Question

Let f be a differentiable function such that ${f}'\left( x \right)=7-\frac{3}{4}\frac{f\left( x \right)}{x}$, (x>0) and $f\left( 1 \right)\ne 4$. Then $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,xf\left( \frac{1}{x} \right)$:
(a) exists and equals 4
(b) does not exist
(c) exist and equals
(d) exists and equals $\frac{4}{7}$

Answer 1

First, before proceeding for this, we must know the following given equation is the linear differential equation of the form as $\frac{dy}{dx}+\frac{3y}{4x}=7$. Then, to get the solution of the above differential equation in the form $\frac{dy}{dx}+Py=Q$, we need a integrating factor(IF) given by the formula as$IF={{e}^{\int{Pdx}}}$. Then, to get the solution of the above differential equation in the form $\frac{dy}{dx}+Py=Q$, we have the form of solution as$y\times IF=\int{Q\times IF}dx+c$. Then, we are required to find the value of $f\left( \frac{1}{x} \right)$, so we replace x by $\frac{1}{x}$ in the above expression and find the value of limit as required.

Complete step by step answer:
In this question, we are supposed to find the value of $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,xf\left( \frac{1}{x} \right)$with given conditions that f be a differentiable function such that ${f}'\left( x \right)=7-\frac{3}{4}\frac{f\left( x \right)}{x}$, (x>0) and $f\left( 1 \right)\ne 4$.
So, before proceeding for this, we must know the following given equation is the linear differential equation of the form as:
$\frac{dy}{dx}+\frac{3y}{4x}=7$
Now, to get the solution of the above differential equation in the form $\frac{dy}{dx}+Py=Q$, we need a integrating factor(IF) given by the formula as:
$IF={{e}^{\int{Pdx}}}$
So, the value of P from the above differential equation is $\frac{3}{4x}$ to get the value of IF as:
$\begin{aligned} & IF={{e}^{\int{\frac{3}{4x}dx}}} \\\ & \Rightarrow IF={{e}^{\frac{3}{4}\ln \left| x \right|}} \\\ & \Rightarrow IF={{x}^{\frac{3}{4}}} \\\ \end{aligned}$
Now, to get the solution of the above differential equation in the form $\frac{dy}{dx}+Py=Q$, we have the form of solution as:
$y\times IF=\int{Q\times IF}dx+c$
Then, by substituting the value of IF and Q, we get:
$y\times {{x}^{\frac{3}{4}}}=\int{7\times {{x}^{\frac{3}{4}}}}dx+c$
Then, by solving the above integration, we get:
$\begin{aligned} & y\times {{x}^{\frac{3}{4}}}=7\frac{{{x}^{\frac{7}{4}}}}{\frac{7}{4}}+c \\\ & \Rightarrow y=4x+c\times {{x}^{\frac{-3}{4}}} \\\ \end{aligned}$
So, we can see clearly that by solving the integral we get the value of f(x) as:
$f\left( x \right)=4x+c\times {{x}^{\frac{-3}{4}}}$
Then, we are required to find the value of $f\left( \frac{1}{x} \right)$, so we replace x by $\frac{1}{x}$ in the above expression and we get:
$f\left( \frac{1}{x} \right)=\frac{4}{x}+c\times {{x}^{\frac{3}{4}}}$
Then, we are required to find the value of $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,xf\left( \frac{1}{x} \right)$ by substituting the value of $f\left( \frac{1}{x} \right)$, we get:
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,xf\left( \frac{1}{x} \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,x\left( \frac{4}{x}+c\times {{x}^{\frac{3}{4}}} \right)$
Then, by substituting the limit in the function, we get:
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,x\left( \frac{4}{x}+c\times {{x}^{\frac{3}{4}}} \right)=4$
So, the value of the limit exists and is equal to 4.
Hence, option (a) is correct.

Note:
Now, to solve these types of the questions we need to know some of the basic formulas for the integration beforehand to get the answer easily and accurately. So, the basic integration formulas are:
$\begin{aligned} & \int{{{x}^{n}}dx=\frac{{{x}^{n+1}}}{n+1}+c} \\\ & \int{\frac{1}{x}dx}=\ln \left| x \right|+c \\\ \end{aligned}$