Year: 2007
Paper: 1
Question Number: 4
Course: LFM Stats And Pure
Section: Polynomials
There were significantly more candidates attempting this paper this year (an increase of nearly 50%), but many found it to be very difficult and only achieved low scores. In particular, the level of algebraic skill required by the questions was often lacking. The examiners' express their concern that this was the case despite a conscious effort to make the paper more accessible than last year's. At this level, the fluent, confident and correct handling of mathematical symbols (and numbers) is necessary and is expected; many good starts to questions soon became unstuck after a simple slip. Graph sketching was usually poor: if future candidates wanted to improve one particular skill, they would be well advised to develop this. There were of course some excellent scripts, full of logical clarity and perceptive insight. It was pleasing to note that the applied questions were more popular this year, and many candidates scored well on at least one of these. It was however surprising how rarely answers to questions such as 5, 9, 10, 11 and 12 began with a diagram. However, the examiners were left with the overall feeling that some candidates had not prepared themselves well for the examination. The use of past papers to ensure adequate preparation is strongly recommended. A student's first exposure to STEP questions can be a daunting, demanding experience; it is a shame if that takes place during a public examination on which so much rides. Further, and fuller, discussion of the solutions to these questions can be found in the Hints and Answers document.
Difficulty Rating: 1516.0
Difficulty Comparisons: 1
Banger Rating: 1530.2
Banger Comparisons: 6
Show that
$x^3-3xbc + b^3 + c^3$ can be written in the form
$\left( x+ b+ c \right) {\rm Q}( x)$,
where ${\rm Q}( x )$ is a quadratic expression.
Show that $2{\rm Q }( x )$ can be written
as the sum of three expressions, each of which is a perfect square.
It is given that the equations $ay^2 + by + c =0$ and $by^2 + cy + a = 0$ have a common root $k$. The coefficients $a$, $b$ and $c$ are real, $a$ and $b$ are both non-zero, and $ac \neq b^2$. Show that
\[
\left( ac - b^2 \right) k = bc - a^2
\]
and determine a similar expression involving $k^2$. Hence show that
\[
\left( ac - b^2 \right) \left(ab-c^2 \right) = \left( bc - a^2 \right)^2
\]
and that $ a^3 -3abc + b^3 +c^3 = 0\,$. Deduce that either $k=1$ or the two equations are identical.\begin{align*}
&& x^3 - 3xbc+b^3 + c^3 &= (x+b+c)(x^2-x(b+c)+b^2+c^2-bc) \\
&&&= \tfrac12(x+b+c)((x-b)^2+(x-c)^2+(b-c)^2) \\
\end{align*}
We must have:
\begin{align*}
&& 0 &= ak^2 + bk+c \tag{1}\\
&&0 &= bk^2+ck+a \tag{2}\\
b*(1)&& 0 &= abk^2 + b^2k+cb \\
a*(2)&& 0 &= abk^2 + ack + a^2 \\
\Rightarrow && 0 &= k(ac-b^2)+a^2-bc \\
\Rightarrow && (ac-b^2)k &= bc-a^2 \\
\\
c*(1) && 0 &= ack^2+bck+c^2 \\
b*(2) && 0 &= b^2k^2+bck+ab \\
\Rightarrow && 0 &= (ac-b^2)k^2 +c^2-ab \\
\Rightarrow && (ac-b^2)k^2 &= ab-c^2 \\
\\
\Rightarrow && \frac{ab-c^2}{ac-b^2} &= k^2 = \left (\frac{bc-a^2}{ac-b^2} \right)^2 \\
\Rightarrow && (ab-c^2)(ac-b^2) &= (bc-a^2)^2 \\
\Rightarrow && a^2bc - ab^3-ac^3+b^2c^2 &= b^2c^2-2a^2bc+a^4 \\
\Rightarrow && 0 &= a^4+ab^3+ac^3-3a^2bc \\
&&&= a(a^3+b^3+c^3-3abc) \\
\underbrace{\Rightarrow}_{a \neq 0} && 0 &= a^3+b^3+c^3-3abc \\
&&&= (a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)
\end{align*}
Therefore $a+b+c = 0$. (Since otherwise $a=b=c$ but $ac \neq b^2$). This means $1$ is a root of our equations. Therefore, either $k = 1$ or they have both roots in common, ie they are the same equation up to a scalar factor. ie $b = la, c = lb, a= lc \Rightarrow l^3 = 1 \Rightarrow l = 1$. Therefore, they are the same equation.
This question was found to be very difficult. The initial factorisation was beyond most candidates, even given the linear factor x + b + c. Anyone who wants to read Mathematics at university must be able to factorise quickly cubic expressions such as this one, and also x3 ± y3. The Hints and Answers document discusses this in more detail. Candidates who progressed to the second part of the question often deduced that ak2 + bk + c = 0 and bk2 + ck + a = 0, but then tried to eliminate k; given that the result they were asked to derive was still in terms of k, this was an unwise strategy.