Solve this


Let $\mathrm{f}:(0, \infty) \rightarrow(0, \infty)$ be a differentiable function such that $f(1)=e$ and

$\lim _{t \rightarrow x} \frac{t^{2} f^{2}(x)-x^{2} f^{2}(t)}{t-x}=0$

If $f(x)=1$, then $x$ is equal to :

  1. $2 \mathrm{e}$

  2. $\frac{1}{2 \mathrm{e}}$

  3. e

  4. $\frac{1}{\mathrm{e}}$

Correct Option: , 4


$L=\operatorname{Lim}_{t \rightarrow x} \frac{t^{2} f^{2}(x)-x^{2} f^{2}(t)}{t-x}$

using L.H. rule

$L=\operatorname{Lim}_{t \rightarrow x} \frac{2 t f^{2}(x)-x^{2} \cdot 2 f^{\prime}(t) \cdot f(t)}{1}$

$\Rightarrow L=2 x f(x)\left(f(x)-x f^{\prime}(x)\right)=0$ (given)

$\Rightarrow f(x)=x f^{\prime}(x) \Rightarrow \int \frac{f^{\prime}(x) d x}{f(x)}=\int \frac{d x}{x}$

$\Rightarrow \ell \mathrm{n}|\mathrm{f}(\mathrm{x})|=\ell \mathrm{n}|\mathrm{x}|+\mathrm{C}$

$\because f(1)=e, x>0, f(x)>0$

$\Rightarrow f(x)=e x, \quad$ if $f(x)=1 \Rightarrow x=\frac{1}{e}$

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