Given, f(x)=sin‌−1(√x2+x+1) The domain of sin‌−1(√x2+x+1) is [−1,1] ∴‌‌−1≤√x2+x+1≤1 Since √x2+x+1 is always, non-negative, so 0≤√x2+x+1≤1 Let g(x)=x2+x+1=(x+‌
1
2
)2+‌
3
4
The minimum value of g(x) is 3∕4, which occurs at x=−‌
1
2
‌∴‌‌x2+x+1≥‌
3
4
‌⇒‌‌√x2+x+1≥√‌
3
4
=‌
√3
2
‌⇒‌‌‌
√3
2
≤√x2+x+1≤1 ‌⇒‌‌sin‌−1(‌
√3
2
)≤sin‌−1(√x2+x+1) ‌⇒‌‌‌
Ï€
3
≤f(x)≤‌
Ï€
2
‌‌≤sin‌−1 Thus, the range of f(x)=sin‌−1(√x2+x+1) is [‌