Question

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

• A. does not exist.
• B. exists and equals $4$.
• C. exists and equals $\frac{4}{7}.$
• D. exists and equals $0.$

Answer 1

The correct option is B exists and equals $4$.

$f^{\prime }\left(x\right)=7-\frac{3}{4}\cdot \frac{f\left(x\right)}{x}\cdots \left(1\right)$
Let $y=f\left(x\right)\therefore f^{\prime }\left(x\right)=\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=7-\frac{3y}{4x}$
$\Rightarrow \frac{dy}{dx}+\frac{3y}{4x}=7$
The above equation is linear differential equation
$I.F.=e^{\int \frac{3}{4x}dx}=x^{\frac{3}{4}}$
$\Rightarrow y\cdot x^{\frac{3}{4}}=7\int x^{\frac{3}{4}}dx$
$\Rightarrow y\cdot x^{\frac{3}{4}}=4x^{\frac{7}{4}}+c$
$\Rightarrow f\left(x\right)=4x+c\cdot x^{-\frac{3}{4}}$
So, $\underset{x\to 0^{+}}{lim}x\cdot f\left(\frac{1}{x}\right)=\underset{x\to 0^{+}}{lim}x(\frac{4}{x}+c{x}^{-\frac{3}{4}})$
$=4$