If h(x) = [ln(x/e)] + [ln(e/x)],where [.] den | vedclass

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(B) Let $t = \ln(x/e)$. Then $h(x) = [t] + [-t]$.Using the property of the greatest integer function,$[t] + [-t] = 0$ if $t$ is an integer,and $[t] +

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$h(x)$ is a periodic function

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(B) Let $t = \ln(x/e)$. Then $h(x) = [t] + [-t]$.Using the property of the greatest integer function,$[t] + [-t] = 0$ if $t$ is an integer,and $[t] + [-t] = -1$ if $t$ is not an integer.Thus,$h(x) = 0$ if $\ln(x/e) = k$ for some integer $k$,which implies $x/e = e^k$,or $x = e^{k+1}$.$h(x) = -1$ if $\ln(x/e) 
otin \mathbb{Z}$.Since $e$ is a transcendental number,$e^{k+1}$ is irrational for all integers $k$ except when $k+1 = 0$ (i.e.,$x=1$,which is rational). Thus,$h(x)=0$ can occur for both rational and irrational $x$.Option $A$ is true (Range is $\{-1, 0\}$).Option $B$ is false because $h(x)$ is not periodic.Option $C$ is true because if $h(x) = -1$,$x$ can be any value such that $\ln(x/e)$ is not an integer,which includes both rational and irrational numbers.Option $D$ is false because if $h(x) = 0$,$x = e^{k+1}$. For $k = -1$,$x = e^0 = 1$,which is rational. Thus,$x$ does not have to be irrational.

If $h(x) = [\ln(x/e)] + [\ln(e/x)]$,where $[.]$ denotes the greatest integer function,then which of the following is false?

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