Let f: R - \0\ (- , 1) be a polynomial of degree 2, satisfying f(x)f ( 1x ) = f(x) + f ( 1x ). If f(K) = -2K, then the sum of squares of all possible values of K is:
- 9
- 1
- 6
- 7
Answer
(3) 6
Solution
Let g(x) = f(x) - 1. Then the given relation becomes
(g(x) + 1) (g ( 1x ) + 1 ) = g(x) + g ( 1x ) + 2
g(x) g ( 1x ) = 1
Let g(x) = ax2 + bx + c (a 0). Then (ax2 + bx + c)(cx2 + bx + a) = x2 for all x.
The coefficient of x4 gives ac = 0, so c = 0; the coefficient of x3 gives ab + bc = 0, so b = 0; and the coefficient of x2 gives a2 = 1. Hence f(x) = 1 x2, and the range (- , 1), f(x) = 1 - x2.
Now f(K) = -2K gives
1 - K2 = -2K
K2 - 2K - 1 = 0 (K = 1 2, both non-zero)
K1 + K2 = 2, K1K2 = -1
K12 + K22 = (K1 + K2)2 - 2K1K2 = 4 + 2 = 6
Answer: (3)