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(B) The given differential equation is $f'(x) + \frac{3}{4x} f(x) = 7$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(

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exists and equals $4$

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(B) The given differential equation is $f'(x) + \frac{3}{4x} f(x) = 7$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{3}{4x}$ and $Q(x) = 7$. The integrating factor $IF = e^{\int P(x) dx} = e^{\int \frac{3}{4x} dx} = e^{\frac{3}{4} \ln x} = x^{3/4}$. The general solution is $f(x) \cdot IF = \int Q(x) \cdot IF dx + c$. $f(x) \cdot x^{3/4} = \int 7 x^{3/4} dx + c = 7 \cdot \frac{x^{7/4}}{7/4} + c = 4x^{7/4} + c$. Thus,$f(x) = 4x + c x^{-3/4}$. Now,we evaluate $\lim_{x 	o 0^+} x f\left(\frac{1}{x}ight)$. $f\left(\frac{1}{x}ight) = 4\left(\frac{1}{x}ight) + c\left(\frac{1}{x}ight)^{-3/4} = \frac{4}{x} + c x^{3/4}$. Therefore,$\lim_{x 	o 0^+} x f\left(\frac{1}{x}ight) = \lim_{x 	o 0^+} x \left(\frac{4}{x} + c x^{3/4}ight) = \lim_{x 	o 0^+} (4 + c x^{7/4}) = 4$.

Let $f$ be a differentiable function such that $f'(x) = 7 - \frac{3}{4} \frac{f(x)}{x}$ for $x > 0$ and $f(1) \neq 4$. Then $\lim_{x \to 0^+} x f\left(\frac{1}{x}\right)$

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