CAIE P2 2024 June — Question 3 8 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2024
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferentiating Transcendental Functions
TypeCalculate area under curve
DifficultyModerate -0.3 This is a straightforward application of standard techniques: differentiation of exponentials (part a), solving an exponential equation (part b), and integration of exponentials. All steps are routine A-level procedures with no novel insight required. The 'show that' in part (b) provides the answer, making it slightly easier than average, though the integration requires careful handling of limits.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.07j Differentiate exponentials: e^(kx) and a^(kx)1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08e Area between curve and x-axis: using definite integrals
\includegraphics{figure_3} The diagram shows the curve with equation \(y = 8e^{-x} - e^{2x}\). The curve crosses the y-axis at the point A and the x-axis at the point B. The shaded region is bounded by the curve and the two axes.
  1. Find the gradient of the curve at A. [3]
  2. Show that the x-coordinate of B is \(\ln 2\) and hence find the area of the shaded region. [5]
Question 3:
AnswerMarks Guidance
3(a)Differentiate to obtain form kex k e2x
1 2M1 Where k k 0, k 8 and k 1.
1 2 1 2
AnswerMarks
Obtain 8ex 2e2xA1
Substitute x0 to obtain –10A1
3
AnswerMarks Guidance
3(b)Attempt to find x-coordinate of B M1
Obtain e3x 8 and hence xln2A1 AG so necessary detail needed.
A0 if decimals used.
Integrate to obtain 8ex 1e2x
AnswerMarks Guidance
2B1
Use limits 0 and ln2 correctly to find areaM1 For integral of form k ex k e2x where k k 0.
3 4 3 4
k 8 and k 1.
1 2
Obtain 5
AnswerMarks Guidance
2A1 OE
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(a) ---
3(a) | Differentiate to obtain form kex k e2x
1 2 | M1 | Where k k 0, k 8 and k 1.
1 2 1 2
Obtain 8ex 2e2x | A1
Substitute x0 to obtain –10 | A1
3
--- 3(b) ---
3(b) | Attempt to find x-coordinate of B | M1 | 8ex e2x 0.
Obtain e3x 8 and hence xln2 | A1 | AG so necessary detail needed.
A0 if decimals used.
Integrate to obtain 8ex 1e2x
2 | B1
Use limits 0 and ln2 correctly to find area | M1 | For integral of form k ex k e2x where k k 0.
3 4 3 4
k 8 and k 1.
1 2
Obtain 5
2 | A1 | OE
5
Question | Answer | Marks | Guidance
\includegraphics{figure_3}

The diagram shows the curve with equation $y = 8e^{-x} - e^{2x}$. The curve crosses the y-axis at the point A and the x-axis at the point B. The shaded region is bounded by the curve and the two axes.

\begin{enumerate}[label=(\alph*)]
\item Find the gradient of the curve at A. [3]
\item Show that the x-coordinate of B is $\ln 2$ and hence find the area of the shaded region. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE P2 2024 Q3 [8]}}
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