CAIE P2 2019 June — Question 1 3 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2019
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLaws of Logarithms
TypeSimplify or prove logarithmic identity
DifficultyModerate -0.8 This is a straightforward application of the quotient law of logarithms (ln a - ln b = ln(a/b)) followed by basic algebraic factorization and cancellation. The factorization is simple (x³-4x = x(x²-4) = x(x-2)(x+2) and x²-2x = x(x-2)), requiring only routine algebraic manipulation with no problem-solving insight needed. Easier than average for A-level.
Spec1.06f Laws of logarithms: addition, subtraction, power rules
1 Show that \(\ln \left( x ^ { 3 } - 4 x \right) - \ln \left( x ^ { 2 } - 2 x \right) \equiv \ln ( x + 2 )\).
Question 1:
\(\ln(x^3 - 4x) - \ln(x^2 - 2x) \equiv \ln(x+2)\)
AnswerMarks Guidance
\(\ln\left(\dfrac{x^3-4x}{x^2-2x}\right)\)M1 Use of \(\ln a - \ln b = \ln\dfrac{a}{b}\)
\(= \ln\left(\dfrac{x(x^2-4)}{x(x-2)}\right)\)M1 Factorising numerator and denominator correctly
\(= \ln\left(\dfrac{(x+2)(x-2)}{(x-2)}\right) = \ln(x+2)\)A1 Correct cancellation leading to given answer 
# Question 1:

$\ln(x^3 - 4x) - \ln(x^2 - 2x) \equiv \ln(x+2)$

| $\ln\left(\dfrac{x^3-4x}{x^2-2x}\right)$ | M1 | Use of $\ln a - \ln b = \ln\dfrac{a}{b}$ |
|---|---|---|
| $= \ln\left(\dfrac{x(x^2-4)}{x(x-2)}\right)$ | M1 | Factorising numerator and denominator correctly |
| $= \ln\left(\dfrac{(x+2)(x-2)}{(x-2)}\right) = \ln(x+2)$ | A1 | Correct cancellation leading to given answer |

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1 Show that $\ln \left( x ^ { 3 } - 4 x \right) - \ln \left( x ^ { 2 } - 2 x \right) \equiv \ln ( x + 2 )$.\\

\hfill \mbox{\textit{CAIE P2 2019 Q1 [3]}}
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