| Direction: For the following questions, choose the correct answer from the codes [a], [b], [c] and [d] defined as follows. |
| Let us define the function as \[{{\cos }^{-1}}\,(\cos \theta )=\theta \]and\[2{{\tan }^{-1}}x=\frac{2x}{1-{{x}^{2}}}\]. |
| Statement I If \[\sin \,[2\,{{\cos }^{-1}}\{\cot \,(2\,{{\tan }^{-1}}x)\}]=0,\] then \[x=\pm 1,\,\,\pm \,(1\pm \sqrt{2})\] |
| Statement II \[\cot \,\,(2\,{{\tan }^{-1}}x)=\frac{1-{{x}^{2}}}{2x}\] |
A) Statement I is true. Statement II is also true and Statement II is the correct explanation of Statement I.
B) Statement I is true. Statement II is also true and Statement II is not the correct explanation of Statement I.
C) Statement I is true. Statement II is false.
D) Statement I is false. Statement II is true.
Correct Answer: A
Solution :
\[\sin \,\theta =0\,\,\,\Rightarrow \,\,\,\theta n\pi \] \[\Rightarrow \] \[2\,{{\cos }^{-1}}[\cot (2\,{{\tan }^{-1}}x)]\,=n\pi \] \[\Rightarrow \] \[{{\cos }^{-1}}\,[\cot \,(2\,{{\tan }^{-1}}x)]=\frac{n\pi }{2}\] \[\Rightarrow \] \[\cot \,(2\,{{\tan }^{-1}}x)=\cos \,\frac{n\pi }{2}\] \[\Rightarrow \] \[\frac{1}{\tan \,\left( {{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}} \right)}=\cos \,\frac{n\pi }{2}\] \[\Rightarrow \] \[\frac{1-{{x}^{2}}}{2x}=\cos \,\frac{n\pi }{2}\] Now, \[\cos \,\frac{n\pi }{2}=0,\] if \[n=1,\,\,3,\,5,...\] \[\therefore \] \[\frac{1-{{x}^{2}}}{2x}=0\,\,\Rightarrow \,\,x=\pm 1\] Now, \[\cos \,\frac{n\pi }{2}=1,\] if \[n=0,4,...\] \[\therefore \] \[\frac{1-{{x}^{2}}}{2x}=1\Rightarrow \,x=-1\,\pm \,\sqrt{2}\] Now, \[\cos \,\frac{n\pi }{2}=-1,\] if \[n=2,\,\,6,...\] \[\therefore \] \[\frac{1-{{x}^{2}}}{2x}=-1\,\,\,\Rightarrow \,\,x=1\pm \sqrt{2}\] Hence, \[x=\pm 1,\,\,\pm 1\pm \sqrt{2}\]
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