:[0,1] 	o R$ be such that $f(xy) = f(x).f(y)$ for all $x,y \in [0,1]$, and $f(0) 
e 0$. If $y = y(x)$ satisfies the differential equation, $\frac{dy}{dx} = f(x)$ with $y(0) = 1$, then $y(1/4) + y(3/4)$ is equal to

Options:

(A) 4

(B) 3

(C) 5

(D) 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to peel back the layers of a problem that looks like a daunting mix of functional equations and calculus, but is actually a beautiful exercise in logical deduction.

Imagine you are a detective, and the functional equation $f(xy) = f(x)f(y)$ is your crime scene. It tells us that the function behaves multiplicatively.

We have a vital clue: $f(0) 
eq 0$. This is not just a piece of information; it is the skeleton key.

If we substitute $x=0$ and $y=0$ into our equation, we get:

$$f(0) = f(0) \cdot f(0) \implies f(0) = [f(0)]^2$$

Since we know $f(0) 
eq 0$, we can divide by it, revealing that $f(0) = 1$. This is our first victory.

### Unmasking the Function

Now that we know $f(0) = 1$, let us see what happens when we set $y=0$ in the original equation. We get $f(x \cdot 0) = f(x) \cdot f(0)$.

This simplifies to $f(0) = f(x) \cdot 1$. Since we already established that $f(0) = 1$, we are left with the elegant result $f(x) = 1$.

The function is not some complex, oscillating curve; it is a simple, horizontal line at height $1$. It is constant for all $x$ in the interval $[0, 1]$.

### The Calculus Bridge

With $f(x) = 1$ firmly in our grasp, the differential equation $\frac{dy}{dx} = f(x)$ becomes trivial. It transforms into:

$$\frac{dy}{dx} = 1$$

This is a statement about the slope of our function $y(x)$. It tells us that the slope is constant and equal to $1$ everywhere.

To find $y(x)$, we integrate both sides:

$$\int dy = \int 1 \, dx \implies y = x + C$$

We are looking at a family of lines with a slope of $1$. We use the initial condition $y(0) = 1$ to find the specific constant.

Substituting $x=0$ and $y=1$ into our equation, we get $1 = 0 + C$, which means $C = 1$. Our unique function is **$y(x) = x + 1$**.

### The Final Calculation

We have arrived at the finish line. We need to evaluate $y(\frac{1}{4}) + y(\frac{3}{4})$.

Using our function $y(x) = x + 1$, we calculate:

$$y\left(\frac{1}{4}ight) = \frac{1}{4} + 1 = \frac{5}{4}$$

$$y\left(\frac{3}{4}ight) = \frac{3}{4} + 1 = \frac{7}{4}$$

Adding these together, we get:

$$\frac{5}{4} + \frac{7}{4} = \frac{12}{4} = 3$$

The result is a clean, satisfying integer. The final answer is **$3$**.

Select Answer:

Visualized Solution

Analyzing the Functional Equation

Evaluating $f (0)$

Finding the General Form of $f (x)$

The Differential Equation

Integration

Finding the Constant of Integration

Evaluating $y (\frac{1}{4})$

Evaluating $y (\frac{3}{4})$

Final Calculation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

Unmasking the Function

The Calculus Bridge

The Final Calculation

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Select Answer:

Visualized Solution

Analyzing the Functional Equation

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

Finding the General Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Differential Equation

Integration

Finding the Constant of Integration

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(41​)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(43​)

Final Calculation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

Unmasking the Function

The Calculus Bridge

The Final Calculation

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