Question

The number of solutions for the equation $2{\sin }^{-1}\sqrt{x^2-x+1}+{\cos }^{-1}\sqrt{x^2-x}=\frac{3\pi }{2}$ is

• A. $1$
• B. $2$
• C. $3$
• D. Infinite

Answer 1

Answer: (B)

$2$

$2{\sin }^{-1}\sqrt{x^2-x+1}+{\cos }^{-1}\sqrt{x^2-x}=\frac{3\pi }{2}$

For existence of domain of ${\sin }^{-1}\sqrt{x^2-x+1}$

$-1\le \sqrt{x^2-x+1}\le 10\le x^2-x+1\le 1{\Rightarrow x}^2-x\le 0x\in [0,1]$

For ${\cos }^{-1}\sqrt{x^2-x}$

$0\le x^2-x\le 1\Rightarrow x^2-x\ge 0x(x-1)\ge 0x\in [-\infty ,0]\cup [1,\infty ]$

Only two points are common in their domains i.e. $0$ and $1$ which also satisfies the given equation.

So option $B$ is correct