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(A) The given curves are $y^2=4x$ and $y=e^{-x/2}$.Let $(x_1, y_1)$ be the point of intersection of both curves.For $y^2=4x$,differentiating with resp

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$2$

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(A) The given curves are $y^2=4x$ and $y=e^{-x/2}$.Let $(x_1, y_1)$ be the point of intersection of both curves.For $y^2=4x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$.Thus,the slope $m_1 = \frac{2}{y_1}$.For $y=e^{-x/2}$,differentiating with respect to $x$ gives $\frac{dy}{dx} = -\frac{1}{2}e^{-x/2} = -\frac{1}{2}y$.Thus,the slope $m_2 = -\frac{y_1}{2}$.The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the slopes: $m_1 m_2 = (\frac{2}{y_1})(-\frac{y_1}{2}) = -1$.Since $1 + m_1 m_2 = 1 - 1 = 0$,the denominator is zero,which implies $	an 	heta = \infty$.Therefore,$	heta = \frac{\pi}{2}$.Finally,$\operatorname{cosec}^2(	heta/2) = \operatorname{cosec}^2(\frac{\pi}{4}) = (\sqrt{2})^2 = 2$.

If the angle between the curves $y^2=4x$ and $y=e^{-x/2}$ is $\theta$,then $\operatorname{cosec}^2(\theta/2)=$

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