If 2y = 3x - 1 is a tangent drawn to the curv | vedclass

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(A) The equation of the tangent is $2y = 3x - 1$,which can be written as $y = \frac{3}{2}x - \frac{1}{2}$.Comparing this with $y = mx + c$,the slope o

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$(1, 0)$

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(A) The equation of the tangent is $2y = 3x - 1$,which can be written as $y = \frac{3}{2}x - \frac{1}{2}$.Comparing this with $y = mx + c$,the slope of the tangent is $m = \frac{3}{2}$.Given the curve $y^2 = ax^3 + b$,we differentiate with respect to $x$:$2y \frac{dy}{dx} = 3ax^2 \implies \frac{dy}{dx} = \frac{3ax^2}{2y}$.At the point $(1, 1)$,the slope is $\frac{dy}{dx} = \frac{3a(1)^2}{2(1)} = \frac{3a}{2}$.Equating the slopes: $\frac{3a}{2} = \frac{3}{2} \implies a = 1$.Since the point $(1, 1)$ lies on the curve $y^2 = ax^3 + b$,we substitute $x=1, y=1, a=1$:$1^2 = 1(1)^3 + b \implies 1 = 1 + b \implies b = 0$.Thus,$(a, b) = (1, 0)$.

If $2y = 3x - 1$ is a tangent drawn to the curve $y^2 = ax^3 + b$ at $(1, 1)$,where $a$ and $b$ are constants,then $(a, b) = $

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