Question

The number of real solutions of $ta{n}^{-1}(\sqrt{x(x+1)}+si{n}^{-1}\sqrt{(x^2+x+1)}=\frac{\pi }{2}$ is

• A. $0$
• B. $1$
• C. $2$
• D. infinite

Answer 1

The correct option is C $2$

From the above expression
$-1\le \sqrt{x^2+x+1}\le 1$ and $x(x+1)\ge 0$
$\Rightarrow 0\le x^2+x+1\le 1and{x}^2+x+1\ge 1$
$\Rightarrow x^2+x+1=1$
$\Rightarrow x^2+x=0$
$\Rightarrow x(x+1)=0$
Hence there will be two solutions. One at $x=-1$ and another at$x=0$.