Question

Let \(f: R \rightarrow R, f(x-f(y))=f(f(y))+x f(y)+f(x)-1 \forall x\), \(y \in R\), if \(f(0)=1\) and \(f^{\prime}(0)=0\), then (A) \(\mathrm{f}(\mathrm{x})=1-\frac{\mathrm{x}^{2}}{2}\) (B) \(f(x)=x^{2}+1\) (C) \(f(x)=\left(\frac{2 x+1}{x+1}\right)\) (D) none of these

Short answer

Given: \(f(x - f(y)) = f(f(y)) + xf(y) + f(x) - 1\), \(f(0) = 1\), and \(f'(0) = 0\) Options: (A) \(f(x) = 1 - \frac{x^2}{2}\) (B) \(f(x) = x^2 + 1\) (C) \(f(x) = \frac{2x + 1}{x + 1}\) (D) None of these Answer: (A) \(f(x) = 1 - \frac{x^2}{2}\)

Step-by-step solution

Substitute x and y = 0 in the functional equation

Substitute x = 0 and y = 0 in the given functional equation: \(f(x - f(y)) = f(f(y)) + xf(y) + f(x) - 1\) and use the given condition \(f(0) = 1\): \(f(0 - f(0)) = f(f(0)) + 0f(0) + f(0) - 1\) \(f(1) = f(1) + 1 - 1\) Thus, \(f(1) = 1\)

Apply the functional equation to f(1)

Substitute x = y = 1 in the functional equation: \(f(1 - f(1)) = f(f(1)) + f(1) + f(1) - 1\) \(f(0) = f(1) + 1 + 1 - 1\) This confirms that \(f(0) = 1\)

Apply the conditions to find f'(x)

We know that \(f(0) = 1\) and \(f'(0) = 0\). Now differentiate both sides of the functional equation w.r.t. x and apply chain rule: \(f'(x - f(y)) (1 - f'(y)) = f'(f(y)) f'(y) + f(y) + f'(x)\) Since we're dealing with \(f'(0) = 0\), we need to find a link between the given information. If we set y = 0, we get: \(f'(x - f(0)) = f'(x) + f'(f(1)) = f'(x) + f'(1)\) From the above equation, we observe that \(f'(x) = - f'(1)\), which implies \(f'(x) = 0\), as we were given \(f'(0) = 0\).

Consider the candidates for f(x)

We have four given options for the function \(f(x)\). We know that \(f(0) = 1\) and \(f'(0) = 0\). Let's test each option with these conditions: (A) \(f(x) = 1 - \frac{x^2}{2}\) : This function satisfies both conditions. (B) \(f(x) = x^2 + 1\) : Since its derivative is \(f'(x) = 2x\), it does not satisfy the condition \(f'(0) = 0\). (C) \(f(x) = \frac{2x + 1}{x + 1}\) : This function doesn't satisfy the condition \(f(0) = 1\). (D) None of these.

Verify the functional equation for the chosen candidate

From Step 4, the only option that satisfies the given initial conditions is (A). Let's verify if the functional equation holds for this function: Choose (A) \(f(x) = 1 - \frac{x^2}{2}\). Then, calculate \(f(x - f(y)) = 1 - \frac{(x - f(y))^2}{2}\), \(f(f(y)) = 1 - \frac{(1 - \frac{y^2}{2})^2}{2}\), and \(x f(y) + f(x) - 1 = x(1 - \frac{y^2}{2}) + (1 - \frac{x^2}{2}) - 1\). By plugging the values of \(f(x - f(y))\), \(f(f(y))\), and \(x f(y) + f(x) - 1\), we can see that both sides of the functional equation hold true for this function. Therefore, the correct answer is (A) \(\mathrm{f}(\mathrm{x})=1-\frac{\mathrm{x}^{2}}{2}\).