Question 13

For (P) and (Q), roots of $x^2 + x + 1 = 0$ are $\omega$ and $\omega^2$.

Since $\alpha + 1 = -\alpha^2$ and $\alpha^3 = 1$:

(P) New roots are $\frac{1}{(-\alpha^2)^{2026}} = \frac{1}{\alpha^{4052}}$.

Since $4052 \equiv 2 \pmod 3$, roots are $\frac{1}{\alpha^2} = \alpha$ and $\frac{1}{\beta^2} = \beta$.

The equation remains $x^2 + x + 1 = 0$. Thus (P) → (1).

(Q) New roots are $\frac{1}{(-\alpha^2)^{2027}} = \frac{-1}{\alpha^{4054}} = \frac{-1}{\alpha} = -\alpha^2$. The roots are $-\omega^2$ and $-\omega$. Sum $= 1$, Product $= 1$.

Equation is $x^2 - x + 1 = 0$. Thus (Q) → (2).

(R) Roots of $x^2 - x + 1 = 0$ are $-\omega, -\omega^2$. Let $\gamma = -\omega$. Then $\gamma - 1 = -\omega - 1 = \omega^2$.

Value $= \frac{1}{(\omega^2)^{2026}} + \frac{1}{(\omega)^{2026}} = \frac{1}{\omega^2} + \frac{1}{\omega} = \omega + \omega^2 = -1$. Thus (R) → (4).

(S) Roots of $x^2 + x - 1 = 0$ satisfy $p+1 = \frac{1}{p}$.

Value $= p^3 + r^3 = (p+r)^3 - 3pr(p+r) = (-1)^3 - 3(-1)(-1) = -1 - 3 = -4$. Thus (S) → (5).

Matching sequence: (P) → (1), (Q) → (2), (R) → (4), (S) → (5).

Correct Option: C