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Question

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

A
2017
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B
2016
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C
2018
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D
2015
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Solution

The correct option is D 2015
Note that

x[0,2015],Q(x)=P(x)(x2015)=0

Q(x)=ax(x1)(x2)(x2015)

P(x)=q(x)+(x2015)

P(x)=ax(x1)(x2)(x2015)+(x2015)

As a=1!, Hence we got a solution
P(2016)=a2016(20161)1+(20162015)=12016!+1

thus
n=2016,a=1n+a=2015


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