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2016 Paper 1 Q3
D: 1500.0 B: 1487.6
curve sketching floor function greatest integer function trigonometric piecewise function discontinuity step function

In this question, \(\lfloor x \rfloor\) denotes the greatest integer that is less than or equal to \(x\), so that (for example) \(\lfloor 2.9 \rfloor = 2\), \(\lfloor 2\rfloor = 2\) and \(\lfloor -1.5 \rfloor = -2\). On separate diagrams draw the graphs, for \(-\pi \le x \le \pi\), of:

(i) \(y = \lfloor x \rfloor\); (ii) \(y=\sin\lfloor x \rfloor\); (iii) \(y = \lfloor \sin x\rfloor\); (iv) \(y= \lfloor 2\sin x\rfloor\).
In each case, you should indicate clearly the value of \(y\) at points where the graph is discontinuous.

Solution:

  1. TikZ diagram
  2. TikZ diagram
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  4. TikZ diagram

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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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2004 Paper 1 Q2
D: 1484.0 B: 1499.3
integration floor function greatest integer function curve sketching summation geometric series piecewise function

The square bracket notation \(\boldsymbol{[} x\boldsymbol{]}\) means the greatest integer less than or equal to \(x\,\). For example, \(\boldsymbol{[}\pi\boldsymbol{]} = 3\,\), \(\boldsymbol{[}\sqrt{24}\,\boldsymbol{]} = 4\,\) and \(\boldsymbol{[}5\boldsymbol{]}=5\,\).

  1. Sketch the graph of \(y = \sqrt{\boldsymbol{[}x\boldsymbol{]}}\) and show that \[ \displaystyle \int^a_0 \sqrt{\boldsymbol{[}x\boldsymbol{]}} \; \mathrm{d}x = \sum^{a-1}_{r=0} \sqrt{r} \] when \(a\) is a positive integer.
  2. Show that $\displaystyle \int^{a}_0 2_{\vphantom{A}}^{\pmb{\boldsymbol {[} } x \pmb{ \boldsymbol{]}} }\; \mathrm{d}x = 2^{a}-1\( when \)a\( is a positive integer.
  3. Determine an expression for \)\displaystyle \int^{a}_0 2_{\vphantom{\dot A}}^{\pmb{\boldsymbol{[} }x \pmb{ \boldsymbol{]}} } \; \mathrm{d}x\( when \)a$ is positive but not an integer.

Solution:

  1. \(\,\)
    TikZ diagram
    \begin{align*} && \int_0^a \sqrt{\boldsymbol{[}x\boldsymbol{]}} \d x &= \sum_{r=0}^{a-1} \int_{x=r}^{x=r+1} \sqrt{\boldsymbol{[}x\boldsymbol{]}} \d x \\ &&&= \sum_{r=0}^{a-1} \int_r^{r+1} \sqrt{r} \d x \\ &&&= \sum_{r=0}^{a-1} \sqrt{r} \\ \end{align*}
  2. \(\,\) \begin{align*} && \int^{a}_0 2^{\boldsymbol {[} x \boldsymbol{]}} \d x &= \sum_{r=0}^{a-1} \int_{x=r}^{x=r+1} 2^{\boldsymbol {[} x \boldsymbol{]}} \d x \\ &&&= \sum_{r=0}^{a-1} \int_{x=r}^{x=r+1} 2^{r} \d x \\ &&&= \sum_{r=0}^{a-1} 2^{r}\\ &&&= 2^{a}-1 \end{align*}
  3. \(\,\) \begin{align*} && \int^{a}_0 2^{\boldsymbol {[} x \boldsymbol{]}} \d x &= \int_0^{\boldsymbol {[} a \boldsymbol{]}} 2^{\boldsymbol {[} x \boldsymbol{]}} \d x + \int_{\boldsymbol {[} a \boldsymbol{]}}^a 2^{\boldsymbol {[} x \boldsymbol{]}} \d x \\ &&&= 2^{ \boldsymbol {[} a \boldsymbol{]}}-1 + (a-\boldsymbol {[} a \boldsymbol{]})2^{\boldsymbol {[} a \boldsymbol{]}} \\ &&&= (a-\boldsymbol {[} a \boldsymbol{]}+1)2^{\boldsymbol {[} a \boldsymbol{]}} -1 \end{align*}

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