Year: 2016
Paper: 2
Question Number: 3
Course: LFM Pure
Section: Differentiation
As in previous years the Pure questions were the most popular of the paper with questions 1, 3 and 7 the most popular of these. The least popular questions of the paper were questions 10, 11, 12 and 13 with fewer than 400 attempts for each of them. There were many examples of solutions in this paper that were insufficiently well explained, given that the answer to be reached had been provided in the question.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1517.4
Banger Comparisons: 3
For each non-negative integer $n$, the polynomial $\f_n$ is defined by
\[ \f_n(x) = 1 + x + \frac{x^2}{2!} + \frac {x^3}{3!} + \cdots + \frac{x^n}{n!} \]
\begin{questionparts}
\item Show that $\f'_{n}(x) = \f_{n-1}(x)\,$ (for $n\ge1$).
\item Show that, if $a$ is a real root of the equation \[\f_n(x)=0\,,\tag{$*$}\] then $a<0$.
\item Let $a$ and $b$ be distinct real roots of $(*)$, for $n\ge2$. Show that $\f_n'(a)\, \f_n'(b)>0\,$ and use a sketch to deduce that $\f_n(c)=0$ for some number $c$ between $a$ and $b$.
Deduce that $(*)$ has at most one real root. How many real roots does $(*)$ have if $n$ is odd? How many real roots does $(*)$ have if $n$ is even?
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& f'_n(x) &= 0 + 1 + \frac{2x}{2!} + \frac{3x^2}{3!} + \cdots + \frac{nx^{n-1}}{n!} \\
&&&= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{n-1}}{(n-1)!} \\
&&&= f_{n-1}(x)
\end{align*}
\item Claim: $f_n(x) > 0$ for all $x > 0$
Proof: (By induction)
Base case: ($n = 1$) $f_1(x) = 1 + x > 1$ therefore $f_1(x) > 0$
Suppose it's true for $n = k$, then consider $f_{k+1}$, if we differentiate it, we find it is increasing on $(0, \infty)$ by our inductive hypothesis. But then $f_{k+1}(0) = 1 > 0$. Therefore $f_{k+1}(x) > 0$ as well.
Therefore by the principle of mathematical induction we are done.
Since $f_n(x) > 0$ for non-negative $x$, if $a$ is a root it must be negative.
\item Suppose $f_n(a) = f_n(b) = 0$ then $f'_n(a) = -\frac{a^n}{n!}$ and $f'_n(b) = -\frac{b^n}{n!}$, but then $f_n'(a) f_n'(b) = \frac{(-a)^n(-b)^n}{(n!)^2} > 0$ since $a < 0, b < 0$.
$_n'(a) f_n'(b)$ is positive, the two gradients must have the same sign (and not be zero). Therefore if they are both increasing, at some point the curve must cross the axis in between. Therefore there is some root $c$ between $a$ and $b$. But then there is also a root between $c$ and $a$ and $c$ and $b$, and very quickly we find more than $n$ roots which is not possivel. Therefore there must be at most $1$ root.
If $n$ is odd there must be exactly one root, since $f_n$ changes sign as $x \to -\infty$ vs $x = 0$.
If $n$ is even then there can't be any roots, since if it crossed the $x$-axis there would be two roots (not possible) and it cannot touch the axis, since $f'_n(a) \neq 0$ unless $a = 0$, and we know $a < 0$
\end{questionparts}
This was the most popular question on the paper, and candidates generally scored well here. The first two parts were relatively straightforward and were answered well. The final part required careful explanation from candidates and many were not able to take the initial step of rewriting the relationship between the function and its derivative in a useful way. Those who did successfully complete this part were often able to identify the number of roots in each of the two cases.