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(C) Given equation is $	an ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{2}$.For the domain,we must have $x(x+1) \ge 0$ and $0 \le x^2+x

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(C) Given equation is $	an ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{2}$.For the domain,we must have $x(x+1) \ge 0$ and $0 \le x^2+x+1 \le 1$.The condition $x^2+x+1 \le 1$ implies $x^2+x \le 0$,which means $x(x+1) \le 0$.Combining $x(x+1) \ge 0$ and $x(x+1) \le 0$,we get $x(x+1) = 0$,so $x=0$ or $x=-1$.If $x=0$,the equation becomes $	an ^{-1} (0) + \sin ^{-1} (1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. This is a solution.If $x=-1$,the equation becomes $	an ^{-1} (0) + \sin ^{-1} (1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. This is also a solution.Thus,there are $2$ real solutions.

The number of real solutions of the equation $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{2}$ is

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