A) 0
B) 1
C) 2
D) \[\infty \]
Correct Answer: C
Solution :
From given function \[{{\tan }^{-1}}\sqrt{x(x+1)}+{{\sin }^{-1}}\sqrt{{{x}^{2}}+x+1}=\frac{\pi }{2},\]it is clear that (1)\[x(x+1)\ge 0\] (\[\because \]Domain of square root function) (2) \[{{x}^{2}}+x+1\ge 0\](\[\because \]Domain of square root function) (3)\[{{x}^{2}}+x+1\le 1\] (\[\because \] Domain of \[{{\sin }^{-1}}\] function) From (1), (2) and (3), \[(0\le {{x}^{2}}+x+1\le 1)\cap ({{x}^{2}}+x\ge 0)\] \[\Rightarrow \]\[(0\le {{x}^{2}}+x+1\le 1)\cap ({{x}^{2}}+x\ge 1)\] \[\Rightarrow \]\[{{x}^{2}}+x+1=1\]\[\Rightarrow \]\[{{x}^{2}}+x=0\]\[\Rightarrow \]\[x=0,x=-1\]
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