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If $y=2\tan^{-1}x+\sin^{-1}\frac{2x}{1+x^2}$ for all x,then _________< y < ___________.

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Toolbox:
  • Put \( x=tan\theta \rightarrow \theta = tan^{-1}x\)
  • \(\large\frac{2tan\theta}{1+tan^2\theta}=sin2\theta\)
  • Principal value of tan is \((-\large\frac{\pi}{2},\large\frac{\pi}{2})\)
Ans : \( (-2\pi \: \: 2\pi)\)
 
Put \( x=tan\theta \Rightarrow \theta = tan^{-1}x\)
\(\Rightarrow\: \large\frac{2x}{1+x^2}=\large\frac{2tan\theta}{1+tan^2\theta}=sin2\theta\)
\(sin^{-1}\large\frac{2x}{1+x^2}=sin^{-1}sin2\theta=2\theta\)
 
Substituting the values in the given exepression we get
\(y=2.tan^{-1}tanx+sin^{-1}\large\frac{2x}{1+x^2}=2\theta+2\theta\)
 
\(\Rightarrow\: y = 4\theta=4\: tan^{-1}x\)
\( tan^{-1}x \: lies \: in \: (-\large\frac{\pi}{2} \: \large\frac{\pi}{2})\)
 
\( \Rightarrow y \in (-2\pi \: \: 2\pi)\)

 

answered Feb 18, 2013 by thanvigandhi_1
edited Mar 16, 2013 by thanvigandhi_1
 
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