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(B) The slope of the tangent is given by $\frac{dy}{dx} = -k \frac{y}{x}$,where $k$ is a constant of proportionality.Separating the variables,we get $

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$4$

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(B) The slope of the tangent is given by $\frac{dy}{dx} = -k \frac{y}{x}$,where $k$ is a constant of proportionality.Separating the variables,we get $\frac{dy}{y} = -k \frac{dx}{x}$.Integrating both sides,we obtain $\ln |y| = -k \ln |x| + C$,which can be written as $y = C x^{-k}$.Given that the curve passes through $(1, 2)$,we have $2 = C(1)^{-k} \Rightarrow C = 2$.Given that the curve passes through $(8, 1)$,we have $1 = 2(8)^{-k} \Rightarrow 8^k = 2 \Rightarrow (2^3)^k = 2^1 \Rightarrow 3k = 1 \Rightarrow k = \frac{1}{3}$.Thus,the equation of the curve is $y = 2 x^{-1/3}$.We need to find $\left| y \left(\frac{1}{8}ight) ight|$.Substituting $x = \frac{1}{8}$ into the equation,$y = 2 \left(\frac{1}{8}ight)^{-1/3} = 2 \left( (2^{-3})^{-1/3} ight) = 2 	imes 2^1 = 4$.Therefore,$\left| y \left(\frac{1}{8}ight) ight| = 4$.

Let a smooth curve $y=f(x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\left(\frac{-y}{x}\right)$. If the curve passes through the points $(1, 2)$ and $(8, 1)$,then $\left| y \left(\frac{1}{8}\right) \right|$ is equal to

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