CAIE P2 2019 November — Question 3 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2019
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCurve Sketching
TypeLogarithmic graph for power law
DifficultyModerate -0.5 This is a standard logarithmic linearization problem requiring students to recognize that ln y = ln k + a ln x gives a straight line with gradient a and intercept ln k. Finding the gradient from two points and exponentiating to find k are routine A-level techniques 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]{9c26457d-4b65-4cd4-a9b9-128aba92dbf4-04_586_734_260_701} The variables \(x\) and \(y\) satisfy the equation \(y = k x ^ { a }\), where \(k\) and \(a\) are constants. The graph of \(\ln y\) against \(\ln x\) is a straight line passing through the points ( \(0.22,3.96\) ) and ( \(1.32,2.43\) ), as shown in the diagram. Find the values of \(k\) and \(a\) correct to 3 significant figures.
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(\ln y = \ln k + a\ln x\)B1 SOI
Equate gradient of line to \(a\)M1
Obtain \(a=-1.39\)A1 OE
Substitute appropriate values into a correct equation to find \(\ln k\)M1
Obtain \(\ln k=4.266...\) and \(k=71.2\)A1 SC1 for gradient \(=-1.39\) and no other relevant working
Total5
*Alternative method 1:*
AnswerMarks Guidance
AnswerMarks Guidance
\(\ln y = \ln k + a\ln x\)B1 SOI
\(3.96=\ln k+0.22a\)M1 For one correct equation
\(2.43=\ln k+1.32a\)M1 For a second correct equation and attempt to solve to find one unknown
Obtain \(a=-1.39\)A1 OE
Obtain \(\ln k=4.266...\) and \(k=71.2\)A1 SC1 for gradient \(=-1.39\) and no other relevant working
*Alternative method 2:*
AnswerMarks Guidance
AnswerMarks Guidance
\(e^{3.96}=k\times 0.22^a\) and \(e^{2.43}=k\times 1.32^a\)B1
Apply a correct method to obtain \(a\)M1
Obtain \(a=-1.39\)A1 OE
Substitute appropriate values into a correct equation to find \(k\)M1
Obtain \(k=71.2\)A1 AWRT 
**Question 3:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ln y = \ln k + a\ln x$ | B1 | SOI |
| Equate gradient of line to $a$ | M1 | |
| Obtain $a=-1.39$ | A1 | OE |
| Substitute appropriate values into a correct equation to find $\ln k$ | M1 | |
| Obtain $\ln k=4.266...$ and $k=71.2$ | A1 | SC1 for gradient $=-1.39$ and no other relevant working |
| **Total** | **5** | |

*Alternative method 1:*

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ln y = \ln k + a\ln x$ | B1 | SOI |
| $3.96=\ln k+0.22a$ | M1 | For one correct equation |
| $2.43=\ln k+1.32a$ | M1 | For a second correct equation and attempt to solve to find one unknown |
| Obtain $a=-1.39$ | A1 | OE |
| Obtain $\ln k=4.266...$ and $k=71.2$ | A1 | SC1 for gradient $=-1.39$ and no other relevant working |

*Alternative method 2:*

| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{3.96}=k\times 0.22^a$ and $e^{2.43}=k\times 1.32^a$ | B1 | |
| Apply a correct method to obtain $a$ | M1 | |
| Obtain $a=-1.39$ | A1 | OE |
| Substitute appropriate values into a correct equation to find $k$ | M1 | |
| Obtain $k=71.2$ | A1 | AWRT |
3\\
\includegraphics[max width=\textwidth, alt={}, center]{9c26457d-4b65-4cd4-a9b9-128aba92dbf4-04_586_734_260_701}

The variables $x$ and $y$ satisfy the equation $y = k x ^ { a }$, where $k$ and $a$ are constants. The graph of $\ln y$ against $\ln x$ is a straight line passing through the points ( $0.22,3.96$ ) and ( $1.32,2.43$ ), as shown in the diagram. Find the values of $k$ and $a$ correct to 3 significant figures.\\

\hfill \mbox{\textit{CAIE P2 2019 Q3 [5]}}
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