Let R be the interior region between the lines 3x−y+1=0 and x+2y−5=0 containing the origin. The set of all values of a , for which the points ( a2,a+1 ) lie in R , is :
(A)(−3,−1)∪(−31,1)
(B)(−3,0)∪(31,1)✓ Correct
(C)(−3,0)∪(32,1)
(D)(−3,−1)∪(31,1)
Answer: (B)
Solution
Let L1:3x−y+1=0 and L2:x+2y−5=0, and P(a2,a+1).
P and the origin lie on the same side of L1 when L1(0)⋅L1(P)>0:
1⋅[3a2−(a+1)+1]>0⇒3a2−a>0⇒a(3a−1)>0
⇒a∈(−∞,0)∪(31,∞).(1)
Similarly for L2: L2(0)⋅L2(P)>0:
(−5)[a2+2(a+1)−5]>0⇒a2+2a−3<0⇒(a+3)(a−1)<0
⇒a∈(−3,1).(2)
Intersecting (1) and (2):
a∈(−3,0)∪(31,1).
JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option