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

1. Solve the equation

$$15 \operatorname { sech } ^ { 2 } x + 7 \tanh x = 13$$ Give your answers in terms of simplified natural logarithms.

9 The function \(\mathrm { f } ( x ) = \ln \left( 1 + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = \mathrm { f } ( x )\). \begin{figure}[h] \end{figure}

1. Show algebraically that the function is even. State how this property relates to the shape of the curve.
2. Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
3. Explain why the function does not have an inverse for the domain \(- 3 \leqslant x \leqslant 3\). The domain of \(\mathrm { f } ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(\mathrm { f } ( x )\) is the function \(\mathrm { g } ( x )\).
4. Sketch the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) on the same axes. State the domain of the function \(\mathrm { g } ( x )\). Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
5. Differentiate \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the connection between this result and your answer to part (ii).

9

1. Sketch the curve \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\), and state the coordinates of any point where the curve crosses an axis.
2. Use the trapezium rule, with 4 strips of width 0.5 , to estimate the area of the region bounded by the curve \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\), the axes, and the line \(x = 2\).
3. The point \(P\) on the curve \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\) has \(y\)-coordinate equal to \(\frac { 1 } { 6 }\). Prove that the \(x\)-coordinate of \(P\) may be written as $$1 + \frac { \log _ { 10 } 3 } { \log _ { 10 } 2 }$$

9 The polynomial \(f ( x )\) is given by $$f ( x ) = x ^ { 3 } + 6 x ^ { 2 } + x - 4 .$$

1. (a) Show that ( \(\mathrm { x } + 1\) ) is a factor of \(\mathrm { f } ( \mathrm { x } )\).
   (b) Hence find the exact roots of the equation \(f ( x ) = 0\).
2. (a) Show that the equation $$2 \log _ { 2 } ( x + 3 ) + \log _ { 2 } x - \log _ { 2 } ( 4 x + 2 ) = 1$$ can be written in the form \(f ( x ) = 0\).
   (b) Explain why the equation $$2 \log _ { 2 } ( x + 3 ) + \log _ { 2 } x - \log _ { 2 } ( 4 x + 2 ) = 1$$ has only one real root and state the exact value of this root.

11 Answer part (iii) of this question on the insert provided.
A hot drink is made and left to cool. The table shows its temperature at ten-minute intervals after it is made. The room temperature is \(22 ^ { \circ } \mathrm { C }\). The difference between the temperature of the drink and room temperature at time \(t\) minutes is \(z ^ { \circ } \mathrm { C }\). The relationship between \(z\) and \(t\) is modelled by $$z = z _ { 0 } 10 ^ { - k t }$$ where \(z _ { 0 }\) and \(k\) are positive constants.

1. Give a physical interpretation for the constant \(z _ { 0 }\).
2. Show that \(\log _ { 10 } z = - k t + \log _ { 10 } z _ { 0 }\).
3. On the insert, complete the table and draw the graph of \(\log _ { 10 } z\) against \(t\). Use your graph to estimate the values of \(k\) and \(z _ { 0 }\).
   Hence estimate the temperature of the drink 70 minutes after it is made. \section*{OXFORD CAMBRIDGE AND RSA EXAMINATIONS} Advanced Subsidiary General Certificate of Education Advanced General Certificate of Education \section*{MEI STRUCTURED MATHEMATICS} Concepts for Advanced Mathematics (C2)
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   Wednesday

9 You are given that \(\log _ { 10 } y = 3 x + 2\).

1. Find the value of \(x\) when \(y = 500\), giving your answer correct to 2 decimal places.
2. Find the value of \(y\) when \(x = - 1\).
3. Express \(\log _ { 10 } \left( y ^ { 4 } \right)\) in terms of \(x\).
4. Find an expression for \(y\) in terms of \(x\). Section B (36 marks)

5. The function \(f\) is defined by $$\mathrm { f } ( x ) \equiv 4 - \ln 3 x , \quad x \in \mathbb { R } , \quad x > 0$$

1. Solve the equation \(\mathrm { f } ( x ) = 0\).
2. Sketch the curve \(y = \mathrm { f } ( x )\). The function g is defined by $$\mathrm { g } ( x ) \equiv \mathrm { e } ^ { 2 - x } , \quad x \in \mathbb { R }$$
3. Show that $$\operatorname { fg } ( x ) = x + a - \ln b$$ where \(a\) and \(b\) are integers to be found.

1. The diagram shows the curve with equation \(y = \ln ( 2 + \cos x ) , x \geq 0\).
The smallest value of \(x\) for which the curve meets the \(x\)-axis is \(a\) as shown.

1. Find the value of \(a\).
2. Use Simpson's rule with four strips of equal width to estimate the area of the region bounded by the curve in the interval \(0 \leq x \leq a\) and the coordinate axes.

2. \includegraphics[max width=\textwidth, alt={}, center]{b124d427-1f9b-4770-95bb-ed79bae5b4fb-1_460_805_587_486} The diagram shows the curve with equation \(y = \frac { 1 } { 2 } \ln 3 x\).

1. Express the equation of the curve in the form \(x = \mathrm { f } ( y )\). The shaded region is bounded by the curve, the coordinate axes and the line \(y = 1\).
2. Find, in terms of \(\pi\) and e, the volume of the solid formed when the shaded region is rotated through four right angles about the \(y\)-axis.

6. \(\quad f ( x ) = 2 x ^ { 2 } + 3 \ln ( 2 - x ) , \quad x \in \mathbb { R } , \quad x < 2\).

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form $$x = 2 - \mathrm { e } ^ { k x ^ { 2 } }$$ where \(k\) is a constant to be found. The root, \(\alpha\), of the equation \(\mathrm { f } ( x ) = 0\) is 1.9 correct to 1 decimal place.
2. Use the iterative formula $$x _ { n + 1 } = 2 - \mathrm { e } ^ { k x _ { n } ^ { 2 } }$$ with \(x _ { 0 } = 1.9\) and your value of \(k\), to find \(\alpha\) correct to 3 decimal places.
   You should show the result of each iteration.
3. Solve the equation \(\mathrm { f } ^ { \prime } ( x ) = 0\).

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

$$\begin{aligned} & \mathrm { f } : x \rightarrow | 2 x - 5 | , \quad x \in \mathbb { R } , \\ & \mathrm {~g} : x \rightarrow \ln ( x + 3 ) , \quad x \in \mathbb { R } , \quad x > - 3 \end{aligned}$$

1. State the range of f .
2. Evaluate fg(-2).
3. Solve the equation $$\operatorname { fg } ( x ) = 3$$ giving your answers in exact form.
4. Show that the equation $$\mathrm { f } ( x ) = \mathrm { g } ( x )$$ has a root, \(\alpha\), in the interval [3,4].
5. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left[ 5 + \ln \left( x _ { n } + 3 \right) \right]$$ with \(x _ { 0 } = 3\), to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 significant figures.
6. Show that your answer for \(x _ { 4 }\) is the value of \(\alpha\) correct to 4 significant figures.

7 The curve \(y = \ln x\) is transformed to the curve \(y = \ln \left( \frac { 1 } { 2 } x - a \right)\) by means of a translation followed by a stretch. It is given that \(a\) is a positive constant.

1. Give full details of the translation and stretch involved.
2. Sketch the graph of \(y = \ln \left( \frac { 1 } { 2 } x - a \right)\).
3. Sketch, on another diagram, the graph of \(y = \left| \ln \left( \frac { 1 } { 2 } x - a \right) \right|\).
4. State, in terms of \(a\), the set of values of \(x\) for which \(\left| \ln \left( \frac { 1 } { 2 } x - a \right) \right| = - \ln \left( \frac { 1 } { 2 } x - a \right)\).

8 The definite integral \(I\) is defined by $$I = \int _ { 0 } ^ { 6 } 2 ^ { x } \mathrm {~d} x$$

1. Use Simpson's rule with 6 strips to find an approximate value of \(I\).
2. By first writing \(2 ^ { x }\) in the form \(\mathrm { e } ^ { k x }\), where the constant \(k\) is to be determined, find the exact value of \(I\).
3. Use the answers to parts (i) and (ii) to deduce that \(\ln 2 \approx \frac { 9 } { 13 }\).

3 The graph shows part of the function \(y = a \ln ( b x )\). \includegraphics[max width=\textwidth, alt={}, center]{2f403099-2813-40d8-a9ae-1f7e64d41f80-2_377_762_900_685} The graph passes through the points \(( 2,0 )\) and \(( 4,1 )\).

1. Show that \(b = \frac { 1 } { 2 }\) and find the exact value of \(a\).
2. Solve the inequality \(| a \ln ( b x ) | < 2\).

3 Fig. 9 shows the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\). The function \(y = \mathrm { f } ( x )\) is given by $$f ( x ) = \ln \left( \frac { 2 x } { 1 + x } \right) , x > 0$$ The curve \(y = \mathrm { f } ( x )\) crosses the \(x\)-axis at P , and the line \(x = 2\) at Q . \begin{figure}[h] \end{figure}

1. Verify that the \(x\)-coordinate of P is 1 . Find the exact \(y\)-coordinate of Q .
2. Find the gradient of the curve at P. [Hint: use \(\ln \frac { a } { b } = \ln a - \ln b\).] The function \(\mathrm { g } ( x )\) is given by $$\mathrm { g } ( x ) = \frac { \mathrm { e } ^ { x } } { 2 - \mathrm { e } ^ { x } } , \quad x < \ln 2 .$$ The curve \(y = \mathrm { g } ( x )\) crosses the \(y\)-axis at the point R .
3. Show that \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\). Write down the gradient of \(y = \mathrm { g } ( x )\) at R .
4. Show, using the substitution \(u = 2 - \mathrm { e } ^ { x }\) or otherwise, that \(\int _ { 0 } ^ { \ln \frac { 4 } { 3 } } \mathrm {~g} ( x ) \mathrm { d } x = \ln \frac { 3 } { 2 }\). Using this result, show that the exact area of the shaded region shown in Fig. 9 is \(\ln \frac { 32 } { 27 }\). [Hint: consider its reflection in \(y = x\).]

3 The function \(f ( x ) = \ln \left( 1 + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = f ( x )\). \begin{figure}[h] \end{figure} (1) Show algebraically that the function is even. State how this property relates to the shape of the curve.
(ii) Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
(iii) Explain why the function does not have an inverse for the domain \(- 3 \leqslant x \leqslant 3\). The domain of \(f ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(f ( x )\) is the function \(g ( x )\),
(iv) Sketch the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) on the same axes. State the domain of the function \(g ( x )\).
Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
(v) Differentiate \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the commection between this result and your answer to part (ii).

6 The functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are defined for the domain \(x > 0\) as follows: $$\mathrm { f } ( x ) = \ln x , \quad \mathrm {~g} ( x ) = x ^ { 3 } .$$ Express the composite function \(\mathrm { fg } ( x )\) in terms of \(\ln x\).
State the transformation which maps the curve \(y = \mathrm { f } ( x )\) onto the curve \(y = \mathrm { fg } ( x )\).

1 Fig. 9 shows the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\). The function \(y = \mathrm { f } ( x )\) is given by $$f ( x ) = \ln \left( \frac { 2 x } { 1 + x } \right) , x > 0$$ The curve \(y = \mathrm { f } ( x )\) crosses the \(x\)-axis at P , and the line \(x = 2\) at Q . \begin{figure}[h] \end{figure}

1. Verify that the \(x\)-coordinate of P is 1 . Find the exact \(y\)-coordinate of Q .
2. Find the gradient of the curve at P . [Hint: use \(\ln \frac { a } { b } = \ln a - \ln b\).] The function \(\mathrm { g } ( x )\) is given by $$\mathrm { g } ( x ) = \frac { \mathrm { e } ^ { x } } { 2 - \mathrm { e } ^ { x } } , \quad x < \ln 2 .$$ The curve \(y = \mathrm { g } ( x )\) crosses the \(y\)-axis at the point R .
3. Show that \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\). Write down the gradient of \(y = \mathrm { g } ( x )\) at R.
4. Show, using the substitution \(u = 2 - \mathrm { e } ^ { x }\) or otherwise, that \(\int _ { 0 } ^ { \ln \frac { 4 } { 3 } } \mathrm {~g} ( x ) \mathrm { d } x = \ln \frac { 3 } { 2 }\). Using this result, show that the exact area of the shaded region shown in Fig. 9 is \(\ln \frac { 32 } { 27 }\).
   [0pt] [Hint: consider its reflection in \(y = x\).]

2 Fig. 8 shows the line \(y = x\) and parts of the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\), where $$\mathrm { f } ( x ) = \mathrm { e } ^ { x - 1 } , \quad \mathrm {~g} ( x ) = 1 + \ln x$$ The curves intersect the axes at the points A and B, as shown. The curves and the line \(y = x\) meet at the point C . \begin{figure}[h] \end{figure}

1. Find the exact coordinates of A and B . Verify that the coordinates of C are \(( 1,1 )\).
2. Prove algebraically that \(\mathrm { g } ( x )\) is the inverse of \(\mathrm { f } ( x )\).
3. Evaluate \(\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x\), giving your answer in terms of e .
4. Use integration by parts to find \(\int \ln x \mathrm {~d} x\). Hence show that \(\int _ { \mathrm { e } ^ { - 1 } } ^ { 1 } \mathrm {~g} ( x ) \mathrm { d } x = \frac { 1 } { \mathrm { e } }\).
5. Find the area of the region enclosed by the lines OA and OB , and the arcs AC and BC .

3 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { 2 x } } { 1 + \mathrm { e } ^ { 2 x } }\). The curve crosses the \(y\)-axis at P . \begin{figure}[h] \end{figure}

1. Find the coordinates of P .
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), simplifying your answer. Hence calculate the gradient of the curve at P .
3. Show that the area of the region enclosed by \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = 1\) is \(\frac { 1 } { 2 } \ln \left( \frac { 1 + \mathrm { e } ^ { 2 } } { 2 } \right)\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { 2 } \left( \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } } { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } } \right)\).
4. Prove algebraically that \(\mathrm { g } ( x )\) is an odd function. Interpret this result graphically.
5. (A) Show that \(\mathrm { g } ( x ) + \frac { 1 } { 2 } = \mathrm { f } ( x )\).
   (B) Describe the transformation which maps the curve \(y = \mathrm { g } ( x )\) onto the curve \(y = \mathrm { f } ( x )\).
   (C) What can you conclude about the symmetry of the curve \(y = \mathrm { f } ( x )\) ?

1 Fig. 7 shows the curve $$y = 2 x - x \ln x , \text { where } x > 0 .$$ The curve crosses the \(x\)-axis at A , and has a turning point at B . The point C on the curve has \(x\)-coordinate 1 . Lines CD and BE are drawn parallel to the \(y\)-axis. \begin{figure}[h] \end{figure}

1. Find the \(x\)-coordinate of A , giving your answer in terms of e .
2. Find the exact coordinates of B .
3. Show that the tangents at A and C are perpendicular to each other.
4. Using integration by parts, show that $$\int x \ln x \mathrm {~d} x = \frac { 1 } { 2 } x ^ { 2 } \ln x - \frac { 1 } { 4 } x ^ { 2 } + c$$ Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines CD and BE .

1 It is given that \(\mathrm { f } ( x ) = \ln ( 3 + x )\).

1. Find the exact values of \(f ( 0 )\) and \(f ^ { \prime } ( 0 )\), and show that \(f ^ { \prime \prime } ( 0 ) = - \frac { 1 } { 9 }\).
2. Hence write down the first three terms of the Maclaurin series for \(\mathrm { f } ( x )\), given that \(- 3 < x \leqslant 3\).

3 It is given that \(\mathrm { F } ( x ) = 2 + \ln x\). The iteration \(x _ { n + 1 } = \mathrm { F } \left( x _ { n } \right)\) is to be used to find a root, \(\alpha\), of the equation \(x = 2 + \ln x\).

1. Taking \(x _ { 1 } = 3.1\), find \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers correct to 5 decimal places.
2. The error \(e _ { n }\) is defined by \(e _ { n } = \alpha - x _ { n }\). Given that \(\alpha = 3.14619\), correct to 5 decimal places, use the values of \(e _ { 2 }\) and \(e _ { 3 }\) to make an estimate of \(\mathrm { F } ^ { \prime } ( \alpha )\) correct to 3 decimal places. State the true value of \(\mathrm { F } ^ { \prime } ( \alpha )\) correct to 4 decimal places.
3. Illustrate the iteration by drawing a sketch of \(y = x\) and \(y = \mathrm { F } ( x )\), showing how the values of \(x _ { n }\) approach \(\alpha\). State whether the convergence is of the 'staircase' or 'cobweb' type.

7.The variable \(y\) is defined by $$y = \ln \left( \sec ^ { 2 } x + \operatorname { cosec } ^ { 2 } x \right) \text { for } 0 < x < \frac { \pi } { 2 } .$$ A student was asked to prove that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = - 4 \cot 2 x .$$ The attempted proof was as follows: $$\begin{aligned} y & = \ln \left( \sec ^ { 2 } x + \operatorname { cosec } ^ { 2 } x \right) \\ & = \ln \left( \sec ^ { 2 } x \right) + \ln \left( \operatorname { cosec } ^ { 2 } x \right) \\ & = 2 \ln \sec x + 2 \ln \operatorname { cosec } x \\ \frac { \mathrm {~d} y } { \mathrm {~d} x } & = 2 \tan x - 2 \cot x \\ & = \frac { 2 \left( \sin ^ { 2 } x - \cos ^ { 2 } x \right) } { \sin x \cos x } \\ & = \frac { - 2 \cos 2 x } { \frac { 1 } { 2 } \sin 2 x } \\ & = - 4 \cot 2 x \end{aligned}$$

1. Identify the error in this attempt at a proof.
2. Give a correct version of the proof.
3. Find and simplify a general relationship between \(p\) and \(q\) ,where \(p\) and \(q\) are variables that depend on \(x\) ,such that the student would obtain the correct result when differentiating \(\ln ( p + q )\) with respect to \(x\) by the above incorrect method.
4. Given that \(p ( x ) = k \sec r x\) and \(q ( x ) = \operatorname { cosec } ^ { 2 } x\) ,where \(k\) and \(r\) are positive integers,find the values of \(k\) and \(r\) such that \(p\) and \(q\) satisfy the relationship found in part(c). \section*{END} Marks for presentation: 7
   TOTAL MARKS: 100

2.The functions \(f\) and \(g\) are defined by $$\begin{array} { l l } \mathrm { f } ( x ) = 2 \sqrt { 1 - \mathrm { e } ^ { - x } } & x \in \mathbb { R } , x \geqslant 0 \\ \mathrm {~g} ( x ) = \ln \left( 4 - x ^ { 2 } \right) & x \in \mathbb { R } , - 2 < x < 2 \end{array}$$

1. 1. Explain why fg cannot be formed as a composite function.
   2. Explain why gf can be formed as a composite function.
2. 1. Find \(\mathrm { gf } ( x )\) ,giving the answer in the form \(\mathrm { gf } ( x ) = a + b x\) ,where \(a\) and \(b\) are constants.
   2. State the domain and range of gf.
3. Sketch the graph of the function gf.
   On your sketch,you should show the coordinates of any points where the graph meets or crosses the coordinate axes. The circle \(C\) with centre \(( 0 , - \ln 9 )\) touches the line with equation \(y = \operatorname { gf } ( x )\) at precisely one point.
4. Find an equation of the circle \(C\) .