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If $\cos^{-1}x\;>\;\sin^{-1}x,$ then \begin{array}{1 1}(A)\quad\frac{1}{\sqrt 2}<x\leq 1& (B)\quad 0\leq x<\frac{1}{\sqrt 2}\\(C)\quad -1\leq x<\frac{1}{\sqrt 2} & (D)\quad x>0\end{array}

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Toolbox:
  • cosx>sinx if \(x\:\in[0,\large\frac{\pi}{4}]\)
Ans -(B)   \(0,\leq\:x<\large\frac{1}{\sqrt 2}  \)
 
\(Because\:If\:x\in \big[ 0, \large\frac{1}{\sqrt2} \big)\;then\: cos^{-1}x\:\in\:(\large\frac{\pi}{4},\frac{\pi}{2}]\)
\(and\: sin^{-1}x\in\:[0,\large\frac{\pi}{4})\)
 
\(\Rightarrow\: cos^{-1}x\: is\:>sin^{-1}x\: in\: the \: interval \: \big[0, \large\frac{1}{\sqrt 2}\big) \)

 

answered Feb 18, 2013 by thanvigandhi_1
edited Mar 17, 2013 by rvidyagovindarajan_1
 

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