Question

Let\[f(x) = \sqrt{\lim_{r \to x} \left( \frac{2r^2 \left[ (f(r))^2 - f(x) f(r) \right]}{r^2 - x^2} - r^3 e^{\frac{f(r)}{r}} \right)}\]be differentiable in \( (-\infty, 0) \cup (0, \infty) \) and \( f(1) = 1 \). Then the value of \( ea \), such that \( f(a) = 0 \), is equal to ______.

Answer 1

Correct Answer: 2

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