Question

Let \(f(x)\) be defined for all \(x \in R\) and the continuous. Let $\mathrm{f}(\mathrm{x}+\mathrm{y})-\mathrm{f}(\mathrm{x}-\mathrm{y})=4 \mathrm{xy} \forall \mathrm{x}, \mathrm{y}=\in \mathrm{R}$ and \(f(0)=0\) then (A) \(\mathrm{f}(\mathrm{x})\) is bounded (B) \(f(x)+f\left(\frac{1}{x}\right)=f\left(x+\frac{1}{x}\right)+2\) (C) \(\mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{f}\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)+2\) (D) none of these

Short answer

In this problem, we were given a functional equation involving an unknown function \(f(x)\) along with an initial condition \(f(0)=0\). We first examined some properties of the functional equation and then found the function to be \(f(x) = 2x^2\). After determining the function, we checked it against the given statements (A, B, and C). We found that statement (B) is correct: \(f(x) + f\left(\frac{1}{x}\right) = f\left(x+\frac{1}{x}\right) + 2\) holds true for the given function.

Step-by-step solution

Identify the given functional equation

We are given the functional equation: \(f(x+y) - f(x-y) = 4xy \forall x, y \in ℝ\)

Investigate properties of the functional equation

Let's plug in \(x=y\) in the given functional equation: \(f(2x) - f(0) = 4x^2\) Since \(f(0)=0\), we have \(f(2x) = 4x^2\) for all \(x \in ℝ\) Now, let's plug in \(x=0\) in the given functional equation: \(f(y) - f(-y) = 0\) This equation implies that \(f(-y) = f(y)\) for all \(y \in ℝ\) These two properties are very useful in understanding the behavior of the function \(f(x)\).

Find the function \(f(x)\)

Since \(f(2x) = 4x^2\) and \(f\) is continuous, we can express \(f(x)\) as a continuous function on the interval (0,2) and also continuous at 0 with the initial condition \(f(0)=0\). So, we can write \(f(x) = 2x^2\) for all \(x \in ℝ\)

Verify the given statements

Now that we have found the function \(f(x) = 2x^2\), we can check it against the given statements: (A) \(f(x)\) is bounded - This statement is incorrect, as \(f(x) = 2x^2\) is unbounded for all \(x \in ℝ\). (B) \(f(x) + f\left(\frac{1}{x}\right) = f\left(x+\frac{1}{x}\right) + 2\) By plugging in the function we found, we get: \(2x^2 + 2\left(\frac{1}{x^2}\right) = 2\left(x+\frac{1}{x}\right)^2 + 2\) Simplifying, we get: \(2x^2 + 2\left(\frac{1}{x}\right)^2 = 2x^2 + 4x + 2 - 4x\left(\frac{1}{x}\right) + 2\) This statement is correct. (C) \(f(x) + f\left(\frac{1}{x}\right) = f\left(x-\frac{1}{x}\right) + 2\) By plugging in the function we found, we get: \(2x^2 + 2\left(\frac{1}{x^2}\right) = 2\left(x-\frac{1}{x}\right)^2 + 2\) Simplifying, we get: \(2x^2 + 2\left(\frac{1}{x}\right)^2 = 2x^2 - 4x + 2 + 4x\left(\frac{1}{x}\right) + 2\) This statement is incorrect. So, the correct answer is statement (B).