Question

Let $p\left(x\right)$ be a polynomial such that $p\left(x\right)–p^{\prime }\left(x\right)=x^n$, where $n$ is a positive integer. Then $p\left(0\right)$ equals

• A. $n!$
• B. $(n-1)!$
• C. $\frac{1}{n!}$
• D. $\frac{1}{(n-1)!}$

Answer 1

The correct option is A $n!$

Let $p\left(x\right)=a_0{x}^n+a_1{x}^{n-1}+a_2{x}^{n-2}+....+a_n=\sum _{r=0}^n{a}_r.x^{n-r}$
$p^{\prime }\left(x\right)=\sum _{r=1}^{n-1}{a}_r.(n-r).x^{n-r-1}$
$p\left(x\right)-p^{\prime }\left(x\right)=a_0.x^n+\sum _{r=1}^n\{a_r-a_{r-1}(n-r+1\left)\right\}{x}^{n-r}=x^n$
From above expression we get $a_0=1$
and $a_r=a_{r-1}(n-r+1)\Rightarrow \frac{a_r}{a_{r-1}}=n-r+1$
Now,
$p\left(0\right)=a_n=\frac{a_n}{a_{n-1}}\frac{a_{n-1}}{a_{n-2}}....\frac{a_2}{a_1}\frac{a_1}{a_0}.a_0=(1\times \mathrm{2......}n)=n!$