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(B) Given differential equation: $\cos \left(\frac{1}{2} \cos ^{-1}\left(e^{-x}\right)\right) d x=\sqrt{e^{2 x}-1} \,d y$. Let $\cos ^{-1}\left(e^{-x}

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$2$

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(B) Given differential equation: $\cos \left(\frac{1}{2} \cos ^{-1}\left(e^{-x}ight)ight) d x=\sqrt{e^{2 x}-1} \,d y$. Let $\cos ^{-1}\left(e^{-x}ight)=	heta$,where $	heta \in[0, \pi]$. Then $\cos 	heta = e^{-x}$. Using the identity $\cos 	heta = 2 \cos^2 \frac{	heta}{2} - 1$,we have $2 \cos^2 \frac{	heta}{2} = 1 + e^{-x} = \frac{e^x + 1}{e^x}$. Thus,$\cos \frac{	heta}{2} = \sqrt{\frac{e^x + 1}{2e^x}}$. Substituting this into the differential equation: $\sqrt{\frac{e^x + 1}{2e^x}} dx = \sqrt{e^{2x} - 1} dy$. Since $\sqrt{e^{2x} - 1} = \sqrt{(e^x - 1)(e^x + 1)}$,we get $\sqrt{\frac{e^x + 1}{2e^x}} dx = \sqrt{e^x - 1} \sqrt{e^x + 1} dy$. Dividing by $\sqrt{e^x + 1}$: $\frac{1}{\sqrt{2e^x}} dx = \sqrt{e^x - 1} dy$,which simplifies to $\frac{dx}{\sqrt{2} \sqrt{e^x(e^x - 1)}} = dy$. Let $e^x = t$,then $e^x dx = dt \Rightarrow dx = \frac{dt}{t}$. So,$\int \frac{dt}{\sqrt{2} t \sqrt{t(t-1)}} = \int dy$. Let $t = \frac{1}{z}$,then $dt = -\frac{1}{z^2} dz$. Substituting: $\int \frac{-dz/z^2}{\sqrt{2} (1/z) \sqrt{1/z^2 - 1/z}} = \int dy \Rightarrow -\int \frac{dz}{\sqrt{2} \sqrt{1-z}} = y + C$. Integrating: $\sqrt{2} \sqrt{1-z} = y + C \Rightarrow \sqrt{2} \sqrt{1 - e^{-x}} = y + C$. At $x=0, y=-1$: $\sqrt{2} \sqrt{1 - 1} = -1 + C \Rightarrow C = 1$. So,$\sqrt{2} \sqrt{1 - e^{-x}} = y + 1$. For the $x$-axis intersection $(\alpha, 0)$,set $y=0$: $\sqrt{2} \sqrt{1 - e^{-\alpha}} = 1$. Squaring both sides: $2(1 - e^{-\alpha}) = 1 \Rightarrow 1 - e^{-\alpha} = \frac{1}{2} \Rightarrow e^{-\alpha} = \frac{1}{2}$. Therefore,$e^{\alpha} = 2$.

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