Let f be a differentiable function such that

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. (1) exists and equals $\frac{4}{7}$.

  2. (2) exists and equals 4 .

  3. (3) does not exist.

  4. (4) exists and equals $0 .$


Correct Option: , 2

Solution:

Let $y \quad f(x)$

$\frac{d y}{d x}+\left(\frac{3}{4 x}\right) y=7$

I.F. $=e^{\int \frac{3}{4 x} d x}=e^{\frac{3}{4} \ln x}=x^{\left(\frac{3}{4}\right)}$

Solution of differential equation

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

$y \cdot x^{\frac{3}{4}}=7 \cdot \frac{x^{\frac{7}{4}}}{\left(\frac{7}{4}\right)}+C=4 x^{\frac{7}{4}}+C$

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

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

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

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