Let f(x)=r xlim2r^2[(f(r))^2-f(x)f(r)]r^2-x^2-r^3e^(f(r))/(r) be differentiable in (-,0) (…

Question

Let $f(x)=\sqrt{\underset{r	o x}{lim}{\frac{2{r}^{2}[(f(r){)}^{2}-f(x)f(r)]}{{r}^{2}-{x}^{2}}-{r}^{3}{e}^{\frac{f(r)}{r}}}}$ 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: 2

Accepted Answer

$\begin{gathered} f(1)=1,f(a)=0 \ {f}^{2}(x)=\underset{r	o x}{lim}(\frac{2{r}^{2}({f}^{2}(r)-f(x)f(r))}{{r}^{2}-{x}^{2}}-{r}^{3}{e}^{\frac{f(r)}{r}}) \ =\underset{r	o x}{lim}(\frac{2{r}^{2}f(r)}{r+x}\frac{(f(r)-f(x))}{r-x}-{r}^{3}{e}^{\frac{f(r)}{r}}) \ {f}^{2}(x)=\frac{2{x}^{2}f(x)}{2x}{f}^{'}(x)-{x}^{3}{e}^{\frac{f(x)}{x}} \ {y}^{2}=xy\frac{dy}{dx}-{x}^{3}{e}^{\frac{y}{x}} \ \frac{y}{x}=\frac{dy}{dx}-\frac{{x}^{2}}{y}{e}^{\frac{y}{x}} \end{gathered}$

Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\begin{gathered} v=v+x\frac{dv}{dx}-\frac{x}{v}{e}^{v} \ \frac{dv}{dx}=\frac{{e}^{v}}{v}\Rightarrow {e}^{-v}vdv=dx \end{gathered}$

Integrating both side

$\begin{gathered} {e}^{v}(x+c)+1+v=0 \ f(1)=1\Rightarrow x=1,y=1 \end{gathered}$

$\begin{gathered} \Rightarrow c=-1-\frac{2}{e} \ {e}^{v}(-1-\frac{2}{e}+x)+1+v=0 \ {e}^{\frac{y}{x}}(-1-\frac{2}{e}+x)+1+\frac{y}{x}=0 \ x=a,y=0\Rightarrow a=\frac{2}{e} \ ae=2 \end{gathered}$

Solutions · 1