Relations and Functions (Class 12) · Question 35

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:

1. 9
2. 1
3. 6
4. 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.

Coefficient of x4: ac = 0 c = 0; coefficient of x3: ab + bc = 0 b = 0; coefficient of x2: a2 = 1

f(x) = 1 x2

Range (- , 1) f(x) = 1 - x2

f(K) = -2K 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)