Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $\infty$

Answer 1

The correct option is A

$0$

Explanation for correct option

The given function is ${\tan }^{-1}\left(\sqrt{x\left(x-1\right)}\right)+{\sin }^{-1}\left(\sqrt{x^2+x+1}\right)=\frac{\pi }{2}$

The function is defined if,

1. $x\left(x-1\right)\ge 0$ [because it is the domain of square root function]
2. $x^2+x+1\ge 0$ [because it is the domain of square root function]
3. $\sqrt{x^2+x+1}\le 1$ [domain of ${\sin }^{-1}$ function]

From $\left(2\right)$ and $\left(3\right)$

$$\begin{gathered} 0\le x^2+x+1\le 1\cap x^2-x\ge 0 \\ =0\le x^2+x+1\le 1\cap x^2-x+1\ge 1 \\ =\varnothing \end{gathered}$$

Therefore no solution exist.

Hence, the correct option is $\left(\mathrm{A}\right)$.