Let $f:(0, \infty) \rightarrow(0, \infty)$ 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:
Correct Option: 1
$\lim _{t \rightarrow x} \frac{t^{2} f^{2}(x)-x^{2} f^{2}(t)}{t-x}=0$
$\Rightarrow \lim _{t \rightarrow x} \frac{2 t f^{2}(x)-2 x^{2} f(t) \cdot f^{\prime}(t)}{1}=0$
Using L'Hospital's rule
$\Rightarrow f(x)=x f^{\prime}(x)$
$\int \frac{f^{\prime}(x)}{f(x)} d x=\int \frac{1}{x} d x$
$\log _{e} f(x)=\log _{e} x+\log _{e} C$
$\Rightarrow f(x)=C x, \quad \because f(1)=e$
$\Rightarrow C=e ;$ so $f(x)=e x$
When $f(x)=1=e x \Rightarrow x=\frac{1}{e}$
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