CAIE P3 2020 June — Question 2 5 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2020
SessionJune
Marks5
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Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
Typeln(y) vs x: find constants from two points
DifficultyModerate -0.3 This is a standard logarithmic linearization problem requiring students to recognize that ln(y) = (1/2)ln(A) + (k/2)x, find the gradient and intercept from two points, then solve for A and k. It's slightly easier than average because the method is routine and well-practiced in P3, though it requires careful algebraic manipulation of the given form.
Spec1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form
2 \includegraphics[max width=\textwidth, alt={}, center]{3149080d-ad1a-4d2e-8e20-eb9977ced619-03_515_901_260_623} The variables \(x\) and \(y\) satisfy the equation \(y ^ { 2 } = A \mathrm { e } ^ { k x }\), where \(A\) and \(k\) are constants. The graph of \(\ln y\) against \(x\) is a straight line passing through the points (1.5, 1.2) and (5.24, 2.7) as shown in the diagram. Find the values of \(A\) and \(k\) correct to 2 decimal places.
Question 2:
Main Method:
AnswerMarks Guidance
AnswerMark Guidance
State or imply \(2\ln y = \ln A + kx\)B1
Substitute values of \(\ln y\) and \(x\), or equate gradient of line to \(k\), and solve for \(k\)M1
Obtain \(k = 0.80\)A1
Solve for \(\ln A\)M1
Obtain \(A = 3.31\)A1
Alternative Method:
AnswerMarks Guidance
AnswerMark Guidance
Obtain two correct equations in \(y\) and \(x\) by substituting \(y\)- and \(x\)-values in the given equationB1
Solve for \(k\)M1
Obtain \(k = 0.80\)A1
Solve for \(A\)M1
Obtain \(A = 3.31\)A1
5 
## Question 2:

**Main Method:**

| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $2\ln y = \ln A + kx$ | B1 | |
| Substitute values of $\ln y$ and $x$, or equate gradient of line to $k$, and solve for $k$ | M1 | |
| Obtain $k = 0.80$ | A1 | |
| Solve for $\ln A$ | M1 | |
| Obtain $A = 3.31$ | A1 | |

**Alternative Method:**

| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain two correct equations in $y$ and $x$ by substituting $y$- and $x$-values in the given equation | B1 | |
| Solve for $k$ | M1 | |
| Obtain $k = 0.80$ | A1 | |
| Solve for $A$ | M1 | |
| Obtain $A = 3.31$ | A1 | |
| | **5** | |
2\\
\includegraphics[max width=\textwidth, alt={}, center]{3149080d-ad1a-4d2e-8e20-eb9977ced619-03_515_901_260_623}

The variables $x$ and $y$ satisfy the equation $y ^ { 2 } = A \mathrm { e } ^ { k x }$, where $A$ and $k$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points (1.5, 1.2) and (5.24, 2.7) as shown in the diagram.

Find the values of $A$ and $k$ correct to 2 decimal places.\\

\hfill \mbox{\textit{CAIE P3 2020 Q2 [5]}}
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