The number of real solutions of $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$ is
The number of real solutions of $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$ is
one
zero
two
infinite
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
$x(x+1) \geq 0...(i)$
$\sin ^{-1} \sqrt{x^2+x+1}$ is defined when
$x(x+1)+1 \leq 1 \text { or } x(x+1) \leq 0...(ii)$ From (i) and (ii),
$x(x+1)=0 \Rightarrow x=0 \text { or }-1$ Hence, number of solutions is 2 .