CAIE P2 2022 November — Question 3 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2022
SessionNovember
Marks5
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Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
Typeln(y) vs ln(x) linear graph
DifficultyModerate -0.3 This is a standard logarithmic linearization question requiring students to recognize that ln(y) = ln(A) + k·ln(x) gives a straight line with gradient k and intercept ln(A). Finding k from two points uses basic coordinate geometry (gradient formula), then substituting to find A. It's slightly easier than average because it's a routine textbook exercise with clear structure and no conceptual surprises.
Spec1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form
3 \includegraphics[max width=\textwidth, alt={}, center]{68f4b2dc-a05d-4061-aaf0-de15cfe186a9-04_714_515_262_804} The variables \(x\) and \(y\) satisfy the equation \(y = A x ^ { k }\), where \(A\) and \(k\) are constants. The graph of \(\ln y\) against \(\ln x\) is a straight line passing through the points ( \(0.56,2.87\) ) and ( \(0.81,3.47\) ), as shown in the diagram. Find the value of \(k\), and the value of \(A\) correct to 2 significant figures.
Question 3:
AnswerMarks Guidance
AnswerMark Guidance
State or imply equation is \(\ln y = \ln A + k \ln x\)B1
Equate \(k\) to gradient of lineM1 or eliminate \(\ln A\) from simultaneous equations using appropriate values
Obtain \(k = 2.4\)A1
Substitute appropriate values to find \(\ln A\)M1
Obtain \(\ln A = 1.526\) and hence \(A = 4.6\)A1 AWRT
Alternative method:
AnswerMarks Guidance
AnswerMark Guidance
State or imply \(e^{3.47} = Ae^{0.81k}\) and \(e^{2.87} = Ae^{0.56k}\)B1
Eliminate \(A\)M1
Obtain \(k = 2.4\)A1
Substitute appropriate values to find \(A\)M1
Obtain \(A = e^{1.526}\) and hence \(A = 4.6\)A1 
## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply equation is $\ln y = \ln A + k \ln x$ | B1 | |
| Equate $k$ to gradient of line | M1 | or eliminate $\ln A$ from simultaneous equations using appropriate values |
| Obtain $k = 2.4$ | A1 | |
| Substitute appropriate values to find $\ln A$ | M1 | |
| Obtain $\ln A = 1.526$ and hence $A = 4.6$ | A1 | AWRT |

**Alternative method:**

| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $e^{3.47} = Ae^{0.81k}$ and $e^{2.87} = Ae^{0.56k}$ | B1 | |
| Eliminate $A$ | M1 | |
| Obtain $k = 2.4$ | A1 | |
| Substitute appropriate values to find $A$ | M1 | |
| Obtain $A = e^{1.526}$ and hence $A = 4.6$ | A1 | |

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3\\
\includegraphics[max width=\textwidth, alt={}, center]{68f4b2dc-a05d-4061-aaf0-de15cfe186a9-04_714_515_262_804}

The variables $x$ and $y$ satisfy the equation $y = A x ^ { k }$, where $A$ and $k$ are constants. The graph of $\ln y$ against $\ln x$ is a straight line passing through the points ( $0.56,2.87$ ) and ( $0.81,3.47$ ), as shown in the diagram.

Find the value of $k$, and the value of $A$ correct to 2 significant figures.\\

\hfill \mbox{\textit{CAIE P2 2022 Q3 [5]}}
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