Question

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

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

Answer 1

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

$f\left(x\right)=(1+{\sec }^{-1}x)(1+{\cos }^{-1}x)$
Domain of ${\sec }^{-1}x$ is $(-\infty ,-1]\cup [1,\infty )$
and domain of ${\cos }^{-1}x$ is $[-1,1].$
Therefore, domain of $f\left(x\right)$ is $\{-1,1\}.$
$f\left(1\right)=(1+0)(1+0)=1$
$f(-1)=(1+\pi )(1+\pi )={(1+\pi )}^2$
Hence, range of $f\left(x\right)$ is $\{1,{(1+\pi )}^2\}$