OCR Further Pure Core 2 2019 June — Question 3 5 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2019
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine intersection with line
DifficultyStandard +0.3 This is a standard Further Maths line intersection problem requiring students to equate parametric forms and solve simultaneous equations. While it involves more algebraic manipulation than a typical A-level question and is from Further Maths content, it follows a well-practiced procedure without requiring geometric insight or novel problem-solving approaches.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms

3
1 \end{array} \right) + \lambda \left( \begin{array} { r } - 2
4
- 2 \end{array} \right)
& l _ { 2 } : \mathbf { r } = \left( \begin{array} { l }

Question 3:
AnswerMarks
3DR 1
3 1
− −
∫(x−1) 2dx=−2(x−1) 2(+c)
N
3  1
N − −
∫ (x−1) 2dx=−2(x−1) 2
5  
5
2 2
− +
N −1 5−1
1
lim =0oe
N→∞ N −1
∞ − 3  2 2 
∫ (x−1) 2dx= lim− + =1
AnswerMarks
5 N→∞ N −1 5−1B1
M1
A1
B1
A1
AnswerMarks
[5]1.1a
2.1
1.1
2.1
AnswerMarks
2.2aConsideration of a finite upper
limit
1
Not just eg =0
AG. Convincing argument
equating improper integral to
AnswerMarks
solutionCan be seen as part of limit of
both terms, but must be explicitly
shown as zero
Calculates cross product of ‘their
r’ and direction vector.
Allow if full method seen or if 2
terms correct
Question 3:
3 | DR 1
3 1
− −
∫(x−1) 2dx=−2(x−1) 2(+c)
N
3  1
N − −
∫ (x−1) 2dx=−2(x−1) 2
5  
5
2 2
− +
N −1 5−1
1
lim =0oe
N→∞ N −1
∞ − 3  2 2 
∫ (x−1) 2dx= lim− + =1
5 N→∞ N −1 5−1 | B1
M1
A1
B1
A1
[5] | 1.1a
2.1
1.1
2.1
2.2a | Consideration of a finite upper
limit
1
Not just eg =0
∞
AG. Convincing argument
equating improper integral to
solution | Can be seen as part of limit of
both terms, but must be explicitly
shown as zero
Calculates cross product of ‘their
r’ and direction vector.
Allow if full method seen or if 2
terms correct
3 \\
1
\end{array} \right) + \lambda \left( \begin{array} { r } 
- 2 \\
4 \\
- 2
\end{array} \right) \\
& l _ { 2 } : \mathbf { r } = \left( \begin{array} { l }

\hfill \mbox{\textit{OCR Further Pure Core 2 2019 Q3 [5]}}