CAIE P2 2015 June — Question 2 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2015
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
Typeln(y) vs x: find constants from two points
DifficultyModerate -0.5 This is a standard logarithmic transformation question requiring students to recognize that ln(y) = ln(A) + px - p gives a linear relationship, then use two points to find the gradient (p) and y-intercept to find A. It involves routine algebraic manipulation and substitution with no novel insight required, making it slightly easier than average.
Spec1.06f Laws of logarithms: addition, subtraction, power rules1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form

2 \includegraphics[max width=\textwidth, alt={}, center]{595e38f4-c52e-4509-8b16-f08e30dec96b-2_456_716_529_712} The variables \(x\) and \(y\) satisfy the equation $$y = A \mathrm { e } ^ { p ( x - 1 ) } ,$$ where \(A\) and \(p\) are constants. The graph of \(\ln y\) against \(x\) is a straight line passing through the points \(( 2,1.60 )\) and \(( 5,2.92 )\), as shown in the diagram. Find the values of \(A\) and \(p\) correct to 2 significant figures.

AnswerMarks Guidance
State or imply that \(\ln y = \ln A + p(x - 1)\)B1
Equate gradient to \(p\) or obtain two equations for \(\ln A\) and \(p\)M1
Obtain \(p = 0.44\)A1
Substitute values correctly, to find value of \(\ln A\)DM1
Obtain \(A = 3.2\)A1 [5]
Alternative:
AnswerMarks Guidance
Obtain an equation either \(e^{1a} = Ae^p\) or \(e^{2a2} = Ae^{4p}\)M1
Obtain both equations correctlyA1
Solve to obtain \(p = 0.44\)A1
Substitute value correctly to find \(A\)DM1
Obtain \(A = 3.2\)A1 [5]
State or imply that $\ln y = \ln A + p(x - 1)$ | B1 |
Equate gradient to $p$ or obtain two equations for $\ln A$ and $p$ | M1 |
Obtain $p = 0.44$ | A1 |
Substitute values correctly, to find value of $\ln A$ | DM1 |
Obtain $A = 3.2$ | A1 | [5]

**Alternative:**
Obtain an equation either $e^{1a} = Ae^p$ or $e^{2a2} = Ae^{4p}$ | M1 |
Obtain both equations correctly | A1 |
Solve to obtain $p = 0.44$ | A1 |
Substitute value correctly to find $A$ | DM1 |
Obtain $A = 3.2$ | A1 | [5]

---
2\\
\includegraphics[max width=\textwidth, alt={}, center]{595e38f4-c52e-4509-8b16-f08e30dec96b-2_456_716_529_712}

The variables $x$ and $y$ satisfy the equation

$$y = A \mathrm { e } ^ { p ( x - 1 ) } ,$$

where $A$ and $p$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points $( 2,1.60 )$ and $( 5,2.92 )$, as shown in the diagram. Find the values of $A$ and $p$ correct to 2 significant figures.

\hfill \mbox{\textit{CAIE P2 2015 Q2 [5]}}