The number of real roots of the equation ^-1x(x+1)+^-1x^2+x+1=4 is

Question

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

Accepted Answer

Domain requires $x(x+1)\ge0$ and $0\le x^2+x+1\le1$. The latter forces $x^2+x\le0$ and $x^2+x\ge-1$ (always true), combined with $x(x+1)\ge0$ forces $x^2+x=0\Rightarrow x=0$ or $x=-1$.

At $x=0$: $	an^{-1}0+\sin^{-1}1=\dfrac{\pi}{2}
e\dfrac{\pi}{4}$.

At $x=-1$: $	an^{-1}0+\sin^{-1}1=\dfrac{\pi}{2}
e\dfrac{\pi}{4}$.

Neither satisfies the equation, so there are 0 real roots.

Solutions · 1