Question

3. F(x) is a polynomial in x of degree 7. The coefficient
of x' in F(x) is 1. F(1) = 5, F(2) = 7. F(3) = 9
F(4) = 11, F(5) = 13, F(6) = 15, F(7) = 17. Find F(8).
(A) 21 (B) 4943 (C) 5059 (D) 5071​

Answer 1

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 1,2,3,4and5 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