Moderate -0.3 This is a standard logarithmic linearization problem requiring students to take ln of both sides, identify gradient and intercept from two points, then solve simultaneous equations. While it involves multiple steps (finding gradient, using point-slope, solving for two unknowns), these are routine A-level techniques with no novel insight required. Slightly easier than average due to being a textbook-style exponential modelling question.
2
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The variables \(x\) and \(y\) satisfy the equation \(y = A \mathrm { e } ^ { ( A - B ) x }\), where \(A\) and \(B\) are constants. The graph of \(\ln y\) against \(x\) is a straight line passing through the points \(( 0.4,3.6 )\) and \(( 2.9,14.1 )\), as shown in the diagram.
Find the values of \(A\) and \(B\) correct to 3 significant figures.
State or imply equation \(\ln y = \ln A + (A-B)x\)
B1
Allow inclusion of \(\ln e\)
Equate \(A-B\) to gradient of line
M1
Obtain \(A-B=4.2\)
A1
Substitute appropriate values to find value of \(\ln A\)
M1
Obtain \(\ln A = 1.92\) and hence \(A=6.82\) and \(B=2.62\)
A1
or greater accuracy
Alternative Method 1:
Answer
Marks
Guidance
Answer
Marks
Guidance
State or imply equation \(\ln y = \ln A + (A-B)x\)
B1
Allow inclusion of \(\ln e\)
Use of coordinates to obtain equation of line \(\frac{\ln y - 3.6}{14.1-3.6} = \frac{x-0.4}{2.9-0.4}\)
M1
condone use of \(y\) in place of \(\ln y\)
Obtain gradient equal to \(4.2\)
A1
Substitute appropriate values to find value of \(\ln A\)
M1
Obtain \(\ln A=1.92\) and hence \(A=6.82\) and \(B=2.62\)
A1
or greater accuracy
Alternative Method 2:
Answer
Marks
Guidance
Answer
Marks
Guidance
State or imply equation \(\ln y = \ln A + (A-B)x\)
B1
Allow inclusion of \(\ln e\)
\(3.6 = \ln A + 0.4(A-B)\)
M1
For one correct equation
\(14.1 = \ln A + 2.9(A-B)\)
A1
For both correct equations
Obtain \(A=6.82\) and \(B=2.62\)
M1A1
For attempt at solution by elimination of \(\ln A\) to obtain both values. SC B1 for \(y=4.2x+1.92\)
Total
5
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply equation $\ln y = \ln A + (A-B)x$ | B1 | Allow inclusion of $\ln e$ |
| Equate $A-B$ to gradient of line | M1 | |
| Obtain $A-B=4.2$ | A1 | |
| Substitute appropriate values to find value of $\ln A$ | M1 | |
| Obtain $\ln A = 1.92$ and hence $A=6.82$ and $B=2.62$ | A1 | or greater accuracy |
**Alternative Method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply equation $\ln y = \ln A + (A-B)x$ | B1 | Allow inclusion of $\ln e$ |
| Use of coordinates to obtain equation of line $\frac{\ln y - 3.6}{14.1-3.6} = \frac{x-0.4}{2.9-0.4}$ | M1 | condone use of $y$ in place of $\ln y$ |
| Obtain gradient equal to $4.2$ | A1 | |
| Substitute appropriate values to find value of $\ln A$ | M1 | |
| Obtain $\ln A=1.92$ and hence $A=6.82$ and $B=2.62$ | A1 | or greater accuracy |
**Alternative Method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply equation $\ln y = \ln A + (A-B)x$ | B1 | Allow inclusion of $\ln e$ |
| $3.6 = \ln A + 0.4(A-B)$ | M1 | For one correct equation |
| $14.1 = \ln A + 2.9(A-B)$ | A1 | For both correct equations |
| Obtain $A=6.82$ and $B=2.62$ | M1A1 | For attempt at solution by elimination of $\ln A$ to obtain both values. SC B1 for $y=4.2x+1.92$ |
| **Total** | **5** | |
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The variables $x$ and $y$ satisfy the equation $y = A \mathrm { e } ^ { ( A - B ) x }$, where $A$ and $B$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points $( 0.4,3.6 )$ and $( 2.9,14.1 )$, as shown in the diagram.
Find the values of $A$ and $B$ correct to 3 significant figures.\\
\hfill \mbox{\textit{CAIE P2 2023 Q2 [5]}}