Use this A-Level Mathematics November 2006 marking scheme for checking answers, revision, and study practice in Zimbabwe.
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Questions: ZIMSEC A-Level Mathematics Paper 1 November 2006 Question Paper
This is the A-Level Mathematics marking scheme for Paper 1 from November 2006. It shows how marks are awarded, so you can mark your own answers to the matching question paper. See the question paper.
A-level Mattenatics Syllabus 9164
Neveruber 2006 Paper 1 Working and Answers
(1) let $f(x)=x^{3}+p x^{2}+q x-81$
By the Factor Theorem, if $(x+1)$ is a factor of $f(x)$, then $f(-1)=0$
so
\[
\begin{array}{l}
0=(-1)^{3}+p(-1)^{2}+q(-1)-81
0=-1+p-q-81 \Rightarrow p-q=82
\end{array}
\]
and if $(x-3)$ is a factor of $f(x), f(3)=0$;
\[
\begin{array}{l}
0=(3)^{3}+p(3)^{2}+q(3)-81
0=27+9 p+3 q-81 \Rightarrow 3 p+q=18
\end{array}
\]
solving simultaneously, $4 p=100 \Rightarrow p=25$
substituting back, $25-q=82 \Rightarrow q=-57$
(2) By the Cosine Rule: $\cos R=\frac{x^{2}+(x+3)^{2}-k^{2}}{2 x(x+3)}$
\[
\cos R=\frac{2 x^{2}+6 x+9-k^{2}}{2 x(x+3)}
\]
If $\cos R=\frac{1}{2}$, then
\[
\begin{array}{c}
\frac{1}{2}=\frac{2 x^{2}+6 x+9-k^{2}}{2 x(x+3)}
x^{2}+3 x=2 x^{2}+6 x+9-k^{2}
0=x^{2}+3 x+9-k^{2}
\end{array}
\]
If the above quadratic equation in $x$ has two identical real roots, then there will be only one possible triangle. This occurs if $b^{2}-4 a c=0$
\[
\begin{aligned}
(3)^{2}-4(1)\left(9-k^{2}\right) | =0
9-36+4 k^{2} | =0
k^{2} | =\frac{27}{4}
k | =\frac{3 \sqrt{3}}{2}
\end{aligned}
\]
( $k$ must be positive because it is the length of one sitce of a triangle)