| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Parametric form dy/dx |
| Difficulty | Standard +0.3 Part (a) is a standard implicit differentiation problem requiring the chain rule for logarithms and product rule, then substitution - routine A-level technique. Part (b) requires finding the parameter value from given coordinates, then applying dy/dx = (dy/dt)/(dx/dt) - also standard parametric differentiation. Both parts are textbook exercises with no novel insight required, making this slightly easier than average overall. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (a) Differentiate \(4\ln y\) to obtain \(\frac{4}{y} \cdot \frac{dy}{dx}\); Differentiate \(6xy\) to obtain \(6y + 6x\frac{dy}{dx}\); Substitute 1 and 1 and solve for \(\frac{dy}{dx}\); Obtain \(-\frac{9}{10}\) or equivalent | B1, B1, M1, A1 | [4] |
| (b) Obtain \(\frac{dx}{dt} = -10r^{-2} - 1\); Obtain derivative of form \(k(2t-1)^{-\frac{1}{2}}\) for \(\frac{dy}{dt}\); Obtain correct \((2t-1)^{-\frac{1}{2}}\); Identify value of \(t\) as 5; Obtain expression for \(\frac{dy}{dx}\) correctly, with numerical value of \(t\) substituted; Obtain \(-\frac{5}{21}\) or exact equivalent | B1, M1, A1, B1, M1, A1 | [6] |
| Answer | Marks | Guidance |
|--------|-------|----------|
| **(a)** Differentiate $4\ln y$ to obtain $\frac{4}{y} \cdot \frac{dy}{dx}$; Differentiate $6xy$ to obtain $6y + 6x\frac{dy}{dx}$; Substitute 1 and 1 and solve for $\frac{dy}{dx}$; Obtain $-\frac{9}{10}$ or equivalent | B1, B1, M1, A1 | [4] |
| **(b)** Obtain $\frac{dx}{dt} = -10r^{-2} - 1$; Obtain derivative of form $k(2t-1)^{-\frac{1}{2}}$ for $\frac{dy}{dt}$; Obtain correct $(2t-1)^{-\frac{1}{2}}$; Identify value of $t$ as 5; Obtain expression for $\frac{dy}{dx}$ correctly, with numerical value of $t$ substituted; Obtain $-\frac{5}{21}$ or exact equivalent | B1, M1, A1, B1, M1, A1 | [6] |\begin{enumerate}[label=(\alph*)]
\item Find the gradient of the curve
$$3\ln x + 4\ln y + 6xy = 6$$
at the point $(1, 1)$. [4]
\item The parametric equations of a curve are
$$x = \frac{10}{t} - t, \quad y = \sqrt{2t - 1}.$$
Find the gradient of the curve at the point $(-3, 3)$. [6]
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2015 Q7 [10]}}