]} and B={y∈R∕y=sin−1(√x2+x+1),x∈A} Since, domain of sin−1(x) is [−1,1] So, domain of sin−1√x2+x+1 is real, so so √x2+x+1∈[−1,1] ⇒√x2+x+1∈[0,1] ( ∵ square root is always non-negative) ⇒x2+x+1≤1 ⇒x2+x≤0 ⇒x(x+1)≤0 So, x∈[−1,0] for set A Thus, A=[−1,0] Now, for set B, since x∈A=[−1,0], the range values for √x2+x+1 for x∈[−1,0] we will find. At x=−1,√(−1)2−1+1\=√1−1+1=1 At x=0,√02+0+1=√0+1=1 At x=−
1
2
,√(
−1
2
)2−
1
2
+1=√
1
4
−
1
2
+1=√
3
4
So, √x2+x+1∈[√
3
4
,1]=[
√3
2
,1] Thus, y=sin−1(√x2+x+1) ∈[sin−1(
√3
2
),sin−1(1)] =[
π
3
,
π
2
] Hence, B=[
π
3
,
π
2
] ⇒ Now, A∩B=[−1,0]∩[
π
3
,
π
2
]=φ So, A∩B≠φ is false. ⇒A∩BC=[−1,0]∩[
π
3
,
π
2
]C and the intersection with any set outside this cannot be [0,1], So, A∩BC=[0,1] is false. ⇒AC is everything except [−1,0]. Since B=(π∕3,π∕2)⊂R∕(−1,0) ⇒AC∩B is true. ⇒A∪B=R−{[−1,0]∪[
π
3
,
π
2
]} So, union of A and B is the complement of their own union. That's impossible. Thus, AC∩B=[
