CAIE P2 2024 March — Question 6 9 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2024
SessionMarch
Marks9
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TopicParametric differentiation
DifficultyStandard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: finding dy/dx using the chain rule, solving ln t = 3 to find point B, and setting dy/dx = 0 to find the minimum. All steps are routine applications of AS/A2 methods with no novel problem-solving required, making it slightly easier than average.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.06d Natural logarithm: ln(x) function and properties1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation
\includegraphics{figure_6} The diagram shows the curve with parametric equations $$x = 1 + \sqrt{t}, \quad y = (\ln t + 2)(\ln t - 3),$$ for \(0 < t < 25\). The curve crosses the \(x\)-axis at the points \(A\) and \(B\) and has a minimum point \(M\).
  1. Show that \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4\ln t - 2}{\sqrt{t}}\). [4]
  2. Find the exact gradient of the curve at \(B\). [2]
  3. Find the exact coordinates of \(M\). [3]
Question 6:
AnswerMarks
6(a)dy
Use product rule to find
AnswerMarks Guidance
dtM1 Or chain rule if expression expanded before
differentiation.
dy 2lnt 1
Obtain = −
AnswerMarks
dt t tA1
dy dx
Divide attempt at by attempt at
AnswerMarks
dt dtM1
dy 4lnt−2
Confirm =
AnswerMarks Guidance
dx tA1 Answer given – necessary detail needed.
4
AnswerMarks Guidance
6(b)State or imply t =e3 at point B B1
−3
AnswerMarks
Obtain gradient 10e 2 or exact equivalentB1
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
6(c)Equate first derivative to zero and obtain value for lnt or t M1
1
AnswerMarks Guidance
Obtain x-coordinate 1+e4A1 Or exact equivalent.
Obtain y-coordinate −25
AnswerMarks Guidance
4A1 Or equivalent.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(a) ---
6(a) | dy
Use product rule to find
dt | M1 | Or chain rule if expression expanded before
differentiation.
dy 2lnt 1
Obtain = −
dt t t | A1
dy dx
Divide attempt at by attempt at
dt dt | M1
dy 4lnt−2
Confirm =
dx t | A1 | Answer given – necessary detail needed.
4
--- 6(b) ---
6(b) | State or imply t =e3 at point B | B1
−3
Obtain gradient 10e 2 or exact equivalent | B1
2
Question | Answer | Marks | Guidance
--- 6(c) ---
6(c) | Equate first derivative to zero and obtain value for lnt or t | M1
1
Obtain x-coordinate 1+e4 | A1 | Or exact equivalent.
Obtain y-coordinate −25
4 | A1 | Or equivalent.
3
Question | Answer | Marks | Guidance
\includegraphics{figure_6}

The diagram shows the curve with parametric equations
$$x = 1 + \sqrt{t}, \quad y = (\ln t + 2)(\ln t - 3),$$
for $0 < t < 25$. The curve crosses the $x$-axis at the points $A$ and $B$ and has a minimum point $M$.

\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4\ln t - 2}{\sqrt{t}}$. [4]
\item Find the exact gradient of the curve at $B$. [2]
\item Find the exact coordinates of $M$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P2 2024 Q6 [9]}}
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