Question

Let $k$ be a positive integer and $f(x)$ be a polynomial with integer coefficients satisfying $2\underset{1}{\overset{x}{\int }}f(t)dt+x^k=xf(x)$, where $x\ge 1$. Find the sum of all possible values of $k$.

Answer 1

Answer: 8

We have $2{\int }_1^{x}f(t)dt+x^k=xf(x)$....(1)
Differentiate both sides of equation (1) with respect to $x$, we get $2f(x)+k{x}^{k-1}=x{f}^{\prime }(x)+f(x)$
$\Rightarrow x{f}^{\prime }(x)-f(x)=k{x}^{k-1}$
$\Rightarrow f^{\prime }(x)-\frac{1}{x}f(x)=kx{x}^{k-2}\left(\frac{dy}{dx}-\frac{1}{x}y=kx{k}^{k-2}\right)$
Integrating factor $=$ L.F. $=\frac{1}{x}$
So, $\frac{f(x)}{x}=k\int x^{k-3}dx\Rightarrow \frac{f(x)}{x}=\frac{k\cdot x^{k-2}}{k-2}+C$
Put $x=1$ in (1), we get $f(1)=1$
Now, $C=1-\frac{k}{k-2}\Rightarrow C=\frac{2}{2-k}$
$\Rightarrow f(x)=\frac{k{x}^{k-1}}{k-2}+\frac{2x}{2-k}$
Since coefficients of $f(x)$ are integer, so
$\frac{k}{k-2}$ and $\frac{2}{2-k}$ are integer $\Rightarrow 2-k=\pm 1,\pm 2\Rightarrow k=0,1,3,4$
But $k\ne 0$ as $k>0$ and at $k=1,3,4,\frac{k}{k-2}$ is also integer hence possible values of $k$ are $1,3,4$
Hence, sum of possible values of $k=1+3+4=8$.