Moderate -0.3 This is a standard logarithmic linearization problem requiring students to take ln of both sides to get ln(y) = ln(A) + (A-B)x, then use two points to find gradient and intercept. It involves routine algebraic manipulation and simultaneous equations, slightly easier than average due to its mechanical nature with no conceptual surprises.
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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 \(B = 2.62\)
M1 A1
For attempt at solution by elimination of \(\ln A\) to obtain both values. SC B1 for \(y = 4.2x + 1.92\)
## 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 $B = 2.62$ | M1 A1 | For attempt at solution by elimination of $\ln A$ to obtain both values. SC B1 for $y = 4.2x + 1.92$ |
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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]}}