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If f(x) is a polynomial of degree n such that f(0)=0,f(1)=12,....,f(n)=nn+1, then the value of f(n+1) is 


A
1 when n is odd
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B
nn+2 when n is even
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C
nn+1 when n is odd
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D
1 when n is even
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Solution

The correct options are
A 1 when n is odd
B nn+2 when n is even
(x+1)f(x)x is a polynomial of degree n+1
(x+1)f(x)x=k(x)(x1)(x2)....(xn).....(i)
Also, 1=k(1)(2)...((n1)      (Putting x=1 in (i))
1=k(1)n+1(n+1)!
k=1(1)n+1(n+1)!
By replacing x with n+1,we get​​​​
(n+2)f(n+1)(n+1)=k[(n+1)!]
(n+2)f(n+1)(n+1)=(1)n+1
f(n+1)=1,if n is odd and nn+2,if n is even.

Mathematics

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