CAIE P1 2020 November — Question 3 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2020
SessionNovember
Marks5
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Mark schemeDownload PDF ↗
TopicQuadratic trigonometric equations
TypeQuartic in sin or cos substitution
DifficultyStandard +0.3 This is a quadratic-in-tan² equation requiring substitution (u = tan²θ), solving a quadratic, then finding angles. It's slightly above average due to the reciprocal term requiring rearrangement to standard form and needing to handle both positive solutions carefully, but remains a standard technique for P1 level with straightforward algebra and no conceptual surprises.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
3 Solve the equation \(3 \tan ^ { 2 } \theta + 1 = \frac { 2 } { \tan ^ { 2 } \theta }\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
Question 3:
AnswerMarks Guidance
\(3\tan^4\theta + \tan^2\theta - 2 (= 0)\)M1 SOI 3-term quartic, condone sign errors for this mark only
\((3\tan^2\theta - 2)(\tan^2\theta + 1)(= 0)\)M1 Attempt to factorise or solve 3-term quadratic in \(\tan^2\theta\)
\(\tan\theta = (\pm)\sqrt{\frac{2}{3}}\) or \((\pm)0.816\) or \((\pm)0.817\)A1 SOI Implied by final answer \(= 39.2°\) after 1st M1 scored
\(39.2°, 140.8°\)A1, A1 FT FT for 2nd solution \(= 180° -\) 1st solution 
## Question 3:

| $3\tan^4\theta + \tan^2\theta - 2 (= 0)$ | M1 | SOI 3-term quartic, condone sign errors for this mark only |
|---|---|---|
| $(3\tan^2\theta - 2)(\tan^2\theta + 1)(= 0)$ | M1 | Attempt to factorise or solve 3-term quadratic in $\tan^2\theta$ |
| $\tan\theta = (\pm)\sqrt{\frac{2}{3}}$ or $(\pm)0.816$ or $(\pm)0.817$ | A1 | SOI Implied by final answer $= 39.2°$ after 1st M1 scored |
| $39.2°, 140.8°$ | A1, A1 FT | FT for 2nd solution $= 180° -$ 1st solution |

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3 Solve the equation $3 \tan ^ { 2 } \theta + 1 = \frac { 2 } { \tan ^ { 2 } \theta }$ for $0 ^ { \circ } < \theta < 180 ^ { \circ }$.\\

\hfill \mbox{\textit{CAIE P1 2020 Q3 [5]}}
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