Question

For ${\sin }^{-1}x+{\cos }^{-1}y+{\sec }^{-1}z\ge t^2-\sqrt{2\pi }t+3\pi$, then the value of $x+y+z$ is equal to,

• A. 1
• B. 0
• C. 2
• D. -1

Answer 1

The correct option is D -1

Let $p=t^2-\sqrt{2\pi }t+3\pi$

$={(t-\sqrt{\frac{\pi }{2}})}^2+\frac{5\pi }{2}\ge \frac{5\pi }{2}$

Also L.H.S $={\sin }^{-1}x+{\sec }^{-1}z+{\cos }^{-1}y\le \frac{5\pi }{2}$

Thus there is only one solution

${\sin }^{-1}x=\frac{\pi }{2},$

${\cos }^{-1}y=\pi ,$

${\sec }^{-1}z=\pi$

$\Rightarrow x=1,y=-1,z=-1\Rightarrow x+y+z=-1$