rehaucesnwr1

2023-02-23

Let f be a differentiable function such that ${f}^{\prime }\left(x\right)=7-\frac{3}{4}\ast \frac{f\left(x\right)}{x},\left(x>0\right)$ and $f\left(1\right)\ne 4$. Then $\underset{x\to {0}^{+}}{lim}x\ast f\left(\frac{1}{x}\right):$
A)does not exist.
B)exists and equals 4.
C)exists and equals 4/7.
D)exists and equals 0.

Ruby Rollins

The right decision is A exists and equals 4.
${f}^{\prime }\left(x\right)=7-\frac{3}{4}\ast \frac{f\left(x\right)}{x}$...(1)
Let
$⇒\frac{dy}{dx}=7-\frac{3y}{4x}$
$⇒\frac{dy}{dx}+\frac{3y}{4x}=7$
An example of a linear differential equation is the one above
$I.F.={e}^{\int \frac{3}{4x}dx}={x}^{\frac{3}{4}}$
$⇒y\ast {x}^{\frac{3}{4}}=7\int {x}^{\frac{3}{4}}dx$
$⇒y\ast {x}^{\frac{3}{4}}=4{x}^{\frac{7}{4}}+c$
$⇒f\left(x\right)=4x+c\ast {x}^{-\frac{3}{4}}$
Now, $\underset{x\to {0}^{+}}{lim}x\ast f\left(\frac{1}{x}\right)=\underset{x\to {0}^{+}}{lim}x\left(\frac{4}{x}+c{x}^{-\frac{3}{4}}\right)=4$

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