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:

Let f: R 0 (-, 1) be a polynomial of degree 2, satisfying f(x)f (… — Solution | JEE Mathematics

MCQMediumJEE 2025Nature of Roots & Formation of Equations

Let $f : \mathbb{R} \setminus \{0\} \to (-\infty, 1)$ be a polynomial of degree $2$ , satisfying $f(x)f\left( \frac{1}{x} \right) = f(x) + f\left( \frac{1}{x} \right)$ . If $f(K) = -2K$ , then the sum of squares of all possible values of $K$ is:

A
$1$
B
$7$
C
$9$
D
$6$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: $f(x)$ is a polynomial of degree $2$ and satisfies

f(x)f\left( \frac{1}{x} \right) = f(x) + f\left( \frac{1}{x} \right)

Also, find the sum of squares of all possible values of $K$ when

f(K) = -2K

Let

f(x) = ax^2 + bx + c, \quad a \ne 0

Then

f\left( \frac{1}{x} \right) = a\frac{1}{x^2} + b\frac{1}{x} + c

Using the given condition,

(ax^2 + bx + c)\left(a\frac{1}{x^2} + b\frac{1}{x} + c\right) = (ax^2 + bx + c) + \left(a\frac{1}{x^2} + b\frac{1}{x} + c\right)

After simplifying, the resulting equation becomes

1 - K^2 = -2K

So,

K^2 - 2K - 1 = 0

Solving the quadratic,

K = \frac{2 \pm \sqrt{(2)^2 - 4(1)(-1)}}{2(1)}

= \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}

Let the roots be $\alpha$ and $\beta$ . Using

\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta

From Vieta's formulas for

K^2 - 2K - 1 = 0

we get

\alpha + \beta = 2, \quad \alpha\beta = -1

Therefore,

\alpha^2 + \beta^2 = 2^2 - 2(-1) = 4 + 2 = 6

Hence, the sum of squares of all possible values of $K$ is $6$ . The correct option is D.

Common mistakes

Assuming the final quadratic in $K$ without using the given functional condition correctly is incorrect. The relation between $f(x)$ and $f\left(\frac{1}{x}\right)$ must first be used to restrict the quadratic before applying $f(K) = -2K$ .
Using the wrong identity for sum of squares is a common error. Do not take $\alpha^2 + \beta^2 = (\alpha + \beta)^2$ . The correct identity is $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ .
Sign mistakes in Vieta's formulas can change the answer. For $K^2 - 2K - 1 = 0$ , the product of roots is $-1$ , not $1$ . Substitute the signs carefully.

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Answer

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