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