CAIE P2 2017 June — Question 5 6 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2017
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
Typeln(y) vs x: find constants from two points
DifficultyModerate -0.3 This is a standard logarithmic transformation question requiring students to convert the exponential equation to linear form (ln y = ln K - 2x ln a), find the gradient and intercept from two points, then solve for K and a. While it involves multiple steps (finding gradient, using point-slope, exponentiating), these are routine techniques practiced extensively in P2 with no novel problem-solving required.
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]{de2f8bf3-fd03-4199-9eb2-c9cbac4d4385-05_551_535_260_806} The variables \(x\) and \(y\) satisfy the equation \(y = \frac { K } { a ^ { 2 x } }\), where \(K\) and \(a\) are constants. The graph of \(\ln y\) against \(x\) is a straight line passing through the points \(( 0.6,1.81 )\) and \(( 1.4,1.39 )\), as shown in the diagram. Find the values of \(K\) and \(a\) correct to 2 significant figures.

Question 5:
AnswerMarks Guidance
AnswerMark Guidance
State or imply \(\ln y = \ln K - 2x\ln a\)B1
EITHER:
Obtain \(-0.525\) as gradient of line(M1
Equate their \(-2\ln a\) to their gradient and solve for \(a\)M1 Allow \(2\ln a =\) their gradient for M1
Obtain \(a = 1.3\)A1
Substitute to find value of \(K\)M1
Obtain \(K = 8.4\)A1)
OR:
Obtain two equations using coordinates correctly(M1
Solve these equations to obtain \(2\ln a\) or equivalentM1
Obtain \(a = 1.3\)A1
Substitute to find value of \(K\)M1
Obtain \(K = 8.4\)A1)
Total:6
## Question 5:

| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $\ln y = \ln K - 2x\ln a$ | B1 | |
| **EITHER:** | | |
| Obtain $-0.525$ as gradient of line | (M1 | |
| Equate their $-2\ln a$ to their gradient and solve for $a$ | M1 | Allow $2\ln a =$ their gradient for **M1** |
| Obtain $a = 1.3$ | A1 | |
| Substitute to find value of $K$ | M1 | |
| Obtain $K = 8.4$ | A1) | |
| **OR:** | | |
| Obtain two equations using coordinates correctly | (M1 | |
| Solve these equations to obtain $2\ln a$ or equivalent | M1 | |
| Obtain $a = 1.3$ | A1 | |
| Substitute to find value of $K$ | M1 | |
| Obtain $K = 8.4$ | A1) | |
| **Total:** | **6** | |
5\\
\includegraphics[max width=\textwidth, alt={}, center]{de2f8bf3-fd03-4199-9eb2-c9cbac4d4385-05_551_535_260_806}

The variables $x$ and $y$ satisfy the equation $y = \frac { K } { a ^ { 2 x } }$, where $K$ and $a$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points $( 0.6,1.81 )$ and $( 1.4,1.39 )$, as shown in the diagram. Find the values of $K$ and $a$ correct to 2 significant figures.\\

\hfill \mbox{\textit{CAIE P2 2017 Q5 [6]}}