Question

Let p(x) be a polynomial such that p(x) - p'(x)=$x^n$ , where n is a positive integer. Then p(0) equals.

• A. n!
• B. (n-1)!
• C.
• D.

Answer 1

The correct option is A n!

$\text{Let}P\left(x\right)=a_0{x}^n+a_1{x}^{n-1}+a_2{x}^{n-2}....a_n={\sum }_{r=0}^n{a}_r{x}^{n-r}{P}^1\left(x\right)={\sum }_{r=0}^{n-1}{a}_r(n-r)x^{n-r-1}P\left(x\right)-P^1\left(x\right)=a_0{x}^n+{\sum }_{r-1}^n\{a_r-a_{r+}(n-r+1\left)\right\}{x}^{n-r}=x^n\text{So},a_0=1\text{and}{a}_r=a_{r-1}(n-r+1)\frac{a_r}{a_{r-1}}=n-r+1\text{Now},P\left(0\right)=a_n=\frac{a_n}{a_{n-1}}.\frac{a_{n-1}}{a_{n-2}}.....\frac{a_2}{a_1}\frac{a_1}{a_0}.a_0=(1\times \mathrm{2......}n)=n!$