Key Concepts and Formulas

• Functional Equations: Equations that define a function through a relationship between its values at different points.
• Differential Equations: Equations involving an unknown function and its derivatives.
• Integration: Finding the integral of a function.
• Initial Conditions: Specific values of the function at a certain point, used to determine the constant of integration in differential equations.

Step-by-Step Solution

Step 1: Determining the Value of $f(0)$

We are given the functional equation $f(xy) = f(x)f(y)$ for all $x, y \in [0, 1]$, and the condition $f(0) \ne 0$. Our goal is to find the value of $f(0)$.

Let $x = 0$ and $y = 0$. Substituting these values into the functional equation yields: $f(0 \cdot 0) = f(0)f(0)$ $f(0) = [f(0)]^2$ Let $k = f(0)$. Then the equation becomes: $k = k^2$ $k^2 - k = 0$ $k(k - 1) = 0$ This gives two possible solutions: $k = 0$ or $k = 1$. Therefore, $f(0) = 0$ or $f(0) = 1$.

Since we are given that $f(0) \ne 0$, we must have $f(0) = 1$.

Step 2: Determining the Explicit Form of $f(x)$

Now that we know $f(0) = 1$, we can use this information to find the general form of $f(x)$.

Let $y = 0$. Substituting this into the functional equation $f(xy) = f(x)f(y)$ gives: $f(x \cdot 0) = f(x)f(0)$ $f(0) = f(x)f(0)$ Since $f(0) = 1$, we have: $1 = f(x) \cdot 1$ $f(x) = 1$ Therefore, $f(x) = 1$ for all $x \in [0, 1]$.

Step 3: Solving the Differential Equation

We are given the differential equation $\frac{dy}{dx} = f(x)$. Substituting $f(x) = 1$ into this equation gives: $\frac{dy}{dx} = 1$

To solve this, we integrate both sides with respect to $x$: $\int dy = \int 1 \, dx$ $y = x + C$ where $C$ is the constant of integration.

Step 4: Applying the Initial Condition

We are given the initial condition $y(0) = 1$. Substituting $x = 0$ and $y = 1$ into the equation $y = x + C$ gives: $1 = 0 + C$ $C = 1$ Therefore, the particular solution is $y(x) = x + 1$.

Step 5: Evaluating the Required Expression

We need to find the value of $y\left(\frac{1}{4}\right) + y\left(\frac{3}{4}\right)$. We have $y(x) = x + 1$. Therefore: $y\left(\frac{1}{4}\right) = \frac{1}{4} + 1 = \frac{5}{4}$ $y\left(\frac{3}{4}\right) = \frac{3}{4} + 1 = \frac{7}{4}$ So, $y\left(\frac{1}{4}\right) + y\left(\frac{3}{4}\right) = \frac{5}{4} + \frac{7}{4} = \frac{12}{4} = 3$

Common Mistakes & Tips

• Ignoring the condition $f(0) \ne 0$ would lead to an incorrect solution for $f(x)$ and consequently for $y(x)$.
• Forgetting to use the initial condition $y(0) = 1$ would result in a general solution instead of a particular solution.
• Always verify that the derived function satisfies the original functional equation.

Summary

We first determined that $f(0) = 1$ using the given condition $f(0) \ne 0$. Then, we found that $f(x) = 1$ for all $x \in [0, 1]$. Next, we solved the differential equation $\frac{dy}{dx} = 1$ to get $y = x + C$. Using the initial condition $y(0) = 1$, we found $C = 1$, so $y(x) = x + 1$. Finally, we evaluated $y\left(\frac{1}{4}\right) + y\left(\frac{3}{4}\right)$ and found it to be equal to 3.

The final answer is \boxed{3}, which corresponds to option (A).