|=√3 √3α+β=±2√3......(1) Given α=2+√3β......(2) From equation (1) and (2),we get =2or−1 So |α|=1or2 (B) f(x)={
−3αx2−2,
x<1
bx+a2,
x≥1
For continuity −3a−2=b+a2......(1) a2+3a+2=−b For differentiability −6a=b 6a=−b a2−3a+2=0 a=1,2 (C)(3−3ω+2ω2)4n+3+(2+3ω−3ω2)4n+3+(−3+2ω+3ω2)4n+3=0 (3−3ω+2ω2)4n+3+(ω(2ω2+3−3ω))4n+3+(ω2(−3ω+2ω2+3))4n+3=0 ⇒(3−3ω+2ω2)4n+3(1+ω4n+ω8n)=0 ⇒n≠3k,k∊N Let a=5−d q=5+d b=5+2d |q−a|=|2d| Given