A) zero
B) one
C) two
D) infinite
Correct Answer: C
Solution :
We have \[{{\tan }^{-1}}\sqrt{x(x+1)}+{{\sin }^{-1}}\sqrt{{{x}^{2}}+x+1}=\frac{\pi }{2}\] \[\Rightarrow \]\[{{\cos }^{-1}}\frac{1}{\sqrt{{{({{x}^{2}}+x)}^{2}}+1}}+{{\sin }^{-1}}\sqrt{{{x}^{2}}+x+1}=\frac{\pi }{2}\] \[\Rightarrow \]\[{{\cos }^{-1}}\frac{1}{\sqrt{{{({{x}^{2}}+x)}^{2}}+1}}={{\cos }^{-1}}\sqrt{{{x}^{2}}+x+1}\] \[\Rightarrow \]\[\frac{1}{\sqrt{{{({{x}^{2}}+x)}^{2}}+1}}=\sqrt{{{x}^{2}}+x+1}\] \[\Rightarrow \]\[1=({{x}^{2}}+x+1)[{{({{x}^{2}}+x)}^{2}}+1]\] \[\Rightarrow \]\[{{({{x}^{2}}+x)}^{3}}+{{({{x}^{2}}+x)}^{2}}+({{x}^{2}}+x)+1=1\] \[\Rightarrow \]\[({{x}^{2}}+x)\{{{({{x}^{2}}+x)}^{2}}+({{x}^{2}}+x)+1\}=0\] \[\Rightarrow \] \[{{x}^{2}}+x=0\Rightarrow x=0,-1\]
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