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

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