| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2016 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Definite integral with trigonometric functions |
| Difficulty | Standard +0.3 Part (i) is straightforward algebraic manipulation using the double angle formula cos 2x = 2cos²x - 1, requiring routine factorization. Part (ii) applies a substitution u = 2x with the result from (i), then integrates 2cos u + 1 over symmetric limits where the cosine term vanishes. While multi-step, this follows a standard 'show then use' pattern with no novel insight required, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits |
| Answer | Marks |
|---|---|
| (ii) | Use cos2x=2cos2 x−1 and attempt factorisation of numerator |
| Answer | Marks |
|---|---|
| Obtain 2π | M1 |
| Answer | Marks |
|---|---|
| A1 | [3] |
Question 5:
--- 5 (i)
(ii) ---
5 (i)
(ii) | Use cos2x=2cos2 x−1 and attempt factorisation of numerator
Obtain (2cosx+1)(cosx+4)
Confirm given result 2cosx+1
Express integrand as 2cos2x+1
Integrate to obtain sin2x+x
Apply limits correctly to integral of form k sin2x+k x
1 2
Obtain 2π | M1
A1
A1
B1
B1
M1
A1 | [3]
[4]\begin{enumerate}[label=(\roman*)]
\item Show that $\frac{\cos 2x + 9\cos x + 5}{\cos x + 4} \equiv 2\cos x + 1$. [3]
\item Hence find the exact value of $\int_{-\pi}^{\pi} \frac{\cos 4x + 9\cos 2x + 5}{\cos 2x + 4} dx$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2016 Q5 [7]}}