if f(x) = 1, tham x is equal to:
Question:

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:

1. (1) $\frac{1}{e}$

2. (2) $2 e$

3. (3) $\frac{1}{2 e}$

4. (4) $e$

Correct Option: 1

Solution:

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

$\Rightarrow \lim _{t \rightarrow x} \frac{2 t f^{2}(x)-2 x^{2} f(t) \cdot f^{\prime}(t)}{1}=0$

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