A) Zero
B) One
C) Two
D) Infinite
Correct Answer: C
Solution :
\[{{\tan }^{-1}}\sqrt{x(x+1)}+{{\sin }^{-1}}\sqrt{{{x}^{2}}+x+1}=\frac{\pi }{2}\] \[{{\tan }^{-1}}\sqrt{x(x+1)}\]is defined when \[x(x+1)\ge 0\] ?..(i) \[{{\sin }^{-1}}\sqrt{{{x}^{2}}+x+1}\]is defined when \[0\le x(x+1)+1\le 1\]or \[0\le x(x+1)\le 0\] ?..(ii) From (i) and (ii), \[x(x+1)=0\] or \[x=0\]and -1. Hence number of solution is 2.
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