A) \[x\in \left[ \frac{\sqrt{2}}{2},1 \right]\]
B) \[x=3\]
C) \[x\in [3,4]\]
D) \[x=0\]
Correct Answer: A
Solution :
[a] \[x=\cos y;\]where \[0\le y\le \pi ,\left| x \right|\le 1\] \[2{{\cos }^{-1}}x={{\sin }^{-1}}(2x\sqrt{1-{{x}^{2}}})\] \[\Rightarrow 2{{\cos }^{-1}}(cosy)=si{{n}^{-1}}(2cosy\,\,\cdot \sqrt{1-{{\cos }^{2}}y})\] \[\Rightarrow 2{{\cos }^{-1}}(cosy)=si{{n}^{-1}}(2cosy.siny)\] \[\Rightarrow 2{{\cos }^{-1}}(cosy)=si{{n}^{-1}}(sin2y)\] \[\Rightarrow {{\sin }^{-1}}(sin2y)=2y\] for \[-\pi /4\le y\le \pi /4\] and \[2{{\cos }^{-1}}(cosy)=2y\]for \[0\le y\le \pi \] Thus, Eq. (i) holds only when, \[y\in [0,\pi /4]\Rightarrow x\in [\sqrt{2}/2,1]\]
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