| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Standard +0.3 This is a standard two-part trigonometric equation question requiring routine manipulation of trig identities (tan²θ = sec²θ - 1, sec²θ = 1/cos²θ) to convert to quadratic form, then solving a straightforward quadratic in cos 2x. The algebraic steps are mechanical and the solving process is textbook standard, making it slightly easier than average for A-level. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| *EITHER:* Uses \(\tan^2 2x = \frac{\sin^2 2x}{\cos^2 2x}\) | (M1 | Replaces \(\tan^2 2x\) by \(\frac{\sin^2 2x}{\cos^2 2x}\), not \(\frac{\sin^2}{\cos^2} 2x\) |
| Uses \(\sin^2 2x = (1 - \cos^2 2x)\) | M1 | Replaces \(\sin^2 2x\) by \((1-\cos^2 2x)\) |
| \(\rightarrow 2\cos^2 2x + 3\cos 2x + 1 = 0\) | A1) | AG. All correct |
| *OR:* \(\tan^2 2x = \sec^2 2x - 1\) | (M1 | Replaces \(\tan^2 2x\) by \(\sec^2 2x - 1\) |
| \(\sec^2 2x = \frac{1}{\cos^2 2x}\), multiply through by \(\cos^2 2x\) and rearrange | M1 | Replaces \(\sec^2 2x\) by \(\frac{1}{\cos^2 2x}\) |
| \(\rightarrow 2\cos^2 2x + 3\cos 2x + 1 = 0\) | A1) | AG. All correct |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos 2x = -\frac{1}{2},\ -1\) | M1 | Uses (i) to get values for \(\cos 2x\). Allow incorrect sign(s). |
| \(2x = 120°, 240°\) or \(2x = 180°\); \(x = 60°\) or \(120°\) | A1 A1 FT | A1 for 60° or 120° FT for 180° — 1st answer |
| or \(x = 90°\) | A1 | Any extra answer(s) in given range only penalise fourth mark so max 3/4. |
| Total: 4 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| *EITHER:* Uses $\tan^2 2x = \frac{\sin^2 2x}{\cos^2 2x}$ | (M1 | Replaces $\tan^2 2x$ by $\frac{\sin^2 2x}{\cos^2 2x}$, not $\frac{\sin^2}{\cos^2} 2x$ |
| Uses $\sin^2 2x = (1 - \cos^2 2x)$ | M1 | Replaces $\sin^2 2x$ by $(1-\cos^2 2x)$ |
| $\rightarrow 2\cos^2 2x + 3\cos 2x + 1 = 0$ | A1) | AG. All correct |
| *OR:* $\tan^2 2x = \sec^2 2x - 1$ | (M1 | Replaces $\tan^2 2x$ by $\sec^2 2x - 1$ |
| $\sec^2 2x = \frac{1}{\cos^2 2x}$, multiply through by $\cos^2 2x$ and rearrange | M1 | Replaces $\sec^2 2x$ by $\frac{1}{\cos^2 2x}$ |
| $\rightarrow 2\cos^2 2x + 3\cos 2x + 1 = 0$ | A1) | AG. All correct |
| **Total: 3** | | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos 2x = -\frac{1}{2},\ -1$ | M1 | Uses (i) to get values for $\cos 2x$. Allow incorrect sign(s). |
| $2x = 120°, 240°$ or $2x = 180°$; $x = 60°$ or $120°$ | A1 A1 FT | A1 for 60° or 120° FT for 180° — 1st answer |
| or $x = 90°$ | A1 | Any extra answer(s) in given range only penalise fourth mark so max 3/4. |
| **Total: 4** | | |5 (i) Show that the equation $\cos 2 x \left( \tan ^ { 2 } 2 x + 3 \right) + 3 = 0$ can be expressed as
$$2 \cos ^ { 2 } 2 x + 3 \cos 2 x + 1 = 0$$
(ii) Hence solve the equation $\cos 2 x \left( \tan ^ { 2 } 2 x + 3 \right) + 3 = 0$ for $0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$.\\
\hfill \mbox{\textit{CAIE P1 2017 Q5 [7]}}