The sketch shows the curve with parametric equations $x = \ln(2t + 3)$ and $y = \dfrac{2t - 3}{2t + 3}$. The curve meets the $y$-axis at the point $A$ and meets the $x$-axis at the point $B$.
(a)[4]
Show that $\dfrac{dy}{dx} = \dfrac{6}{2t + 3}$.
(b)[2]
Find the gradient of the curve at $A$.
(c)[2]
Find the gradient of the curve at $B$.
Worked solution & mark scheme
This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Obtain $\dfrac{dx}{dt}=\dfrac{2}{2t+3}$” …