The range of the function, f(x)=(1+ ^ -1 x)(1+ ^ -1 x) is: - Vidyadip

MCQ

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

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

✓

Answer

Correct option: D.

$[1.(1+\pi)^2]$

$\text{f(x)}=(1+\sec^{-1}(\text{x}))(1+\cos^{-1}(\text{x}))$

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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