Use this ZIMSEC A-Level Mathematics Paper 1 June 2004 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.
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Questions: ZIMSEC A-Level Mathematics Paper 1 June 2004 Question Paper
This is the ZIMSEC A-Level Mathematics marking scheme for Paper 1 from June 2004. It shows how marks are awarded, so you can mark your own answers to the matching question paper. See the question paper.
June 2004 Paper 1 A-Level Mallematics ZIMSEC Sylabus 9164 WORKING AND ANSWERS
(1) Let $f(x)=x^{4}+a x^{2}+3 x+6$
if $(x+2)$ is a factor of $f(x)$, when $f(-2)=0$
\[
\begin{aligned}
(0) | =(-2)^{4}+a(-2)^{2}+3(-2)+6
0 | =16+4 a-6+6 \Rightarrow a=-4
\end{aligned}
\]
(2)
\[
\begin{array}{l}
\left(9+3 x^{2}\right)^{\frac{1}{2}} \equiv\left(9\left(1+\frac{x^{2}}{3}\right)\right)^{\frac{1}{2}} \equiv 3\left(1+\frac{x^{2}}{3}\right)^{\frac{1}{2}}
=3\left\{1+\left(\frac{1}{3}\right)\left(\frac{x^{2}}{3}\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}\left(\frac{x^{2}}{3}\right)^{2}+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}\left(\frac{x^{2}}{3}\right)^{3}+\ldots\right\}
\end{array}
\]
so the coefficient of $x^{6}$ is $\frac{3 \times\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!\times 3^{3}}=\frac{1}{144}$
(3)
\[
\begin{array}{l}
\frac{d r}{d t}=3 \quad V=\frac{4 \pi r^{3}}{3} \Rightarrow \frac{d V}{d r}=4 \pi r^{2}
\frac{d V}{d t}=\frac{d V}{d r} \times \frac{d r}{d t}
\frac{d V}{d t}=4 \pi r^{2} \times 3=12 \pi r^{2}
\end{array}
\]
when $r=9, \frac{d N}{d t}=12 \pi(9)^{2}$
so the rate of ticrease of the volume of the sphere when the vadius is 9 cm is $972 \pi \mathrm{~cm}^{3} / \mathrm{s}$
(4) $0,432 i=0,43+\frac{21}{10000}+\frac{21}{1000000}+\frac{21}{100000000}+\ldots$
The terms after 0,43 make a GP. with $a=\frac{21}{10000}$ and $r=\frac{1}{100}$ and an infinote number of terms
\[
S_{\infty}=\frac{a}{1-r}=\frac{21}{10000} \times \frac{100}{99}=\frac{7}{3300}
\]
so $0,4321=0,43+\frac{7}{3300}$