Use this ZIMSEC A-Level Mathematics Paper 2 November 2010 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.
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Questions: ZIMSEC A-Level Mathematics Paper 2 November 2010 Question Paper
This is the ZIMSEC A-Level Mathematics marking scheme for Paper 2 from November 2010. It shows how marks are awarded, so you can mark your own answers to the matching question paper. See the question paper.
TO REVIEREND
NDLOV4
Att
H.O.D
MATHS
CHENGATO HIGH SCHOOL
mr munyikale
2
2010.
I
1
\[
\begin{array}{l}
\mathrm{A}+\mathrm{B}=45^{\circ}
\tan (\mathrm{A}+\mathrm{B})=\tan 45^{\circ}
\frac{\tan \mathrm{A}+\tan \mathrm{B}}{1-\tan \mathrm{A} \tan \mathrm{~B}}=1
\tan \mathrm{~A}+\tan \mathrm{B}=1-\tan \mathrm{A} \tan \mathrm{~B}
\tan \mathrm{~A}+\tan \mathrm{B}+\tan \mathrm{A} \tan \mathrm{~B}=1
\tan 22 \frac{1}{2}^{\circ}+\tan 22 \frac{1}{2}^{\circ}+\tan ^{2} 22 \frac{1}{2}^{\circ}=1
\tan ^{2} 22 \frac{1}{2}^{\circ}+2 \tan 22 \frac{1}{2}^{\circ}-1=0
\tan 22 \frac{1}{2}^{\circ}=\frac{-2 \pm \sqrt{8}}{2}
=\quad \frac{-2}{2}+2 \sqrt{2}
=\quad \sqrt{2}-1
\end{array}
\]
c.a.o
For applying addition formula
M1A1-
(AG)
A1
B1
M1
A1
[6]
2 Verify statement true for $\mathrm{n}=2 .(\mathrm{n}>1)$
i.e. $7^{2}-6(2)-1=49-12-1=36$
B1
Assuming true for $\mathrm{n}=\mathrm{k}$ and stating $7^{\mathrm{k}}-6 \mathrm{k}-1=36 \mathrm{p}(p \in N)$
B1
Correct method to prove validity for $\mathrm{n}=\mathrm{k}+1$
e.g. $7^{\mathrm{k}+1}-6(\mathrm{k}+1)=7.7^{\mathrm{k}}-6 \mathrm{k}-7$
B1
For substituting $7^{\mathrm{k}}$ by $36 \mathrm{p}+6 \mathrm{k}+1$
M1
Correct reduction to obtain $36(7 \mathrm{p}+\mathrm{k})$
A1
Give correct conclusion
A1
[6]
3
(a) $e^{-2 x}=1-2 x+\frac{4 x^{2}}{2!}-\frac{8 x^{3}}{3!}$
M1
\[
=\quad 1-2 x+2 x^{2}-\frac{4 x^{3}}{3}
\]
A1
\[
\sin 3 x=3 x-\frac{27 x^{3}}{3!}=3 x-\frac{9 x^{3}}{2}
\]
M1A1 [4]