CAIE P2 2020 June — Question 4 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2020
SessionJune
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 take logs of both sides to get ln y = ln A - 2p ln x, recognize this as a straight line with gradient -2p and intercept ln A, then use two points to find the gradient and substitute to find the intercept. While it requires multiple steps and understanding of the log-linear transformation technique, it's a routine textbook exercise with no novel problem-solving required.
Spec1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form
4 \includegraphics[max width=\textwidth, alt={}, center]{ad833f8c-80de-42ae-a186-93091a6fdf1e-06_659_828_262_660} The variables \(x\) and \(y\) satisfy the equation \(y = A x ^ { - 2 p }\), where \(A\) and \(p\) are constants. The graph of \(\ln y\) against \(\ln x\) is a straight line passing through the points \(( - 0.68,3.02 )\) and \(( 1.07 , - 1.53 )\), as shown in the diagram. Find the values of \(A\) and \(p\).
Question 4:
AnswerMarks Guidance
AnswerMark Guidance
State or imply equation is \(\ln y = \ln A - 2p\ln x\)B1
Equate gradient of line to \(-2p\)M1
Obtain \(-2p = -2.6\) and hence \(p = 1.3\)A1
Substitute appropriate values to find \(\ln A\)M1
Obtain \(\ln A = 1.252\) and hence \(A = 3.5\)A1
Alternative method:
AnswerMarks Guidance
AnswerMark Guidance
State or imply equation is \(\ln y = \ln A - 2p\ln x\)B1
Substitute given coordinates to obtain 2 simultaneous equations and solve to obtain \(3.5p\)M1
Obtain \(3.5p = 4.55\) and hence \(p = 1.3\)A1
Substitute appropriate values to find \(\ln A\)M1
Obtain \(\ln A = 1.252\) and hence \(A = 3.5\)A1 
## Question 4:

| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply equation is $\ln y = \ln A - 2p\ln x$ | B1 | |
| Equate gradient of line to $-2p$ | M1 | |
| Obtain $-2p = -2.6$ and hence $p = 1.3$ | A1 | |
| Substitute appropriate values to find $\ln A$ | M1 | |
| Obtain $\ln A = 1.252$ and hence $A = 3.5$ | A1 | |

**Alternative method:**

| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply equation is $\ln y = \ln A - 2p\ln x$ | B1 | |
| Substitute given coordinates to obtain 2 simultaneous equations and solve to obtain $3.5p$ | M1 | |
| Obtain $3.5p = 4.55$ and hence $p = 1.3$ | A1 | |
| Substitute appropriate values to find $\ln A$ | M1 | |
| Obtain $\ln A = 1.252$ and hence $A = 3.5$ | A1 | |

---
4\\
\includegraphics[max width=\textwidth, alt={}, center]{ad833f8c-80de-42ae-a186-93091a6fdf1e-06_659_828_262_660}

The variables $x$ and $y$ satisfy the equation $y = A x ^ { - 2 p }$, where $A$ and $p$ are constants. The graph of $\ln y$ against $\ln x$ is a straight line passing through the points $( - 0.68,3.02 )$ and $( 1.07 , - 1.53 )$, as shown in the diagram.

Find the values of $A$ and $p$.\\

\hfill \mbox{\textit{CAIE P2 2020 Q4 [5]}}
This paper (1 questions)
View full paper
Q4 5