rehaucesnwr1

2023-02-23

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

A)does not exist.

B)exists and equals 4.

C)exists and equals 4/7.

D)exists and equals 0.

A)does not exist.

B)exists and equals 4.

C)exists and equals 4/7.

D)exists and equals 0.

Ruby Rollins

Beginner2023-02-24Added 5 answers

The right decision is A exists and equals 4.

${f}^{\prime}(x)=7-\frac{3}{4}\ast \frac{f(x)}{x}$...(1)

Let $y=f(x)\text{}\text{}\text{}\text{}\therefore {f}^{\prime}(x)=\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=7-\frac{3y}{4x}$

$\Rightarrow \frac{dy}{dx}+\frac{3y}{4x}=7$

An example of a linear differential equation is the one above

$I.F.={e}^{\int \frac{3}{4x}dx}={x}^{\frac{3}{4}}$

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

$\Rightarrow y\ast {x}^{\frac{3}{4}}=4{x}^{\frac{7}{4}}+c$

$\Rightarrow f(x)=4x+c\ast {x}^{-\frac{3}{4}}$

Now, $\underset{x\to {0}^{+}}{lim}x\ast f(\frac{1}{x})=\underset{x\to {0}^{+}}{lim}x(\frac{4}{x}+c{x}^{-\frac{3}{4}})=4$

${f}^{\prime}(x)=7-\frac{3}{4}\ast \frac{f(x)}{x}$...(1)

Let $y=f(x)\text{}\text{}\text{}\text{}\therefore {f}^{\prime}(x)=\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=7-\frac{3y}{4x}$

$\Rightarrow \frac{dy}{dx}+\frac{3y}{4x}=7$

An example of a linear differential equation is the one above

$I.F.={e}^{\int \frac{3}{4x}dx}={x}^{\frac{3}{4}}$

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

$\Rightarrow y\ast {x}^{\frac{3}{4}}=4{x}^{\frac{7}{4}}+c$

$\Rightarrow f(x)=4x+c\ast {x}^{-\frac{3}{4}}$

Now, $\underset{x\to {0}^{+}}{lim}x\ast f(\frac{1}{x})=\underset{x\to {0}^{+}}{lim}x(\frac{4}{x}+c{x}^{-\frac{3}{4}})=4$