CAIE P2 2021 November — Question 3 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2021
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
Typey vs ln(x) linear graph
DifficultyStandard +0.3 This is a standard linearization problem requiring students to take logarithms of both sides to get y = (ln k/ln a) + (1/ln a)·ln x, then use two points to find the gradient and intercept. It involves routine algebraic manipulation and simultaneous equations, slightly easier than average due to its formulaic nature.
Spec1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form
3 \includegraphics[max width=\textwidth, alt={}, center]{6294c4f4-70a9-4b81-87e0-20e2cc24dd27-05_606_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
Total5 
## 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 | |
| **Total** | **5** | |

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3\\
\includegraphics[max width=\textwidth, alt={}, center]{6294c4f4-70a9-4b81-87e0-20e2cc24dd27-05_606_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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