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(A) Given the curve $y = ax^3 + b$ and the point $A(2, 3)$ lies on the curve,we have: $3 = a(2)^3 + b \implies 8a + b = 3$  (Equation $1$). The slope

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$\frac{1}{3}$

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(A) Given the curve $y = ax^3 + b$ and the point $A(2, 3)$ lies on the curve,we have: $3 = a(2)^3 + b \implies 8a + b = 3$  (Equation $1$). The slope of the tangent to the curve at any point $x$ is given by the derivative $\frac{dy}{dx} = 3ax^2$. At $x = 2$,the slope is $m = 3a(2)^2 = 12a$. The equation of the tangent is given as $y = 4x - 5$,which is in the form $y = mx + c$,so the slope $m = 4$. Equating the slopes: $12a = 4 \implies a = \frac{4}{12} = \frac{1}{3}$. Substitute $a = \frac{1}{3}$ into Equation $1$: $8(\frac{1}{3}) + b = 3$ $\frac{8}{3} + b = 3$ $b = 3 - \frac{8}{3} = \frac{9 - 8}{3} = \frac{1}{3}$. Thus,$b = \frac{1}{3}$.

The equation of the tangent at $A(2, 3)$ to the curve $y = ax^3 + b$ is $y = 4x - 5$. Then $b =$

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