question_answer

Solve the following equation. \[{{\tan }^{-1}}\sqrt{{{x}^{2}}+x}+{{\sin }^{-1}}\sqrt{{{x}^{2}}+x+1}=\frac{\pi }{2}\]

Answer:

We have, \[{{\tan }^{-1}}\sqrt{{{x}^{2}}+x}+{{\sin }^{-1}}\sqrt{{{x}^{2}}+x+1}=\frac{\pi }{2}\] This equation holds, if             \[{{x}^{2}}+x\ge 0\] and \[0\le {{x}^{2}}+x+1\le 1\] Now,     \[{{x}^{2}}+x\ge 0\] and \[0\le {{x}^{2}}+x+1\le 1\] \[\Rightarrow \]   \[{{x}^{2}}+x\ge 0\] and \[{{x}^{2}}+x+1\le 1\]                         \[[\because {{x}^{2}}+x+1>0\,\,\text{for}\,\,\text{all}\,\,x]\] \[\Rightarrow \]   \[{{x}^{2}}+x\ge 0\] and \[{{x}^{2}}+x\le 0\] \[\Rightarrow \]   \[{{x}^{2}}+x=0\] \[\Rightarrow \]   \[x(x+1)=0\] \[\Rightarrow \]   \[x=0,\,\,-\,1\] Hence, \[x=0,\,\,-\,1\] are the solution of the given equation.