Answers_-_Lockdown_Pure_Maths_Test_42_-_Trockers.pdf. Free PDF download (635 KB) from the ExamineAI past papers library.
A free account is needed to download. Sign in or create one
Questions: Lockdown Pure Maths Test 51 Question Paper
This is the A-Level Pure Mathematics marking scheme from September 2020. It shows how marks are awarded, so you can mark your own answers to the matching question paper. See the question paper.
Upper $6{ }^{\text {th }}$
LOCKDOWN PURE MATHS TEST 42
Friday 11 September 2020
ANSWERS
35 MARKS
NB: If a numerical answer cannot be given exactly, and the accuracy required is not specified in the question, then in the case of an angle it should be given to the nearest degree, and in other cases it should be given correct to $\mathbf{2}$ significant figures.
1. (i) Given that $x$ and $y$ are real, find an expression for the real part of the complex number $\frac{z-8 i}{z-6}$, where $z=x+i y$.
[4]
\[
\begin{aligned}
\frac{z-8 i}{z-6} | =\frac{x+i y-8 i}{x+i y-6} | =\frac{x+i(y-8)}{(x-6)+i y} | =\frac{[x+i(y-8)][(x-6)-i y]}{[(x-6)+i y][(x-6)-i y]} | =\frac{x(x-6)-i x y+i(y-8)(x-6)+y(y-8)}{(x-6)^{2}+y^{2}} | =\frac{[x(x-6)+y(y-8)]+i[(y-8)(x-6)-x y]}{(x-6)^{2}+y^{2}} | =\frac{[x(x-6)+y(y-8)]}{(x-6)^{2}+y^{2}}+\frac{[(y-8)(x-6)-x y]}{(x-6)^{2}+y^{2}} i
\end{aligned}
\]
∴ The real part is $\frac{x(x-6)+y(y-8)}{(x-6)^{2}+y^{2}}$
It is going to be HARD, but hard is not IMPOSSIBLE.
Page 1
(C) Tarakino N.P. 0772978155/0717267175
ntarakino@gmail.com