JEE Advanced 2026 — Paper 1 — MATH-IV solution

JEE Advanced2026Paper 1MATH-IV

Match each entry in List-I to the correct entry in List-II and choose the correct option.

List-I	List-II
(P) If $α$ and $β$ are the distinct roots of the equation $x^{2} + x + 1 = 0$ , then the quadratic equation with roots $\frac{1}{( α + 1 ) ^{2026}}$ and $\frac{1}{( β + 1 ) ^{2026}}$ is	(1) $x^{2} + x + 1 = 0$
(Q) If $α$ and $β$ are the distinct roots of the equation $x^{2} + x + 1 = 0$ , then the quadratic equation with roots $\frac{1}{( α + 1 ) ^{2027}}$ and $\frac{1}{( β + 1 ) ^{2027}}$ is	(2) $x^{2} - x + 1 = 0$
(R) If $γ$ and $δ$ are the distinct roots of the equation $x^{2} - x + 1 = 0$ , then the value of $\frac{1}{( γ - 1 ) ^{2026}} + \frac{1}{( δ - 1 ) ^{2026}}$ is	(3) $x^{2} + x - 1 = 0$
(S) If $p$ and $r$ are the distinct roots of the equation $x^{2} + x - 1 = 0$ , then the value of $\frac{1}{( p + 1 ) ^{3}} + \frac{1}{( r + 1 ) ^{3}}$ is	(4) $- 1$
	(5) $- 4$

A
(P) $\to$ (1), (Q) $\to$ (2), (R) $\to$ (5), (S) $\to$ (4)
B
(P) $\to$ (3), (Q) $\to$ (1), (R) $\to$ (4), (S) $\to$ (5)
C
(P) $\to$ (1), (Q) $\to$ (2), (R) $\to$ (4), (S) $\to$ (5)
D
(P) $\to$ (2), (Q) $\to$ (3), (R) $\to$ (5), (S) $\to$ (4)

Solution

Let $ω = e^{2 π i /3}$ , a primitive cube root of unity, so $ω^{3} = 1$ and $1 + ω + ω^{2} = 0$ .

(P) Roots of $x^{2} + x + 1 = 0$ are $ω, ω^{2}$ . Then $ω + 1 = - ω^{2}$ , so $(ω + 1)^{2026} = (- ω^{2})^{2026} = ω^{4052} = ω^{4052 mod 3} = ω^{2}$ . Similarly $(ω^{2} + 1)^{2026} = (- ω)^{2026} = ω^{2026 mod 3} = ω$ .

So the new roots are $\frac{1}{ω ^{2}} = ω$ and $\frac{1}{ω} = ω^{2}$ , giving the equation $x^{2} + x + 1 = 0$ . P $\to$ 1.

(Q) $(ω + 1)^{2027} = (- ω^{2})^{2027} = - ω^{4054} = - ω$ and $(ω^{2} + 1)^{2027} = (- ω)^{2027} = - ω^{2027} = - ω^{2}$ . New roots: $- 1/ ω = - ω^{2}$ and $- 1/ ω^{2} = - ω$ . Sum $= - (ω + ω^{2}) = 1$ , product $= ω^{3} = 1$ , giving $x^{2} - x + 1 = 0$ . Q $\to$ 2.

(R) Roots of $x^{2} - x + 1 = 0$ are $- ω$ and $- ω^{2}$ (since these satisfy $x^{2} - x + 1 = 0$ ). So $γ - 1 = - ω - 1 = ω^{2}$ and $δ - 1 = - ω^{2} - 1 = ω$ . Therefore $\frac{1}{( γ - 1 ) ^{2026}} + \frac{1}{( δ - 1 ) ^{2026}} = ω^{- 4052} + ω^{- 2026} = ω + ω^{2} = - 1$ . R $\to$ 4.

(S) For $x^{2} + x - 1 = 0$ : $p + r = - 1$ , $p r = - 1$ . Then $(p + 1) (r + 1) = p r + (p + r) + 1 = - 1$ . Also $(p + 1)^{3} + (r + 1)^{3} = (p^{3} + r^{3}) + 3 (p^{2} + r^{2}) + 3 (p + r) + 2$ .

Compute $p^{2} + r^{2} = (p + r)^{2} - 2 p r = 1 + 2 = 3$ . And $p^{3} + r^{3} = (p + r)^{3} - 3 p r (p + r) = - 1 - 3 (- 1) (- 1) = - 1 - 3 = - 4$ .

So $(p + 1)^{3} + (r + 1)^{3} = - 4 + 9 - 3 + 2 = 4$ .

$\frac{1}{( p + 1 ) ^{3}} + \frac{1}{( r + 1 ) ^{3}} = \frac{( p + 1 ) ^{3} + ( r + 1 ) ^{3}}{[( p + 1 ) ( r + 1 ) ] ^{3}} = \frac{4}{( - 1 ) ^{3}} = - 4$ . S $\to$ 5.

$∴$ The correct option is (C).

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