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First page of ZIMSEC A-Level Pure Mathematics Paper 1 November 2017 Marking Scheme

ZIMSEC A-Level Pure Mathematics Paper 1 November 2017 Marking Scheme

Use this ZIMSEC A-Level Pure Mathematics Paper 1 November 2017 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.

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Questions: ZIMSEC A-Level Pure Mathematics Paper 1 November 2017 Question Paper

Exam board
ZIMSEC
Subject
Pure Mathematics
Level
A-Level
Document
Marking Scheme
Paper
Paper 1
Session
November
Year
2017
File
PDF, 4.4 MB

About this marking scheme

This is the ZIMSEC A-Level Pure Mathematics marking scheme for Paper 1 from November 2017. It shows how marks are awarded, so you can mark your own answers to the matching question paper. See the question paper.

More Pure Mathematics A-Level 2017 papers

Other years: Pure Mathematics A-Level

Text from the first page

$1 \omega=1+2 i$ i $\mu=3-i$
(a)
\[
\text { (1) } \begin{aligned}
\mu+\omega  |  =3-i+1+2 i   |  =4+i
\end{aligned}
\]
(II)
\[
\begin{aligned}
\mu w  |  =(3-i)(1+2 i)   |  =3+6 i-i-2 i^{2}   |  =5+5 i
\end{aligned}
\]
(b) $\arg (\mu \omega)=\tan ^{-1}(5 / 5)=45^{\circ}$
2.
\[
\begin{array}{ll}
f(x)=3 x-1 ;  |  x \in R
h(x)=2 x+5 ;  |  x \in R .
\end{array}
\]
\[
\begin{array}{rl}
h f(x)  |  =h(3 x-1)   |  =2(3 x-1)+5   |  =6 x+3
f h(x)  |  =f(2 x+5)   |  =3(2 x+5)-1   |  =6 x+14
f h(x)=2  |  h f(x)
\Rightarrow 6 x+14  |  =2(6 x+3)
6 x+14  |  =12 x+6
8  |  =6 x
x  |  =\frac{4}{3}
\end{array}
\]

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