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JEE Advanced Mathematics Quadratic Equation 2026 JEE Advanced 2026 (Paper 1)

JEE Advanced Mathematics Question (2026) — Solution

Question

Match each entry in List-I to the correct entry in List-II and choose the correct option. List-I List-II (P) If and are the distinct roots of the equation x^2 + x + 1 = 0, then the quadratic equation with roots 1 ( +1)^ 2026 and 1 ( +1)^ 2026 is (1) x^2 + x + 1 = 0 (Q) If and are the distinct roots of the equation x^2 + x + 1 = 0, then the quadratic equation with roots 1 ( +1)^ 2027 and 1 ( +1)^ 2027 is (2) x^2 - x + 1 = 0 (R) If and are the distinct roots of the equation x^2 - x + 1 = 0, then the value of 1 ( -1)^ 2026 + 1 ( -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 1 (p+1)^3 + 1 (r+1)^3 is (4) -1 (5) -4

Options

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

Answer

C. (P) (1), (Q) (2), (R) (4), (S) (5)

Step-by-step solution

For (P): Since and are roots of x^2 + x + 1 = 0, we have ^2 + + 1 = 0 + 1 = - ^2 and ^3 = 1. 1 ( +1)^ 2026 = 1 (- ^2)^ 2026 = 1 ^ 4052 Since 4052 = 3 1350 + 2, ^ 4052 = ( ^3)^ 1350 ^2 = ^2. Thus, 1 ^2 = ^3 ^2 = . Similarly, 1 ( +1)^ 2026 = . The quadratic equation with roots and is the original equation x^2 + x + 1 = 0. So, (P) (1). For (Q): 1 ( +1)^ 2027 = 1 (- ^2)^ 2027 = - 1 ^ 4054 Since 4054 = 3 1351 + 1, ^ 4054 = . Thus, - 1 = - . Similarly, the other root is - . Sum of roots = - - = -( + ) = -(-1) = 1. Product of roots = (- )(- ) = = 1. The quadratic equation is x^2 - x + 1 = 0. So, (Q) (2). For (R): Since and are roots of x^2 - x + 1 = 0, we have ^2 - + 1 = 0 - 1 = ^2 and ^3 = -1. 1 ( -1)^ 2026 = 1 ( ^2)^ 2026 = 1 ^ 4052 ^ 4052 = ( ^3)^ 1350 ^2 = (-1)^ 1350 ^2 = ^2. Thus, 1 ^2 = ^3 = - . Similarly, the second term is - . The sum is - - = -( + ) = -(1) = -1. So, (R) (4). For (S): Since p and r are roots of x^2 + x - 1 = 0, we have x^2 + x = 1 x(x+1) = 1 x+1 = 1 x . 1 (p+1)^3 + 1 (r+1)^3 = p^3 + r^3 We know p+r = -1 and pr = -1. p^3 + r^3 = (p+r)^3 - 3pr(p+r) = (-1)^3 - 3(-1)(-1) = -1 - 3 = -4. So, (S) (5). Answer: (P) (1), (Q) (2), (R) (4), (S) (5)

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