Use this ZIMSEC A-Level Pure Mathematics Paper 1 November 2023 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.
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This is the ZIMSEC A-Level Pure Mathematics marking scheme for Paper 1 from November 2023. It shows how marks are awarded, so you can mark your own answers.
A' LEVEL: PUREMATHS.
N2023 P. 1: MS.
Af $y=16 x-17$ is a tangent to
$y=3 x^{2}+4 x-5$.
Then $3 x^{2}+4 x-5=16 x-17$.
$3 x^{2}-12 x+12=0$
$x^{2}-4 x+4=0$
$\Rightarrow$ Discriminant $=0$.
$(-4)^{2}-4(1)(4)=16-16$
$\therefore \quad \operatorname{Shoman} \quad=0$.
(2) $\quad 4 \operatorname{Sin}\left(x+\frac{\pi}{6}\right)-8 \operatorname{Cos}\left(x-\frac{\pi}{6}\right)$
$=4\left[\operatorname{Sin} x \operatorname{Cos} \frac{\pi}{6}+\operatorname{Cos} x \operatorname{Sin} \frac{\pi}{6}\right]-8\left[\operatorname{Cos} x \operatorname{Cos} \frac{\pi}{6}+\operatorname{Sin} x \operatorname{Sin} \frac{\pi}{6}\right.$
$=4\left(\frac{\sqrt{3}}{2} \operatorname{Sin} x+\frac{1}{2} \operatorname{Cos} x\right)-8\left(\frac{\sqrt{3}}{2} \operatorname{Cos} x+\frac{1}{2} \operatorname{Sin} x\right)$
$=2 \sqrt{3} \operatorname{Sin} x+2 \operatorname{Cos} x-4 \sqrt{3} \operatorname{Cos} x-4 \operatorname{Sin} x$.
$=2 \sqrt{3} x+2\left(1-\frac{x^{2}}{2}\right)-4 \sqrt{3}\left(1-\frac{x^{2}}{2}\right)-4 x$.
$=2 \sqrt{3} x+2-x^{2}-4 \sqrt{3}+2 \sqrt{3} x^{2}-4 x$.
$=2 \sqrt{3} x^{2}-x^{2}+2 \sqrt{3} x-4 x+2-4 \sqrt{3}$.
$=(2 \sqrt{3}-1) x^{2}+(2 \sqrt{3}-4) x+2-4 \sqrt{3}$.