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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)âˆ’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 2015Note that∀x∈[0,2015],Q(x)=P(x)−(x−2015)=0⟹Q(x)=ax(x−1)(x−2)…(x−2015)⟹P(x)=q(x)+(x−2015)⟹P(x)=ax(x−1)(x−2)…(x−2015)+(x−2015) As a=1!, Hence we got a solutionP(2016)=a⋅2016⋅(2016−1)⋅…⋅1+(2016−2015)=1⋅2016!+1thusn=2016,a=−1⟹n+a=2015

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