A) \[0\]
B) \[1,\,\,-1\]
C) \[0,\,\,\frac{1}{2}\]
D) none of these
Correct Answer: C
Solution :
Given \[{{\sin }^{-1}}x+{{\sin }^{-1}}(1-x)={{\cos }^{-1}}x\] \[\Rightarrow \]\[{{\sin }^{-1}}x+{{\sin }^{-1}}(1-x)={{\sin }^{-1}}\sqrt{1-{{x}^{2}}}\] \[\Rightarrow \] \[\sin [{{\sin }^{-1}}x+{{\sin }^{-1}}(1-x)]\] \[\Rightarrow \] \[\sin [{{\sin }^{-1}}[x\sqrt{1-{{(1-x)}^{2}}}\] \[\Rightarrow \] \[+(1-x)\sqrt{1-{{x}^{2}}}]=\sin {{\sin }^{-1}}(\sqrt{1-{{x}^{2}}})\] \[\Rightarrow \] \[x\sqrt{1-{{(1-x)}^{2}}}+(1-x)\sqrt{1-{{x}^{2}}}=\sqrt{1-{{x}^{2}}}\] \[\Rightarrow \] \[x\sqrt{2x-{{x}^{2}}}=\sqrt{1-{{x}^{2}}}x\] \[\Rightarrow \] \[x(\sqrt{2x-{{x}^{2}}}-\sqrt{1-{{x}^{2}}})=0\] \[\Rightarrow \] \[x=0\]or\[2x-{{x}^{2}}=1-{{x}^{2}}\] \[\Rightarrow \] \[x=0\]or\[x=\frac{1}{2}\]
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