CAIE P3 2019 November — Question 6 8 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2019
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferentiating Transcendental Functions
TypeIntegration by parts
DifficultyStandard +0.3 Part (i) is a straightforward quotient rule application to verify a standard result. Part (ii) is a routine integration by parts with cosec²x, requiring recognition that ∫cosec²x = -cot x and careful evaluation at limits. This is a standard textbook exercise testing technique rather than problem-solving, slightly easier than average due to the guided structure.
Spec1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07l Derivative of ln(x): and related functions1.08i Integration by parts

6
  1. By differentiating \(\frac { \cos x } { \sin x }\), show that if \(y = \cot x\) then \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } ^ { 2 } x\).
  2. Show that \(\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 2 } \pi } x \operatorname { cosec } ^ { 2 } x \mathrm {~d} x = \frac { 1 } { 4 } ( \pi + \ln 4 )\). \(7 \quad\) Two lines \(l\) and \(m\) have equations \(\mathbf { r } = a \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } + \lambda ( \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } )\) and \(\mathbf { r } = 2 \mathbf { i } + \mathbf { j } + 2 \mathbf { k } + \mu ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } )\) respectively, where \(a\) is a constant. It is given that the lines intersect.

Question 6(i):
AnswerMarks Guidance
AnswerMark Guidance
Use correct quotient ruleM1
Obtain \(\frac{dy}{dx} = -\cosec^2 x\) correctlyA1 AG
Question 6(ii):
AnswerMarks Guidance
AnswerMark Guidance
Integrate by parts and reach \(ax\cot x + b\int \cot x\,dx\)\*M1
Obtain \(-x\cot x + \int \cot x\,dx\)A1 OE
State \(\pm\ln\sin x\) as integral of \(\cot x\)M1
Obtain complete integral \(-x\cot x + \ln\sin x\)A1 OE
Use correct limits correctlyDM1 \(0 + 0 + \frac{\pi}{4} - \ln\frac{1}{\sqrt{2}}\)
Obtain \(\frac{1}{4}(\pi + \ln 4)\) following full and exact workingA1 AG
## Question 6(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct quotient rule | M1 | |
| Obtain $\frac{dy}{dx} = -\cosec^2 x$ correctly | A1 | AG |

## Question 6(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| Integrate by parts and reach $ax\cot x + b\int \cot x\,dx$ | \*M1 | |
| Obtain $-x\cot x + \int \cot x\,dx$ | A1 | OE |
| State $\pm\ln\sin x$ as integral of $\cot x$ | M1 | |
| Obtain complete integral $-x\cot x + \ln\sin x$ | A1 | OE |
| Use correct limits correctly | DM1 | $0 + 0 + \frac{\pi}{4} - \ln\frac{1}{\sqrt{2}}$ |
| Obtain $\frac{1}{4}(\pi + \ln 4)$ following full and exact working | A1 | AG |
6 (i) By differentiating $\frac { \cos x } { \sin x }$, show that if $y = \cot x$ then $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } ^ { 2 } x$.\\

(ii) Show that $\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 2 } \pi } x \operatorname { cosec } ^ { 2 } x \mathrm {~d} x = \frac { 1 } { 4 } ( \pi + \ln 4 )$.\\

$7 \quad$ Two lines $l$ and $m$ have equations $\mathbf { r } = a \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } + \lambda ( \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } )$ and $\mathbf { r } = 2 \mathbf { i } + \mathbf { j } + 2 \mathbf { k } + \mu ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } )$ respectively, where $a$ is a constant. It is given that the lines intersect.\\

\hfill \mbox{\textit{CAIE P3 2019 Q6 [8]}}