Let α, ß, \(\gamma\) are roots of P1(x) and α, ß, \(\delta\) are roots of P2(x).
∴ α + ß + \(\gamma\) = 2020
and α + ß + \(\delta\) = 2021
⇒ \(\delta-\gamma\) = 1
Also αß + ß\(\gamma\) + \(\gamma\)α = b1
αß + ß\(\delta\) + \(\delta\)α = b2
⇒ ß(\(\delta\) - \(\gamma\)) + α(\(\delta\) - \(\gamma\)) = b2 - b1
⇒ ß + α = b2 - b1
Also, αß\(\gamma\) = -C1
αß\(\gamma\) = -C2
⇒ αß = C1 - C2
Given that p1(x) q1(x)+ p2(x)q2(x) = x2 - 3x + 2
⇒ (x - α) (x - ß) (x - γ)q1 (x) + (x - α) (x - ß) (x - δ )q2(x) = (x - 1) (x - 2)
⇒ (x - α) (x - ß) ((x - γ )q1(x) + (x - δ)q2(x)) = (x - 1) (x - 2)
∃ q1(x), q2(x) such that (x - γ)q1(x) + (x - δ) q2(x) = 1
α = 1, ß = 2
Therefore, r = 2017
δ = 2018
∴ p1(x) = (x- 1) (x- 2) (x - 2017)
p2(x) = (x - 1)(x - 2) (x - 2018)
p1(3) = 2 x 1 x 2014 = -4032
p2(1) = 0
p1(3) + p2(1) = -4032
⇒ p1(3) + p2(1) + 4032 = 0