1.02l Modulus function: notation, relations, equations and inequalities

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

1. Sketch the graph of \(y = f(x)\), showing the coordinates of the points where the graph cuts the axes. [2]
2. On a separate diagram, sketch the graph of \(y = f(2x)\), showing the coordinates of the points where the graph cuts the axes. [2]
3. Given that a solution of the equation \(f(x) = \frac{1}{2}x\) is \(x = 4\), find the two possible values of \(a\). [4]

$$f(x) = x^2 - 2x - 3, \quad x \in \mathbb{R}, x \geq 1.$$

1. Find the range of \(f\). [1]
2. Write down the domain and range of \(f^{-1}\). [2]
3. Sketch the graph of \(f^{-1}\), indicating clearly the coordinates of any point at which the graph intersects the coordinate axes. [4]

Given that \(g(x) = |x - 4|, x \in \mathbb{R}\),

1. find an expression for \(gf(x)\). [2]
2. Solve \(gf(x) = 8\). [5]

\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = e^{-x} - 1\).

1. Copy Fig. 1 and on the same axes sketch the graph of \(y = \frac{1}{2}|x - 1|\). Show the coordinates of the points where the graph meets the axes. [2]

The \(x\)-coordinate of the point of intersection of the graphs is \(\alpha\).

1. Show that \(x = \alpha\) is a root of the equation \(x + 2e^{-x} - 3 = 0\). [3]
2. Show that \(-1 < \alpha < 0\). [2]

The iterative formula \(x_{n+1} = -\ln[\frac{1}{2}(3 - x_n)]\) is used to solve the equation \(x + 2e^{-x} - 3 = 0\).

1. Starting with \(x_0 = -1\), find the values of \(x_1\) and \(x_2\). [2]
2. Show that, to 2 decimal places, \(\alpha = -0.58\). [2]

\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = f(x), -1 \leq x \leq 3\). The curve touches the \(x\)-axis at the origin \(O\), crosses the \(x\)-axis at the point \(A(2, 0)\) and has a maximum at the point \(B(\frac{4}{3}, 1)\). In separate diagrams, show a sketch of the curve with equation

1. \(y = f(x + 1)\), [3]
2. \(y = |f(x)|\), [3]
3. \(y = f(|x|)\), [4]

marking on each sketch the coordinates of points at which the curve

1. has a turning point,
2. meets the \(x\)-axis.

The functions \(f\) and \(g\) are defined by $$f: x \mapsto |x - a| + a, \quad x \in \mathbb{R},$$ $$g: x \mapsto 4x + a, \quad x \in \mathbb{R},$$ where \(a\) is a positive constant.

1. On the same diagram, sketch the graphs of \(f\) and \(g\), showing clearly the coordinates of any points at which your graphs meet the axes. [5]
2. Use algebra to find, in terms of \(a\), the coordinates of the point at which the graphs of \(f\) and \(g\) intersect. [3]
3. Find an expression for \(fg(x)\). [2]
4. Solve, for \(x\) in terms of \(a\), the equation $$fg(x) = 3a.$$ [3]

Find the set of values of \(x\) for which $$|x^2 - 4| > 3x.$$ [5]

Find the set of values for which $$|x - 1| > 6x - 1.$$ [5]

Find the complete set of values of \(x\) for which $$|x^2 - 2| > 2x.$$ [7]

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

1. Sketch the graph of \(y = f(x)\), showing the coordinates of the points where the graph cuts the axes. [2]
2. On a separate diagram, sketch the graph of \(y = f(2x)\), showing the coordinates of the points where the graph cuts the axes. [2]
3. Given that a solution of the equation f(x) = \(\frac{1}{2}x\) is \(x = 4\), find the two possible values of \(a\). [4]

The function f is given by $$f: x \mapsto \ln(3x - 6), \quad x \in \mathbb{R}, \quad x > 2$$

1. Find \(f^{-1}(x)\). [3]
2. Write down the domain of \(f^{-1}\) and the range of \(f^{-1}\). [2]
3. Find, to 3 significant figures, the value of \(x\) for which f(x) = 3. [2]

The function g is given by $$g: x \mapsto \ln|3x - 6|, \quad x \in \mathbb{R}, \quad x \neq 2$$

1. Sketch the graph of \(y = g(x)\). [3]
2. Find the exact coordinates of all the points at which the graph of \(y = g(x)\) meets the coordinate axes. [3]

The functions f and g are defined by $$\text{f}: x \alpha |x - a| + a, \quad x \in \mathbb{R},$$ $$\text{g}: x \alpha 4x + a, \quad x \in \mathbb{R}.$$ where \(a\) is a positive constant.

1. On the same diagram, sketch the graphs of f and g, showing clearly the coordinates of any points at which your graphs meet the axes. [5]
2. Use algebra to find, in terms of \(a\), the coordinates of the point at which the graphs of f and g intersect. [3]
3. Find an expression for fg(x). [2]
4. Solve, for \(x\) in terms of \(a\), the equation $$\text{fg}(x) = 3a.$$ [3]

\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = e^{-x} - 1\).

1. Copy Fig. 1 and on the same axes sketch the graph of \(y = \frac{1}{2}|x - 1|\). Show the coordinates of the points where the graph meets the axes. [2]

The \(x\)-coordinate of the point of intersection of the graph is \(\alpha\).

1. Show that \(x = \alpha\) is a root of the equation \(x + 2e^{-x} - 3 = 0\). [3]
2. Show that \(-1 < \alpha < 0\). [2]

The iterative formula \(x_{n+1} = -\ln[\frac{1}{2}(3 - x_n)]\) is used to solve the equation \(x + 2e^{-x} - 3 = 0\).

1. Starting with \(x_0 = -1\), find the values of \(x_1\) and \(x_2\). [2]
2. Show that, to 2 decimal places, \(\alpha = -0.58\). [2]

f(x) = \(x^2 - 2x - 3\), \(x \in \mathbb{R}\), \(x \geq 1\).

1. Find the range of f. [1]
2. Write down the domain and range of \(f^{-1}\). [2]
3. Sketch the graph of \(f^{-1}\), indicating clearly the coordinates of any point at which the graph intersects the coordinate axes. [4]

Given that g(x) = \(|x - 4|\), \(x \in \mathbb{R}\),

1. find an expression for gf(x). [2]
2. Solve gf(x) = 8. [5]

Find the exact solutions of the equation \(|6x - 1| = |x - 1|\). [4]