| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Polynomial division before integration |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring standard techniques: finding a tangent using differentiation of ln, algebraic manipulation to find a constant in partial fractions, and integration using the result. All steps are routine A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) State derivative is \(k(3x - 2)\) where \(k = 3.1\), or \(\frac{1}{3}\) | M1 | |
| State correct derivative \(\frac{3}{(3x-2)}\) | A1 | |
| Form the equation of the tangent at the point where \(x = 1\) | M1 | |
| Obtain answer \(y = 3x - 3\), or equivalent | A1 | [4] |
| (b) (i) Carry out a complete method for finding \(A\) | M1 | |
| Obtain \(A = 4\) | A1 | [2] |
| (ii) Integrate and obtain term \(2x\) | B1 | |
| Obtain second term of the form \(a\ln(3x - 2)\) | M1 | |
| Obtain second term \(\frac{4}{3}\ln(3x - 2)\) | A1√ | |
| Substitute limits correctly | M1 | |
| Obtain given answer following full and correct working | A1 | [5] |
**(a)** State derivative is $k(3x - 2)$ where $k = 3.1$, or $\frac{1}{3}$ | M1 |
State correct derivative $\frac{3}{(3x-2)}$ | A1 |
Form the equation of the tangent at the point where $x = 1$ | M1 |
Obtain answer $y = 3x - 3$, or equivalent | A1 | [4]
**(b) (i)** Carry out a complete method for finding $A$ | M1 |
Obtain $A = 4$ | A1 | [2]
**(ii)** Integrate and obtain term $2x$ | B1 |
Obtain second term of the form $a\ln(3x - 2)$ | M1 |
Obtain second term $\frac{4}{3}\ln(3x - 2)$ | A1√ |
Substitute limits correctly | M1 |
Obtain given answer following full and correct working | A1 | [5]8
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the tangent to the curve $y = \ln ( 3 x - 2 )$ at the point where $x = 1$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the value of the constant $A$ such that
$$\frac { 6 x } { 3 x - 2 } \equiv 2 + \frac { A } { 3 x - 2 }$$
\item Hence show that $\int _ { 2 } ^ { 6 } \frac { 6 x } { 3 x - 2 } \mathrm {~d} x = 8 + \frac { 8 } { 3 } \ln 2$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2009 Q8 [11]}}