CAIE P2 2014 November — Question 2 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2014
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
Typeln(y) vs x: find constants from two points
DifficultyModerate -0.3 This is a standard logarithmic transformation question requiring students to recognize that ln(y) = ln(a) + x·ln(b) gives a linear relationship, then use two points to find the gradient and intercept. The arithmetic is straightforward with no conceptual challenges beyond the core technique, making it slightly easier than average.
Spec1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form

2 \includegraphics[max width=\textwidth, alt={}, center]{72d50061-ead5-466a-96fc-2203438d1407-2_654_693_532_724} The variables \(x\) and \(y\) satisfy the equation \(y = a \left( b ^ { x } \right)\), where \(a\) and \(b\) are constants. The graph of \(\ln y\) against \(x\) is a straight line passing through the points ( \(0.75,1.70\) ) and ( \(1.53,2.18\) ), as shown in the diagram. Find the values of \(a\) and \(b\) correct to 2 decimal places.

Question 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
State or imply \(\ln y = \ln a + x\ln b\)B1
Equate \(\ln b\) to numerical gradient of lineM1
Obtain \(b = 1.85\)A1
Substitute to find value of \(a\)M1
Obtain \(a = 3.45\)A1 [5]
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply $\ln y = \ln a + x\ln b$ | B1 | |
| Equate $\ln b$ to numerical gradient of line | M1 | |
| Obtain $b = 1.85$ | A1 | |
| Substitute to find value of $a$ | M1 | |
| Obtain $a = 3.45$ | A1 | [5] |

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2\\
\includegraphics[max width=\textwidth, alt={}, center]{72d50061-ead5-466a-96fc-2203438d1407-2_654_693_532_724}

The variables $x$ and $y$ satisfy the equation $y = a \left( b ^ { x } \right)$, where $a$ and $b$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points ( $0.75,1.70$ ) and ( $1.53,2.18$ ), as shown in the diagram. Find the values of $a$ and $b$ correct to 2 decimal places.

\hfill \mbox{\textit{CAIE P2 2014 Q2 [5]}}