1.06d Natural logarithm: ln(x) function and properties

1. The functions f and g are defined by

$$\mathrm { f } : x \mapsto \mathrm { e } ^ { x } + 2 \quad x \in \mathbb { R }$$ $$\mathrm { g } : x \mapsto \ln x \quad x > 0$$

1. State the range of f .
2. Find \(\mathrm { fg } ( x )\), giving \(y\) our answer in its simplest form.
3. Find the exact value of \(x\) for which \(\mathrm { f } ( 2 x + 3 ) = 6\)
4. Find \(\mathrm { f } ^ { - 1 }\) stating its domain.
5. On the same axes sketch the curves with equation \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\), giving the coordinates of all the points where the curves cross the axes.

3. $$\mathrm { f } ( x ) = \frac { x ^ { 2 } } { 4 } + \ln ( 2 x ) , \quad x > 0$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be rewritten as $$x = \frac { 1 } { 2 } \mathrm { e } ^ { - \frac { 1 } { 4 } x ^ { 2 } }$$ The equation \(\mathrm { f } ( x ) = 0\) has a root near 0.5
2. Starting with \(x _ { 1 } = 0.5\) use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { - \frac { 1 } { 4 } x _ { n } ^ { 2 } }$$ to calculate the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
3. Using a suitable interval, show that 0.473 is a root of \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.

9. $$\mathrm { f } ( x ) = 2 \ln ( x ) - 4 , \quad x > 0 , \quad x \in \mathbb { R }$$

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. Find the exact solutions of the equation \(| \mathrm { f } ( x ) | = 4\) $$\mathrm { g } ( x ) = \mathrm { e } ^ { x + 5 } - 2 , \quad x \in \mathbb { R }$$
3. Find \(\mathrm { gf } ( x )\), giving your answer in its simplest form.
4. Hence, or otherwise, state the range of gf.

9. A rare species of mammal is being studied. The population \(P\), \(t\) years after the study started, is modelled by the formula $$P = \frac { 900 \mathrm { e } ^ { \frac { 1 } { 4 } t } } { 3 \mathrm { e } ^ { \frac { 1 } { 4 } t } - 1 } , \quad t \in \mathbb { R } , \quad t \geqslant 0$$ Using the model,

1. calculate the number of mammals at the start of the study,
2. calculate the exact value of \(t\) when \(P = 315\) Give your answer in the form \(a \ln k\), where \(a\) and \(k\) are integers to be determined.
3. 1. Find \(\frac { \mathrm { d } P } { \mathrm {~d} t }\)
   2. Hence find the value of \(\frac { \mathrm { d } P } { \mathrm {~d} t }\) when \(t = 8\), giving your answer to 2 decimal places.

1. The functions \(f\) and \(g\) are defined by

$$\begin{array} { l l } \mathrm { f } : x \mapsto \mathrm { e } ^ { - x } + 2 , & x \in \mathbb { R } \\ \mathrm {~g} : x \mapsto 2 \ln x , & x > 0 \end{array}$$

1. Find \(\mathrm { fg } ( x )\), giving your answer in its simplest form.
2. Find the exact value of \(x\) for which \(\mathrm { f } ( 2 x + 3 ) = 6\)
3. Find \(\mathrm { f } ^ { - 1 }\), stating its domain.
4. On the same axes, sketch the curves with equation \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\), giving the coordinates of all the points where the curves cross the axes.

1. The function \(f\) is defined by

$$\mathrm { f } : x \mapsto \ln ( 4 - 2 x ) , \quad x < 2 \quad \text { and } \quad x \in \mathbb { R } .$$

1. Show that the inverse function of f is defined by $$\mathrm { f } ^ { - 1 } : x \mapsto 2 - \frac { 1 } { 2 } \mathrm { e } ^ { x }$$ and write down the domain of \(\mathrm { f } ^ { - 1 }\).
2. Write down the range of \(\mathrm { f } ^ { - 1 }\).
3. In the space provided on page 16, sketch the graph of \(y = f ^ { - 1 } ( x )\). State the coordinates of the points of intersection with the \(x\) and \(y\) axes. The graph of \(y = x + 2\) crosses the graph of \(y = f ^ { - 1 } ( x )\) at \(x = k\). The iterative formula $$x _ { n + 1 } = - \frac { 1 } { 2 } e ^ { x _ { n } } , x _ { 0 } = - 0.3$$ is used to find an approximate value for \(k\).
4. Calculate the values of \(x _ { 1 }\) and \(x _ { 2 }\), giving your answers to 4 decimal places.
5. Find the value of \(k\) to 3 decimal places.

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

5. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = ( 8 - x ) \ln x , \quad x > 0$$ The curve cuts the \(x\)-axis at the points \(A\) and \(B\) and has a maximum turning point at \(Q\), as shown in Figure 1.

1. Write down the coordinates of \(A\) and the coordinates of \(B\).
2. Find f'(x).
3. Show that the \(x\)-coordinate of \(Q\) lies between 3.5 and 3.6
4. Show that the \(x\)-coordinate of \(Q\) is the solution of $$x = \frac { 8 } { 1 + \ln x }$$ To find an approximation for the \(x\)-coordinate of \(Q\), the iteration formula $$x _ { n + 1 } = \frac { 8 } { 1 + \ln x _ { n } }$$ is used.
5. Taking \(x _ { 0 } = 3.55\), find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\). Give your answers to 3 decimal places.

7. For the constant \(k\), where \(k > 1\), the functions f and g are defined by $$\begin{aligned} & \mathrm { f } : x \mapsto \ln ( x + k ) , \quad x > - k , \\ & \mathrm {~g} : x \mapsto | 2 x - k | , \quad x \in \mathbb { R } . \end{aligned}$$

1. On separate axes, sketch the graph of f and the graph of g . On each sketch state, in terms of \(k\), the coordinates of points where the graph meets the coordinate axes.
2. Write down the range of f.
3. Find \(\mathrm { fg } \left( \frac { k } { 4 } \right)\) in terms of \(k\), giving your answer in its simplest form. The curve \(C\) has equation \(y = \mathrm { f } ( x )\). The tangent to \(C\) at the point with \(x\)-coordinate 3 is parallel to the line with equation \(9 y = 2 x + 1\).
4. Find the value of \(k\).

4. The function \(f\) is defined by $$\mathrm { f } : x \mapsto 4 - \ln ( x + 2 ) , \quad x \in \mathbb { R } , x \geqslant - 1$$

1. Find \(\mathrm { f } ^ { - 1 } ( x )\).
2. Find the domain of \(\mathrm { f } ^ { - 1 }\). The function \(g\) is defined by $$\mathrm { g } : x \mapsto \mathrm { e } ^ { x ^ { 2 } } - 2 , \quad x \in \mathbb { R }$$
3. Find \(\mathrm { fg } ( x )\), giving your answer in its simplest form.
4. Find the range of fg.

6. The functions \(f\) and \(g\) are defined by $$\begin{aligned} & \mathrm { f } : x \mapsto \mathrm { e } ^ { x } + 2 , \quad x \in \mathbb { R } \\ & \mathrm {~g} : x \mapsto \ln x , \quad x > 0 \end{aligned}$$

1. State the range of f.
2. Find \(\mathrm { fg } ( x )\), giving your answer in its simplest form.
3. Find the exact value of \(x\) for which \(\mathrm { f } ( 2 x + 3 ) = 6\)
4. Find \(\mathrm { f } ^ { - 1 }\), the inverse function of f , stating its domain.
5. On the same axes sketch the curves with equation \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\), giving the coordinates of all the points where the curves cross the axes.

1. (a) On the same diagram, sketch and clearly label the graphs with equations

$$y = \mathrm { e } ^ { x } \quad \text { and } \quad y = 10 - x$$ Show on your sketch the coordinates of each point at which the graphs cut the axes.
(b) Explain why the equation \(\mathrm { e } ^ { x } - 10 + x = 0\) has only one solution.
(c) Show that the solution of the equation $$\mathrm { e } ^ { x } - 10 + x = 0$$ lies between \(x = 2\) and \(x = 3\) (d) Use the iterative formula $$x _ { n + 1 } = \ln \left( 10 - x _ { n } \right) , \quad x _ { 1 } = 2$$ to calculate the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\).
Give your answers to 4 decimal places.

2. Given that $$\mathrm { f } ( x ) = \ln x , \quad x > 0$$ sketch on separate axes the graphs of

1. \(\quad y = \mathrm { f } ( x )\),
2. \(y = | \mathrm { f } ( x ) |\),
3. \(y = - \mathrm { f } ( x - 4 )\). Show, on each diagram, the point where the graph meets or crosses the \(x\)-axis. In each case, state the equation of the asymptote.

6. The function f is defined by $$\mathrm { f } : x \rightarrow \mathrm { e } ^ { 2 x } + k ^ { 2 } , \quad x \in \mathbb { R } , \quad k \text { is a positive constant. }$$

1. State the range of f .
2. Find \(\mathrm { f } ^ { - 1 }\) and state its domain. The function g is defined by $$g : x \rightarrow \ln ( 2 x ) , \quad x > 0$$
3. Solve the equation $$\mathrm { g } ( x ) + \mathrm { g } \left( x ^ { 2 } \right) + \mathrm { g } \left( x ^ { 3 } \right) = 6$$ giving your answer in its simplest form.
4. Find \(\mathrm { fg } ( x )\), giving your answer in its simplest form.
5. Find, in terms of the constant \(k\), the solution of the equation $$\mathrm { fg } ( x ) = 2 k ^ { 2 }$$

6. \begin{figure}[h] \end{figure} Figure 1 is a sketch showing part of the curve with equation \(y = 2 ^ { x + 1 } - 3\) and part of the line with equation \(y = 17 - x\). The curve and the line intersect at the point \(A\).

1. Show that the \(x\) coordinate of \(A\) satisfies the equation $$x = \frac { \ln ( 20 - x ) } { \ln 2 } - 1$$
2. Use the iterative formula $$x _ { n + 1 } = \frac { \ln \left( 20 - x _ { n } \right) } { \ln 2 } - 1 , \quad x _ { 0 } = 3$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.
3. Use your answer to part (b) to deduce the coordinates of the point \(A\), giving your answers to one decimal place.

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = g ( x )\), where $$\mathrm { g } ( x ) = \left| 4 \mathrm { e } ^ { 2 x } - 25 \right| , \quad x \in \mathbb { R }$$ The curve cuts the \(y\)-axis at the point \(A\) and meets the \(x\)-axis at the point \(B\). The curve has an asymptote \(y = k\), where \(k\) is a constant, as shown in Figure 1

1. Find, giving each answer in its simplest form,
   1. the \(y\) coordinate of the point \(A\),
   2. the exact \(x\) coordinate of the point \(B\),
   3. the value of the constant \(k\). The equation \(\mathrm { g } ( x ) = 2 x + 43\) has a positive root at \(x = \alpha\)
2. Show that \(\alpha\) is a solution of \(x = \frac { 1 } { 2 } \ln \left( \frac { 1 } { 2 } x + 17 \right)\) The iteration formula $$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( \frac { 1 } { 2 } x _ { n } + 17 \right)$$ can be used to find an approximation for \(\alpha\)
3. Taking \(x _ { 0 } = 1.4\) find the values of \(x _ { 1 }\) and \(x _ { 2 }\) Give each answer to 4 decimal places.
4. By choosing a suitable interval, show that \(\alpha = 1.437\) to 3 decimal places. \includegraphics[max width=\textwidth, alt={}, center]{d3ba2776-eedb-48f0-834f-41aa454afba3-07_2258_47_315_37}

7. (a) Sketch the curve with equation \(y = \ln x\).
(b) Show that the tangent to the curve with equation \(y = \ln x\) at the point ( \(\mathrm { e } , 1\) ) passes through the origin.
(c) Use your sketch to explain why the line \(y = m x\) cuts the curve \(y = \ln x\) between \(x = 1\) and \(x = \mathrm { e }\) if \(0 < m < \frac { 1 } { \mathrm { e } }\). Taking \(x _ { 0 } = 1.86\) and using the iteration \(x _ { n + 1 } = \mathrm { e } ^ { \frac { 1 } { 3 } x _ { n } }\),
(d) calculate \(x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 }\) and \(x _ { 5 }\), giving your answer to \(x _ { 5 }\) to 3 decimal places. The root of \(\ln x - \frac { 1 } { 3 } x = 0\) is \(\alpha\).
(e) By considering the change of sign of \(\ln x - \frac { 1 } { 3 } x\) over a suitable interval, show that your answer for \(x _ { 5 }\) is an accurate estimate of \(\alpha\), correct to 3 decimal places.

5. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation \(y = \mathrm { f } ( x ) , x \in \mathbb { R }\). The curve meets the \(x\)-axis at \(P ( p , 0 )\) and meets the \(y\)-axis at \(Q ( 0 , q )\).

1. On separate diagrams, sketch the curve with equation
   1. \(y = | \mathrm { f } ( x ) |\),
   2. \(y = 3 \mathrm { f } \left( \frac { 1 } { 2 } x \right)\). In each case show, in terms of \(p\) or \(q\), the coordinates of points at which the curve meets the axes. Given that \(\mathrm { f } ( x ) = 3 \ln ( 2 x + 3 )\),
2. state the exact value of \(q\),
3. find the value of \(p\),
4. find an equation for the tangent to the curve at \(P\).

4. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve with equation \(y = x ^ { 3 } \ln \left( x ^ { 2 } + 2 \right) , x \geqslant 0\).
The finite region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis and the line \(x = \sqrt { } 2\). The table below shows corresponding values of \(x\) and \(y\) for \(y = x ^ { 3 } \ln \left( x ^ { 2 } + 2 \right)\).

1. Complete the table above giving the missing values of \(y\) to 4 decimal places.
2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the area of \(R\), giving your answer to 2 decimal places.
3. Use the substitution \(u = x ^ { 2 } + 2\) to show that the area of \(R\) is $$\frac { 1 } { 2 } \int _ { 2 } ^ { 4 } ( u - 2 ) \ln u \mathrm {~d} u$$
4. Hence, or otherwise, find the exact area of \(R\).

1. Find the gradient of the curve with equation

$$\ln y = 2 x \ln x , \quad x > 0 , y > 0$$ at the point on the curve where \(x = 2\). Give your answer as an exact value.

7. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve with equation \(y = x ^ { \frac { 1 } { 2 } } \ln 2 x\).
The finite region \(R\), shown shaded in Figure 3, is bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 4\)

1. Use the trapezium rule, with 3 strips of equal width, to find an estimate for the area of \(R\), giving your answer to 2 decimal places.
2. Find \(\int x ^ { \frac { 1 } { 2 } } \ln 2 x \mathrm {~d} x\).
3. Hence find the exact area of \(R\), giving your answer in the form \(a \ln 2 + b\), where \(a\) and \(b\) are exact constants.

2. The curve \(C\) has equation $$3 ^ { x - 1 } + x y - y ^ { 2 } + 5 = 0$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at the point \(( 1,3 )\) on the curve \(C\) can be written in the form \(\frac { 1 } { \lambda } \ln \left( \mu \mathrm { e } ^ { 3 } \right)\), where \(\lambda\) and \(\mu\) are integers to be found.

5. \begin{figure}[h] \end{figure}

1. Find \(\int \frac { \ln x } { x ^ { 2 } } \mathrm {~d} x\) Figure 3 shows a sketch of part of the curve with equation $$y = \frac { 3 + 2 x + \ln x } { x ^ { 2 } } \quad x > 0.5$$ The finite region \(R\), shown shaded in Figure 3, is bounded by the curve, the line with equation \(x = 2\), the \(x\)-axis and the line with equation \(x = 4\)
2. Use the answer to part (a) to find the exact area of \(R\), writing your answer in simplest form.

6. A curve \(C\) has equation $$y = x ^ { \sin x } \quad x > 0 \quad y > 0$$

1. Find, by firstly taking natural logarithms, an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
2. Hence show that the \(x\) coordinates of the stationary points of \(C\) are solutions of the equation $$\tan x + x \ln x = 0$$

1. Solve the equation

$$18 \cosh x + 14 \sinh x = 11 + \mathrm { e } ^ { x }$$ Give your answers in the form \(\ln a\), where \(a\) is rational.