JEE Main 2021 — Mathematics Trigonometry

We have,

$${\mathrm{tan}}^{-1}\sqrt{{x}^{2}+x}+{\mathrm{sin}}^{-1}\sqrt{{x}^{2}+x+1}=\frac{\pi }{4}$$tan−1x2+x​+sin−1x2+x+1​=4π​

For equation to be defined,

for ${\mathrm{tan}}^{-1}\sqrt{x(x+1)}$tan−1x(x+1)​,

$${x}^{2}+x\geq 0$$x2+x≥0

$$\Rightarrow {x}^{2}+x+1\geq 1....(1)$$⇒x2+x+1≥1....(1)

And, from domain of ${\mathrm{sin}}^{-1}\sqrt{{x}^{2}+x+1}$sin−1x2+x+1​, ${x}^{2}+x+1\leq 1....(2)$x2+x+1≤1....(2)

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

$${x}^{2}+x=0$$x2+x=0

$$\Rightarrow x=0,-1$$⇒x=0,−1

at $x=0$x=0, ${\mathrm{tan}}^{-1}\sqrt{x(x+1)}+{\mathrm{sin}}^{-1}\sqrt{{x}^{2}+x+1}=0+\frac{\pi }{2}\neq \frac{\pi }{4}$tan−1x(x+1)​+sin−1x2+x+1​=0+2π​=4π​

and at $x=-1$x=−1, ${\mathrm{tan}}^{-1}\sqrt{x(x+1)}+{\mathrm{sin}}^{-1}\sqrt{{x}^{2}+x+1}=0+\frac{\pi }{2}\neq \frac{\pi }{4}$tan−1x(x+1)​+sin−1x2+x+1​=0+2π​=4π​

None of these values satisfy

$\therefore$∴ Number of roots$=0$=0