JEE Main 2021 — Inverse Trigonometric Functions Question with Solution

We have,

${\tan }^{-1}\sqrt{x^2+x}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{4}$

For equation to be defined,

for ${\tan }^{-1}\sqrt{x\left(x+1\right)}$, 

$x^2+x\ge 0$

$\Rightarrow x^2+x+1\ge 1 ....\left(1\right)$

And, from domain of ${\sin }^{-1}\sqrt{x^2+x+1}$, $x^2+x+1\le 1 ....\left(2\right)$

$\therefore$ from $\left(1\right) \& \left(2\right)$ only possibility that the equation is defined is

$x^2+x=0$

$\Rightarrow x=0, -1$

at $x=0$, ${\tan }^{-1}\sqrt{x\left(x+1\right)}+{\sin }^{-1}\sqrt{x^2+x+1}=0+\frac{\mathrm{\pi }}{2}\ne \frac{\pi }{4}$

and at $x=-1$, ${\tan }^{-1}\sqrt{x\left(x+1\right)}+{\sin }^{-1}\sqrt{x^2+x+1}=0+\frac{\mathrm{\pi }}{2}\ne \frac{\pi }{4}$

None of these values satisfy

$\therefore$ Number of roots$=0$