Edexcel C34 2014 January — Question 2 6 marks

Exam BoardEdexcel
ModuleC34 (Core Mathematics 3 & 4)
Year2014
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeSolve equation with tan(θ ± α)
DifficultyStandard +0.3 This question requires recognizing the tan addition formula, then solving a linear equation in tan(2x), followed by routine inverse tan calculations with period consideration. While it tests formula recognition and careful angle arithmetic, it's a standard textbook application with no novel insight required, making it slightly easier than average.
Spec1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals
2. Solve, for \(0 \leqslant x \leqslant 270 ^ { \circ }\), the equation $$\frac { \tan 2 x + \tan 50 ^ { \circ } } { 1 - \tan 2 x \tan 50 ^ { \circ } } = 2$$ Give your answers in degrees to 2 decimal places.
(6) \includegraphics[max width=\textwidth, alt={}, center]{5b698944-41ac-4072-b5e1-c580b7752c39-05_104_95_2613_1786}
Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{\tan 2x + \tan 50°}{1 - \tan 2x \tan 50°} = 2 \Rightarrow \tan(2x + 50°) = 2\)M1A1 M1: Uses compound angle identity \(\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\) to write in form \(\tan(2x \pm 50°) = 2\). Accept sign error. A1: Correct form \(\tan(2x + 50°) = 2\)
\(\Rightarrow 2x + 50° = 63.43°,\ (243.43°,\ 423.43°)\)— —
\(\Rightarrow x = \text{awrt } 6.72°\) or \(96.72°\) or \(186.72°\)dM1, A1 dM1: Correct order of operations to find one solution. Moves from \(\tan(2x \pm 50°) = 2 \Rightarrow 2x \pm 50° = \arctan 2 \Rightarrow x = ...\). Dependent on first M1. A1: One correct answer, usually awrt \(6.72°\)
\(\Rightarrow 2x + 50° = 243.43°\ (423.43°) \Rightarrow x = ..\)dM1 Uses correct order of operations to find a second solution. Can be scored by \(2x \pm 50° = 180 + \text{their } 63\) or \(360 + \text{their } 63\). Dependent on first M1
\(x = \text{awrt } 6.72°, 96.72°, 186.72°\)A1 All three answers in range. Extra solutions in range withhold last A mark. Ignore solutions outside \(0 \leq x \leq 270°\)
Alt 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\tan 2x = \frac{2 - \tan 50°}{1 + 2\tan 50°} = (0.239...)\)M1A1 M1: Cross multiplies, collects terms in \(\tan 2x\), makes \(\tan 2x\) subject. A1: Accept \(\tan 2x = \frac{2 - \tan 50°}{1 + 2\tan 50°}\) or decimal equivalent \(\text{awrt } 0.239\)
\(2x = 13.435° \Rightarrow x = \text{awrt } 6.72°\)dM1, A1 dM1: Correct order of operations for one solution. Dependent on first M1. A1: One correct solution usually awrt \(6.72°\)
\(\Rightarrow 2x° = 193.435°\ (373.435°) \Rightarrow x = ..\)dM1 Uses correct order of operations for another solution. By \(2x = 180 + \text{their } 13.4\) or \(360 + \text{their } 13.4\), or \(90 + \text{their } 6.7\) or \(180 + \text{their } 6.7\)
\(x = \text{awrt } 6.72°, 96.72°, 186.72°\)A1 All three answers in range. Extra solutions withhold last A mark. Ignore solutions outside \(0 \leq x \leq 270°\)
Alt 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Uses \(\tan 2x = \frac{2\tan x}{1 - \tan^2 x}\); \((2 - \tan 50)\tan^2 x + (2 + 4\tan 50)\tan x + (\tan 50 - 2) = 0\)M1A1 —
\(\tan x = \text{awrt } 0.118,\ -8.49\)dM1, A1 —
\(x = 6.72°, 96.72°, 186.72°\)dM1, A1 — 
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\tan 2x + \tan 50°}{1 - \tan 2x \tan 50°} = 2 \Rightarrow \tan(2x + 50°) = 2$ | M1A1 | M1: Uses compound angle identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ to write in form $\tan(2x \pm 50°) = 2$. Accept sign error. A1: Correct form $\tan(2x + 50°) = 2$ |
| $\Rightarrow 2x + 50° = 63.43°,\ (243.43°,\ 423.43°)$ | — | — |
| $\Rightarrow x = \text{awrt } 6.72°$ or $96.72°$ or $186.72°$ | dM1, A1 | dM1: Correct order of operations to find one solution. Moves from $\tan(2x \pm 50°) = 2 \Rightarrow 2x \pm 50° = \arctan 2 \Rightarrow x = ...$. Dependent on first M1. A1: One correct answer, usually awrt $6.72°$ |
| $\Rightarrow 2x + 50° = 243.43°\ (423.43°) \Rightarrow x = ..$ | dM1 | Uses correct order of operations to find a second solution. Can be scored by $2x \pm 50° = 180 + \text{their } 63$ or $360 + \text{their } 63$. Dependent on first M1 |
| $x = \text{awrt } 6.72°, 96.72°, 186.72°$ | A1 | All three answers in range. Extra solutions in range withhold last A mark. Ignore solutions outside $0 \leq x \leq 270°$ |

**Alt 1:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan 2x = \frac{2 - \tan 50°}{1 + 2\tan 50°} = (0.239...)$ | M1A1 | M1: Cross multiplies, collects terms in $\tan 2x$, makes $\tan 2x$ subject. A1: Accept $\tan 2x = \frac{2 - \tan 50°}{1 + 2\tan 50°}$ or decimal equivalent $\text{awrt } 0.239$ |
| $2x = 13.435° \Rightarrow x = \text{awrt } 6.72°$ | dM1, A1 | dM1: Correct order of operations for one solution. Dependent on first M1. A1: One correct solution usually awrt $6.72°$ |
| $\Rightarrow 2x° = 193.435°\ (373.435°) \Rightarrow x = ..$ | dM1 | Uses correct order of operations for another solution. By $2x = 180 + \text{their } 13.4$ or $360 + \text{their } 13.4$, or $90 + \text{their } 6.7$ or $180 + \text{their } 6.7$ |
| $x = \text{awrt } 6.72°, 96.72°, 186.72°$ | A1 | All three answers in range. Extra solutions withhold last A mark. Ignore solutions outside $0 \leq x \leq 270°$ |

**Alt 2:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$; $(2 - \tan 50)\tan^2 x + (2 + 4\tan 50)\tan x + (\tan 50 - 2) = 0$ | M1A1 | — |
| $\tan x = \text{awrt } 0.118,\ -8.49$ | dM1, A1 | — |
| $x = 6.72°, 96.72°, 186.72°$ | dM1, A1 | — |

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2. Solve, for $0 \leqslant x \leqslant 270 ^ { \circ }$, the equation

$$\frac { \tan 2 x + \tan 50 ^ { \circ } } { 1 - \tan 2 x \tan 50 ^ { \circ } } = 2$$

Give your answers in degrees to 2 decimal places.\\
(6)\\

\includegraphics[max width=\textwidth, alt={}, center]{5b698944-41ac-4072-b5e1-c580b7752c39-05_104_95_2613_1786}\\

\hfill \mbox{\textit{Edexcel C34 2014 Q2 [6]}}
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