Let be a polynomial of degree 2, satisfying . If f(K) = –2K, then the sum of squares of all possible values of K is : [2025]

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Solution

Q.
Let $f : R - {0} \to (- \infty, 1)$ be a polynomial of degree 2, satisfying $f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x})$ . If f(K) = –2K, then the sum of squares of all possible values of K is : [2025]

1 1

2 6

3 9

4 7

Ans.
(2)

We have, $f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x})$

$\Rightarrow f (x) f (\frac{1}{x}) - f (x) = f (\frac{1}{x})$

$\Rightarrow f (x) = \frac{f (\frac{1}{x})}{f (\frac{1}{x}) - 1}$ ... (i)

Also, $f (x) f (\frac{1}{x}) - f (\frac{1}{x}) = f (x)$

$\Rightarrow f (\frac{1}{x}) = \frac{f (x)}{f (x) - 1}$ ...(ii)

On multiplying (i) and (ii), we get

$f (x) f (\frac{1}{x}) = \frac{f (\frac{1}{x}) f (x)}{{f (\frac{1}{x}) - 1} {f (x) - 1}}$

$\Rightarrow (f (\frac{1}{x}) - 1) (f (x) - 1) = 1$ ... (iii)

Since, f(x) is polynomial function, so f(x) – 1 and $f (\frac{1}{x}) - 1$ are reciprocal of each other. Also x and $\frac{1}{x}$ are reciprocal of each other.

So, (iii) hold only if, $f (x) - 1 = \pm x^{n}, n \in R$

$∴ f (x) = 1 \pm x^{n}$

As, f(x) is a polynomial of degree 2 and range $\in (- \infty, 1)$

$∴ f (x) = 1 - x^{2}$

Now, $f (K) = - 2 K \Rightarrow 1 - k^{2} = - 2 K \Rightarrow K^{2} - 2 K - 1 = 0$

Let its roots i.e., possible values of K be $α$ and $β$

$∴ α^{2} + β^{2} = {(α + β)}^{2} - 2 α β = 4 - 2 (- 1) = 6$ .

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