CAIE P2 2014 November — Question 5 8 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2014
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeFind stationary points and nature
DifficultyStandard +0.3 This is a straightforward multi-part question requiring standard techniques: differentiate using chain rule, solve a simple trigonometric equation for stationary points, then integrate. The trigonometric manipulations are routine (using double angle formula cos 2x = 2cos²x - 1 makes the derivative factorise easily), and the integration is direct. Slightly above average difficulty due to the exact value requirement and multi-step nature, but all techniques are standard A-level fare with no novel insight required.
Spec1.05l Double angle formulae: and compound angle formulae1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits

5 \includegraphics[max width=\textwidth, alt={}, center]{c703565b-8aa8-424b-9684-6592d4effdf8-2_554_689_1354_726} The diagram shows part of the curve $$y = 2 \cos x - \cos 2 x$$ and its maximum point \(M\). The shaded region is bounded by the curve, the axes and the line through \(M\) parallel to the \(y\)-axis.
  1. Find the exact value of the \(x\)-coordinate of \(M\).
  2. Find the exact value of the area of the shaded region.

(i)
AnswerMarks Guidance
Differentiate to obtain \(-2\sin x + 2\sin 2x\) or equivalentB1
Use \(\sin 2x = 2\sin x \cos x\) or equivalentB1
Equate first derivative to zero and solve for \(x\)M1
Obtain \(\frac{1}{3}\pi\)A1 [4]
(ii)
AnswerMarks Guidance
Integrate to obtain form \(k_1 \sin x + k_2 \sin 2x\)M1
Obtain correct \(2\sin x - \frac{1}{2}\sin 2x\)A1
Apply limits 0 and their answer from part (i)M1
Obtain \(\frac{3}{4}\sqrt{3}\) or exact equivalentA1 [4]
**(i)**

Differentiate to obtain $-2\sin x + 2\sin 2x$ or equivalent | B1 |
Use $\sin 2x = 2\sin x \cos x$ or equivalent | B1 |
Equate first derivative to zero and solve for $x$ | M1 |
Obtain $\frac{1}{3}\pi$ | A1 | [4]

**(ii)**

Integrate to obtain form $k_1 \sin x + k_2 \sin 2x$ | M1 |
Obtain correct $2\sin x - \frac{1}{2}\sin 2x$ | A1 |
Apply limits 0 and their answer from part (i) | M1 |
Obtain $\frac{3}{4}\sqrt{3}$ or exact equivalent | A1 | [4]
5\\
\includegraphics[max width=\textwidth, alt={}, center]{c703565b-8aa8-424b-9684-6592d4effdf8-2_554_689_1354_726}

The diagram shows part of the curve

$$y = 2 \cos x - \cos 2 x$$

and its maximum point $M$. The shaded region is bounded by the curve, the axes and the line through $M$ parallel to the $y$-axis.\\
(i) Find the exact value of the $x$-coordinate of $M$.\\
(ii) Find the exact value of the area of the shaded region.

\hfill \mbox{\textit{CAIE P2 2014 Q5 [8]}}