Question

Find the range of
$f(x) = {cos}^{ - 1} \sqrt{ log_{[x]}( \frac{|x|}{x} ) }$
, where [.] denotes the greatest integer function ​

Answer 1

Step-by-step explanation:

We have,

$f(x) = { \cos}^{ - 1} \sqrt{ log_{[x]} \bigg( \frac{|x|}{x} \bigg) }$

Here, f(x) is only defined for $x\in(0,1)\:\cup\:[2,\infty)$, because,

$[x] > 0 \: , \: \frac{ |x| }{x} > 0 \: \: \: and \: \: \: [x] \not = 1$

$\implies x > 0 \: , \: x > 0 \: \: \: and \: \: \: x \not \in [1 ,2)$

$\implies x \in (0 ,1) \cup [2 , \infty )$

But, if $x>0$ , then $\frac{|x|}{x}=1$

So,

$f(x) = { \cos}^{ - 1} \sqrt{ log_{[x]} ( 1 ) } \: \: \: \: , \: x \in(0 ,1) \: \cup \: [2, \infty )$

$\implies \: f(x) = { \cos}^{ - 1} \sqrt{ 0} \: \: \: \: \:$

$\implies \: f(x) = { \cos}^{ - 1} ( 0) \: \: \: \: \:$

$\implies \: f(x) = \frac{\pi}{2} \: \: \: \: \:$

So,

Range of f(x) $\bold{\in\bigg\{\frac{\pi}{2}\bigg\}}$

Answer 2

answer is in the picture