1.07l Derivative of ln(x): and related functions

10 \includegraphics[max width=\textwidth, alt={}, center]{9cdb00a6-1e86-4185-bb73-ed3ecab981ba-4_634_937_696_603} The diagram shows part of the curve \(y = \sqrt { } \left( 9 - 2 x ^ { 2 } \right)\). The point \(P ( 2,1 )\) lies on the curve and the normal to the curve at \(P\) intersects the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\).

1. Show that \(B\) is the mid-point of \(A P\). The shaded region is bounded by the curve, the \(y\)-axis and the line \(y = 1\).
2. Find, showing all necessary working, the exact volume obtained when the shaded region is rotated through \(360 ^ { \circ }\) about the \(\boldsymbol { y }\)-axis. {www.cie.org.uk} after the live examination series. }

10 A curve has equation \(y = \frac { 1 } { 2 } ( 4 x - 3 ) ^ { - 1 }\). The point \(A\) on the curve has coordinates \(\left( 1 , \frac { 1 } { 2 } \right)\).

1. (a) Find and simplify the equation of the normal through \(A\).
   (b) Find the \(x\)-coordinate of the point where this normal meets the curve again.
2. A point is moving along the curve in such a way that as it passes through \(A\) its \(x\)-coordinate is decreasing at the rate of 0.3 units per second. Find the rate of change of its \(y\)-coordinate at \(A\).

10 \includegraphics[max width=\textwidth, alt={}, center]{d5d94eb8-7f41-4dff-b503-8be4f20e21b7-16_648_823_262_660} The diagram shows part of the curve \(y = 2 ( 3 x - 1 ) ^ { - \frac { 1 } { 3 } }\) and the lines \(x = \frac { 2 } { 3 }\) and \(x = 3\). The curve and the line \(x = \frac { 2 } { 3 }\) intersect at the point \(A\).

1. Find, showing all necessary working, the volume obtained when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
2. Find the equation of the normal to the curve at \(A\), giving your answer in the form \(y = m x + c\).

5 \includegraphics[max width=\textwidth, alt={}, center]{2d6fc4c5-70ec-4cd8-9b48-59d5ce0e39b7-08_575_618_262_762} The diagram shows the curve with equation \(y = \frac { 3 x + 2 } { \ln x }\). The curve has a minimum point \(M\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that the \(x\)-coordinate of \(M\) satisfies the equation \(x = \frac { 3 x + 2 } { 3 \ln x }\). [3]
2. Use the equation in part (a) to show by calculation that the \(x\)-coordinate of \(M\) lies between 3 and 4.
3. Use an iterative formula, based on the equation in part (a), to find the \(x\)-coordinate of \(M\) correct to 5 significant figures. Give the result of each iteration to 7 significant figures.

6 \includegraphics[max width=\textwidth, alt={}, center]{388d7076-636c-417d-84cb-e6e2a3e9a6a0-08_451_1086_260_525} The diagram shows the curve with equation $$y = ( \ln x ) ^ { 2 } - 2 \ln x$$ The curve crosses the \(x\)-axis at the points \(A\) and \(B\), and has a minimum point \(M\).

1. Find the exact value of the gradient of the curve at each of the points \(A\) and \(B\).
2. Find the exact \(x\)-coordinate of \(M\).

6 \includegraphics[max width=\textwidth, alt={}, center]{61df367d-741f-4906-8ab9-2f32e8711aa6-08_451_1086_260_525} The diagram shows the curve with equation $$y = ( \ln x ) ^ { 2 } - 2 \ln x$$ The curve crosses the \(x\)-axis at the points \(A\) and \(B\), and has a minimum point \(M\).

1. Find the exact value of the gradient of the curve at each of the points \(A\) and \(B\).
2. Find the exact \(x\)-coordinate of \(M\).

1 Given that \(y = \frac { \ln x } { x ^ { 2 } }\), find the exact value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(x = \mathrm { e }\).

2 A curve has equation \(y = \frac { 2 + 3 \ln x } { 1 + 2 x }\).
Find the equation of the tangent to the curve at the point \(\left( 1 , \frac { 2 } { 3 } \right)\). Give your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

6 \includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-10_417_700_310_685} The diagram shows the curve with equation \(y = \frac { \ln ( 2 x + 1 ) } { x + 3 }\). The curve has a maximum point \(M\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Show that the \(x\)-coordinate of \(M\) satisfies the equation \(x = \frac { x + 3 } { \ln ( 2 x + 1 ) } - 0.5\).
3. Show by calculation that the \(x\)-coordinate of \(M\) lies between 2.5 and 3.0 .
4. Use an iterative formula based on the equation in part (b) to find the \(x\)-coordinate of \(M\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

4 A curve has equation $$3 x ^ { 2 } - y ^ { 2 } - 4 \ln ( 2 y + 3 ) = 26$$ Find the equation of the tangent to the curve at the point \(( 3 , - 1 )\).

2 A curve has equation \(y = 7 + 4 \ln ( 2 x + 5 )\).
Find the equation of the tangent to the curve at the point ( \(- 2,7\) ), giving your answer in the form \(y = m x + c\).

7 \includegraphics[max width=\textwidth, alt={}, center]{389df578-e7a7-4d19-9416-5e580d107717-10_456_598_269_762} The diagram shows the curve with equation \(y = \frac { 2 \ln x } { 3 x + 1 }\). The curve crosses the \(x\)-axis at the point \(A\) and has a maximum point \(B\). The shaded region is bounded by the curve and the lines \(x = 3\) and \(y = 0\).

1. Find the gradient of the curve at \(A\).
2. Show by calculation that the \(x\)-coordinate of \(B\) lies between 3.0 and 3.1.
3. Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 decimal places.

7 \includegraphics[max width=\textwidth, alt={}, center]{1cd04df5-3fe3-4573-b880-d49262afd16a-10_456_598_269_762} The diagram shows the curve with equation \(y = \frac { 2 \ln x } { 3 x + 1 }\). The curve crosses the \(x\)-axis at the point \(A\) and has a maximum point \(B\). The shaded region is bounded by the curve and the lines \(x = 3\) and \(y = 0\).

1. Find the gradient of the curve at \(A\).
2. Show by calculation that the \(x\)-coordinate of \(B\) lies between 3.0 and 3.1.
3. Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 decimal places.

6 The equation of a curve is \(y = \frac { 1 } { 1 + \tan x }\).

1. Show, by differentiation, that the gradient of the curve is always negative.
2. Use the trapezium rule with 2 intervals to estimate the value of $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { 1 } { 1 + \tan x } \mathrm {~d} x$$ giving your answer correct to 2 significant figures.
3. \includegraphics[max width=\textwidth, alt={}, center]{a31a4b4e-83a6-47d9-9679-3471b3da1b6e-3_556_802_1384_708} The diagram shows a sketch of the curve for \(0 \leqslant x \leqslant \frac { 1 } { 4 } \pi\). State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (ii).

6 \includegraphics[max width=\textwidth, alt={}, center]{08210e25-0f0e-405b-b72d-1bf989689b0a-3_641_865_264_641} The diagram shows the part of the curve \(y = \frac { \ln x } { x }\) for \(0 < x \leqslant 4\). The curve cuts the \(x\)-axis at \(A\) and its maximum point is \(M\).

1. Write down the coordinates of \(A\).
2. Show that the \(x\)-coordinate of \(M\) is e, and write down the \(y\)-coordinate of \(M\) in terms of e.
3. Use the trapezium rule with three intervals to estimate the value of $$\int _ { 1 } ^ { 4 } \frac { \ln x } { x } \mathrm {~d} x$$ correct to 2 decimal places.
4. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (iii).

7

1. Differentiate \(\ln ( 2 x + 3 )\).
2. Hence, or otherwise, show that $$\int _ { - 1 } ^ { 3 } \frac { 1 } { 2 x + 3 } \mathrm {~d} x = \ln 3$$
3. Find the quotient and remainder when \(4 x ^ { 2 } + 8 x\) is divided by \(2 x + 3\).
4. Hence show that $$\int _ { - 1 } ^ { 3 } \frac { 4 x ^ { 2 } + 8 x } { 2 x + 3 } d x = 12 - 3 \ln 3$$

8

1. Find the equation of the tangent to the curve \(y = \ln ( 3 x - 2 )\) at the point where \(x = 1\).
2. 1. Find the value of the constant \(A\) such that $$\frac { 6 x } { 3 x - 2 } \equiv 2 + \frac { A } { 3 x - 2 }$$
   2. Hence show that \(\int _ { 2 } ^ { 6 } \frac { 6 x } { 3 x - 2 } \mathrm {~d} x = 8 + \frac { 8 } { 3 } \ln 2\).

5 Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(x = 4\) in each of the following cases:

1. \(y = x \ln ( x - 3 )\),
2. \(y = \frac { x - 1 } { x + 1 }\).

6 The curve \(y = 4 x ^ { 2 } \ln x\) has one stationary point.

1. Find the coordinates of this stationary point, giving your answers correct to 3 decimal places.
2. Determine whether this point is a maximum or a minimum point.

1 Find the gradient of the curve $$y = 3 e ^ { 4 x } - 6 \ln ( 2 x + 3 )$$ at the point for which \(x = 0\).

8 \includegraphics[max width=\textwidth, alt={}, center]{6295873e-7db4-4e7e-8dcd-912ad9c41675-10_643_414_260_863} The diagram shows the curve with equation $$y = 3 x ^ { 2 } \ln \left( \frac { 1 } { 6 } x \right) .$$ The curve crosses the \(x\)-axis at the point \(P\) and has a minimum point \(M\).

1. Find the gradient of the curve at the point \(P\).
2. Find the exact coordinates of the point \(M\).

4 \includegraphics[max width=\textwidth, alt={}, center]{873a104f-e2e2-49bb-b943-583769728fbb-06_355_839_260_653} The diagram shows the curve with equation \(y = \frac { 5 \ln x } { 2 x + 1 }\). The curve crosses the \(x\)-axis at the point \(P\) and has a maximum point \(M\).

1. Find the gradient of the curve at the point \(P\).
2. Show that the \(x\)-coordinate of the point \(M\) satisfies the equation \(x = \frac { x + 0.5 } { \ln x }\).
3. Use an iterative formula based on the equation in part (ii) to find the \(x\)-coordinate of \(M\) correct to 4 significant figures. Show the result of each iteration to 6 significant figures.

2 A curve has equation \(y = 3 \ln ( 2 x + 9 ) - 2 \ln x\).

1. Find the \(x\)-coordinate of the stationary point.
2. Determine whether the stationary point is a maximum or minimum point.

10 \includegraphics[max width=\textwidth, alt={}, center]{0f081749-4fe0-46e3-96c2-466e69cf49d3-4_620_894_338_687} The function f is defined by \(\mathrm { f } ( x ) = ( \ln x ) ^ { 2 }\) for \(x > 0\). The diagram shows a sketch of the graph of \(y = \mathrm { f } ( x )\). The minimum point of the graph is \(A\). The point \(B\) has \(x\)-coordinate e .

1. State the \(x\)-coordinate of \(A\).
2. Show that \(\mathrm { f } ^ { \prime \prime } ( x ) = 0\) at \(B\).
3. Use the substitution \(x = \mathrm { e } ^ { u }\) to show that the area of the region bounded by the \(x\)-axis, the line \(x = \mathrm { e }\), and the part of the curve between \(A\) and \(B\) is given by $$\int _ { 0 } ^ { 1 } u ^ { 2 } \mathrm { e } ^ { u } \mathrm {~d} u .$$
4. Hence, or otherwise, find the exact value of this area.

8 The equation of a curve is \(y = \ln x + \frac { 2 } { x }\), where \(x > 0\).

1. Find the coordinates of the stationary point of the curve and determine whether it is a maximum or a minimum point.
2. The sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 2 } { 3 - \ln x _ { n } }$$ with initial value \(x _ { 1 } = 1\), converges to \(\alpha\). State an equation satisfied by \(\alpha\), and hence show that \(\alpha\) is the \(x\)-coordinate of a point on the curve where \(y = 3\).
3. Use this iterative formula to find \(\alpha\) correct to 2 decimal places, showing the result of each iteration.