1.02g Inequalities: linear and quadratic in single variable

7. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the graph with equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = \frac { 1 } { 2 } | 2 x + 7 | - 10$$

1. State the coordinates of the vertex, V, of the graph.
2. Solve, using algebra, $$\frac { 1 } { 2 } | 2 x + 7 | - 10 \geqslant \frac { 1 } { 3 } x + 1$$
3. Sketch the graph with equation $$y = | \mathrm { f } ( x ) |$$ stating the coordinates of the local maximum point and each local minimum point.

6. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the graph with equation $$y = | 3 x - 5 a | - 2 a$$ where \(a\) is a positive constant.
The graph

• cuts the \(y\)-axis at the point \(P\)
• cuts the \(x\)-axis at the points \(Q\) and \(R\)
• has a minimum point at \(S\)
• Find, in simplest form in terms of \(a\), the values of \(x\) for which

$$| 3 x - 5 a | - 2 a = | x - 2 a |$$

6. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the graph \(y = \mathrm { f } ( x )\), where $$f ( x ) = 3 | x - 2 | - 10$$ The vertex of the graph is at point \(P\), shown in Figure 2.

1. Find the coordinates of \(P\)
2. Find \(\mathrm { ff } ( 0 )\)
3. Solve the inequality $$3 | x - 2 | - 10 < 5 x + 10$$
4. Solve the equation $$\mathrm { f } ( | x | ) = 0$$

1. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the graph with equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = 2 | x - 5 | + 10$$ The point \(P\), shown in Figure 1, is the vertex of the graph.

1. State the coordinates of \(P\)
2. Use algebra to solve $$2 | x - 5 | + 10 > 6 x$$ (Solutions relying on calculator technology are not acceptable.)
3. Find the point to which \(P\) is mapped, when the graph with equation \(y = \mathrm { f } ( x )\) is transformed to the graph with equation \(y = 3 \mathrm { f } ( x - 2 )\)

2. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the graph with equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = | 3 x - 13 | + 5 \quad x \in \mathbb { R }$$ The vertex of the graph is at point \(P\), as shown in Figure 1.

1. State the coordinates of \(P\).
2. 1. State the range of f .
   2. Find the value of ff(4)
3. Solve, using algebra and showing your working, $$16 - 2 x > | 3 x - 13 | + 5$$ The graph with equation \(y = \mathrm { f } ( x )\) is transformed onto the graph with equation \(y = a \mathrm { f } ( x + b )\) The vertex of the graph with equation \(y = a \mathrm { f } ( x + b )\) is \(( 4,20 )\) Given that \(a\) and \(b\) are constants,
4. find the value of \(a\) and the value of \(b\).

1. In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve with equation $$y = | 2 - 4 \ln ( x + 1 ) | \quad x > k$$ where \(k\) is a constant.
Given that the curve

• has an asymptote at \(x = k\)
• cuts the \(y\)-axis at point \(A\)
• meets the \(x\)-axis at point \(B\) as shown in Figure 2,
  1. state the value of \(k\)
  2. 1. find the \(y\) coordinate of \(A\)
     2. find the exact \(x\) coordinate of \(B\)
  3. Using algebra and showing your working, find the set of values of \(x\) such that

$$| 2 - 4 \ln ( x + 1 ) | > 3$$

2. $$f ( x ) = 1 - \frac { 3 } { x + 2 } + \frac { 3 } { ( x + 2 ) ^ { 2 } } , x \neq - 2$$

1. Show that \(\mathrm { f } ( x ) = \frac { x ^ { 2 } + x + 1 } { ( x + 2 ) ^ { 2 } } , x \neq - 2\).
2. Show that \(x ^ { 2 } + x + 1 > 0\) for all values of \(x\).
3. Show that \(\mathrm { f } ( x ) > 0\) for all values of \(x , x \neq - 2\).

5. (a) Sketch the graph with equation $$y = | 4 x - 3 |$$ stating the coordinates of any points where the graph cuts or meets the axes. Find the complete set of values of \(x\) for which
(b) $$| 4 x - 3 | > 2 - 2 x$$ (c) $$| 4 x - 3 | > \frac { 3 } { 2 } - 2 x$$

2.

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), writing your answer as a single fraction in its simplest form.
2. Hence find the set of values of \(x\) for which \(\frac { \mathrm { d } y } { \mathrm {~d} x } < 0\) 2. $$y = \frac { 4 x } { x ^ { 2 } + 5 }$$

3. Use algebra to obtain the set of values of \(x\) for which $$\left| x ^ { 2 } + x - 2 \right| < \frac { 1 } { 2 } ( x + 5 )$$

3. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve \(C _ { 1 }\) with equation $$y = \frac { 4 x } { 4 - | x | }$$ and the curve \(C _ { 2 }\) with equation $$y = x ^ { 2 } - 8 x$$ For \(x > 0 , C _ { 1 }\) has equation \(y = \frac { 4 x } { 4 - x }\)

1. Use algebra to show that \(C _ { 1 }\) touches \(C _ { 2 }\) at a point \(P\), stating the coordinates of \(P\)
2. Hence or otherwise, using algebra, solve the inequality $$x ^ { 2 } - 8 x > \frac { 4 x } { 4 - | x | }$$

1. In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
Use algebra to determine the set of values of \(x\) for which $$\frac { x ^ { 2 } - 9 } { | x + 8 | } > 6 - 2 x$$

1. Using algebra, solve the inequality

$$\frac { 1 } { x + 2 } > 2 x + 3$$

2. Use algebra to find the set of values of \(x\) for which $$\frac { 6 } { x - 3 } \leqslant x + 2$$

1. Using algebra, find the set of values of \(x\) for which

$$\frac { x } { x + 2 } < \frac { 2 } { x + 5 }$$

2. Use algebra to find the set of values of \(x\) for which $$\left| x ^ { 2 } - 9 \right| < | 1 - 2 x |$$

1. Use algebra to find the set of values of \(x\) for which

$$\frac { x - 4 } { ( x + 3 ) } \leqslant \frac { 5 } { x ( x + 3 ) }$$

3. Use algebra to obtain the set of values of \(x\) for which $$\left| \frac { x ^ { 2 } + 3 x + 10 } { x + 2 } \right| < 7 - x$$

1. Use algebra to find the set of values of \(x\) for which

$$\left| 2 x ^ { 2 } + x - 3 \right| > 3 ( 1 - x )$$ [Solutions based entirely on graphical or numerical methods are not acceptable.] \includegraphics[max width=\textwidth, alt={}, center]{0d44aec7-a6e8-47fc-a215-7c8c4790e93f-21_2647_1840_118_111}

1. (a) Use algebra to determine the set of values of \(x\) for which

$$x - 5 < \frac { 9 } { x + 3 }$$ (b) Hence, or otherwise, determine the set of values of \(x\) for which $$x - 5 < \frac { 9 } { | x + 3 | }$$

1. In this question you must show all stages of your working.

\section*{Solutions relying on calculator technology are not acceptable.} Given that $$\frac { x + 2 } { x + 4 } \leqslant \frac { x } { k ( x - 1 ) }$$ where \(k\) is a positive constant,

1. show that $$( x + 4 ) ( x - 1 ) \left( p x ^ { 2 } + q x + r \right) \leqslant 0$$ where \(p , q\) and \(r\) are expressions in terms of \(k\) to be determined.
2. Hence, or otherwise, determine the values for \(x\) for which $$\frac { x + 2 } { x + 4 } \leqslant \frac { x } { 3 ( x - 1 ) }$$

1. In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
Use algebra to determine the values of \(x\) for which $$\frac { x + 1 } { ( x - 3 ) ( x + 2 ) } \leqslant 1 - \frac { 2 } { x - 3 }$$

Find the set of values of \(x\) for which \(\frac { x ^ { 2 } } { x - 2 } > 2 x\).
(Total 6 marks)

5. Using algebra, find the set of values of \(x\) for which \(2 x - 5 > \frac { 3 } { x }\).

5. Solve the inequality \(\frac { 1 } { 2 x + 1 } > \frac { x } { 3 x - 2 }\).