Question

For ${\sin }^{-1}x+{\cos }^{-1}y+{\sec }^{-1}z\ge t^2-\sqrt{2\pi }t+3\pi$, then the principal value of ${\cos }^{-1}\left(\cos 5t^2\right)$ is,

• A. $\frac{3\pi }{2}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{2\pi }{3}$

Answer 1

The correct option is B $\frac{\pi }{2}$

As $max\left\{si{n}^{-1}x\right\}=\frac{\pi }{2}max\left\{co{s}^{-1}y\right\}=\pi max\left\{se{c}^{-1}z\right\}=\pi$

We have
$si{n}^{-1}x+co{s}^{-1}y+se{c}^{-1}z>\frac{\pi }{2}+\pi +\pi =\frac{5\pi }{2}$

Gives
$\frac{5\pi }{2}\ge t^2-\sqrt{2\pi }t+3\pi \Rightarrow t^2-\sqrt{2\pi }t+\frac{\pi }{2}\le 0\Rightarrow {(t-\sqrt{\frac{\pi }{2}})}^2\le 0\Rightarrow t\le \sqrt{\frac{\pi }{2}}$

Therefore,

$co{s}^{-1}\left[cos{\left(5\left(\sqrt{\frac{\pi }{2}}\right)\right)}^2\right]=co{s}^{-1}\left[cos\left(\frac{5\pi }{2}\right)\right]$

$=co{s}^{-1}\left[cos(2\pi +\frac{\pi }{2})\right]=\frac{\pi }{2}$