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(B) The function is $y = \ln^2 x - 1$. The $4^{th}$ quadrant corresponds to the region where $y  0$.Setting $y = 0$,we get $\ln^2 x = 1$,

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$\frac{4}{e}$

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(B) The function is $y = \ln^2 x - 1$. The $4^{th}$ quadrant corresponds to the region where $y  0$.Setting $y = 0$,we get $\ln^2 x = 1$,so $\ln x = \pm 1$,which gives $x = e$ or $x = 1/e$.For $x \in (1/e, e)$,$\ln x$ lies between $-1$ and $1$,so $\ln^2 x The area $A$ is given by $\int_{1/e}^{e} |\ln^2 x - 1| dx = \int_{1/e}^{e} (1 - \ln^2 x) dx$.Using integration by parts for $\int \ln^2 x dx = x \ln^2 x - 2x \ln x + 2x$,we have:$\int (1 - \ln^2 x) dx = x - (x \ln^2 x - 2x \ln x + 2x) = -x \ln^2 x + 2x \ln x - x$.Evaluating from $1/e$ to $e$:At $x = e$: $-e(1)^2 + 2e(1) - e = -e + 2e - e = 0$.At $x = 1/e$: $-(1/e)(-1)^2 + 2(1/e)(-1) - (1/e) = -1/e - 2/e - 1/e = -4/e$.Thus,$A = 0 - (-4/e) = 4/e$.

The area enclosed by the graph of the function $y = \ln^2 x - 1$ lying in the $4^{th}$ quadrant is:

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