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 equal to :

• A. 1
• B. 2
• C. 3
• D. None

Answer 1

The correct option is B 2

In eqn. $ta{n}^{-1}\sqrt{x^2+x}+si{n}^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$
We must have $x^2+x\ge 0$ and $0\le x^2+x+1\le 1$
$\Rightarrow x^2+x\ge 0$ and $x^2+x\le 0$
$\Rightarrow x=0$ and $x=-1$ are the only point in the domain of the equation.
For $x=0,L.H.S.=ta{n}^{-1}0+si{n}^{-1}1=\frac{\pi }{2}$
and for $x=-1,L.H.S.=ta{n}^{-1}0+si{n}^{-1}1=\frac{\pi }{2}$
Hence there are only two solutions $x=0$ and $x=-1$