| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | y vs ln(x) linear graph |
| Difficulty | Moderate -0.3 This is a straightforward logarithmic transformation question requiring students to rearrange x = A(3^(-y)) into the form y = m·ln(x) + c, then use a given point to find A. The algebraic manipulation is routine (take ln of both sides, use log laws), and part (b) is simple substitution. Slightly easier than average as it's a standard textbook exercise with no problem-solving insight required. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(\ln x = \ln A - y\ln 3\) | B1 | \(\left(y = -\frac{1}{\ln 3}\ln x + \frac{\ln A}{\ln 3}\right)\) |
| State that the graph of \(y\) against \(\ln x\) has an equation that is *linear* in \(y\) and \(\ln x\), or has an equation of the standard form \(y = mx + c\) and is thus a straight line | B1 | Must be a correct statement. Accept if the 2 equations are written side by side with no comment. An equation with \(y\ln 3\) should be compared with the form \(py + q\ln x = c\) |
| State that the gradient is \(-\frac{1}{\ln 3}\) | B1 | OE. Exact answer required. ISW after a correct statement. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(\ln x = 0\), \(y = 1.3\) and use correct method to solve for \(A\) | M1 | \((\ln A = 1.3\ln 3)\). Follow *their* equation in \(y\) and \(\ln x\). Must be substituting \(\ln x = 0\), not \(x = 0\). \(\ln 0\) 'used' in the solution scores M0A0. |
| Obtain answer \(A = 4.17\) only | A1 | Must be 2 d.p. as specified in question |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $\ln x = \ln A - y\ln 3$ | B1 | $\left(y = -\frac{1}{\ln 3}\ln x + \frac{\ln A}{\ln 3}\right)$ |
| State that the graph of $y$ against $\ln x$ has an equation that is *linear* in $y$ and $\ln x$, or has an equation of the standard form $y = mx + c$ and is thus a straight line | B1 | Must be a correct statement. Accept if the 2 equations are written side by side with no comment. An equation with $y\ln 3$ should be compared with the form $py + q\ln x = c$ |
| State that the gradient is $-\frac{1}{\ln 3}$ | B1 | OE. Exact answer required. ISW after a correct statement. |
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## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $\ln x = 0$, $y = 1.3$ and use correct method to solve for $A$ | M1 | $(\ln A = 1.3\ln 3)$. Follow *their* equation in $y$ and $\ln x$. Must be substituting $\ln x = 0$, not $x = 0$. $\ln 0$ 'used' in the solution scores M0A0. |
| Obtain answer $A = 4.17$ only | A1 | Must be 2 d.p. as specified in question |
---3 The variables $x$ and $y$ satisfy the equation $x = A \left( 3 ^ { - y } \right)$, where $A$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Explain why the graph of $y$ against $\ln x$ is a straight line and state the exact value of the gradient of the line.\\
It is given that the line intersects the $y$-axis at the point where $y = 1.3$.
\item Calculate the value of $A$, giving your answer correct to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2021 Q3 [5]}}