Use this ZIMSEC A-Level Pure Mathematics Paper 2 June 2024 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.
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Questions: ZIMSEC A-Level Pure Mathematics Paper 2 June 2024 Question Paper
This is the ZIMSEC A-Level Pure Mathematics marking scheme for Paper 2 from June 2024. It shows how marks are awarded, so you can mark your own answers to the matching question paper. See the question paper.
JUNE 2024 P2 PURE SUGGESTED MARKING GMDE,
1. a) $y \& F$
\[
\begin{array}{l}
y=k f
0,45=k(6)
k=\frac{3}{40}
y=\frac{3}{40} F
\end{array}
\]
b) when $y=0,2$
\[
\begin{aligned}
0,2 | =\frac{3}{40} \mathrm{~F}
\frac{0,2}{3 / 40} | =\mathrm{F}
F | =\frac{8}{3} \text { Newtons }
\end{aligned}
\]
2. (a)
\[
\begin{aligned}
f \circ g | =f(x)=\frac{x-1}{x-1-1}
f \circ f | =f(x-1)=x-2
h \circ f | =h(x-1) | =x-1+1 | =x
\text { hoh } | =h(x+1) | =x+1 | =x+2
\end{aligned}
\]
| \multicolumn{1}{c}{9} | \multicolumn{1}{c}{$f$} | \multicolumn{1}{c}{$h$}
0 | $x$ | $x-1$ | $x+1$
$x$ | $x$ | $x-1$ | $x+1$
$x-1$ | $x-1$ | $x-2$ | $x$
$x+1$ | $x+1$ | $x$ | $x+2$
b) (1) $e=x$
(II)
element | $x$ | $x-1$ | $x+1$
hiverse | $x$ | $x+1$ | $x-1$
(iii) It is not closed because for \& hok \&S hence it is not a group.
3.(a)Sheded area
⇒ Area of Rhombus - Area of Sector OADC
Area of Rhombus
\[
\Rightarrow \quad r^{2} \sin 2 \theta
\]
Area of sector
\[
\begin{array}{r}
\Rightarrow \frac{1}{2} r^{2} \theta
=r^{2} \theta
\end{array}
\]
shaded area
\[
\begin{array}{r}
\Rightarrow r^{2} \sin 2 \theta-r^{2} \theta
=r^{2}(\sin 2 \theta-\theta)
\end{array}
\]
b)
\[
\begin{aligned}
\text { Area } | =8^{2}(\sin 2 \times \pi / 6-\pi / 6) | =32 \sqrt{3}-\frac{32}{3} \pi
\end{aligned}
\]
\[
\begin{aligned}
\text { 5.(a) } x^{3}+x^{2}+2 x+2 | \equiv(x+1)\left(x^{2}+2\right)
\text { LHS } |
x^{3}+x^{2}+2 x+2 | =x^{2}(x+1)+2(x+1) | =\left(x^{2}+2\right)(x+1) | =\text { RHS }
\end{aligned}
\]
(b)
\[
\text { now } \begin{aligned}
\frac{x^{2}+2 x+2}{\left(x^{2}+2\right)(x+1)} | =\frac{A}{x+1}+\frac{B x+C}{x^{2}+2}
x^{3}+2 x+2=A\left(x^{2}+2\right)+(B x+C)(x+1) | =1-\left(\frac{1}{3(x+1}+\frac{2 x+4}{3\left(x^{2}+2\right)}\right)
A=\frac{1}{3} ; B | =\frac{2}{3} ; C=\frac{4}{3(x+1)}-\frac{2 x+4}{3\left(x^{2}+2\right)}
\end{aligned}
\]
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