Question

Let $P\left(x\right)$ be a polynomial with degree $2016$ and leading co-efficient unity such that $P\left(0\right)=2015,P\left(1\right)=2014,P\left(2\right)\left(2013\right).......P\left(2015\right)=0$ then the value of $P\left(2016\right)=(!n)-a$ where n and a are natural number then value of $(n+a)$

• A. $2017$
• B. $2016$
• C. $2018$
• D. $2015$

Answer 1

The correct option is D $2015$

Note that

$\forall x\in [0,2015],Q\left(x\right)=P\left(x\right)-(x-2015)=0$

$⟹Q\left(x\right)=ax(x-1)(x-2)\dots (x-2015)$

$⟹P\left(x\right)=q\left(x\right)+(x-2015)$

$⟹P\left(x\right)=ax(x-1)(x-2)\dots (x-2015)+(x-2015)$

As $a=1!$, Hence we got a solution

$P\left(2016\right)=a\cdot 2016\cdot (2016-1)\cdot \dots \cdot 1+(2016-2015)=1\cdot 2016!+1$

thus

$n=2016,a=-1⟹n+a=2015$