Question

If $\theta ={\sin }^{-1}x+{\cos }^{-1}x-{\tan }^{-1}x,x\ge 0$ then the smallest interval in which $\theta$ lies is

• A. $\frac{\pi }{6}\le \theta \le \frac{\pi }{4}$
• B. $\frac{\pi }{6}\le \theta \le \frac{\pi }{3}$
• C. $\frac{\pi }{4}\le \theta \le \frac{\pi }{2}$
• D. $0\le \theta \le \frac{\pi }{6}$

Answer 1

The correct option is C

$\frac{\pi }{4}\le \theta \le \frac{\pi }{2}$

${\sin }^{-1}x+{\cos }^{-1}x$ is defined for $xϵ[-1,1]\&{\tan }^{-1}x$ for $xϵR$

But, given $x\ge 0$

$\therefore \theta ={\sin }^{-1}x+{\cos }^{-1}x-{\tan }^{-1}x,0\le x\le 1$

$\Rightarrow \theta =\frac{\pi }{2}-{\tan }^{-1}x$

Now $0\le x\le 1$

$\Rightarrow 0\le {\tan }^{-1}x\le \frac{\pi }{4}$

$\Rightarrow 0\le \frac{\pi }{2}-\theta \le \frac{\pi }{4}$

$\Rightarrow \frac{-\pi }{2}\le -\theta \le \frac{-\pi }{4}$

$=\frac{\pi }{4}\le \theta \le \frac{\pi }{2}$