Year: 2016
Paper: 3
Question Number: 8
Course: LFM Pure
Section: Simultaneous equations
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
The function f satisfies, for all $x$, the equation
\[
\f(x) + (1- x)\f(-x) = x^2\, .
\]
Show that $\f(-x) + (1 + x)\f(x) = x^2\,$.
Hence find $\f(x)$ in terms of $x$. You should
verify that your function satisfies the original equation.
\item
The function ${\rm K}$ is defined, for $x\ne 1$, by
\[{\rm K}(x) = \dfrac{x+1}{x-1}\,.\]
Show that, for $x\ne1$,
${\rm K(K(}x)) =x\,$.
The function g satisfies the equation
\[
\g(x)+ x\, \g\Big(\frac{ x+1 }{x-1}\Big)
= x \ \ \ \ \ \ \ \ \ \ \
( x\ne 1)
\,.
\]
Show
that, for $x\ne1$, $\g(x)= \dfrac{2x}{x^2+1}\,$.
\item
Find $\h(x)$, for $x\ne0$, $x\ne1$, given that
\[
\h(x)+ \h\Big(\frac 1 {1-x}\Big)= 1-x -\frac1{1-x}
\ \ \ \ \ \ (
x\ne0, \ \ x\ne1 )
\,.
\]
\end{questionparts}\begin{questionparts}
\item $\,$ Let $P(x)$ mean the proposition that $f(x) + (1-x)f(-x) = x^2$ so
\begin{align*}
P(x): && f(x) + (1-x)f(-x) &= x^2 \\
P(-x): && f(-x)+(1+x)f(x) &= (-x)^2 = x^2 \\
\Rightarrow && f(x)+(1-x)\left (x^2-(1+x)f(x) \right) &= x^2 \\
\Rightarrow && f(x) \left (1 -(1-x^2) \right) &= x^2 + (x-1)x^2 \\
\Rightarrow && f(x)x^2 &= x^3 \\
\Rightarrow && f(x) &= x
\end{align*}
Notice that $x + (1-x)(-x) = x^2$ so it does satisfy the functional equation.
\item Let $K(x) = \frac{x+1}{x-1}$ if $x \neq 1$ so
\begin{align*}
&& K(K(x)) &= \frac{K(x)+1}{K(x)-1} \\
&&&= \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} \\
&&&= \frac{\frac{2x}{x-1}}{\frac{2}{x-1}} \\
&&&= x
\end{align*}
Let $Q(x)$ denote the proposition that $g(x) + xg(K(x)) = x$ so
\begin{align*}
Q(x): && g(x) + xg(K(x)) &= x \\
Q(K(x)): && g(K(x)) + K(x)g(x) &= K(x) \\
\Rightarrow && g(x) +xK(x)[1-g(x)] &= x \\
\Rightarrow && g(x)[1-xK(x)] &= x(1-K(x)) \\
\Rightarrow && g(x) \frac{x-1-x^2-x}{x-1} &= \frac{-2x}{x-1} \\
\Rightarrow && g(x) &= \frac{2x}{x^2+1}
\end{align*}.
And notice that $\frac{2x}{x^2+1} + x \frac{2\frac{x+1}{x-1}}{\left( \frac{x+1}{x-1}\right)^2+1} = \frac{2x}{x^2+1} + \frac{2x(x^2-1)}{2x^2+2} = x$
\item Consider $H(x) = \frac{1}{1-x}$ then notice that $H(H(x)) = \frac{1}{1-\frac{1}{1-x}} = \frac{x-1}{x}$ and $H^3(x) = \frac{\frac{1}{1-x}-1}{\frac{1}{1-x}} = 1-(1-x) = x$. So
So letting $S(x)$ be the statement that $h(x) + h(H(x)) = 1 - x - \frac{1}{1-x}$ we have
\begin{align*}
S(x): && h(x) + h(H(x)) &= 1 - x - H(x) \\
S(H(x)): && h(H(x)) + h(H^2(x)) &= 1 - H(x) - H^2(x) \\
S(H^2(x)): && h(H^2(x)) + h(x) &= 1 - H^2(x) - x \\
S(x) - S(H(x)) + S(H^2(x)): && 2h(x) &= 1 - 2x \\
\Rightarrow && h(x)& = \frac12 - x
\end{align*}
and notice that $\frac12 -x +\frac12 - \frac{1}{1-x} = 1 - x - \frac{1}{1-x}$ so it does satisfy the equation.
\end{questionparts}
Attempted by nearly as many as attempted question 1, it was marginally more successful, and a good number achieved full marks. Generally, the idea of repeatedly applying a function to create a cycle was well-spotted. However, candidates did sometimes fall down trying to find f(x) in (ii) and some substituted the given f(x) rather than finding it. In part (iii), some stopped having made the first substitution and so could not find the solution. Also, some guessed the solution for part (iii) but, of course, this did not do the full job.