Question

The number of real roots of the equation ${\tan }^{-1}\sqrt{x(x+1)}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{4}$ is

• A. $2$
• B. $0$
• C. $1$
• D. $4$

Answer 1

The correct option is B $0$

${\tan }^{-1}\sqrt{x(x+1)}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{4}$
Finding domain
$x(x+1)\ge 0\dots \left(1\right)$
and $0\le x^2+x+1\le 1$
$\Rightarrow -1\le x^2+x\le 0\dots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we have
$x^2+x=0\Rightarrow x=0,–1$
When $x=0$ or $–1,$
LHS $=\frac{\pi }{2}\ne \frac{\pi }{4}\Rightarrow$ No solution