Question

Let $k$ be a positive integer and $f ( x )$ be a polynomial with integer coefficients satisfying $2 \int\limits_1^x f(t) d t+x^k=x f(x)$, where $x \geq 1$. Find the sum of all possible values of $k$.

Answer 1

We have $2 \int_1^x f(t) d t+x^k=x f(x)$....(1)
Differentiate both sides of equation (1) with respect to $x$, we get $2 f(x)+k x^{k-1}=x f^{\prime}(x)+f(x)$
$\Rightarrow xf ^{\prime}( x )- f ( x )= kx ^{ k -1}$
$\Rightarrow f ^{\prime}( x )-\frac{1}{ x } f ( x )= kx x ^{ k -2} \left(\frac{ dy }{ dx }-\frac{1}{ x } y = kx k ^{ k -2}\right)$
Integrating factor $=$ L.F. $=\frac{1}{x}$
So, $\frac{f(x)}{x}=k \int x^{k-3} d x \Rightarrow \frac{f(x)}{x}=\frac{k \cdot x^{k-2}}{k-2}+C$
Put $x=1$ in (1), we get $f(1)=1$
Now, $C =1-\frac{ k }{ k -2} \Rightarrow C =\frac{2}{2- k }$
$\Rightarrow f ( x )=\frac{ kx ^{ k -1}}{ k -2}+\frac{2 x }{2- k }$
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 $k \neq 0$ 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 $k=1+3+4=8$.