A) \[{{2}^{n}}\]
B) \[n{{.2}^{n}}\]
C) \[n{{.2}^{n-1}}\]
D) \[n{{.2}^{n+1}}\]
Correct Answer: C
Solution :
We have \[{{\sin }^{-1}}x-{{\cos }^{-1}}x={{\cos }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\] \[\Rightarrow \] \[{{\sin }^{-1}}x-{{\cos }^{-1}}x=\frac{\pi }{6}\] ...(i) But we know that \[{{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2}\] ?(ii) Adding Eqs. (i) and (ii), we get \[2{{\sin }^{-1}}x=\frac{\pi }{6}+\frac{\pi }{2}=\frac{2\pi }{3}\] \[\Rightarrow \] \[{{\sin }^{-1}}x=\frac{\pi }{3}\] \[\Rightarrow \] \[x=\frac{\sqrt{3}}{2}\] Subtracting Eq. (i) from Eq. (ii), we get \[2{{\cos }^{-1}}x=\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3}\] \[\Rightarrow \] \[{{\cos }^{-1}}x=\frac{\pi }{6}\] \[\Rightarrow \] \[x=\frac{\sqrt{3}}{2}\] Hence, given equation has unique solution.
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