Step 1: Evaluate \( f^2(x) \):
\[ f^2(x) = \lim_{r \to x} \frac{(2x^2f(r))^2 - f(x)f(r)x}{x^2 - r^2} \cdot \frac{r - x}{r} \]
Simplifying this expression using L'Hôpital's Rule and differentiating the terms with respect to \( r \), we eventually get:
\[ f^2(x) = 2x f(x)f'(x) - xe^{x} \]
Step 2: Rewrite the equation:
We now have the differential equation:
\[ f(x)^2 = x f(x)f'(x) - xe^{x} \]
Step 3: Substitute \( y = f(x) \):
This substitution gives \( y = f(x) \), so that \( \frac{dy}{dx} = f'(x) \), and the equation becomes:
\[ y^2 = xy \frac{dy}{dx} - xe^x \]
Step 4: Separate variables and simplify:
Let \( y = vx \), so that \( \frac{dy}{dx} = v + x \frac{dv}{dx} \). Substitute into the equation:
\[ v^2x^2 = x e^{x}(v + x \frac{dv}{dx}) - xe^x \]
Step 5: Solve the resulting differential equation:
By separating variables and integrating both sides, we obtain:
\[ \int v^2 \, dv = \int \frac{dx}{x} \]
Step 6: Integrate:
Integrating both sides, we get:
\[ e^v = \ln|x| + c \]
Step 7: Apply initial condition:
Given \( f(1) = 1 \), substitute \( x = 1 \) and \( y = 1 \) to find \( c \):
\[ e^1 = \ln 1 + c = c = 2 \]
Step 8: Find \( a \) such that \( f(a) = 0 \):
When \( y = 0 \), we solve for \( v = 0 \):
\[ a = -\frac{2}{e} \]
Step 9: Calculate \( ea \):
\[ ea = e \cdot -\frac{2}{e} = -2 \]
Thus, \( ea = 2 \).
The Correct Answer is : 2
We are given a function \( f(x) \) with the conditions \( f(1) = 1 \) and \( f(a) = 0 \). We are tasked with finding the value of \( a \), starting from the following equation: \[ f^2(x) = \lim_{r \to x} \left( \frac{2r^2(f^2(r)) - f(r)f(r)}{r^2 - x^2} - r^3 e^{f(r)} \right) \] This leads us through various steps of limits, derivatives, and integrations.
Start with the expression: \[ f^2(x) = 2x^2 f(x) \cdot \frac{2}{x} \cdot f'(x) - x^3 e^{f(x)} \] We start simplifying the derivative expression: \[ y^2 = x y \frac{dy}{dx} - x^3 \cdot e^{y/x} \] where we perform differentiation by using the chain rule.
By substitution, we get: \[ y = \frac{dy}{dx} \quad \text{and we solve for} \quad \frac{dy}{dx} = \frac{x^2}{y e^{y/x}} \] This relationship allows us to rewrite: \[ \frac{dx}{dy} = e^{-v} v \, dv = dx \]
Next, we integrate both sides of the equation: \[ e^v(x + c) + 1 + v = 0 \] Using the initial condition \( f(1) = 1 \), we set \( x = 1 \) and \( y = 1 \), which gives: \[ c = -1 - \frac{2}{e} \]
Substituting into the expression, we calculate: \[ x = a, \, y = 0 \quad \Rightarrow \quad a = \frac{2}{e} \]
The value of \( a \) is: \[ \boxed{2} \]
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____. 