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Tuesday, May 5, 2015

Problem from Indian Statistical Institute: B.Stat. (Hons.)

Let P(x) be a polynomial of degree 11 such that P(x)=1x+1, for x=0,1,2,,11.
Find the value of P(12)
Solution: Let f(x)=(x+1)P(x)1, clearly f(x) is a polynomial of degree 12.
Now for x{0,1,2,,11}, f(x)=(x+1)P(x)1=x+1x+11=11=0
This shows that f(x) vanishes at the points x=0,1,2,,11.
f(x) being of degree 12, the above statement assures that x=0,1,2,,11. are the possible roots of f(x)
Therefore f(x)=(x+1)P(x)1=a0(x0)(x1)(x2)(x11)
Letting x=1 in the above equality, we have 1=a0(1)(2)(3)(12)a0=112!
Therefore f(x)=(x+1)P(x)1=112!(x0)(x1)(x2)(x11)
Now, letting x=12 we have 13P(12)1=112!(12)(11)(10)(1)=12!12!=1
13P(12)1=1P(12)=0

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