Question

The number of value(s) of $x$ for which ${\tan }^{-1}\sqrt{x^2+x}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$ is

• A. \N
• B. $2$
• C. $3$
• D. $\infty$

Answer 1

The correct option is B $2$

We have,
${\tan }^{-1}\sqrt{x^2+x}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$
This equation holds, if
$x^2+x\ge 0\cdots \left(1\right)$
and $0\le x^2+x+1\le 1\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we get
$x^2+x=0$
$\Rightarrow x=0,-1$
Clearly, both these values satisfy the given equation.
Hence, $x=0,-1$ are the solutions of the given equation.