Area enclosed by the curves y = ln x,y = ln|x | vedclass

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(D) Let us analyze the curves: $1$. $y = \ln x$ is defined for $x > 0$. $2$. $y = \ln|x|$ is defined for $x 
eq 0$. For $x > 0$,$y = \ln x$. For $x <

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cannot be determined

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(D) Let us analyze the curves: $1$. $y = \ln x$ is defined for $x > 0$. $2$. $y = \ln|x|$ is defined for $x 
eq 0$. For $x > 0$,$y = \ln x$. For $x $3$. $y = |\ln x|$ is defined for $x > 0$. $4$. $y = |\ln|x||$ is defined for $x 
eq 0$. Upon plotting these curves,we observe that for $x > 0$,the curves are $y = \ln x$,$y = \ln x$,$y = |\ln x|$,and $y = |\ln x|$. Specifically,for $x \in (0, 1)$,$\ln x For $x \in (1, \infty)$,$\ln x > 0$,so $|\ln x| = \ln x$. However,the question asks for the area enclosed by these four curves. Since $y = \ln x$ and $y = \ln|x|$ are identical for $x > 0$,and $y = |\ln x|$ and $y = |\ln|x||$ are identical for $x > 0$,the region is not bounded in the standard sense because the curves coincide or extend to infinity without forming a closed finite region in the Cartesian plane. Thus,the area cannot be determined as a finite value.

Area enclosed by the curves $y = \ln x$,$y = \ln|x|$,$y = |\ln x|$,and $y = |\ln|x||$ is equal to

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