Question

Let $f:\left(0,\infty \right)\to \left(0,\infty \right)$ be a differentiable function such that $f\left(1\right)=e$ and $\underset{t\to x}{\lim }\frac{t^2{f}^2\left(x\right)-x^2{f}^2\left(t\right)}{t-x}=0$.if $f\left(x\right)=1$then $x$ is equal to.

• A. $e$
• B. $2e$
• C. $\frac{1}{e}$
• D. $\frac{1}{2e}$

Answer 1

The correct option is C

$\frac{1}{e}$

Explanation for the correct options:

Find the value of $x$.

Step 1: Given data

$\underset{t\to x}{\lim }\frac{t^2{f}^2\left(x\right)-x^2{f}^2\left(t\right)}{t-x}=0$

$f:\left(0,\infty \right)\to \left(0,\infty \right)$

$f\left(1\right)=e$

$f\left(x\right)=1$

Consider the given equation as

$$\begin{gathered} \underset{t\to x}{\Rightarrow \lim }\frac{t^2{f}^2\left(x\right)-x^2{f}^2\left(t\right)}{t-x}=0 \\ \Rightarrow \frac{x^2{f}^2\left(x\right)-x^2{f}^2\left(x\right)}{x-x}=\frac{0}{0} \end{gathered}$$

Step 2: Using the L 'Hospitality rule

$$\begin{gathered} \Rightarrow \underset{t\to x}{\lim }\frac{2t{f}^2\left(x\right)-x^{2}2f\left(t\right)f\text{'}\left(t\right)}{1}=0 \\ \Rightarrow 2x{f}^2\left(x\right)-x^{2}2f\left(x\right)f\text{'}\left(x\right)=0 \\ \Rightarrow 2x{f}^2\left(x\right)=x^{2}2f\left(x\right)f\text{'}\left(x\right) \\ \Rightarrow f\left(x\right)=xf\text{'}\left(x\right) \\ \Rightarrow \frac{f\text{'}\left(x\right)}{f\left(x\right)}=\frac{1}{x} \end{gathered}$$

Step 3: Integrate the above Equation

$$\begin{gathered} \Rightarrow \int \frac{f\text{'}\left(x\right)}{f\left(x\right)}dx=\int \frac{1}{x}dx \\ \Rightarrow \log \left(f\left(x\right)\right)=\log x+\log e^c \\ \Rightarrow \log \left(f\left(x\right)\right)=\log \left(x{e}^c\right) \\ \Rightarrow f\left(x\right)=x{e}^c \\ \Rightarrow f\left(1\right)=1e^c\left[\text{putting x=1; given},f\left(1\right)=e\right] \\ \Rightarrow f\left(1\right)=e \\ \Rightarrow c=1 \end{gathered}$$

Then,

$$\begin{gathered} \Rightarrow f\left(x\right)=xe \\ \Rightarrow f\left(x\right)=1[givenf(x)=1] \\ \Rightarrow xe=1 \\ \Rightarrow x=\frac{1}{e} \end{gathered}$$

Hence, the correct answer is option C.