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Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
-2\leq [sin^{-1}x]<3
f(x) = sin^{-1}(x)
\Rightarrow -1\leq x\leq 1
\frac{-\pi }{2}\leq sin^{-1}x\leq \frac{\pi }{2}
[\frac{-\pi }{2}]\leq [sin^{-1}x]\leq [\frac{\pi }{2}]
-2\leq [sin^{-1}x]\leq 1
sin^{-1}x = -2, -1, 0, 1
Final Solution:
x\in [-1, 1]
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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