The curve is defined by $\frac{dy}{dx}=kx^3+\frac{2}{x^2}$, where $k$ is a constant. It passes through $S(2,20)$, and the gradient at $S$ is $\frac{65}{2}$.
(a)[1]
Find $k$.
(b)[5]
The coordinates of a point $T$ on the curve are $(1,t)$. Find $t$.
Worked solution & mark scheme
This 6-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Therefore $k=4$.” …