CAIE P1 2019 March — Question 2 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2019
SessionMarch
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind tangent given derivative expression
DifficultyModerate -0.3 This is a straightforward integration problem requiring students to integrate f'(x) to find f(x), apply the constant of integration using one point, then use the second point to find k. It involves standard techniques (integrating polynomials) with no conceptual challenges, making it slightly easier than average but still requiring multiple steps for the 5 marks.
Spec1.08a Fundamental theorem of calculus: integration as reverse of differentiation

A curve with equation \(y = f(x)\) passes through the points \((0, 2)\) and \((3, -1)\). It is given that \(f'(x) = kx^2 - 2x\), where \(k\) is a constant. Find the value of \(k\). [5]

Question 2:
AnswerMarks Guidance
2y= 1 kx3 −x2 (+c )
3M1A1 Attempt integration for M mark
Sub (0, 2)DM1 Dep on c present. Expect c = 2
Sub (3, ‒1) → −1=9k−9+theircDM1
k = 2/3A1
5
AnswerMarks Guidance
QuestionAnswer Marks
Question 2:
2 | y= 1 kx3 −x2 (+c )
3 | M1A1 | Attempt integration for M mark
Sub (0, 2) | DM1 | Dep on c present. Expect c = 2
Sub (3, ‒1) → −1=9k−9+theirc | DM1
k = 2/3 | A1
5
Question | Answer | Marks | Guidance
A curve with equation $y = f(x)$ passes through the points $(0, 2)$ and $(3, -1)$. It is given that $f'(x) = kx^2 - 2x$, where $k$ is a constant. Find the value of $k$. [5]

\hfill \mbox{\textit{CAIE P1 2019 Q2 [5]}}