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

10 \includegraphics[max width=\textwidth, alt={}, center]{2718ebbb-29e3-46f7-8d8d-ec7d526483f8-3_458_920_1144_609} The diagram shows the curve \(y = \frac { \ln x } { x ^ { 2 } }\) and its maximum point \(M\). The curve cuts the \(x\)-axis at \(A\).

1. Write down the \(x\)-coordinate of \(A\).
2. Find the exact coordinates of \(M\).
3. Use integration by parts to find the exact area of the shaded region enclosed by the curve, the \(x\)-axis and the line \(x = \mathrm { e }\).

2 Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in each of the following cases:

1. \(y = \ln ( 1 + \sin 2 x )\),
2. \(y = \frac { \tan x } { x }\).

5 The curve with equation $$6 \mathrm { e } ^ { 2 x } + k \mathrm { e } ^ { y } + \mathrm { e } ^ { 2 y } = c$$ where \(k\) and \(c\) are constants, passes through the point \(P\) with coordinates \(( \ln 3 , \ln 2 )\).

1. Show that \(58 + 2 k = c\).
2. Given also that the gradient of the curve at \(P\) is - 6 , find the values of \(k\) and \(c\).

2 The curve \(y = \frac { \ln x } { x ^ { 3 } }\) has one stationary point. Find the \(x\)-coordinate of this point.

6 \includegraphics[max width=\textwidth, alt={}, center]{436d891d-92ee-4076-8369-db756d413979-2_435_597_1516_776} The diagram shows the curves \(y = \mathrm { e } ^ { 2 x - 3 }\) and \(y = 2 \ln x\). When \(x = a\) the tangents to the curves are parallel.

1. Show that \(a\) satisfies the equation \(a = \frac { 1 } { 2 } ( 3 - \ln a )\).
2. Verify by calculation that this equation has a root between 1 and 2 .
3. Use the iterative formula \(a _ { n + 1 } = \frac { 1 } { 2 } \left( 3 - \ln a _ { n } \right)\) to calculate \(a\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places.

4 The curve with equation \(y = \frac { ( \ln x ) ^ { 2 } } { x }\) has two stationary points. Find the exact values of the coordinates of these points.

4 Find the exact coordinates of the point on the curve \(y = \frac { x } { 1 + \ln x }\) at which the gradient of the tangent is equal to \(\frac { 1 } { 4 }\).

4 The equation of a curve is \(y = \frac { 1 + \mathrm { e } ^ { - x } } { 1 - \mathrm { e } ^ { - x } }\), for \(x > 0\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) is always negative.
2. The gradient of the curve is equal to - 1 when \(x = a\). Show that \(a\) satisfies the equation \(\mathrm { e } ^ { 2 a } - 4 \mathrm { e } ^ { a } + 1 = 0\). Hence find the exact value of \(a\).

10 \includegraphics[max width=\textwidth, alt={}, center]{e26f21c5-3776-4c86-8440-6959c5e37486-18_337_529_260_808} The diagram shows the curve \(y = ( \ln x ) ^ { 2 }\). The \(x\)-coordinate of the point \(P\) is equal to e, and the normal to the curve at \(P\) meets the \(x\)-axis at \(Q\).

1. Find the \(x\)-coordinate of \(Q\).
2. Show that \(\int \ln x \mathrm {~d} x = x \ln x - x + c\), where \(c\) is a constant.
3. Using integration by parts, or otherwise, find the exact value of the area of the shaded region between the curve, the \(x\)-axis and the normal \(P Q\).

4 Show that \(\int _ { 1 } ^ { 4 } x ^ { - \frac { 3 } { 2 } } \ln x \mathrm {~d} x = 2 - \ln 4\).

9 \includegraphics[max width=\textwidth, alt={}, center]{8d134c65-af23-4508-acef-49b6ab49e374-3_504_910_625_614} The diagram shows the curve \(y = \frac { \ln x } { \sqrt { } x }\) and its maximum point \(M\). The curve cuts the \(x\)-axis at the point \(A\).

1. State the coordinates of \(A\).
2. Find the exact value of the \(x\)-coordinate of \(M\).
3. Using integration by parts, show that the area of the shaded region bounded by the curve, the \(x\)-axis and the line \(x = 4\) is equal to \(8 \ln 2 - 4\).

5

1. By differentiating \(\frac { 1 } { \cos x }\), show that if \(y = \sec x\) then \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec x \tan x\).
2. Show that \(\frac { 1 } { \sec x - \tan x } \equiv \sec x + \tan x\).
3. Deduce that \(\frac { 1 } { ( \sec x - \tan x ) ^ { 2 } } \equiv 2 \sec ^ { 2 } x - 1 + 2 \sec x \tan x\).
4. Hence show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { 1 } { ( \sec x - \tan x ) ^ { 2 } } \mathrm {~d} x = \frac { 1 } { 4 } ( 8 \sqrt { } 2 - \pi )\).

5 The curve with equation \(y = \mathrm { e } ^ { - 2 x } \ln ( x - 1 )\) has a stationary point when \(x = p\).

1. Show that \(p\) satisfies the equation \(x = 1 + \exp \left( \frac { 1 } { 2 ( x - 1 ) } \right)\), where \(\exp ( x )\) denotes \(\mathrm { e } ^ { x }\).
2. Verify by calculation that \(p\) lies between 2.2 and 2.6.
3. Use an iterative formula based on the equation in part (i) to determine \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

6

1. By differentiating \(\frac { \cos x } { \sin x }\), show that if \(y = \cot x\) then \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } ^ { 2 } x\).
2. Show that \(\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 2 } \pi } x \operatorname { cosec } ^ { 2 } x \mathrm {~d} x = \frac { 1 } { 4 } ( \pi + \ln 4 )\). \(7 \quad\) Two lines \(l\) and \(m\) have equations \(\mathbf { r } = a \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } + \lambda ( \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } )\) and \(\mathbf { r } = 2 \mathbf { i } + \mathbf { j } + 2 \mathbf { k } + \mu ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } )\) respectively, where \(a\) is a constant. It is given that the lines intersect.

5 The curve with equation \(y = x ^ { 2 } \ln x\), where \(x > 0\), has one stationary point.

1. Find the \(x\)-coordinate of this point, giving your answer in terms of e .
2. Determine whether this point is a maximum or a minimum point.

6 The curve with equation \(y = x \ln x\) has one stationary point.

1. Find the exact coordinates of this point, giving your answers in terms of e .
2. Determine whether this point is a maximum or a minimum point.

3 \includegraphics[max width=\textwidth, alt={}, center]{322eb555-d40a-460c-8c71-5780f5772bcd-2_535_1041_573_552} The diagram shows the curve \(y = x - 2 \ln x\) and its minimum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Use the trapezium rule with three intervals to estimate the value of $$\int _ { 2 } ^ { 5 } ( x - 2 \ln x ) \mathrm { d } x$$ giving your answer correct to 2 decimal places.
3. 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).

1 Find the gradient of the curve \(y = \ln ( 5 x + 1 )\) at the point where \(x = 4\).

8

1. By differentiating \(\frac { 1 } { \cos \theta }\), show that if \(y = \sec \theta\) then \(\frac { \mathrm { d } y } { \mathrm {~d} \theta } = \tan \theta \sec \theta\).
2. Hence show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} \theta ^ { 2 } } = a \sec ^ { 3 } \theta + b \sec \theta$$ giving the values of \(a\) and \(b\).
3. Find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \left( 1 + \tan ^ { 2 } \theta - 3 \sec \theta \tan \theta \right) d \theta$$

5 \includegraphics[max width=\textwidth, alt={}, center]{9bbcee46-c5b8-4836-a4b4-f317bf8b1c0a-2_556_844_1731_648} The diagram shows the curve \(y = \frac { 4 \ln x } { x ^ { 2 } + 1 }\) and its stationary point \(M\). The \(x\)-coordinate of \(M\) is \(m\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence show that \(m = \mathrm { e } ^ { 0.5 \left( 1 + m ^ { - 2 } \right) }\).
2. Use an iterative formula based on the equation in part (i) to find the value of \(m\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

3 A curve has equation \(y = \frac { 3 + 2 \ln x } { 1 + \ln x }\). Find the exact gradient of the curve at the point for which \(y = 4\).

7 Let \(\mathrm { f } ( x ) = \frac { \cos x } { 1 + \sin x }\).

1. Show that \(\mathrm { f } ^ { \prime } ( x ) < 0\) for all \(x\) in the interval \(- \frac { 1 } { 2 } \pi < x < \frac { 3 } { 2 } \pi\).
2. Find \(\int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 1 } { 2 } \pi } \mathrm { f } ( x ) \mathrm { d } x\). Give your answer in a simplified exact form.

3 Find the exact value of $$\int _ { 1 } ^ { 4 } x ^ { \frac { 3 } { 2 } } \ln x \mathrm {~d} x$$

10

1. By writing \(y = \sec ^ { 3 } \theta\) as \(\frac { 1 } { \cos ^ { 3 } \theta }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} \theta } = 3 \sin \theta \sec ^ { 4 } \theta\).
2. The variables \(x\) and \(\theta\) satisfy the differential equation $$\left( x ^ { 2 } + 9 \right) \sin \theta \frac { d \theta } { d x } = ( x + 3 ) \cos ^ { 4 } \theta$$ It is given that \(x = 3\) when \(\theta = \frac { 1 } { 3 } \pi\).
   Solve the differential equation to find the value of \(\cos \theta\) when \(x = 0\). Give your answer correct to 3 significant figures.
   If you use the following page to complete the answer to any question, the question number must be clearly shown.

7

1. Given that \(y = \ln ( \ln x )\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { x \ln x }$$ The variables \(x\) and \(t\) satisfy the differential equation $$x \ln x + t \frac { \mathrm {~d} x } { \mathrm {~d} t } = 0$$ It is given that \(x = \mathrm { e }\) when \(t = 2\).
2. Solve the differential equation obtaining an expression for \(x\) in terms of \(t\), simplifying your answer.
3. Hence state what happens to the value of \(x\) as \(t\) tends to infinity.