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(C) The given differential equation is $x 	an \left(\frac{y}{x}ight) d y = \left(y 	an \left(\frac{y}{x}ight)-xight) d x$.Rearranging the term

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$\frac{1}{8}(\pi-1)$

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(C) The given differential equation is $x 	an \left(\frac{y}{x}ight) d y = \left(y 	an \left(\frac{y}{x}ight)-xight) d x$.Rearranging the terms,we get $	an \left(\frac{y}{x}ight)(x d y - y d x) = -x d x$.Dividing both sides by $x^2$,we have $	an \left(\frac{y}{x}ight) d\left(\frac{y}{x}ight) = -\frac{1}{x} d x$.Integrating both sides,we get $\ln \left| \sec \left(\frac{y}{x}ight) ight| = -\ln |x| + C$,which simplifies to $\ln \left| x \sec \left(\frac{y}{x}ight) ight| = C$.Given $y(1/2) = \pi/6$,we have $\ln \left| \frac{1}{2} \sec \left( \frac{\pi/6}{1/2} ight) ight| = \ln \left| \frac{1}{2} \sec \left( \frac{\pi}{3} ight) ight| = \ln \left| \frac{1}{2} \cdot 2 ight| = \ln(1) = 0$. Thus,$C=0$.So,$\sec \left(\frac{y}{x}ight) = \frac{1}{x}$,which implies $\cos \left(\frac{y}{x}ight) = x$,or $y = x \cos^{-1}(x)$.The required area is $A = \int_{0}^{1/\sqrt{2}} x \cos^{-1}(x) d x$.Using integration by parts,$A = \left[ \frac{x^2}{2} \cos^{-1}(x) ight]_{0}^{1/\sqrt{2}} - \int_{0}^{1/\sqrt{2}} \frac{x^2}{2} \left( -\frac{1}{\sqrt{1-x^2}} ight) d x$.$A = \left( \frac{1}{4} \cdot \frac{\pi}{4} - 0 ight) + \frac{1}{2} \int_{0}^{1/\sqrt{2}} \frac{x^2}{\sqrt{1-x^2}} d x$.Substituting $x = \sin 	heta$,$d x = \cos 	heta d 	heta$,we get $\frac{1}{2} \int_{0}^{\pi/4} \frac{\sin^2 	heta \cos 	heta}{\cos 	heta} d 	heta = \frac{1}{2} \int_{0}^{\pi/4} \frac{1-\cos 2	heta}{2} d 	heta = \frac{1}{4} \left[ 	heta - \frac{\sin 2	heta}{2} ight]_{0}^{\pi/4} = \frac{1}{4} \left( \frac{\pi}{4} - \frac{1}{2} ight) = \frac{\pi}{16} - \frac{1}{8}$.Total area $A = \frac{\pi}{16} + \frac{\pi}{16} - \frac{1}{8} = \frac{2\pi}{16} - \frac{1}{8} = \frac{\pi-1}{8}$.

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