Use this ZIMSEC A-Level Pure Mathematics Paper 2 June 2019 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.
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Questions: ZIMSEC A-Level Pure Mathematics Paper 2 June 2019 Question Paper
This is the ZIMSEC A-Level Pure Mathematics marking scheme for Paper 2 from June 2019. It shows how marks are awarded, so you can mark your own answers to the matching question paper. See the question paper.
(1)
1)
\[
\text { Let } \begin{array}{l}
\frac{2 x}{(x-1)(x+1)^{2}} \equiv \frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}
\equiv A(x+1)^{2}+B(x-1)(x+1)+C(x-1)
(x-1)(x+1)^{2}
\Rightarrow 2 x=A(x+1)^{2}+B(x-1)(x+1)+C(x-1)
\end{array}
\]
when $x=1$ when $x=-1$ Companiry Coefficiest
\[
\begin{array}{lll}
4 A=2 | -2 C=-2 | \text { of } x^{2}
\Rightarrow A=1 / 2 | \underline{C C 1} | 0=A+B | & | & -B=-1 / 2
\end{array}
\]
Thus $\frac{2 x}{(x-1)(x+1)^{2}} \equiv \frac{1}{2(x-1)}-\frac{1}{2(x+1)}+\frac{1}{(x+1)^{2}}$
2(1)
\[
\begin{array}{c}
\text { Let } \sqrt{6} \operatorname{Cs} \theta-\sqrt{2} \sin \theta \equiv R \operatorname{Cos}(\theta+\alpha)
\quad \equiv R \operatorname{Cos} \theta \operatorname{Cos} \theta-R \operatorname{Sin} \theta \operatorname{Sin} \alpha
\Rightarrow \quad \begin{array}{rlr}
R \operatorname{Cos} \alpha=\sqrt{6} | \quad R \operatorname{Sin} \alpha=\sqrt{2}
\operatorname{Cos} \alpha=\frac{\sqrt{6}}{R} | \quad \operatorname{Sin} \alpha=\frac{\sqrt{2}}{R}
\end{array}
\frac{\sqrt{2} \int_{\frac{R}{2}}^{R}}{\sqrt{t}}
\end{array}
\]
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