Let f: R-{0} →(-∞, 1) be a polynomial of degree 2 , satisfying f(x) f(1/x)=f(x)+f(1/x). If f( K)=-2 K, then the sum of squares of all possible values of K is :

Solution

Maths sir · 5 steps

Add 1 to both sides and it factors as (f(x)-1)(f(1/x)-1)=1, which pins f down completely.

1. Factor it. Bring everything to one side and add 1: f(x)f(1/x) - f(x) - f(1/x) + 1 = 1, which is (f(x)-1)(f(1/x)-1) = 1. Let g(x) = f(x) - 1 = ax² + bx + c with a ≠ 0, so g(x) g(1/x) = 1.
2. Clear the fractions. Multiply g(x) g(1/x) = 1 by x². Since x² g(1/x) = cx² + bx + a, this gives a polynomial identity.

   (ax²+bx+c)(cx²+bx+a) = x²

3. Compare coefficients. The x⁴ term gives ac = 0, and a ≠ 0, so c = 0. The left side becomes abx³ + (a²+b²)x² + abx, so ab = 0 gives b = 0, and then a² = 1. Hence f(x) = 1 ± x².
4. Use the codomain. 1 + x² ≥ 1, which is outside (-∞, 1). So f(x) = 1 - x², and this is less than 1 for every x ≠ 0.
5. So. 1 - K² = -2K gives K² - 2K - 1 = 0, so K = 1 ± √2. The roots have sum 2 and product -1, so the sum of squares is 2² - 2(-1) = 6, option (c).

The idea to keep. For f(x)f(1/x) = f(x) + f(1/x), add 1 and factor. The polynomial solutions are f(x) = 1 ± xⁿ. Then use the codomain to choose the sign.

Answer from the NTA final answer key. Solution by PrepWiser.