Question

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

• A. $0$
• B. Infinite
• C. $2$
• D. $4$

Answer 1

The correct option is C $2$

The maximum value of the function is obtained when the value of $\sqrt{x^2-x+1}$ is equal to $1$

and value of $\sqrt{x^2-x}$ is equal to $0$

${\sin }^{-1}\sqrt{x},{\cos }^{-1}\sqrt{x}$ are defined for $x\le 1$ and $x\ge 0$
$\therefore \sqrt{x^2-x+1}\le 1$ and $\sqrt{x^2-x}\ge 0$
$\Rightarrow x^2-x\le 0$ and $x^2-x\ge 0$

Thus, reducing the equation to

$\Rightarrow x^2-x=0$
$\Rightarrow x=1,0$
$\therefore$ there are two solutions, both satisfies the equation.