# Solve this following

Question:

Let $f$ be a differentiable function such that

$f^{\prime}(x)=7-\frac{3}{4} \frac{f(x)}{x},(x>0)$ and $f(1) \neq 4$

Then $\lim _{x \rightarrow 0^{+}} x f\left(\frac{1}{x}\right)$ :

1. Exists and equals 4

2. Does not exist

3. Exist and equals 0

4. Exists and equals $\frac{4}{7}$

Correct Option: 1

Solution:

$f^{\prime}(x)=7-\frac{3}{4} \frac{f(x)}{x} \quad(x>0)$

Given $f(1) \neq 4 \quad \lim _{x \rightarrow 0^{+}} x f\left(\frac{1}{x}\right)=?$

$\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{3}{4} \frac{\mathrm{y}}{\mathrm{x}}=7$ (This is LDE)

$I F=e^{\int \frac{3}{4 x} d x}=e^{\frac{3}{4} \ln |x|}=x^{\frac{3}{4}}$

$y \cdot x^{\frac{3}{4}}=\int 7 \cdot x^{\frac{3}{4}} d x$

$y \cdot x^{\frac{3}{4}}=7 \cdot \frac{x^{\frac{7}{4}}}{\frac{7}{4}}+C$

$f(x)=4 x+C \cdot x^{-\frac{3}{4}}$

$f\left(\frac{1}{x}\right)=\frac{4}{x}+C \cdot x^{\frac{3}{4}}$

$\lim _{x \rightarrow 0^{+}} x f\left(\frac{1}{x}\right)=\lim _{x \rightarrow 0^{+}}\left(4+C \cdot x^{\frac{7}{4}}\right)=4$

$\therefore$ Option (1)