1.

Suppose f(x) is a polynomial of degree 5 and with leading coefficient 2009.Suppose further f(1)=1,f(2)=3,f(3)=5,f(4)=7,f(5)=9.What is the value of f(6).

Answer» Let, `f(x) = g(x)+2x-1`
Then, `f(1) = g(2)+2-1 = 1=> g(1) = 0`
`f(2) = g(2)+4-1 = 3=> g(2) = 0`
`f(3) = g(3)+6-1 = 5=> g(3) = 0`
`f(4) = g(4)+8-1 = 7=> g(4) = 0`
`f(5) = g(5)+10-1 = 9=> g(5) = 0`
From the above, we can see that `1,2,3,4 and 5` are the roots of `g(x)`.
So, we can write,
`f(x) = 2009(x-1)(x-2)(x-3)(x-4)(x-5) +2x-1`
`:. f(6) = 2009**5**4**3**2**1+12-1=241091`


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