Use this ZIMSEC A-Level Pure Mathematics Paper 1 June 2025 marking scheme for checking answers, revision, and exam preparation in Zimbabwe.
A free account is needed to download. Sign in or create one
This is the ZIMSEC A-Level Pure Mathematics marking scheme for Paper 1 from June 2025. It shows how marks are awarded, so you can mark your own answers. The paper code is 6024/1.
DE MANUE SOLUTIONS
SUGGESTTED MARKING GUIDE
PURE MATHEMATICS JUNE 2025
PAPER 1
6024/1
0717256148/0774459409
Constructive comments in form of presentation, omission and errors are welcome . Feel free to contact me so that we improve the document together.
1 Find the series expansion of the function $\frac{3-x}{3+2 x}$ up to and including the term in $x^{3}$. State the range of the values of $x$ for which the expansion is valid.
[6]
SOLUTION
\[
\begin{array}{c}
(3-x)(3+2 x)^{-1}
3^{-1}(3-x)\left(1+\frac{2 x}{3}\right)^{-1}
=1+n x+\frac{n(n-1)(x)^{2}}{2!}+\frac{n(n-1)(n-2)(x)^{3}}{3!} \ldots
\frac{1}{3}(3-x)\left[1+\left(\frac{2 x}{3}\right)(-1)+\frac{-1(-1-1)\left(\frac{2 x}{3}\right)^{2}}{2!}+\frac{\left.-1(-1-1)(-1-2)\left(\frac{2 x}{3}\right)^{3}\right]}{3!}\right]
=\left(1-\frac{x}{3}\right)\left[1-\frac{2 x}{3}+\frac{4 x^{2}}{9}-\frac{8 x^{3}}{27}\right]
=1-\frac{2 x}{3}+\frac{4 x^{2}}{9}-\frac{8 x^{3}}{27}-\frac{x}{3}+\frac{2 x^{2}}{9}-\frac{4 x^{3}}{27}
=\mathbf{1}-\boldsymbol{x}+\frac{\mathbf{2} x^{2}}{\mathbf{3}}-\frac{\mathbf{4} x^{3}}{\mathbf{9}}
-1<|x|<1
-\frac{\mathbf{3}}{\mathbf{2}}<\boldsymbol{x}<\frac{\mathbf{3}}{\mathbf{2}}
\end{array}
\]
2 Solve the equation $\frac{3^{x}\left(9^{x}\right)^{2}}{81}=27^{x}$.
[4]
E.ZIVANAYI 0717256148/0774459409(SIR MANUE)zivanayiemmanuel@gmail.com