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First page of Lockdown Pure Maths Test 31 Answers Marking Scheme

Lockdown Pure Maths Test 31 Answers Marking Scheme

Use this A-Level Pure Mathematicsugust 2020 marking scheme for checking answers, revision, and study practice in Zimbabwe.

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Questions: Trockers A-Level Pure Mathematicsugust 2020 Question Paper

Subject
Pure Mathematics
Level
A-Level
Document
Marking Scheme
Session
August
Year
2020
File
PDF, 515 KB

About this marking scheme

This is the A-Level Pure Mathematics marking scheme from August 2020. It shows how marks are awarded, so you can mark your own answers to the matching question paper. See the question paper.

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Text from the first page

Upper $6{ }^{\text {th }}$
LOCKDOWN PURE MATHS TEST 31
Thursday 27 August 2020
ANSWERS

30 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. Find the possible solutions of the quadratic equation $x^{2}+(3-m) x+5=m^{2}$, where $m$ is a constant, given that the equation has repeated roots.

[7]
\[
\begin{array}{l}
x^{2}+(3-m) x+5=m^{2}
\Rightarrow x^{2}+(3-m) x+\left(5-m^{2}\right)=0
\end{array}
\]

Given that the equation has repeated roots:
\[
\begin{array}{l}
\Rightarrow b^{2}-4 a c=0
\Rightarrow(3-m)^{2}-4(1)\left(5-m^{2}\right)=0
\Rightarrow 9-6 m+m^{2}-20+4 m^{2}=0
\Rightarrow 5 m^{2}-6 m-11=0
\Rightarrow 5 m^{2}+5 m-11 m-11=0
\Rightarrow 5 m(m+1)-11(m+1)=0
\Rightarrow(5 m-11)(m+1)=0
\Rightarrow 5 m-11=0 \text { or } m+1=0
\Rightarrow m=\frac{11}{5} \text { or } m=-1
\end{array}
\]

But $x^{2}+(3-m) x+\left(5-m^{2}\right)=0$

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