November 2004 ZIMSEC Syllabus 9164 A-Level Mathematics Paper 1 Working and Answers Marking Scheme

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This is the ZIMSEC A-Level Mathematics marking scheme for Paper 1 from November 2004. It shows how marks are awarded, so you can mark your own answers to the matching question paper. See the question paper.

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Text from the first page

November 2004 ZIMSEC Syllabus 9164 A-Level Mathematics Paper 1 WORKING AND ANSWERS
(1) Let $f(x)=x^{3}+a x^{2}+b x-9$
if $(x-1)$ is a factor of $f(x)$ then $f(1)=0$
\[
\begin{array}{c}
f(1)=(1)^{3}+a(1)^{2}+b(1)-9=0
a+b=8
\end{array}
\]
if $(x+3)$ is a factor of $f(x)$ then $f(-3)=0$
\[
\begin{array}{c}
f(-3)=(-3)^{3}+a(-3)^{2}+b(-3)-9=0
-27+9 a-3 b-9=0
9 a-3 b=36
3 a-b=12
\end{array}
\]
solving sinultaneously, $4 a=20 \Rightarrow a=5$
\[
\Rightarrow b=3
\]
(2) $z=4-2 i$
(i) $|z|=\sqrt{4^{2}+(-2)^{2}}=\sqrt{20}=2 \sqrt{5}$
$\arg z=\arctan \left(\frac{-2}{4}\right)=-27^{\circ}$ (to the nearest degree)
(ii)
\[
\begin{aligned}
\frac{z}{\bar{z}}=\frac{4-2 i}{4+2 i}  |  =\frac{(4-2 i)(4-2 i)}{(4+2 i)(4-2 i)}   |  =\frac{16-16 i+4 i^{2}}{16-4 i^{2}}   |  =\frac{12-16 i}{20}   |  =\frac{3}{5}-\frac{4}{5} i
\end{aligned}
\]
(3) (1)
\[
\begin{aligned}
\overrightarrow{A B}  |  =\underline{b}-\underline{a}   |  =\binom{2}{5}-\binom{3}{-1}=\binom{-1}{6}
\overrightarrow{B C}  |  =\frac{c}{b}-\underline{b}   |  =\binom{8}{3}-\binom{2}{5}=\binom{6}{-2}
\end{aligned}
\]