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

zero one two infinite

two

The given equation holds if $\tan^{-1}\sqrt{x(x+1)}+\cos^{-1}\frac{1}{\sqrt{x^2+x+1}}$ $⇒\cos^{-1}\frac{1}{\sqrt{x^2+x+1}}=\cos^{-1}\sqrt{x^2+x+1}$ ⇒ x2 + x + 1 = 1 or x(x + 1) = 0 ⇒ x = 0, –1. Hence (C) is the correct answer.