Year: 2012
Paper: 2
Question Number: 2
Course: LFM Stats And Pure
Section: Polynomials
There were just over 1000 entries for paper II this year, almost exactly the same number as last year. Overall, the paper was found marginally easier than its predecessor, which means that it was pitched at exactly the level intended and produced the hoped-for outcomes. Almost 50 candidates scored 100 marks or more, with more than 400 gaining at least half marks on the paper. At the lower end of the scale, around a quarter of the entry failed to score more than 40 marks. It was pleasing to note that the advice of recent years, encouraging students not to make attempts at lots of early parts to questions but rather to spend their time getting to grips with the six that can count towards their paper total, was more obviously being heeded in 2012 than I can recall being the case previously. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones. Questions 1 and 2 were the most popular questions, although each drew only around 800 "hits" – fewer than usual. Questions 3 – 5 & 8 were almost as popular (around 700), with Q6 attracting the interest of under 450 candidates and Q7 under 200. Q9 was the most popular applied question – and, as it turned out, the most successfully attempted question on the paper – with very little interest shown in the rest of Sections B or C.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1530.0
Banger Comparisons: 4
If $\p(x)$ and $\q(x)$ are polynomials of degree $m$ and $n$, respectively, what is the degree of $\p(\q(x))$?
\begin{questionparts}
\item The polynomial $\p(x)$ satisfies
\[
\p(\p(\p(x)))- 3 \p(x)= -2x\,
\]
for all $x$.
Explain carefully why $\p(x)$ must be of degree 1, and find all polynomials that satisfy this equation.
\item Find all polynomials that satisfy
\[
2\p(\p(x)) +3 [\p(x)]^2 -4\p(x) =x^4
\]
for all $x$.
\end{questionparts}If $\p(x)$ and $\q(x)$ are polynomials of degree $m$ and $n$, $\p(\q(x))$ has degree $mn$.
\begin{questionparts}
\item Suppose $\p(\p(\p(x)))- 3 \p(x)= -2x$, and suppose $p(x)$ has degree $n = \geq 2$, then $\p(\p(\p(x)))$ has degree $n^3$ and so the left hand side has degree higher than $1$ and the right hand side is degree $1$. Therefore $\p(x)$ is degree $1$ or $0$. If $p(x) = c$ then $c^3-3c = -2x$ but the LHS doesn't depend on $x$ which is also a contradiction. Therefore $\p(x)$ is degree $1$. Suppose $\p(x) = ax+b$ then:
\begin{align*}
&& -2x &= \p(\p(\p(x))) - 3\p(x) \\
&&&= \p(\p(ax+b)) - 3(ax+b) \\
&&&= \p(a(ax+b)+b) - 3ax -3b \\
&&&= a(a^2x+ab+b) + b - 3ax - 3b \\
&&&= (a^3-3a)x + b(a^2+a-2) \\
\Rightarrow &&& \begin{cases} a^3-3a&=-2 \\
b(a^2+a-2) &= 0\end{cases} \\
\Rightarrow &&& \begin{cases} a^3-3a+2 = 0 \\
b = 0, a = 1, a = -2\end{cases} \\
\Rightarrow &&& \begin{cases} (a-1)(a^2+a-2) = 0 \\
b = 0, a = 1, a = -2\end{cases} \\
\Rightarrow && (a,b) &= (1, b), (-2,b)
\end{align*}
\item Suppose $2\p(\p(x)) +3 [\p(x)]^2 -4\p(x) =x^4$ and let $\deg \p(x) = n$, then LHS has degree $\max(n^2,2n,n)$ and the right hand side has degree $4$. Therefore $\p(x)$ must have degree $2$.
Let $\p(x) = ax^2 + bx + c$, then, considering the coefficient of $x^4$ in $2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)$ we will have $2a^3+3a^2=1 \Rightarrow 2a^3+3a^2-1 = (a+1)^2(2a-1) \Rightarrow a = -1, a=\frac12$.
Consider the coefficient of $x^3$ in $2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)$ we have $4a^2b+6ab = 0 \Rightarrow 2ab(2a+3) = 0$ Since $a = -1, \frac12$ this means $b = 0$.
Consider the constant coefficient in $2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)$ (using $b = 0$). $2ac^2+c+3c^2-4c = 0 \Rightarrow c(2ac+3c-3) = 0$. Therefore $c = 0$ or $a = -1, c = 3, a = \frac12, c = \frac34$, so our possible polynomials are:
$\p(x) = -x^2, \frac12x^2, -x^2+3, \frac12x^2+\frac34$
\end{questionparts}
This turned out to be the second most popular question and the highest scoring of the pure questions. Explanations apart, most candidates held their nerve remarkably well to produce careful algebra leading to correct answers. The added "trap" in part (i) – in that each answer contained an arbitrary constant – caught many out.