CAIE P2 2022 June — Question 1 4 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2022
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
Typeln(y) vs x: find constants from two points
DifficultyModerate -0.3 This is a straightforward logarithmic transformation question requiring students to take ln of both sides of an exponential equation and identify gradient/intercept. Part (a) is routine algebraic manipulation (ln 4^(2x-a) = (2x-a)ln 4 = 2ln 4 · x - a ln 4, so gradient is 2ln 4 = ln 16). Part (b) requires using the y-intercept to find a. While it involves multiple steps, each step follows standard A-level techniques with no novel problem-solving required, making it slightly easier than average.
Spec1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form

1 \includegraphics[max width=\textwidth, alt={}, center]{ed12a4fb-e3bf-4d00-ad09-9ba5be941dd5-02_654_396_258_872} The variables \(x\) and \(y\) satisfy the equation \(y = 4 ^ { 2 x - a }\), where \(a\) is an integer. As shown in the diagram, the graph of \(\ln y\) against \(x\) is a straight line passing through the point \(( 0 , - 20.8 )\), where the second coordinate is given correct to 3 significant figures.
  1. Show that the gradient of the straight line is \(\ln 16\).
  2. Determine the value of \(a\).

Question 1:
Part (a)
AnswerMarks Guidance
AnswerMarks Guidance
State or imply equation is \(\ln y = (2x - a)\ln 4\)B1 OE. Do not condone poor use of brackets
State gradient is \(2\ln 4\) and confirm \(\ln 16\)B1 AG – necessary detail needed
Total2
Part (b)
AnswerMarks Guidance
AnswerMarks Guidance
Substitute for \(\ln y\) and attempt value of \(a\)M1 Allow if \(\ln y = 2x - a\ln 4\)
Obtain \(a = 15\)A1 Integer answer required, but condone 15.0
Total2
## Question 1:

### Part (a)

| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply equation is $\ln y = (2x - a)\ln 4$ | B1 | OE. Do not condone poor use of brackets |
| State gradient is $2\ln 4$ and confirm $\ln 16$ | B1 | AG – necessary detail needed |
| **Total** | **2** | |

### Part (b)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute for $\ln y$ and attempt value of $a$ | M1 | Allow if $\ln y = 2x - a\ln 4$ |
| Obtain $a = 15$ | A1 | Integer answer required, but condone 15.0 |
| **Total** | **2** | |

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1\\
\includegraphics[max width=\textwidth, alt={}, center]{ed12a4fb-e3bf-4d00-ad09-9ba5be941dd5-02_654_396_258_872}

The variables $x$ and $y$ satisfy the equation $y = 4 ^ { 2 x - a }$, where $a$ is an integer. As shown in the diagram, the graph of $\ln y$ against $x$ is a straight line passing through the point $( 0 , - 20.8 )$, where the second coordinate is given correct to 3 significant figures.
\begin{enumerate}[label=(\alph*)]
\item Show that the gradient of the straight line is $\ln 16$.
\item Determine the value of $a$.
\end{enumerate}

\hfill \mbox{\textit{CAIE P2 2022 Q1 [4]}}