CAIE P3 2023 June — Question 1 3 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2023
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
TypeQuadratic in exponential form
DifficultyStandard +0.3 This is a standard quadratic-in-exponential problem requiring substitution (let y = e^{2x}), rearranging to quadratic form, solving, then taking logarithms. It's slightly above routine due to the negative exponent requiring multiplication by e^{2x}, but follows a well-practiced technique with no conceptual surprises.
Spec1.06g Equations with exponentials: solve a^x = b
1 Solve the equation $$3 \mathrm { e } ^ { 2 x } - 4 \mathrm { e } ^ { - 2 x } = 5$$ Give the answer correct to 3 decimal places.
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(3(e^{2x})^2 - 5(e^{2x}) - 4 = 0\)B1 OE Form 3 term quadratic in \(e^{2x}\)
\(e^{2x} = \dfrac{5 \pm \sqrt{73}}{6}, \quad x = \dfrac{1}{2}\ln\left(\dfrac{5+\sqrt{73}}{6}\right)\)M1 Use correct method to solve for \(x\)
\(x = 0.407\)A1 Only
Total: 3 
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $3(e^{2x})^2 - 5(e^{2x}) - 4 = 0$ | **B1** | OE Form 3 term quadratic in $e^{2x}$ |
| $e^{2x} = \dfrac{5 \pm \sqrt{73}}{6}, \quad x = \dfrac{1}{2}\ln\left(\dfrac{5+\sqrt{73}}{6}\right)$ | **M1** | Use correct method to solve for $x$ |
| $x = 0.407$ | **A1** | Only |
| | **Total: 3** | |

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1 Solve the equation

$$3 \mathrm { e } ^ { 2 x } - 4 \mathrm { e } ^ { - 2 x } = 5$$

Give the answer correct to 3 decimal places.\\

\hfill \mbox{\textit{CAIE P3 2023 Q1 [3]}}
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