Question

The number of real solutions of the equation ${\tan }^{-1}\sqrt{x(x+1)}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$ is

• A. one
• B. four
• C. two
• D. infinitely many

Answer 1

Answer: (C)

Given,
${\tan }^{-1}\sqrt{x(x+1)}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$
$\Rightarrow co{s}^{-1}\frac{1}{\sqrt{{\left(x^2+x\right)}^2+1}}$
$=\frac{\pi }{2}-si{n}^{-1}\sqrt{x^2+x+1}$
$\Rightarrow co{s}^{-1}\frac{1}{\sqrt{{\left(x^2+x\right)}^2+1}}=co{s}^{-1}\sqrt{x^2+x+1}$
$\Rightarrow \frac{1}{\sqrt{{\left(x^2+x\right)}^2+1}}=\sqrt{x^2+x+1}$
$\Rightarrow 1=\left(x^2+x+1\right)\left[{\left(x^2+x\right)}^2+1\right]$
$\Rightarrow {\left(x^2+x\right)}^3+{\left(x^2+x\right)}^2+\left(x^2+x\right)+1=1$
$\Rightarrow \left(x^2+x\right)\left[{\left(x^2+x\right)}^2+\left(x^2+x\right)+1\right]=0$
$\Rightarrow x^2+x=0$
$\Rightarrow x=0,-1$
Hence, both values of x satisfies the given equation.