| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Solve using given identity |
| Difficulty | Standard +0.3 This is a 'hence' question following previous parts that likely factorize the cubic or provide roots. Students substitute cos θ with known values and solve basic inverse trig equations. Slightly easier than average since the hard work (solving the cubic) is done in earlier parts, requiring only routine application of inverse cosine and angle finding within the given range. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks |
|---|---|
| 7(a) | Show 8 × (–7)3 + 54 × (–7)2 – 17 × (–7) – 21 = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Factors must be stated again in (b) to collect marks there | B1 | No errors allowed. |
| Answer | Marks |
|---|---|
| 7(b) | Commence division and reach partial quotient of the form 8x2 ± 2x |
| Answer | Marks | Guidance |
|---|---|---|
| = ± 2 or C = –3 | M1 | Condone no visible working. |
| Answer | Marks | Guidance |
|---|---|---|
| Stating (x + 7)(8x2 – 2x – 3) is sufficient | A1 | Division can terminate with 0 or −3x – 21 stated once or twice. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 7(c) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | M1 | (x + 7) (8x2 − 2x – 3) = (x + 7)(4x – 3)(2x + 1) = 0 |
| Answer | Marks |
|---|---|
| Obtain one answer, e.g. 𝜃 = 120° | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (condone 319°) and no others in the interval | A1 | Accept more accurate answers. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 7:
--- 7(a) ---
7(a) | Show 8 × (–7)3 + 54 × (–7)2 – 17 × (–7) – 21 = 0
This is sufficient if no errors seen.
[ – 2744 + 2646 +119 – 21 = 0]
Or complete division of 8x3 + 54x2 – 17x – 21 by x + 7 to get
quotient 8x2 – 2x – 3 and remainder of 0
Or state (x + 7)(8x2 – 2x – 3) is sufficient
Factors must be stated again in (b) to collect marks there | B1 | No errors allowed.
Correct division:
8x2 −2x −3 .
x + 7 8x3 + 54x2 − 17x − 21
8x3 + 56x2 .
− 2x2 −17x
− 2x2 −14x .
− 3x − 21
− 3x − 21
1
--- 7(b) ---
7(b) | Commence division and reach partial quotient of the form 8x2 ± 2x
or
8x3 + 54x2 – 17x – 21 = (x + 7)(Ax2 + Bx + C) and reach A = 8 and B
= ± 2 or C = –3 | M1 | Condone no visible working.
Obtain quotient 8x2 – 2x – 3 with no errors seen
Stating (x + 7)(8x2 – 2x – 3) is sufficient | A1 | Division can terminate with 0 or −3x – 21 stated once or twice.
The working of division and finding quotient may be seen in (a)
but results required here to collect marks.
2
Question | Answer | Marks | Guidance
--- 7(c) ---
7(c) | 1
Solve quadratic from (b) to obtain a value for 𝜃 = cos1
2
3
or cos1
4 | M1 | (x + 7) (8x2 − 2x – 3) = (x + 7)(4x – 3)(2x + 1) = 0
2 496 1 3
x = cos and .
16 2 4
Obtain one answer, e.g. 𝜃 = 120° | A1
Obtain three further answers, e.g. 𝜃 = 240°, 41.4° and 318.6°
(condone 319°) and no others in the interval | A1 | Accept more accurate answers.
Answers in radians, maximum 2/3.
3
Question | Answer | Marks | Guidance\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Hence solve the equation
$$8 \cos^3 \theta + 54 \cos^2 \theta - 17 \cos \theta - 21 = 0,$$
for $0° \leqslant \theta \leqslant 360°$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q7 [3]}}