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First page of ZIMSEC O-Level Mathematics Paper 1 June 2005 Marking Scheme

ZIMSEC O-Level Mathematics Paper 1 June 2005 Marking Scheme

Use this ZIMSEC O-Level Mathematics Paper 1 June 2005 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.

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Questions: ZIMSEC O-Level Mathematics Paper 1 June 2005 Question Paper

Exam board
ZIMSEC
Subject
Mathematics
Level
O-Level
Document
Marking Scheme
Paper
Paper 1
Session
June
Year
2005
File
PDF, 7.9 MB

About this marking scheme

This is the ZIMSEC O-Level Mathematics marking scheme for Paper 1 from June 2005. It shows how marks are awarded, so you can mark your own answers to the matching question paper. See the question paper.

More Mathematics O-Level 2005 papers

Other years: Mathematics O-Level

Text from the first page

- Shadrech Blessing JUNE 2005
1)
\[
\begin{array}{l}
A\left(m^{2} ; 2 m\right)
B\left(3 m^{2} ; 6 m\right)
\end{array}
\]

Nkhoma
\[
\text { Midpoint of } \begin{aligned}
A B  |  =\left(\frac{m^{2}+3 m^{2}}{2} ; \frac{2 m+6 m}{2}\right)   |  =\left(2 m^{2} ; 4 m\right)
\end{aligned}
\]
\[
y=2 x-8 .
\]
when $y=0$
\[
\begin{array}{l}
0=2 x-8
8=2 x
x=4
\therefore \frac{4 m-0}{2 m^{2}-4}=2
4 m=4 m^{2}-8
4 m^{2}-4 m-8=0
4 m^{2}-8 m+4 m-8=0
4 m(m-2)+4(m-2)=0
(4 m+4)(m-2)=0
\end{array}
\]
$4 m+4=0 \quad$ or $\quad m-2=0$
$m=-1 \quad m=2$.
\[
\therefore m=-1 \text { or } 2 \text {. }
\]

$(3)$