Calculus – Functional Equations – JEE Main 28 Jan 2025 Shift 2

Solution:

The standard functional equation $f(x)f(1/x) = f(x) + f(1/x)$ for a polynomial has solutions of the form $f(x) = 1 \pm x^n$.

Given that $f(x)$ is a polynomial of degree 2, $n=2$.
So, $f(x) = 1 + x^2$ or $f(x) = 1 – x^2$.

The range is given as $(-\infty, 1)$.
If $f(x) = 1 + x^2$, the range is $[1, \infty)$, which does not match.
If $f(x) = 1 – x^2$, the range is $(-\infty, 1]$, which matches the codomain condition.
So, $f(x) = 1 – x^2$.

We are given $f(K) = -2K$.
$$1 – K^2 = -2K$$
$$K^2 – 2K – 1 = 0$$

Let the roots be $K_1$ and $K_2$.
Sum of roots ($K_1 + K_2$) = 2.
Product of roots ($K_1 K_2$) = -1.

We need the sum of squares of all possible values of K:
$$K_1^2 + K_2^2 = (K_1 + K_2)^2 – 2K_1 K_2$$
$$= (2)^2 – 2(-1)$$
$$= 4 + 2 = 6$$

Ans. (2)