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Find the real solution of the equation\[\tan^{-1}\sqrt{x(x+1)}+\sin^{-1}\sqrt{x^2+x+1}=\frac{\pi}{2}\]

$\begin{array}{1 1} x=0\;or\;1 \\ x=0\;or\;-1 \\ x=1\;or\;-1\\ \text{Solution does not exist} \end{array} $

Can you answer this question?
 
 

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Toolbox:
  • Domain of \(sin^{-1}x=[-1,1]\)
  • Domaim of \(\sqrt{x}\) is x should be \(\geq\:0\)
\( tan^{-1}\sqrt{x^2+x}\:exist\:only\:if \:x^2+x\geq\:0\)
and \( \: sin^{-1} \sqrt{x^2+x+1} \) exist only if \( -1 \leq\:\sqrt{x^2+x+1} \leq 1\)
 
\(\Rightarrow\) \( x^2+x \geq 0 \: and \: 0 \leq x^2+x+1\leq 1\)
\(\Rightarrow\:x^2+x\geq\:0\:and\:x^2+x+1\leq\:1 \)
 
This is possible only if \(x^2+x=0\)
solving which we get \( x=0, -1 \)
 
Both the values satisfy the given equation.

 

answered Feb 19, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 
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