JEE Advanced 2020 Full Test 6 Paper 1

x2+px+2=0<rs and x2+rx+s=0<pq
r+s=−p .....(1)
rs=q ......(2)
p+q=−r ........(3)
pq=s ........(4)
(2)×(4)⇒ pqrs =qs⇒pr=1 .......(5)
(1)+(3)⇒p+q+r+s=−p−r
⇒2(p+r)+q+s=0 ......(6)
Now p is a root of x2+rx+s=0⇒p2+rp+s=0
And r is a root of x2+px+q=0⇒r2+pr+q=0
Adding those two
⇒(p+r)2−2(p+r)=0⇒p+r=0,p+r=2
Also pr=1⇒p=1,r=1
From (4)pq=s⇒q=s
∴q+s=−2(p+r)
⇒q+s=−4⇒q=s=−2
∴4×1−2+2−2=2