Question

The number of integral solutions of $si{n}^{-1}\sqrt{4x-x^2-3}+ta{n}^{-1}\sqrt{x^2-3x+2}=\frac{\pi }{2}$ is

• A. zero
• B. infinite
• C. four
• D. None of these

Answer 1

The correct option is D None of these

Given equation is ${\sin }^{-1}\sqrt{4x-x^2-3}+{\tan }^{-1}\sqrt{x^2-3x+2}=\frac{\pi }{2}$

${\tan }^{-1}\sqrt{x^2-3x+2}=\frac{\pi }{2}-{\sin }^{-1}\sqrt{4x-x^2-3}$
$\Rightarrow {\tan }^{-1}\sqrt{x^2-3x+2}={\cos }^{-1}\sqrt{4x-x^2-3}$
Since, $\sqrt{x^2-3x+2}\ge 0$
$\Rightarrow {\tan }^{-1}\sqrt{x^2-3x+2}<\frac{\pi }{2}$
Also, $\sqrt{4x-x^2-3}\ge 0$
$\Rightarrow 0<{\cos }^{-1}\sqrt{4x-x^2-3}\le \frac{\pi }{2}$