Let C be the curve y = x^3 (where x takes all | vedclass

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(A) Let the point $A$ be $(t, t^3)$ and $B$ be $(T, T^3)$ on the curve $y = x^3$.The slope of the tangent at $A$ is given by $\frac{dy}{dx} = 3x^2$. A

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$4$

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(A) Let the point $A$ be $(t, t^3)$ and $B$ be $(T, T^3)$ on the curve $y = x^3$.The slope of the tangent at $A$ is given by $\frac{dy}{dx} = 3x^2$. At $x = t$,the slope $m_A = 3t^2$.The equation of the tangent at $A$ is $y - t^3 = 3t^2(x - t)$,which simplifies to $y = 3t^2x - 2t^3$.Since this tangent meets the curve $y = x^3$ at $B(T, T^3)$,we have $T^3 = 3t^2T - 2t^3$.Rearranging gives $T^3 - 3t^2T + 2t^3 = 0$.Since $T = t$ is a root (tangent point),we can factor out $(T - t)^2$:$(T - t)^2(T + 2t) = 0$.Thus,$T = -2t$ is the coordinate of the point $B$.The gradient at $B$ is $m_B = 3T^2 = 3(-2t)^2 = 3(4t^2) = 12t^2$.We are given $m_B = K \cdot m_A$,so $12t^2 = K(3t^2)$.Therefore,$K = \frac{12t^2}{3t^2} = 4$.

Let $C$ be the curve $y = x^3$ (where $x$ takes all real values). The tangent at $A(t, t^3)$ meets the curve again at $B(T, T^3)$. If the gradient at $B$ is $K$ times the gradient at $A$,then $K$ is equal to

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