Question

If $f$ is a real valued function satisfying $2\underset{1}{\overset{x}{\int }}f\left(t\right)dt+x^k=xf\left(x\right)$, where $x\ge 1$ and $f\left(1\right)=1$, then $f\left(x\right)$ cannot be
(where $k$ is an positive integer)

• A. $3x^2-2x$
• B. $2x-1$
• C. $2x^3+x$
• D. $x(2\ln x+1)$

Answer 1

The correct option is C $2x^3+x$

$2\underset{1}{\overset{x}{\int }}f\left(t\right)dt+x^k=xf\left(x\right)$
Differentiating both sides of w.r.t. x, we get
$\Rightarrow 2f\left(x\right)+k{x}^{k-1}=x{f}^{\prime }\left(x\right)+f\left(x\right)\Rightarrow f^{\prime }\left(x\right)-\frac{f\left(x\right)}{x}=k{x}^{k-2}\Rightarrow \frac{dy}{dx}-\frac{y}{x}=k{x}^{k-2}$
Integrating factor
$=e^{-\int 1/xdx}=e^{-\ln x}=\frac{1}{x}$
So, the solution of the differential equation is
$\frac{y}{x}=\int k{x}^{k-3}dx+C$

When $k=2$, then
$\frac{y}{x}=\int \frac{k}{x}dx+C\Rightarrow \frac{y}{x}=k\ln x+C$
Putting $x=1$
$\Rightarrow C=1$
Therefore at $k=2$, the function becomes
$f\left(x\right)=2x\ln x+x$

When $k\ne 2$, then
$\Rightarrow \frac{y}{x}=\frac{k{x}^{k-2}}{k-2}+C$
Putting $x=1$
$\Rightarrow 1=\frac{k}{k-2}+C\Rightarrow C=\frac{2}{2-k}\Rightarrow f\left(x\right)=\frac{k{x}^{k-1}}{k-2}+\frac{2x}{2-k}$
For $k=1$,
$f\left(x\right)=2x-1$
For $k=3$,
$f\left(x\right)=3x^2-2x$
For $k=4$,
$f\left(x\right)=2x^3-x$