If y = tan^{-1} left( frac{ln(e/x^2)}{ln(ex^2 | vedclass

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(C) Given $y = 	an^{-1} \left( \frac{\ln e - \ln x^2}{\ln e + \ln x^2} ight) + 	an^{-1} \left( \frac{3 + 2 \ln x}{1 - 6 \ln x} ight)$.Since $\ln

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(C) Given $y = 	an^{-1} \left( \frac{\ln e - \ln x^2}{\ln e + \ln x^2} ight) + 	an^{-1} \left( \frac{3 + 2 \ln x}{1 - 6 \ln x} ight)$.Since $\ln e = 1$ and $\ln x^2 = 2 \ln x$,we have $y = 	an^{-1} \left( \frac{1 - 2 \ln x}{1 + 2 \ln x} ight) + 	an^{-1} \left( \frac{3 + 2 \ln x}{1 - 6 \ln x} ight)$.Using the formula $	an^{-1} A + 	an^{-1} B = 	an^{-1} \left( \frac{A+B}{1-AB} ight)$,let $A = \frac{1 - 2 \ln x}{1 + 2 \ln x}$ and $B = \frac{3 + 2 \ln x}{1 - 6 \ln x}$.Then $A + B = \frac{(1 - 2 \ln x)(1 - 6 \ln x) + (3 + 2 \ln x)(1 + 2 \ln x)}{(1 + 2 \ln x)(1 - 6 \ln x)} = \frac{1 - 8 \ln x + 12(\ln x)^2 + 3 + 8 \ln x + 4(\ln x)^2}{(1 + 2 \ln x)(1 - 6 \ln x)} = \frac{4 + 16(\ln x)^2}{(1 + 2 \ln x)(1 - 6 \ln x)}$.And $1 - AB = 1 - \frac{(1 - 2 \ln x)(3 + 2 \ln x)}{(1 + 2 \ln x)(1 - 6 \ln x)} = \frac{(1 + 2 \ln x)(1 - 6 \ln x) - (3 - 4 \ln x - 4(\ln x)^2)}{(1 + 2 \ln x)(1 - 6 \ln x)} = \frac{1 - 4 \ln x - 12(\ln x)^2 - 3 + 4 \ln x + 4(\ln x)^2}{(1 + 2 \ln x)(1 - 6 \ln x)} = \frac{-2 - 8(\ln x)^2}{(1 + 2 \ln x)(1 - 6 \ln x)}$.Thus,$\frac{A+B}{1-AB} = \frac{4(1 + 4(\ln x)^2)}{-2(1 + 4(\ln x)^2)} = -2$.So $y = 	an^{-1}(-2)$,which is a constant.Therefore,$\frac{dy}{dx} = 0$ and $\frac{d^2y}{dx^2} = 0$.

$LIST I$	$LIST II$
$P$. The range of $f$ is	$1$. $(-\infty, \frac{1}{1-e}] \cup [\frac{e}{e-1}, \infty)$
$Q$. The range of $g$ contains	$2$. $(0, 1)$
$R$. The domain of $f$ contains	$3$. $[-\frac{1}{2}, \frac{1}{2}]$
$S$. The domain of $g$ is	$4$. $(-\infty, 0) \cup (0, \infty)$
	$5$. $(-\infty, \frac{e}{e-1}]$
	$6$. $(-\infty, 0) \cup (\frac{1}{2}, \frac{e}{e-1}]$

If $y = \tan^{-1} \left( \frac{\ln(e/x^2)}{\ln(ex^2)} \right) + \tan^{-1} \left( \frac{3 + 2 \ln x}{1 - 6 \ln x} \right)$,then $\frac{d^2y}{dx^2} =$

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