Year: 2016
Paper: 3
Question Number: 3
Course: LFM Pure
Section: Integration
A substantially larger number of candidates took the paper this year: 14% more than in 2015. However, the mean score was virtually identical to that in 2015. Five questions were very popular, with two being attempted by in excess of 90% of the candidates, but once again, all questions were attempted by significant numbers, with only one dipping under 10% attempting it, and every question was answered perfectly by at least one candidate. Most candidates kept to six sensible attempts, although some did several more scoring weakly overall, except in six outstanding cases that earned very high marks.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
\begin{questionparts}
\item
Given that
\[
\int
\frac {x^3-2}{(x+1)^2}\, \e ^x \d x =
\frac{\P(x)}{Q(x)}\,\e^x + \text{constant}
\,,
\]
where $\P(x)$and $Q(x)$ are polynomials,
show that $Q(x)$ has a factor of $x + 1$.
Show also that the degree of $\P(x)$ is exactly one more than the degree of $Q(x)$, and find $\P(x)$ in the case $Q(x) =x+1$.
\item
Show that there are no polynomials $\P(x)$ and $Q(x)$ such that
\[
\int \frac 1 {x+1} \, \, \e^x \d x
=
\frac{\P(x)}{Q(x)}\,\e^x +\text{constant}
\,.
\]
You need consider only the case when $\P(x)$ and $Q(x)$ have no common factors.
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& \int
\frac {x^3-2}{(x+1)^2}\, \e ^x \d x &=
\frac{\P(x)}{Q(x)}\,\e^x + \text{constant} \\
\underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{x^3-2}{(x+1)^2}e^x &= \frac{P'(x)Q(x)-Q'(x)P(x)}{Q(x)^2}e^x + \frac{P(x)}{Q(x)}e^x \\
\Rightarrow && \frac{x^3-2}{(x+1)^2} &= \frac{(P(x)+P'(x))Q(x)-Q'(x)P(x)}{Q(x)^2} \\
\Rightarrow && Q(x)^2(x^3-2) &= ((P(x)+P'(x))Q(x)-Q'(x)P(x))(x+1)^2 \\
\Rightarrow && Q(-1) &= 0 \\
\Rightarrow && x+1 &\mid Q(x)
\end{align*}
We have $\frac{x^3-2}{(x+1)^2}$ has degree $1$ (plus some remainder term). Therefore
\begin{align*}
1 &= \deg \l (P(x)+P'(x))Q(x)-Q'(x)P(x)\r - 2 \deg Q(x) \\
&= \deg P(x) + \deg Q(x) - 2 \deg Q(x) \\
&= \deg P(x) - \deg Q(x)
\end{align*}
as required.
Suppose $Q(x) = x+1, P(x) = ax^2+bx+c$ then
\begin{align*}
&& \frac{x^3-2}{(x+1)^2} &= \frac{(P(x)+P'(x))(x+1)-P(x)}{(x+1)^2} \\
\Rightarrow && x^3-2 &= (P(x)+P'(x))(x+1) - P(x) \\
\Rightarrow && x^3-2 &= (ax^2+bx+c+2ax+b)(x+1) - (ax^2+bx+c) \\
&&&= a x^3+ x^2 (2 a + b) + x (2 a + b + c)+b \\
\Rightarrow && a &= 1 \\
&& b &= -2 \\
&& c &= 0
\end{align*}
So $P(x) = x^2-2x$
\item \begin{align*}
&& \int \frac1{x+1}e^x \d x &= \frac{P(x)}{Q(x)}e^x + c \\
\Rightarrow && \frac{1}{x+1} e^x &= \frac{P'(x)Q(x)-Q'(x)P(x)}{Q(x)^2}e^x + \frac{P(x)}{Q(x)}e^x \\
\Rightarrow && \frac{1}{x+1} &= \frac{(P(x)+P'(x))Q(x)-Q'(x)P(x)}{Q(x)^2}
\end{align*}
Therefore $Q(-1) = 0$ and so $x +1 \mid Q(x)$.
Considering degrees, we must have that $P(x)$ has degree $1$ less than $Q(x)$. Consider also the number of factors of $x+1$ in the numerator and denominator. Since $P(x)$ and $Q(x)$ have no common factors, the $Q(x)$ could have $q$ factors and $P(x)$ must have none. The denominator therefore has $2q$ factors and the numerator must have $q-1$ factors (coming from $Q'(x)$), we must have $2q = (q-1) + 1$, but that implies $q = 0$. Contradiction!
\end{align*}
\end{questionparts}
Marginally less popular than question 2, this was very slightly better attempted. In both parts, candidates successfully equated the differential of the expressions on the right of the equations to the expression to be integrated, with the exponential function cancelled. In the first part, many obtained P(x) in the given case of Q(x), but the attempted proofs that the degree of P(x) is one more than that of Q(x) and for part (ii) that no such polynomials exist led to many illogical steps. Many would have benefited by multiplying up by denominators and using the remainder/factor theorem, rather than attempting arguments based on degrees of rational expressions. Some subverted part (i) by successfully integrating the first expression.