Question

The range of the function $f\left(x\right)={\sec }^{-1}\left(x\right)+{\tan }^{-1}\left(x\right)$, is

• A. $(0,\pi )$
• B. $(\frac{-\pi }{2},\frac{3\pi }{2})$
• C. $(0,\frac{3\pi }{4}]$
• D. None

Answer 1

The correct option is A $(0,\pi )$

Given, $f\left(x\right)={\sec }^{-1}\left(x\right)+{\tan }^{-1}\left(x\right)$

Domain of ${\sec }^{-1}\left(x\right)=(-\infty ,-1]\cup [1,\infty )$

Domain of ${\tan }^{-1}\left(x\right)=(-\infty ,\infty )$

Thus $D_f=(-\infty ,\infty )-(-1,1)$

$f(-\infty )=\frac{\pi }{2}+\frac{-\pi }{2}=0$

$f(-1)=\pi +\frac{-\pi }{4}=\frac{3\pi }{4}$

$f\left(1\right)=0+\frac{\pi }{4}=\frac{\pi }{4}$

$f\left(\infty \right)=\frac{\pi }{2}+\frac{\pi }{2}=\pi$

Therefore, $f\left(x\right)$ ranges from open interval $0$ to $\pi$.

$\text{Range}\left(f\right)=(0,\pi )$