Sketch single modulus graph

1 Candidates answer on the Question Paper.
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[0pt] The number of marks is given in brackets [ ] at the end of each question or part question.
The total number of marks for this paper is 50. 1

1. Solve the inequality \(| 2 x - 7 | < | 2 x - 9 |\).
2. Hence find the largest integer \(n\) satisfying the inequality \(| 2 \ln n - 7 | < | 2 \ln n - 9 |\).

2

1. Sketch the graph of \(y = | 2 x + 3 |\).
2. Solve the inequality \(3 x + 8 > | 2 x + 3 |\).

1

1. Sketch the graph of \(\mathrm { y } = | \mathrm { x } - 2 \mathrm { a } |\), where \(a\) is a positive constant.
2. Solve the inequality \(2 \mathrm { x } - 3 \mathrm { a } < | \mathrm { x } - 2 \mathrm { a } |\).

1

1. Sketch the graph of \(y = | x - 2 |\).
2. Solve the inequality \(| x - 2 | < 3 x - 4\).

2

1. Sketch the graph of \(y = | 2 x - 3 |\).
2. Solve the inequality \(| 2 x - 3 | < 3 x + 2\).

1

1. Sketch the graph of \(y = | 2 x + 1 |\).
2. Solve the inequality \(3 x + 5 < | 2 x + 1 |\).

1

1. Sketch the graph of \(y = | 4 x - 2 |\).
2. Solve the inequality \(1 + 3 x < | 4 x - 2 |\).

3. The function \(g\) is defined by $$\mathrm { g } : x \mapsto | 8 - 2 x | , \quad x \in \mathbb { R } , \quad x \geqslant 0$$

1. Sketch the graph with equation \(y = \mathrm { g } ( x )\), showing the coordinates of the points where the graph cuts or meets the axes.
2. Solve the equation $$| 8 - 2 x | = x + 5$$ The function f is defined by $$\mathrm { f } : x \mapsto x ^ { 2 } - 3 x + 1 , \quad x \in \mathbb { R } , \quad 0 \leqslant x \leqslant 4$$
3. Find fg(5).
4. Find the range of f. You must make your method clear.

5. Sketch the graph of \(y = \ln | x |\), stating the coordinates of any points of intersection with the axes.

4. The function \(f\) is defined by $$f : x \mapsto | 2 x - 5 | , \quad x \in \mathbb { R }$$

1. Sketch the graph with equation \(y = \mathrm { f } ( x )\), showing the coordinates of the points where the graph cuts or meets the axes.
2. Solve \(\mathrm { f } ( x ) = 15 + x\). The function \(g\) is defined by $$g : x \mapsto x ^ { 2 } - 4 x + 1 , \quad x \in \mathbb { R } , \quad 0 \leqslant x \leqslant 5$$
3. Find fg(2).
4. Find the range of g.

2. Given that $$\mathrm { f } ( x ) = 2 \mathrm { e } ^ { x } - 5 , \quad x \in \mathbb { R }$$

1. sketch, on separate diagrams, the curve with equation
   1. \(y = \mathrm { f } ( x )\)
   2. \(y = | \mathrm { f } ( x ) |\) On each diagram, show the coordinates of each point at which the curve meets or cuts the axes. On each diagram state the equation of the asymptote.
2. Deduce the set of values of \(x\) for which \(\mathrm { f } ( x ) = | \mathrm { f } ( x ) |\)
3. Find the exact solutions of the equation \(| \mathrm { f } ( x ) | = 2\)

4. The function f is defined by $$\mathrm { f } ( x ) \equiv x ^ { 2 } - 2 a x , \quad x \in \mathbb { R }$$ where \(a\) is a positive constant.

1. Showing the coordinates of any points where the graph meets the axes, sketch the graph of \(y = | \mathrm { f } ( x ) |\). The function \(g\) is defined by $$\mathrm { g } ( x ) \equiv 3 a x , \quad x \in \mathbb { R } .$$
2. Find \(\mathrm { fg } ( \mathrm { a } )\) in terms of \(a\).
3. Solve the equation $$\operatorname { gf } ( x ) = 9 a ^ { 3 }$$

1．（a）Sketch the graph of the curve with equation $$y = | \ln ( 2 x + 5 ) | \quad x > - \frac { 5 } { 2 }$$ On your sketch you should clearly state the equations of any asymptotes and mark the coordinates of points where the curve meets the coordinate axes．
（b）Solve the equation \(| \ln ( 2 x + 5 ) | = \ln 9\)

4 Sketch the graph of \(y = | 2 x - 3 |\).

8 A curve has equation $$y = \frac { 5 - 4 x } { 1 + x }$$ 8

1. Sketch the curve. \includegraphics[max width=\textwidth, alt={}, center]{a155b39a-6835-4d62-a481-41ef822bbd5f-10_1205_1219_886_360} 8
2. Hence sketch the graph of \(y = \left| \frac { 5 - 4 x } { 1 + x } \right|\).
   [0pt] [1 mark] \includegraphics[max width=\textwidth, alt={}, center]{a155b39a-6835-4d62-a481-41ef822bbd5f-11_1203_1202_641_331}

4

1. Sketch the graph of \(y = | 8 - 2 x |\).
2. Solve the equation \(| 8 - 2 x | = 4\).
3. Solve the inequality \(| 8 - 2 x | > 4\).

4
7 \(q\) 3 The line \(L\) has equation \(2 x + 3 y = 7\) Which one of the following is perpendicular to \(L\) ?
Tick one box. $$\begin{aligned} & 2 x - 3 y = 7 \\ & 3 x + 2 y = - 7 \\ & 2 x + 3 y = - \frac { 1 } { 7 } \\ & 3 x - 2 y = 7 \end{aligned}$$ □
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□ 4 Sketch the graph of \(y = | 2 x + a |\), where \(a\) is a positive constant. Show clearly where the graph intersects the axes. \includegraphics[max width=\textwidth, alt={}, center]{d9149857-5f94-4fa5-a6d8-550c0c07fefb-03_1001_1002_450_520}

7 \(q\) 3 The line \(L\) has equation \(2 x + 3 y = 7\) Which one of the following is perpendicular to \(L\) ?
Tick one box. $$\begin{aligned} & 2 x - 3 y = 7 \\ & 3 x + 2 y = - 7 \\ & 2 x + 3 y = - \frac { 1 } { 7 } \\ & 3 x - 2 y = 7 \end{aligned}$$ □
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□ 4 Sketch the graph of \(y = | 2 x + a |\), where \(a\) is a positive constant. Show clearly where the graph intersects the axes. \includegraphics[max width=\textwidth, alt={}, center]{d9149857-5f94-4fa5-a6d8-550c0c07fefb-03_1001_1002_450_520} 5 Show that, for small values of \(x\), the graph of \(y = 5 + 4 \sin \frac { x } { 2 } + 12 \tan \frac { x } { 3 }\) can be approximated by a straight line.
6 (b) Use the quotient rule to show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { x - 2 } { ( 2 x - 2 ) ^ { \frac { 3 } { 2 } } }\) 6 (a) State the maximum possible domain of f . \(6 \quad\) A function f is defined by \(\mathrm { f } ( x ) = \frac { x } { \sqrt { 2 x - 2 } }\) $$\begin{gathered} \text { Do not write } \\ \text { outside the } \\ \text { box } \end{gathered}$$ 6 (a)
6 (c) Show that the graph of \(y = \mathrm { f } ( x )\) has exactly one point of inflection.
6 (d) Write down the values of \(x\) for which the graph of \(y = \mathrm { f } ( x )\) is convex.
7 (a) Given that \(\log _ { a } y = 2 \log _ { a } 7 + \log _ { a } 4 + \frac { 1 } { 2 }\), find \(y\) in terms of \(a\).
7 (b) When asked to solve the equation $$2 \log _ { a } x = \log _ { a } 9 - \log _ { a } 4$$ a student gives the following solution: $$\begin{aligned} & 2 \log _ { a } x = \log _ { a } 9 - \log _ { a } 4 \\ & \Rightarrow 2 \log _ { a } x = \log _ { a } \frac { 9 } { 4 } \\ & \Rightarrow \log _ { a } x ^ { 2 } = \log _ { a } \frac { 9 } { 4 } \\ & \Rightarrow x ^ { 2 } = \frac { 9 } { 4 } \\ & \therefore x = \frac { 3 } { 2 } \text { or } - \frac { 3 } { 2 } \end{aligned}$$ Explain what is wrong with the student's solution.