domain for sin inverse (2x√1-x^2) working rule
Answer (1)
10th Nov, 2020
Hello Aspirant,
I think there domain will be -1<x<1
Since value in sin-1 can only be between -1 and 1, we get the conditions
- -1 < x < 1
- -1/2 < x√1-x2 < 1/2
Taking case 2, square it. We get x2(1-x2) <1/4
Replace x2 with t. We get t2-t+1/4>0.
The lhs is a square , thus is greater than zero for all t . Therefore for all x .
The intersection of the two ranges gives -1<x<1
I hope this will help you.
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