| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2018 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Definite integral with logarithmic form |
| Difficulty | Standard +0.3 Part (i) requires standard quotient rule differentiation and solving a trigonometric equation within a given domain—routine calculus techniques. Part (ii) is a straightforward reverse chain rule integration (recognizing the derivative of the denominator in the numerator) followed by solving a logarithmic equation. Both parts are standard textbook exercises requiring competent technique but no novel insight, making this slightly easier than average. |
| Spec | 1.07k 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)1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| 7(i) | Use correct quotient or product rule | M1 |
| Obtain correct derivative in any form | A1 | ( ) |
| Answer | Marks | Guidance |
|---|---|---|
| Equate numerator to zero | M1 | |
| Use cos2x+sin2x=1 and solve for sin x | M1 | −6sinx−3=0 ⇒ sinx =…. |
| Obtain coordinates x=−π /6 and y= 3 ISW | A1 + A1 | From correct working. No others in range |
| Answer | Marks | Guidance |
|---|---|---|
| 7(ii) | State indefinite integral of the form k ln (2 + sin x) | M1* |
| Answer | Marks | Guidance |
|---|---|---|
| 3 ln (2 + sin a) – 3 ln 2 = 1 | A1 | or equivalent |
| Use correct method to solve for a | M1(dep*) | Allow for a correct method to solve an incorrect equation, so |
| Answer | Marks | Guidance |
|---|---|---|
| Obtain answer a = 0.913 or better | A1 | Ignore additional solutions. Must be in radians. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 7:
--- 7(i) ---
7(i) | Use correct quotient or product rule | M1
Obtain correct derivative in any form | A1 | ( )
dy −3sinx 2+sinx −3cosxcosx
=
( )2
dx 2+sinx
Condone invisible brackets if recovery implied later.
Equate numerator to zero | M1
Use cos2x+sin2x=1 and solve for sin x | M1 | −6sinx−3=0 ⇒ sinx =….
Obtain coordinates x=−π /6 and y= 3 ISW | A1 + A1 | From correct working. No others in range
SR: A candidate who only states the numerator of the derivative,
but justifies this, can have full marks. Otherwise they score
M0A0M1M1A0A0
6
--- 7(ii) ---
7(ii) | State indefinite integral of the form k ln (2 + sin x) | M1*
Substitute limits correctly, equate result to 1 and obtain
3 ln (2 + sin a) – 3 ln 2 = 1 | A1 | or equivalent
Use correct method to solve for a | M1(dep*) | Allow for a correct method to solve an incorrect equation, so
long as that equation has a solution.
( )
1+ 1 2 sina=e1 3 ⇒a=sin−1 2 e1 3 −1
Can be implied by 52.3°
Obtain answer a = 0.913 or better | A1 | Ignore additional solutions. Must be in radians.
4
Question | Answer | Marks | GuidanceA curve has equation $y = \frac{3 \cos x}{2 + \sin x}$, for $-\frac{1}{2}\pi \leqslant x \leqslant \frac{1}{2}\pi$.
\begin{enumerate}[label=(\roman*)]
\item Find the exact coordinates of the stationary point of the curve. [6]
\item The constant $a$ is such that $\int_0^a \frac{3 \cos x}{2 + \sin x} \, dx = 1$. Find the value of $a$, giving your answer correct to 3 significant figures. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2018 Q7 [10]}}