Question
The range of the function, $\text{f(x)}=(1+\sec^{-1}\text{x})(1+\cos^{-1}\text{x})$ is:
  1. $(-\infty,\infty)$
  2. $(-\infty,0]\cup[4.\infty)$
  3. $\big\{0,(1+\pi^2)\big\}$
  4. $[1.(1+\pi)^2]$

Answer

  1. $[1.(1+\pi)^2]$
Solution:
$\text{f(x)}=(1+\sec^{-1}(\text{x}))(1+\cos^{-1}(\text{x}))$
 
Here the limiting component is $\cos−1(\text{x}),$ since the domain of $\cos−1(\text{x}),$ is [−1, 1].
Therefore,
$\text{f}(1)=(1+0)(1+0)$
$=1$
$\text{f}(−1)=(1+\pi(1+\pi)$
$=(1+\pi)^2 $
Hence range of $\text{f(x)}=[1,(1+\pi)^2]$

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