A) R
B) \[\left[ -1,1 \right]\]
C) \[\left[ 0,1 \right]\]
D) \[\phi \]
Correct Answer: C
Solution :
[c] \[\left| {{\sin }^{-1}}x \right|+\left| {{\cos }^{-1}}x \right|=\frac{\pi }{2}\] \[\Rightarrow \,\,\,\,\left| {{\sin }^{-1}}x \right|+{{\cos }^{-1}}x=\frac{\pi }{2}\] (\[\therefore {{\cos }^{-1}}x\] is always non-negative) \[\Rightarrow \left| {{\sin }^{-1}}x \right|={{\sin }^{-1}}x\] \[\Rightarrow {{\sin }^{1}}x\ge 0\Rightarrow 0\le {{\sin }^{-1}}x\le \frac{\pi }{2}\] \[\Rightarrow 0\le x\le 1\]
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