Question

Let $f$ be a differentiable function such that $f^{\prime }\left(x\right)=7-\frac{3}{4}\frac{f\left(x\right)}{x},(x>0)$ and $f\left(1\right)\ne 4$.
Then $\underset{x\to 0^{+}}{lim}xf\left(\frac{1}{x}\right)$:

• A. Exists and equals 4
• B. Does not exist
• C. Exist and equals
• D. Exists and equals $\frac{4}{7}$

Answer 1

The correct option is A Exists and equals 4

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

Given $f\left(1\right)\ne 4$ $\underset{x\to 0^{+}}{lim}xf\left(\frac{1}{x}\right)=?$

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

IF $=e^{\int \frac{3}{4x}dx}=e^{\frac{3}{4}ln|x|}=X^{\frac{3}{4}}$

$y.x^{\frac{3}{4}}=\int 7.x^{\frac{3}{4}}dx$

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

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

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

$\underset{x\to 0^{+}}{lim}xf\left(\frac{1}{x}\right)=\underset{x\to 0^{+}}{lim}(4+C.x^{\frac{7}{4}})=4$