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Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
[cos^{-1}x]>0
f(x) = cos^{-1}x
\Rightarrow -1\leq x\leq 1
0\leq cos^{-1}x\leq \pi
[0]\leq [cos^{-1}x]\leq [\pi]
0\leq [cos^{-1}x]\leq 3
[cos^{-1}x] =0, 1, 2, 3
Value of x for:
[cos^{-1}x] =0
\Rightarrow cos1< x\leq 1
So final solutions is:
x\in [-1, cos1)
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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