Question

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

• A. $(-\infty ,\infty )$
• B. $(-\infty ,0]\cup [4,\infty )$
• C. $\{0,{(1+\pi )}^2\}$
• D. $[1,{(1+\pi )}^2]$

Answer 1

The correct option is D $[1,{(1+\pi )}^2]$

$f\left(x\right)=(1+se{c}^{-1}(x\left)\right)(1+co{s}^{-1}(x\left)\right)$
Here the limiting component is $co{s}^{-1}\left(x\right)$, since the domain of $co{s}^{-1}\left(x\right)$ is $[-1,1]$.
Therefore,
$f\left(1\right)=(1+0)(1+0)$
$=1$
$f(-1)=(1+\pi )(1+\pi )$
$=(1+\pi {)}^2$
Hence range of $f\left(x\right)=[1,(1+\pi {)}^2]$