The angle between the curve 2y = e^{-x/2} and | vedclass

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(D) The curve is given by $2y = e^{-x/2}$.At the point of intersection with the $y$-axis,we set $x = 0$.Substituting $x = 0$ into the equation: $2y =

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$4$

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(D) The curve is given by $2y = e^{-x/2}$.At the point of intersection with the $y$-axis,we set $x = 0$.Substituting $x = 0$ into the equation: $2y = e^0 = 1$,so $y = 1/2$.The point of intersection is $(0, 1/2)$.Differentiating the curve with respect to $x$: $2 \frac{dy}{dx} = -\frac{1}{2} e^{-x/2}$.At $x = 0$,the slope of the tangent to the curve is $m_1 = \frac{dy}{dx} = -\frac{1}{4}$.The $y$-axis is a vertical line,so its slope $m_2$ is undefined (or can be considered as $\infty$).The angle $	heta$ between a curve with slope $m$ and the $y$-axis is given by $	an 	heta = |\frac{1}{m}|$.Here,$m = -1/4$,so $	an 	heta = |\frac{1}{-1/4}| = 4$.Given that the angle is $	an^{-1}(k)$,we have $	an^{-1}(k) = 	an^{-1}(4)$.Therefore,$k = 4$.

The angle between the curve $2y = e^{-x/2}$ and the $y$-axis is $\tan^{-1}(k)$,then $k = $

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