CAIE P3 2009 June — Question 10 11 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2009
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeFinding maximum/minimum on curve
DifficultyStandard +0.8 This is a multi-part question requiring differentiation with product and chain rules to find a maximum, then a non-trivial trigonometric substitution (x = sin θ) with careful manipulation of sin²2θ, and finally integration using double-angle formulas. While the techniques are standard for P3/C4, the execution requires careful algebraic manipulation across multiple steps, placing it moderately above average difficulty.
Spec1.07n Stationary points: find maxima, minima using derivatives1.08h Integration by substitution

10 \includegraphics[max width=\textwidth, alt={}, center]{0f73e750-18a0-49ad-b4cb-fd6d14f0789e-4_424_713_262_715} The diagram shows the curve \(y = x ^ { 2 } \sqrt { } \left( 1 - x ^ { 2 } \right)\) for \(x \geqslant 0\) and its maximum point \(M\).
  1. Find the exact value of the \(x\)-coordinate of \(M\).
  2. Show, by means of the substitution \(x = \sin \theta\), that the area \(A\) of the shaded region between the curve and the \(x\)-axis is given by $$A = \frac { 1 } { 4 } \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sin ^ { 2 } 2 \theta \mathrm {~d} \theta$$
  3. Hence obtain the exact value of \(A\).

Question 10:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
EITHER Use product and chain ruleM1
Obtain correct derivative in any formA1
OR Square and differentiate LHS by chain rule and RHS by product rule or as powersM1
Obtain correct result in any formA1
Set \(\dfrac{dy}{dx}\) equal to zero and make reasonable attempt to solve for \(x \neq 0\)M1
Obtain answer \(x = \sqrt{\tfrac{2}{3}}\), or exact equivalent, correctlyA1 [4 marks]
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
State or imply \(dx = \cos\theta\, d\theta\) or \(\dfrac{dx}{d\theta} = \cos\theta\)B1
Substitute for \(x\) and \(dx\) throughout the integral \(\int y\, dx\)M1
Obtain the given form correctly with no errors seenA1 [3 marks]
Part (iii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Attempt integration and reach indefinite integral of the form \(a\theta + b\sin 4\theta\), where \(ab \neq 0\)M1*
Obtain indefinite integral \(\tfrac{1}{8}\theta - \tfrac{1}{32}\sin 4\theta\), or equivalentA1
Substitute limits correctlyM1(dep*)
Obtain exact answer \(\tfrac{1}{16}\pi\)A1 [4 marks]
[Working to carry out the change of limits is needed for the A mark in (ii) but, if omitted, can be earned retrospectively if it is seen in part (iii)]
# Question 10:

## Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| **EITHER** Use product and chain rule | M1 | |
| Obtain correct derivative in any form | A1 | |
| **OR** Square and differentiate LHS by chain rule and RHS by product rule or as powers | M1 | |
| Obtain correct result in any form | A1 | |
| Set $\dfrac{dy}{dx}$ equal to zero and make reasonable attempt to solve for $x \neq 0$ | M1 | |
| Obtain answer $x = \sqrt{\tfrac{2}{3}}$, or exact equivalent, correctly | A1 | **[4 marks]** |

## Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply $dx = \cos\theta\, d\theta$ or $\dfrac{dx}{d\theta} = \cos\theta$ | B1 | |
| Substitute for $x$ and $dx$ throughout the integral $\int y\, dx$ | M1 | |
| Obtain the given form correctly with no errors seen | A1 | **[3 marks]** |

## Part (iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt integration and reach indefinite integral of the form $a\theta + b\sin 4\theta$, where $ab \neq 0$ | M1* | |
| Obtain indefinite integral $\tfrac{1}{8}\theta - \tfrac{1}{32}\sin 4\theta$, or equivalent | A1 | |
| Substitute limits correctly | M1(dep*) | |
| Obtain exact answer $\tfrac{1}{16}\pi$ | A1 | **[4 marks]** |
| [Working to carry out the change of limits is needed for the A mark in **(ii)** but, if omitted, can be earned retrospectively if it is seen in part **(iii)**] | | |
10\\
\includegraphics[max width=\textwidth, alt={}, center]{0f73e750-18a0-49ad-b4cb-fd6d14f0789e-4_424_713_262_715}

The diagram shows the curve $y = x ^ { 2 } \sqrt { } \left( 1 - x ^ { 2 } \right)$ for $x \geqslant 0$ and its maximum point $M$.\\
(i) Find the exact value of the $x$-coordinate of $M$.\\
(ii) Show, by means of the substitution $x = \sin \theta$, that the area $A$ of the shaded region between the curve and the $x$-axis is given by

$$A = \frac { 1 } { 4 } \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sin ^ { 2 } 2 \theta \mathrm {~d} \theta$$

(iii) Hence obtain the exact value of $A$.

\hfill \mbox{\textit{CAIE P3 2009 Q10 [11]}}