Let f be a polynomial function such that 2(f(x))= ( 2 (2+ 23+ 29+ ) ) 3 (1+ f(x)f ( 1x ) ), x>0 and f(6)=37. Then n=110f(n) is equal to .
Answer
395
Solution
2 + 23 + 29 + = 21 - 13 = 3, so
2 f(x) = 2 3 3 (1 + f(x)f(1/x) ) = 2 (1 + f(x)f(1/x) )
f(x) = 1 + f(x)f(1/x) f(x) f ( 1x ) = f(x) + f ( 1x )
A polynomial satisfying this is f(x) = 1 xk, and f(6) = 37 gives
1 + 6k = 37 k = 2 f(x) = x2 + 1
∴ n=110f(n) = n=110n2 + 10 = 10 11 216 + 10 = 385 + 10 = 395
Answer: 395