JEE Challenger
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MathMedAdv 2026 • Paper 1

Matching List Quadratic Equations with Transformed Roots

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

List-IList-II(P) If α and β are the distinct roots of the equation x2+x+1=0, then the quadratic equation with roots 1(α+1)2026 and 1(β+1)2026 is(1) x2+x+1=0(Q) If α and β are the distinct roots of the equation x2+x+1=0, then the quadratic equation with roots 1(α+1)2027 and 1(β+1)2027 is(2) x2−x+1=0(R) If γ and δ are the distinct roots of the equation x2−x+1=0, then the value of 1(γ−1)2026+1(δ−1)2026 is(3) x2+x−1=0(S) If p and r are the distinct roots of the equation x2+x−1=0, then the value of 1(p+1)3+1(r+1)3 is(4) −1(5) −4\begin{array}{ll} \text{\textbf{List-I}} & \text{\textbf{List-II}} \\ \text{(P) If } \alpha \text{ and } \beta \text{ are the distinct roots of the equation } x^2 + x + 1 = 0\text{, then the quadratic equation with roots } \frac{1}{(\alpha+1)^{2026}} \text{ and } \frac{1}{(\beta+1)^{2026}} \text{ is} & \text{(1) } x^2 + x + 1 = 0 \\ \text{(Q) If } \alpha \text{ and } \beta \text{ are the distinct roots of the equation } x^2 + x + 1 = 0\text{, then the quadratic equation with roots } \frac{1}{(\alpha+1)^{2027}} \text{ and } \frac{1}{(\beta+1)^{2027}} \text{ is} & \text{(2) } x^2 - x + 1 = 0 \\ \text{(R) If } \gamma \text{ and } \delta \text{ are the distinct roots of the equation } x^2 - x + 1 = 0\text{, then the value of } \frac{1}{(\gamma-1)^{2026}} + \frac{1}{(\delta-1)^{2026}} \text{ is} & \text{(3) } x^2 + x - 1 = 0 \\ \text{(S) If } p \text{ and } r \text{ are the distinct roots of the equation } x^2 + x - 1 = 0\text{, then the value of } \frac{1}{(p+1)^3} + \frac{1}{(r+1)^3} \text{ is} & \text{(4) } -1 \\ & \text{(5) } -4 \end{array}

Options

A

(P) →\rightarrow (1), (Q) →\rightarrow (2), (R) →\rightarrow (5), (S) →\rightarrow (4)

B

(P) →\rightarrow (3), (Q) →\rightarrow (1), (R) →\rightarrow (4), (S) →\rightarrow (5)

C

(P) →\rightarrow (1), (Q) →\rightarrow (2), (R) →\rightarrow (4), (S) →\rightarrow (5)

Correct
D

(P) →\rightarrow (2), (Q) →\rightarrow (3), (R) →\rightarrow (5), (S) →\rightarrow (4)

Topics & Concepts

Complex Numbers and Quadratic EquationsQuadratic EquationsLocation of Roots

Step-by-Step Solution

To solve the matching list, we analyze each part individually:

Part (P): The roots of x2+x+1=0x^2 + x + 1 = 0 are the non-real cube roots of unity, α=ω\alpha = \omega and β=ω2\beta = \omega^2. Since 1+ω+ω2=01 + \omega + \omega^2 = 0, we have α+1=−ω2\alpha + 1 = -\omega^2 and β+1=−ω\beta + 1 = -\omega. Evaluating the new roots: 1(α+1)2026=1(−ω2)2026=1ω4052=1ω2=ω\frac{1}{(\alpha+1)^{2026}} = \frac{1}{(-\omega^2)^{2026}} = \frac{1}{\omega^{4052}} = \frac{1}{\omega^2} = \omega 1(β+1)2026=1(−ω)2026=1ω2026=1ω=ω2\frac{1}{(\beta+1)^{2026}} = \frac{1}{(-\omega)^{2026}} = \frac{1}{\omega^{2026}} = \frac{1}{\omega} = \omega^2 The transformed roots are ω\omega and ω2\omega^2, so the required quadratic equation is x2+x+1=0x^2 + x + 1 = 0. Thus, (P) →\rightarrow (1).

Part (Q): Using α+1=−ω2\alpha + 1 = -\omega^2 and β+1=−ω\beta + 1 = -\omega: 1(α+1)2027=1(−ω2)2027=−1ω4054=−1ω=−ω2\frac{1}{(\alpha+1)^{2027}} = \frac{1}{(-\omega^2)^{2027}} = -\frac{1}{\omega^{4054}} = -\frac{1}{\omega} = -\omega^2 1(β+1)2027=1(−ω)2027=−1ω2027=−1ω2=−ω\frac{1}{(\beta+1)^{2027}} = \frac{1}{(-\omega)^{2027}} = -\frac{1}{\omega^{2027}} = -\frac{1}{\omega^2} = -\omega The sum of the new roots is (−ω)+(−ω2)=1(-\omega) + (-\omega^2) = 1 and their product is (−ω)(−ω2)=1(-\omega)(-\omega^2) = 1. The quadratic equation is x2−x+1=0x^2 - x + 1 = 0. Thus, (Q) →\rightarrow (2).

Part (R): , The roots of x2−x+1=0x^2 - x + 1 = 0 are γ=−ω\gamma = -\omega and δ=−ω2\delta = -\omega^2. Thus, γ−1=−ω−1=ω2\gamma - 1 = -\omega - 1 = \omega^2 and δ−1=−ω2−1=ω\delta - 1 = -\omega^2 - 1 = \omega. 1(γ−1)2026+1(δ−1)2026=1(ω2)2026+1ω2026=1ω2+1ω=ω+ω2=−1\frac{1}{(\gamma-1)^{2026}} + \frac{1}{(\delta-1)^{2026}} = \frac{1}{(\omega^2)^{2026}} + \frac{1}{\omega^{2026}} = \frac{1}{\omega^2} + \frac{1}{\omega} = \omega + \omega^2 = -1 Thus, (R) →\rightarrow (4).

Part (S): Since pp and rr are roots of x2+x−1=0x^2 + x - 1 = 0, we have p2+p=1  ⟹  p(p+1)=1  ⟹  1p+1=pp^2 + p = 1 \implies p(p+1) = 1 \implies \frac{1}{p+1} = p. Similarly, 1r+1=r\frac{1}{r+1} = r. Using p+r=−1p + r = -1 and pr=−1pr = -1: 1(p+1)3+1(r+1)3=p3+r3=(p+r)3−3pr(p+r)=(−1)3−3(−1)(−1)=−1−3=−4\frac{1}{(p+1)^3} + \frac{1}{(r+1)^3} = p^3 + r^3 = (p+r)^3 - 3pr(p+r) = (-1)^3 - 3(-1)(-1) = -1 - 3 = -4 Thus, (S) →\rightarrow (5).

Conclusion: (P) →\rightarrow (1), (Q) →\rightarrow (2), (R) →\rightarrow (4), (S) →\rightarrow (5)

The correct option is C.

Matching List Quadratic Equations with Transformed Roots | Mathematics PYQ Solution - JEE Challenger