CAIE P2 2024 November — Question 6 7 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2024
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric differentiation
TypeParametric curve crosses axis, find gradient there
DifficultyStandard +0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = dy/dt ÷ dx/dt) with exponential functions. Part (a) involves routine differentiation and algebraic simplification. Part (b) requires finding t when x=0 (which gives e^{2t}=2, so t=ln√2) and substituting. While it involves multiple steps and some algebraic manipulation, it follows a standard template with no novel problem-solving required, making it slightly easier than average.
Spec1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07s Parametric and implicit differentiation

A curve has parametric equations $$x = \frac{e^{2t} - 2}{e^{2t} + 1}, \quad y = e^{3t} + 1.$$
  1. Find an expression for \(\frac{dy}{dx}\) in terms of \(t\). [4]
  2. Find the exact gradient of the curve at the point where the curve crosses the \(y\)-axis. [3]

Question 6:

AnswerMarks Guidance
6(a)Differentiate x using quotient rule or correct equivalent *M1
(e2t +1)2e2t −(e2t −2)2e2t
Obtain or equivalent
AnswerMarks
(e2t +1)2A1
dy
Attempt expression for in terms of t
AnswerMarks
dxDM1
Obtain 1et(e2t +1)2 or (unsimplified) equivalent
AnswerMarks Guidance
2A1 No fractions within fractions.
(e2t +1)2e2t −(e2t −2)2e2t
Attempt to simplify
must be seen.
4

AnswerMarks
6(b)Identify t = 1ln2 at point where curve crosses y-axis
2B1
dy
Substitute non-zero value of t in their expression for and attempt simplification
AnswerMarks
dxM1
Obtain 9 2 or exact equivalent
AnswerMarks
2A1
3
AnswerMarks Guidance
QuestionAnswer Marks
Question 6:
--- 6(a) ---
6(a) | Differentiate x using quotient rule or correct equivalent | *M1
(e2t +1)2e2t −(e2t −2)2e2t
Obtain or equivalent
(e2t +1)2 | A1
dy
Attempt expression for in terms of t
dx | DM1
Obtain 1et(e2t +1)2 or (unsimplified) equivalent
2 | A1 | No fractions within fractions.
(e2t +1)2e2t −(e2t −2)2e2t
Attempt to simplify
must be seen.
4
--- 6(b) ---
6(b) | Identify t = 1ln2 at point where curve crosses y-axis
2 | B1
dy
Substitute non-zero value of t in their expression for and attempt simplification
dx | M1
Obtain 9 2 or exact equivalent
2 | A1
3
Question | Answer | Marks | Guidance
A curve has parametric equations
$$x = \frac{e^{2t} - 2}{e^{2t} + 1}, \quad y = e^{3t} + 1.$$

\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\frac{dy}{dx}$ in terms of $t$. [4]

\item Find the exact gradient of the curve at the point where the curve crosses the $y$-axis. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P2 2024 Q6 [7]}}