CAIE P2 2015 June — Question 6 9 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2015
SessionJune
Marks9
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TopicStandard Integrals and Reverse Chain Rule
TypeDefinite integral with trigonometric functions
DifficultyStandard +0.3 This question requires using the double angle identity cos(2x) = 1 - 2sin²(x) to rewrite the equation, then performing standard integration of sin(x) and cos(2x). Part (i) involves solving a quadratic in sin(x). While it requires multiple steps and knowledge of trigonometric identities, these are routine A-level techniques with clear pathways and no novel problem-solving required.
Spec1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05l Double angle formulae: and compound angle formulae1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals

6 \includegraphics[max width=\textwidth, alt={}, center]{3b217eb4-3bd3-4800-a913-749754bf109f-3_401_586_817_778} The diagram shows part of the curve with equation $$y = 4 \sin ^ { 2 } x + 8 \sin x + 3$$ and its point of intersection \(P\) with the \(x\)-axis.
  1. Find the exact \(x\)-coordinate of \(P\).
  2. Show that the equation of the curve can be written $$y = 5 + 8 \sin x - 2 \cos 2 x$$ and use integration to find the exact area of the shaded region enclosed by the curve and the axes.

AnswerMarks Guidance
(i)Solve three-term quadratic equation for \(\sin x\) M1
Obtain at least \(\sin x = -\frac{1}{2}\) and no errors seenA1
Obtain \(x = \frac{\pi}{6}\)A1 [3]
(ii)State \(\sin^2 x = \frac{1}{4} - \frac{1}{4}\cos 2x\) B1
Obtain given \(5 + 8\sin x - 2\cos 2x\) with necessary detail seenB1
Integrate to obtain expression of form \(ax + b\cos x + c\sin 2x\)M1
Obtain correct \(5x - 8\cos x - \sin 2x\)A1
Apply limits 0 and their \(x\)-value correctlyM1 depM
Obtain \(\frac{35\pi}{6} + \frac{7\sqrt{3}}{2} + 8\) or exact equivalentA1 [6]
(i) | Solve three-term quadratic equation for $\sin x$ | M1 |
| Obtain at least $\sin x = -\frac{1}{2}$ and no errors seen | A1 |
| Obtain $x = \frac{\pi}{6}$ | A1 | [3]

(ii) | State $\sin^2 x = \frac{1}{4} - \frac{1}{4}\cos 2x$ | B1 |
| Obtain given $5 + 8\sin x - 2\cos 2x$ with necessary detail seen | B1 |
| Integrate to obtain expression of form $ax + b\cos x + c\sin 2x$ | M1 |
| Obtain correct $5x - 8\cos x - \sin 2x$ | A1 |
| Apply limits 0 and their $x$-value correctly | M1 depM |
| Obtain $\frac{35\pi}{6} + \frac{7\sqrt{3}}{2} + 8$ or exact equivalent | A1 | [6]

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6\\
\includegraphics[max width=\textwidth, alt={}, center]{3b217eb4-3bd3-4800-a913-749754bf109f-3_401_586_817_778}

The diagram shows part of the curve with equation

$$y = 4 \sin ^ { 2 } x + 8 \sin x + 3$$

and its point of intersection $P$ with the $x$-axis.\\
(i) Find the exact $x$-coordinate of $P$.\\
(ii) Show that the equation of the curve can be written

$$y = 5 + 8 \sin x - 2 \cos 2 x$$

and use integration to find the exact area of the shaded region enclosed by the curve and the axes.

\hfill \mbox{\textit{CAIE P2 2015 Q6 [9]}}
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