Use this ZIMSEC A-Level Mathematics Paper 1 June 1997 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.
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This is the ZIMSEC A-Level Mathematics marking scheme for Paper 1 from June 1997. It shows how marks are awarded, so you can mark your own answers.
Answers
Eime 1997 Paper) A-Leval Mattionatics ( 9200 )
(1)
\[
\begin{aligned}
S_{n} | =\frac{n}{2}(2 a+(n-1) d)
964 | =\frac{241}{2}(2 a+240(0,1))
964 | =\frac{241}{2}(2 a+24)
a | =-8
\end{aligned}
\]
(2). Area of sector $P O B=\frac{\theta}{2 \pi} \times \pi r^{2}=\frac{r^{2} \theta}{2}$
Area of sector $P O A=\frac{\pi r^{2}}{2}-\frac{r^{2} \theta}{2}=\frac{r^{2}(\pi-\theta)}{2}$
Aren of triangle $P O A=\frac{1}{2} x y \sin \alpha$
\[
\begin{array}{l}
=\frac{1}{2} r^{2} \sin (18 \theta-\theta)
=\frac{r^{2} \sin \theta}{2}
\end{array}
\]
Area of shaded segment $=\frac{r^{2}(\pi-\theta)}{2}-\frac{r^{2} \sin \theta}{2}$
\[
=\frac{r^{2}(\pi-\theta-\sin \theta)}{2}
\]
Area of sector $P O B=2 \times$ shaded segment
\[
\begin{aligned}
\frac{r^{2} \theta}{2} | =2 \times \frac{r^{2}(\pi-\theta-\sin \theta)}{-2}
\theta | =2 \pi-2 \theta-2 \sin \theta
3 \theta | =2(\pi-\sin \theta)
\end{aligned}
\]