CAIE P2 2021 November — Question 3 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2021
SessionNovember
Marks5
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Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
Typey vs ln(x) linear graph
DifficultyModerate -0.3 This is a standard logarithmic linearization problem requiring students to take logarithms of both sides, identify the gradient and intercept from two points, then solve for the constants. It involves routine algebraic manipulation and straight-line calculations with no novel insight required, making it slightly easier than average.
Spec1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form
3 \includegraphics[max width=\textwidth, alt={}, center]{83d0697c-b133-47da-a745-dfdafa7dbf10-05_604_933_258_605} The variables \(x\) and \(y\) satisfy the equation \(a ^ { y } = k x\), where \(a\) and \(k\) are constants. The graph of \(y\) against \(\ln x\) is a straight line passing through the points \(( 1.03,6.36 )\) and \(( 2.58,9.00 )\), as shown in the diagram. Find the values of \(a\) and \(k\), giving each value correct to 2 significant figures.
Question 3:
AnswerMarks Guidance
AnswerMark Guidance
State or imply equation is \(y\ln a = \ln k + \ln x\)B1
Equate gradient of line to \(\frac{1}{\ln a}\)M1 Or eliminate \(\ln k\) from simultaneous equations
Obtain \(\frac{1}{\ln a} = \frac{2.64}{1.55}\) or equivalent and hence \(a = 1.8\)A1 AWRT
Substitute appropriate values to find \(\ln k\)M1
Obtain \(\ln k = 2.7...\) and hence \(k = 15\)A1 AWRT
Alternative: \(a^{6.36} = ke^{1.03}\) and \(a^9 = ke^{2.58}\)B1 For both
Elimination of \(k\) to obtain equation in \(a\) only \(\left(a^{2.64} = e^{1.55}\right)\)M1 Must have previous B1
Use of a correct method to obtain \(a\)M1 Allow for \(a = e^{0.59}\)
\(a = 1.8\)A1
\(k = 15\)A1
5 
## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply equation is $y\ln a = \ln k + \ln x$ | B1 | |
| Equate gradient of line to $\frac{1}{\ln a}$ | M1 | Or eliminate $\ln k$ from simultaneous equations |
| Obtain $\frac{1}{\ln a} = \frac{2.64}{1.55}$ or equivalent and hence $a = 1.8$ | A1 | AWRT |
| Substitute appropriate values to find $\ln k$ | M1 | |
| Obtain $\ln k = 2.7...$ and hence $k = 15$ | A1 | AWRT |
| **Alternative:** $a^{6.36} = ke^{1.03}$ and $a^9 = ke^{2.58}$ | B1 | For both |
| Elimination of $k$ to obtain equation in $a$ only $\left(a^{2.64} = e^{1.55}\right)$ | M1 | Must have previous B1 |
| Use of a correct method to obtain $a$ | M1 | Allow for $a = e^{0.59}$ |
| $a = 1.8$ | A1 | |
| $k = 15$ | A1 | |
| | **5** | |

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3\\
\includegraphics[max width=\textwidth, alt={}, center]{83d0697c-b133-47da-a745-dfdafa7dbf10-05_604_933_258_605}

The variables $x$ and $y$ satisfy the equation $a ^ { y } = k x$, where $a$ and $k$ are constants. The graph of $y$ against $\ln x$ is a straight line passing through the points $( 1.03,6.36 )$ and $( 2.58,9.00 )$, as shown in the diagram.

Find the values of $a$ and $k$, giving each value correct to 2 significant figures.\\

\hfill \mbox{\textit{CAIE P2 2021 Q3 [5]}}
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