A) only\[2/3\]
B) only\[-2/3\]
C) \[-2/3\]or \[2/3\]
D) neither 2/3 nor \[-2/3\]
Correct Answer: C
Solution :
We have \[\cos (2{{\sin }^{-1}}x)=\frac{1}{9}\] \[\Rightarrow \] \[\cos ({{\sin }^{-1}}2x\sqrt{1-{{x}^{2}}})=\frac{1}{9}\] \[\Rightarrow \] \[\cos \,{{\cos }^{-1}}\sqrt{1-4{{x}^{2}}(1-{{x}^{2}})}=\frac{1}{9}\] \[\Rightarrow \] \[\sqrt{{{(1-2{{x}^{2}})}^{2}}}=\frac{1}{9}\] \[\Rightarrow \] \[1-2{{x}^{2}}=\frac{1}{9}\Rightarrow 2{{x}^{2}}=1-\frac{1}{9}\] \[\Rightarrow \] \[{{x}^{2}}=\frac{4}{9}\Rightarrow x=\pm \frac{2}{3}\]
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