Let f-x- - -1-x- -where - denotes the greatest integer function- then which of the following is not true-Range of f is -1-displaystyleunderset-xrightarrow n-lim -f-x- exists- every integer nf is an odd functionf is an even

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Toppr Admin · Accepted Answer

F-x-1-x-1 -xA0- since- -x- is integerTherefore- f-x-x2208-1and f-x-1 is an even function-xA0-And limx-x2192-nf-x-1 -xA0- for every integer n

Let $f (x) = (1)^{[x]}$ (where [.] denotes the greatest integer function), then which of the following is not true?
Range of $f$ is ${1}$
$lim x \to n f (x)$ exists, for every integer $n$
$f$ is an odd function
$f$ is an even function

$f (x) = (1)^{[x]} = 1$ since, $[x]$ is integer
Therefore, $f (x) \in 1$
and $f (x) = 1$ is an even function
And ${lim}_{x \to n} f (x) = 1$ for every integer $n$

Let f(x)=(1)[x] (where [.] denotes the greatest integer function), then which of the following is not true?Range of f is {1}limx→nf(x) exists, for every integer nf is an odd functionf is an even function

f(x)=(1)[x]=1 since, [x] is integerTherefore, f(x)∈1and f(x)=1 is an even function And limx→nf(x)=1 for every integer n

Let $f (x) = (1)^{[x]}$ (where [.] denotes the greatest integer function), then which of the following is not true?
Range of $f$ is ${1}$
$lim x \to n f (x)$ exists, for every integer $n$
$f$ is an odd function
$f$ is an even function

$f (x) = (1)^{[x]} = 1$ since, $[x]$ is integer
Therefore, $f (x) \in 1$
and $f (x) = 1$ is an even function
And ${lim}_{x \to n} f (x) = 1$ for every integer $n$