CAIE P2 2014 June — Question 5 6 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2014
SessionJune
Marks6
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TopicExponential Equations & Modelling
Typeln(y) vs x: find constants from two points
DifficultyStandard +0.3 This is a standard logarithmic linearization problem requiring students to take ln of both sides to get ln(y) = ln(K) + px·ln(2), identify gradient and intercept from two points, then solve for K and p. It's slightly easier than average as it's a routine textbook exercise with clear steps: find gradient, find intercept, apply exponential/logarithm rules. The calculation is straightforward with no conceptual surprises.
Spec1.06f Laws of logarithms: addition, subtraction, power rules1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form

5 \includegraphics[max width=\textwidth, alt={}, center]{22ba6cc7-7375-434e-9eaa-d536684dd727-2_583_597_1457_772} The variables \(x\) and \(y\) satisfy the equation \(y = K \left( 2 ^ { p x } \right)\), where \(K\) and \(p\) are constants. The graph of \(\ln y\) against \(x\) is a straight line passing through the points ( \(1.35,1.87\) ) and ( \(3.35,3.81\) ), as shown in the diagram. Find the values of \(K\) and \(p\) correct to 2 decimal places.
[0pt] [6]

AnswerMarks Guidance
State or imply \(\ln y = \ln K + p x \ln 2\)B1
Obtain at least one of:B1
\(1.87 = \ln K + 1.35 p\ln 2\), \(3.81 = \ln K + 3.35 p\ln 2\), \(p\ln 2 = \frac{3.81 - 1.87}{3.35 - 1.35}\)
or equivalents
Solve equation(s) to find one constant, dependent on previous B1M1
Obtain \(p = 1.40\)A1
Substitute to attempt value of \(K\)DM1
Obtain \(\ln K = 0.5605\) and hence \(K = 1.75\)A1 [6]
| State or imply $\ln y = \ln K + p x \ln 2$ | B1 |
| Obtain at least one of: | B1 |
| $1.87 = \ln K + 1.35 p\ln 2$, $3.81 = \ln K + 3.35 p\ln 2$, $p\ln 2 = \frac{3.81 - 1.87}{3.35 - 1.35}$ | |
| or equivalents | |
| Solve equation(s) to find one constant, dependent on previous B1 | M1 |
| Obtain $p = 1.40$ | A1 |
| Substitute to attempt value of $K$ | DM1 |
| Obtain $\ln K = 0.5605$ and hence $K = 1.75$ | A1 | [6] |
5\\
\includegraphics[max width=\textwidth, alt={}, center]{22ba6cc7-7375-434e-9eaa-d536684dd727-2_583_597_1457_772}

The variables $x$ and $y$ satisfy the equation $y = K \left( 2 ^ { p x } \right)$, where $K$ and $p$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points ( $1.35,1.87$ ) and ( $3.35,3.81$ ), as shown in the diagram. Find the values of $K$ and $p$ correct to 2 decimal places.\\[0pt]
[6]

\hfill \mbox{\textit{CAIE P2 2014 Q5 [6]}}
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