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 A1 when n is odd Bnn+2 when n is even (x+1)f(x)−x is a polynomial of degree n+1 ⇒(x+1)f(x)−x=k(x)(x−1)(x−2)....(x−n).....(i) Also, 1=k(−1)(−2)...((−n−1)(Puttingx=−1in(i)) ⇒1=k(−1)n+1(n+1)! ⇒k=1(−1)n+1(n+1)! Byreplacingxwithn+1,weget ⇒(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.