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2013 Paper 1 Q2
D: 1500.0 B: 1487.3
curve sketching floor function piecewise function equation solving necessary and sufficient conditions greatest integer function rational function

In this question, \(\lfloor x \rfloor\) denotes the greatest integer that is less than or equal to \(x\), so that \(\lfloor 2.9 \rfloor = 2 = \lfloor 2.0 \rfloor\) and \(\lfloor -1.5 \rfloor = -2\). The function \(\f\) is defined, for \(x\ne0\), by \(\f(x) = \dfrac{\lfloor x \rfloor}{x}\,\).

  1. Sketch the graph of \(y=\f(x)\) for \(-3\le x \le 3\) (with \(x\ne0\)).
  2. By considering the line \(y= \frac7{12}\) on your graph, or otherwise, solve the equation \(\f(x) = \frac7 {12}\,\). Solve also the equations \(\f(x) =\frac{17}{24}\) and \(\f(x) = \frac{4 }{3 }\,\).
  3. Find the largest root of the equation \(\f(x) =\frac9{10}\,\).
Give necessary and sufficient conditions, in the form of inequalities, for the equation \(\f(x) =c\) to have exactly \(n\) roots, where \(n\ge1\).

Solution:

  1. TikZ diagram
  2. Notice that there are no solutions when \(x < 0\) since \(f(x) \geq 1\) in that region. Suppose \(x = n + \epsilon, 0 < \epsilon < 1\), then \(f(x) = \frac{n}{n+\epsilon}\), ie \(12n = 7n + 7 \epsilon \Rightarrow 5 n = 7\epsilon \Rightarrow \epsilon = \frac{5}{7}n \Rightarrow n < \frac75\), so \(n = 1 ,\epsilon = \frac57, x = \frac{12}5\). \begin{align*} && \frac{17}{24} &= f(x) \\ \Rightarrow && 17n + 17 \epsilon &= 24 n \\ \Rightarrow && 17 \epsilon &= 7 n \\ \Rightarrow && n &< \frac{17}{7} \\ \Rightarrow && n &= 1, 2 \\ \Rightarrow && x &= \frac{24}{17}, \frac{48}{17} \end{align*}. For \(f(x) = \frac{4}{3}\) we notice that \(x < 0\), so let \(x = -n +\epsilon\), ie \begin{align*} && \frac43 &= f(x) \\ \Rightarrow && \frac43 &= \frac{-n}{-n+\epsilon} \\ \Rightarrow && 4\epsilon &= n \\ \Rightarrow && n &= 1,2,3 \\ \Rightarrow && x &= -\frac{5}{4}, -\frac{3}{2}, -\frac{9}{4} \end{align*}
  3. \begin{align*} && \frac9{10} &= f(x) \\ \Rightarrow && 9n + 9 \epsilon &= 10 n \\ \Rightarrow && 9 \epsilon &= n \\ \Rightarrow && n < 9 \end{align} So largest will be when \(n = 8, \epsilon = \frac{8}{9}\), ie \(\frac{80}{9}\)
If \(c < 1\) \begin{align*} && c &= \frac{k}{k + \epsilon} \\ \Rightarrow && \frac{c}{1-c} \epsilon &= k \end{align*} For this to have exactly \(n\) solutions, we need \(n < \frac{c}{1-c} \leq n+1\). If \(c > 1\) \begin{align*} && c &= \frac{-k}{-k+\epsilon} \\ \Rightarrow && c \epsilon &= (c-1) k \\ \Rightarrow && \frac{c}{c-1} \epsilon &= k \end{align*} Therefore for there to be exactly \(n\) solutions we need \(n < \frac{c}{c-1} \leq n+1\)

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2003 Paper 1 Q2
D: 1500.0 B: 1484.0
proof algebraic manipulation equation solving proof by contradiction factorisation necessary and sufficient conditions

The first question on an examination paper is: Solve for \(x\) the equation \(\displaystyle \frac 1x = \frac 1 a + \frac 1b \;.\) where (in the question) \(a\) and \(b\) are given non-zero real numbers. One candidate writes \(x=a+b\) as the solution. Show that there are no values of \(a\) and \(b\) for which this will give the correct answer. The next question on the examination paper is: Solve for \(x\) the equation \(\displaystyle \frac 1x = \frac 1 a + \frac 1b +\frac 1c \;.\) where (in the question) \(a\,\), \(b\) and \(c\) are given non-zero numbers. The candidate uses the same technique, giving the answer as \(\displaystyle x = a + b +c \;.\) Show that the candidate's answer will be correct if and only if \(a\,\), \(b\) and \(c\) satisfy at least one of the equations \(a+b=0\,\), \(b+c=0\) or \(c+a=0\,\).

Solution: Suppose \begin{align*} && \frac{1}{a+b} &= \frac{1}{a} + \frac{1}{b} \\ \Rightarrow && ab &= b(a+b)+a(a+b) \\ &&&= (a+b)^2 \\ \Rightarrow && 0 &= a^2+ab + b^2 \\ &&&= \tfrac12 (a^2+(a+b)^2+b^2) \end{align*} Which clearly has no solution for non-zero \(a,b\). Suppose \begin{align*} && \frac{1}{a+b+c} &= \frac1a + \frac1b+\frac1c \\ \Leftrightarrow && abc &= (a+b+c)(bc+ca+ab) \\ \Leftrightarrow && 0 &= (a+b+c)(bc+ca+ab) - abc \\ &&&= (a+b)(b+c)(c+a) \end{align*} Therefore it is true iff \(a+b = 0\) or \(b+c=0\) or \(c+a =0\) as required.

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1997 Paper 1 Q4
D: 1500.0 B: 1500.0
modulus function piecewise linear case analysis equation solving absolute value equation

Find all the solutions of the equation \[|x+1|-|x|+3|x-1|-2|x-2|=x+2.\]

Solution: Case 1: \(x \leq -1\) \begin{align*} && -1-x+x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && -x-2 &= x + 2 \\ \Leftrightarrow && x = -2 \end{align*} Case \(-1 < x \leq 0\): \begin{align*} && x+1+x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && x&= x + 2 \\ \end{align*} No solutions Case \(0 < x \leq 1\): \begin{align*} && x+1-x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && -x&= x + 2 \\ \end{align*} No solutions Case \(1 < x \leq 2\): \begin{align*} && x+1-x+3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && 5x-6&= x + 2 \\ \Leftrightarrow && x = 2 \end{align*} Case \(2 < x\): \begin{align*} && x+1-x+3(x-1)-2(x-2) &= x + 2 \\ \Leftrightarrow && x+2&= x + 2 \\ \end{align*} Therefore the solutions are \(x \in \{-2\} \cup [2, \infty)\)

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