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

Answer: (A)

$n!$

Let $P\left(x\right)=a_0{x}^n+a_1{x}^{n-1}+a_2{x}^{n-2}....a_n=\sum _{x=0}^n{a}_r{x}^{n-r}$

$P^{\prime }\left(x\right)=\sum _{x=0}^{n-1}{a}_r(n-r)x^{n-r-1}$

$P\left(x\right)-P^{\prime }\left(x\right)=a_0{x}^n+\sum _{x=1}^n\{a_r-a_{r-1}(n-r+1\left)\right\}{x}^{n-r}=x^n$

So, $a_0=1$ and $a_r=a_{r-1}(n-r+1)$

$\frac{a_r}{a_{r-1}}=n-r+1$

Now, $P\left(0\right)=a_n=\frac{a_n}{a_{n-1}}\cdot \frac{a_{n-1}}{a_{n-2}}....\frac{a_2}{a_1}\cdot \frac{a_1}{a_0}.a_0$

$=(1\times 2....n)=n!$