CAIE P2 2024 November — Question 1 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2024
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLaws of Logarithms
TypeLinear relationship between log variables
DifficultyModerate -0.3 Part (a) is a straightforward application of logarithms to linearize an exponential equation (take ln of both sides, rearrange to y = mx + c form), requiring only routine manipulation. Part (b) involves calculating gradient from two points and solving simultaneous equations, which are standard techniques. The question requires multiple steps but no novel insight or challenging problem-solving.
Spec1.02c Simultaneous equations: two variables by elimination and substitution1.06c Logarithm definition: log_a(x) as inverse of a^x1.06f Laws of logarithms: addition, subtraction, power rules

The variables \(x\) and \(y\) satisfy the equation \(a^{2y} = e^{3x+k}\), where \(a\) and \(k\) are constants. The graph of \(y\) against \(x\) is a straight line.
  1. Use logarithms to show that the gradient of the straight line is \(\frac{3}{2\ln a}\). [1]
  2. Given that the straight line passes through the points \((0.4, 0.95)\) and \((3.3, 3.80)\), find the values of \(a\) and \(k\). [4]

Question 1:

AnswerMarks
1(a)3
State or imply 2ylna=3x+k and conclude that gradient is
AnswerMarks Guidance
2lnaB1 AG – necessary detail needed.
1

AnswerMarks
1(b)3
Equate to gradient of line
AnswerMarks
2lnaM1
3 2.85 29
Obtain = or equivalent and hence obtain a=4.6or a=e19
AnswerMarks Guidance
2lna 2.9A1 Allow greater accuracy.
Substitute appropriate values to find value of kM1
Obtain k=1.7A1
Alternative Method for Question 1(b)
0.95(2lna)=3(0.4)+k
Obtain
AnswerMarks Guidance
or a1.9 =e1.2+kM1 OE
3.80(2lna)=3(3.3)+k
Obtain
AnswerMarks Guidance
or a7.6 =e9.9+kM1 OE
29
AnswerMarks Guidance
Obtain a=4.6or a=e19A1 Allow greater accuracy.
Obtain k=1.7A1
4
AnswerMarks Guidance
QuestionAnswer Marks
Question 1:
--- 1(a) ---
1(a) | 3
State or imply 2ylna=3x+k and conclude that gradient is
2lna | B1 | AG – necessary detail needed.
1
--- 1(b) ---
1(b) | 3
Equate to gradient of line
2lna | M1
3 2.85 29
Obtain = or equivalent and hence obtain a=4.6or a=e19
2lna 2.9 | A1 | Allow greater accuracy.
Substitute appropriate values to find value of k | M1
Obtain k=1.7 | A1
Alternative Method for Question 1(b)
0.95(2lna)=3(0.4)+k
Obtain
or a1.9 =e1.2+k | M1 | OE
3.80(2lna)=3(3.3)+k
Obtain
or a7.6 =e9.9+k | M1 | OE
29
Obtain a=4.6or a=e19 | A1 | Allow greater accuracy.
Obtain k=1.7 | A1
4
Question | Answer | Marks | Guidance
The variables $x$ and $y$ satisfy the equation $a^{2y} = e^{3x+k}$, where $a$ and $k$ are constants. The graph of $y$ against $x$ is a straight line.

\begin{enumerate}[label=(\alph*)]
\item Use logarithms to show that the gradient of the straight line is $\frac{3}{2\ln a}$. [1]

\item Given that the straight line passes through the points $(0.4, 0.95)$ and $(3.3, 3.80)$, find the values of $a$ and $k$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P2 2024 Q1 [5]}}