Question

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

• 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. \(P \to (2),\ Q \to (3),\ R \to (5),\ S \to (4)\)

Hint:

For cube roots of unity: \[ 1+\omega+\omega^2=0 \] and: \[ \omega^3=1 \]

Answer 1

The Correct Option is A

Step 1: Solve part (P).
Roots of: \[ x^2+x+1=0 \] are cube roots of unity: \[ \alpha=\omega,\qquad \beta=\omega^2 \] Now: \[ 1+\omega=-\omega^2 \] Thus: \[ \frac1{(\alpha+1)^{2026}} = \frac1{(-\omega^2)^{2026}} = \omega \] Similarly: \[ \frac1{(\beta+1)^{2026}} = \omega^2 \] Hence required quadratic equation is: \[ x^2+x+1=0 \] Therefore: \[ (P)\to(1) \]

Step 2: Solve part (Q).
Now exponent is: \[ 2027 \] Thus: \[ \frac1{(\alpha+1)^{2027}} = -\omega^2 \] and: \[ \frac1{(\beta+1)^{2027}} = -\omega \] Their sum: \[ \omega+\omega^2=-1 \] Hence: \[ (-\omega)+(-\omega^2)=1 \] Product: \[ (-\omega)(-\omega^2)=1 \] Thus equation becomes: \[ x^2-x+1=0 \] Therefore: \[ (Q)\to(2) \]

Step 3: Solve part (R).
Roots of: \[ x^2-x+1=0 \] are: \[ \gamma,\delta \] Using: \[ \gamma-1=-\delta, \qquad \delta-1=-\gamma \] Hence: \[ \frac1{(\gamma-1)^{2026}} + \frac1{(\delta-1)^{2026}} = \frac1{\delta^{2026}} + \frac1{\gamma^{2026}} \] Since: \[ \gamma^6=\delta^6=1 \] and: \[ 2026\equiv4\pmod6 \] we get: \[ = \gamma^2+\delta^2 \] Now: \[ \gamma+\delta=1, \qquad \gamma\delta=1 \] Thus: \[ \gamma^2+\delta^2 = (\gamma+\delta)^2-2\gamma\delta \] \[ =1-2 \] \[ =-1 \] Therefore: \[ (R)\to(4) \]

Step 4: Solve part (S).
Roots of: \[ x^2+x-1=0 \] are: \[ p,r \] Now: \[ (p+1)(r+1) = pr+p+r+1 \] Using: \[ p+r=-1, \qquad pr=-1 \] \[ (p+1)(r+1)=-1-1+1 \] \[ =-1 \] Also: \[ \frac1{(p+1)^3}+\frac1{(r+1)^3} = \frac{(p+1)^3+(r+1)^3}{[(p+1)(r+1)]^3} \] Denominator: \[ =(-1)^3=-1 \] Now: \[ (p+1)+(r+1)=1 \] and: \[ (p+1)(r+1)=-1 \] Thus: \[ (a^3+b^3)=(a+b)^3-3ab(a+b) \] \[ =1^3-3(-1)(1) \] \[ =4 \] Hence: \[ \frac4{-1}=-4 \] Therefore: \[ (S)\to(5) \]

Step 5: Identify the correct option.
Hence: \[ \boxed{\mathrm{(A)}} \]