| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2017 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find constant from definite integral |
| Difficulty | Moderate -0.3 Part (a) requires expanding the product and integrating standard trigonometric functions (sin²x and sinx·cosx), which are routine C3/C4 techniques. Part (b) is a straightforward reverse problem using logarithm integration rules. Both parts are standard textbook exercises with no problem-solving insight required, making this slightly easier than average. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain \(2 - 2\cos 2x\) as part of integrand | B1 | |
| Obtain \(3\sin 2x\) as part of integrand | B1 | Allow second B1 for writing \(\int 6\sin x\cos x\, dx = 6\!\left(\frac{1}{2}\sin^2 x\right)\), M1 may then be implied by subsequent work |
| Integrate to obtain form \(k_1 x + k_2\sin 2x + k_3\cos 2x\) | M1 | |
| Obtain \(2x - \sin 2x - \frac{3}{2}\cos 2x\) or \(2x - \sin 2x + 3\sin^2 x\) | A1 | |
| Apply limits to obtain \(\frac{1}{2}\pi + \frac{1}{2}\) | A1 | |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Integrate to obtain \(2\ln(3x+2)\) | B1 | Allow \(\frac{6}{3}\ln(3x+2)\) for B1 |
| Use at least one relevant logarithm property | \*M1 | |
| Obtain \(\dfrac{3a+2}{2} = 7\) or \(\dfrac{(3a+2)^2}{4} = 49\) or equivalent without ln | A1 | |
| Solve relevant equation to find \(a\) | DM1 | Dep on \*M1, allow for \(49 = (3a+2)^2\) OE or correct working involving \((3a+2)\) |
| Obtain \(a = 4\) only | A1 | |
| Total | 5 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $2 - 2\cos 2x$ as part of integrand | B1 | |
| Obtain $3\sin 2x$ as part of integrand | B1 | Allow second B1 for writing $\int 6\sin x\cos x\, dx = 6\!\left(\frac{1}{2}\sin^2 x\right)$, M1 may then be implied by subsequent work |
| Integrate to obtain form $k_1 x + k_2\sin 2x + k_3\cos 2x$ | M1 | |
| Obtain $2x - \sin 2x - \frac{3}{2}\cos 2x$ or $2x - \sin 2x + 3\sin^2 x$ | A1 | |
| Apply limits to obtain $\frac{1}{2}\pi + \frac{1}{2}$ | A1 | |
| **Total** | **5** | |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Integrate to obtain $2\ln(3x+2)$ | B1 | Allow $\frac{6}{3}\ln(3x+2)$ for B1 |
| Use at least one relevant logarithm property | \*M1 | |
| Obtain $\dfrac{3a+2}{2} = 7$ or $\dfrac{(3a+2)^2}{4} = 49$ or equivalent without ln | A1 | |
| Solve relevant equation to find $a$ | DM1 | Dep on \*M1, allow for $49 = (3a+2)^2$ OE or correct working involving $(3a+2)$ |
| Obtain $a = 4$ only | A1 | |
| **Total** | **5** | |6
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of $\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sin x ( 4 \sin x + 6 \cos x ) \mathrm { d } x$.
\item Given that $\int _ { 0 } ^ { a } \frac { 6 } { 3 x + 2 } \mathrm {~d} x = \ln 49$, find the value of the positive constant $a$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2017 Q6 [10]}}