CAIE P3 2023 November — Question 3 4 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2023
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
Typeln(y) vs x: find constants from two points
DifficultyModerate -0.5 This is a standard textbook exercise on exponential models and logarithmic linearization. Students take ln of both sides to get ln(y) = ln(a) + x·ln(b), recognize this as a linear relationship, find the gradient to get ln(b), then use a point to find ln(a). It requires multiple steps but follows a well-practiced procedure with no novel insight needed—slightly easier than average due to its routine nature.
Spec1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form

3 \includegraphics[max width=\textwidth, alt={}, center]{ce3c4a9c-bf83-4d28-96e2-ef31c3673dea-04_860_451_264_833} The variables \(x\) and \(y\) are related by the equation \(y = a b ^ { x }\), where \(a\) and \(b\) are constants. The diagram shows the result of plotting \(\ln y\) against \(x\) for two pairs of values of \(x\) and \(y\). The coordinates of these points are \(( 1,3.7 )\) and \(( 2.2,6.46 )\). Use this information to find the values of \(a\) and \(b\).

Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
State or imply that \(\ln y = \ln a + x \ln b\)B1
Carry out a completely correct method for finding \(\ln a\) or \(\ln b\)M1 \(3.7 = \ln a + \ln b\) and \(6.46 = \ln a + 2.2\ln b\) leading to \(\ln a = 1.4\), \(\ln b = 2.3\)
Obtain value \(a = 4.06\)A1
Obtain value \(b = 9.97\)A1 SC B1 for \(a = e^{1.4}\) and \(b = e^{2.3}\)
Alternative Method:
AnswerMarks Guidance
AnswerMarks Guidance
\(e^{3.7} = ab^1\) and \(e^{6.46} = ab^{2.2}\)B1
Divide to obtain \(e^{2.76} = b^{1.2}\) and state or imply \(2.76 = 1.2\ln b\)M1
Obtain value \(a = 4.06\)A1
Obtain value \(b = 9.97\)A1
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply that $\ln y = \ln a + x \ln b$ | B1 | |
| Carry out a completely correct method for finding $\ln a$ or $\ln b$ | M1 | $3.7 = \ln a + \ln b$ and $6.46 = \ln a + 2.2\ln b$ leading to $\ln a = 1.4$, $\ln b = 2.3$ |
| Obtain value $a = 4.06$ | A1 | |
| Obtain value $b = 9.97$ | A1 | SC B1 for $a = e^{1.4}$ and $b = e^{2.3}$ |

**Alternative Method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{3.7} = ab^1$ and $e^{6.46} = ab^{2.2}$ | B1 | |
| Divide to obtain $e^{2.76} = b^{1.2}$ and state or imply $2.76 = 1.2\ln b$ | M1 | |
| Obtain value $a = 4.06$ | A1 | |
| Obtain value $b = 9.97$ | A1 | |

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3\\
\includegraphics[max width=\textwidth, alt={}, center]{ce3c4a9c-bf83-4d28-96e2-ef31c3673dea-04_860_451_264_833}

The variables $x$ and $y$ are related by the equation $y = a b ^ { x }$, where $a$ and $b$ are constants. The diagram shows the result of plotting $\ln y$ against $x$ for two pairs of values of $x$ and $y$. The coordinates of these points are $( 1,3.7 )$ and $( 2.2,6.46 )$.

Use this information to find the values of $a$ and $b$.\\

\hfill \mbox{\textit{CAIE P3 2023 Q3 [4]}}