1.07n Stationary points: find maxima, minima using derivatives

1. The curve \(C\) has equation

$$y = \arccos \left( \frac { 1 } { 2 } x \right) \quad - 2 \leqslant x \leqslant 2$$

1. Show that \(C\) has no stationary points. The normal to \(C\), at the point where \(x = 1\), crosses the \(x\)-axis at the point \(A\) and crosses the \(y\)-axis at the point \(B\). Given that \(O\) is the origin,
2. show that the area of the triangle \(O A B\) is \(\frac { 1 } { 54 } \left( p \sqrt { 3 } + q \pi + r \sqrt { 3 } \pi ^ { 2 } \right)\) where \(p\), \(q\) and \(r\) are integers to be determined.
   (5)

10 In this question you must show detailed reasoning. Fig. C2.2 indicates that the curve \(\mathrm { y } = \frac { 4 \mathrm { x } ( \pi - \mathrm { x } ) } { \pi ^ { 2 } } - \sin \mathrm { x }\) has a stationary point near \(x = 3\).

• Verify that the \(x\)-coordinate of this stationary point is between 2.6 and 2.7.
• Show that this stationary point is a maximum turning point.

13 Answer part (i) of this question on the insert provided. The insert shows the graph of \(y = \frac { 1 } { x }\).

1. On the insert, on the same axes, plot the graph of \(y = x ^ { 2 } - 5 x + 5\) for \(0 \leqslant x \leqslant 5\).
2. Show algebraically that the \(x\)-coordinates of the points of intersection of the curves \(y = \frac { 1 } { x }\) and \(y = x ^ { 2 } - 5 x + 5\) satisfy the equation \(x ^ { 3 } - 5 x ^ { 2 } + 5 x - 1 = 0\).
3. Given that \(x = 1\) at one of the points of intersection of the curves, factorise \(x ^ { 3 } - 5 x ^ { 2 } + 5 x - 1\) into a linear and a quadratic factor. Show that only one of the three roots of \(x ^ { 3 } - 5 x ^ { 2 } + 5 x - 1 = 0\) is rational.

12 In this question you must show detailed reasoning. Fig. 12 shows part of the graph of \(y = x ^ { 2 } + \frac { 1 } { x ^ { 2 } }\). \begin{figure}[h] \end{figure} The tangent to the curve \(\mathrm { y } = \mathrm { x } ^ { 2 } + \frac { 1 } { \mathrm { x } ^ { 2 } }\) at the point \(\left( 2 , \frac { 17 } { 4 } \right)\) meets the \(x\)-axis at A and meets the \(y\)-axis at B . O is the origin.

1. Find the exact area of the triangle OAB .
2. Use calculus to prove that the complete curve has two minimum points and no maximum point. \section*{END OF QUESTION PAPER}

8 In this question you must show detailed reasoning.

1. Use differentiation to find the coordinates of the stationary point on the curve with equation \(y = 2 x ^ { 2 } - 3 x - 2\).
2. Use the second derivative to determine the nature of the stationary point.
3. Show by shading on a sketch the region defined by the inequality \(y \geqslant 2 x ^ { 2 } - 3 x - 2\), indicating clearly whether the boundary is included or not.
4. Solve the inequality \(2 x ^ { 2 } - 3 x - 2 > 0\) using set notation for your answer.

5 In this question you must show detailed reasoning.

1. Find the coordinates of the two stationary points on the graph of \(y = 15 - x ^ { 2 } - \frac { 16 } { x ^ { 2 } }\).
2. Show that both these stationary points are maximum points.

3 On a particular voyage, a ship sails 500 km at a constant speed of \(v \mathrm {~km} / \mathrm { h }\). The cost for the voyage is \(\pounds R\) per hour. The total cost of the voyage is \(\pounds T\).

1. Show that \(T = \frac { 500 R } { v }\). The running cost is modelled by the following formula. $$R = 270 + \frac { v ^ { 3 } } { 200 }$$ The ship's owner wishes to sail at a speed that will minimise the total cost for the voyage. It is given that the graph of \(T\) against \(v\) has exactly one stationary point, which is a minimum.
2. Find the speed that gives the minimum value of \(T\).
3. Find the minimum value of the total cost.

8 \includegraphics[max width=\textwidth, alt={}, center]{e3942549-bfc0-432a-bf49-7d01d44af01a-6_533_524_246_772} The diagram shows a container which consists of a cylinder with a solid base and a hemispherical top. The radius of the cylinder is \(r \mathrm {~cm}\) and the height is \(h \mathrm {~cm}\). The container is to be made of thin plastic. The volume of the container is \(45 \pi \mathrm {~cm} ^ { 3 }\).

1. Show that the surface area of the container, \(A \mathrm {~cm} ^ { 2 }\), is given by $$A = \frac { 5 } { 3 } \pi r ^ { 2 } + \frac { 90 \pi } { r } .$$ [The volume of a sphere is \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\) and the surface area of a sphere is \(S = 4 \pi r ^ { 2 }\).]
2. Use calculus to find the minimum surface area of the container, justifying that it is a minimum.
3. Suggest a reason why the manufacturer would wish to minimise the surface area.

2 A curve has equation \(y = a x ^ { 4 } + b x ^ { 3 } - 2 x + 3\).

1. Given that the curve has a stationary point where \(x = 2\), show that \(16 a + 6 b = 1\).
2. Given also that this stationary point is a point of inflection, determine the values of \(a\) and \(b\).

4 The curve \(y = 2 x ^ { 3 } + 3 x ^ { 2 } - k x + 4\) has a stationary point where \(x = 2\).

1. Determine the value of the constant \(k\).
2. Determine whether this stationary point is a maximum or a minimum point.

5 The diagram shows an open-topped water tank with a horizontal rectangular base and four vertical faces. The base has width \(x\) metres and length \(2 x\) metres, and the height of the tank is \(h\) metres. \includegraphics[max width=\textwidth, alt={}, center]{33da89e2-f74f-4d5a-8bbd-ceaa728b6c34-4_403_410_477_792} The combined internal surface area of the base and four vertical faces is \(54 \mathrm {~m} ^ { 2 }\).

1. 1. Show that \(x ^ { 2 } + 3 x h = 27\).
   2. Hence express \(h\) in terms of \(x\).
   3. Hence show that the volume of water, \(V \mathrm {~m} ^ { 3 }\), that the tank can hold when full is given by $$V = 18 x - \frac { 2 x ^ { 3 } } { 3 }$$
2. 1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} x }\).
   2. Verify that \(V\) has a stationary value when \(x = 3\).
3. Find \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} x ^ { 2 } }\) and hence determine whether \(V\) has a maximum value or a minimum value when \(x = 3\).
   (2 marks)

2 The curve with equation \(y = x ^ { 4 } - 32 x + 5\) has a single stationary point, \(M\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Hence find the \(x\)-coordinate of \(M\).
3. 1. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
   2. Hence, or otherwise, determine whether \(M\) is a maximum or a minimum point.
4. Determine whether the curve is increasing or decreasing at the point on the curve where \(x = 0\).

4 A model helicopter takes off from a point \(O\) at time \(t = 0\) and moves vertically so that its height, \(y \mathrm {~cm}\), above \(O\) after time \(t\) seconds is given by $$y = \frac { 1 } { 4 } t ^ { 4 } - 26 t ^ { 2 } + 96 t , \quad 0 \leqslant t \leqslant 4$$

1. Find:
   1. \(\frac { \mathrm { d } y } { \mathrm {~d} t }\);
      (3 marks)
   2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\).
      (2 marks)
2. Verify that \(y\) has a stationary value when \(t = 2\) and determine whether this stationary value is a maximum value or a minimum value.
   (4 marks)
3. Find the rate of change of \(y\) with respect to \(t\) when \(t = 1\).
4. Determine whether the height of the helicopter above \(O\) is increasing or decreasing at the instant when \(t = 3\).

3 Two numbers, \(x\) and \(y\), are such that \(3 x + y = 9\), where \(x \geqslant 0\) and \(y \geqslant 0\). It is given that \(V = x y ^ { 2 }\).

1. Show that \(V = 81 x - 54 x ^ { 2 } + 9 x ^ { 3 }\).
2. 1. Show that \(\frac { \mathrm { d } V } { \mathrm {~d} x } = k \left( x ^ { 2 } - 4 x + 3 \right)\), and state the value of the integer \(k\).
   2. Hence find the two values of \(x\) for which \(\frac { \mathrm { d } V } { \mathrm {~d} x } = 0\).
3. Find \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} x ^ { 2 } }\).
4. 1. Find the value of \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} x ^ { 2 } }\) for each of the two values of \(x\) found in part (b)(ii).
   2. Hence determine the value of \(x\) for which \(V\) has a maximum value.
   3. Find the maximum value of \(V\).

3 The curve with equation \(y = x ^ { 5 } + 20 x ^ { 2 } - 8\) passes through the point \(P\), where \(x = - 2\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Verify that the point \(P\) is a stationary point of the curve.
3. 1. Find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at the point \(P\).
   2. Hence, or otherwise, determine whether \(P\) is a maximum point or a minimum point.
4. Find an equation of the tangent to the curve at the point where \(x = 1\).

6 A curve \(C\) is defined for \(x > 0\) by the equation \(y = x + 1 + \frac { 4 } { x ^ { 2 } }\) and is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{c16d94a6-52f2-4bf3-acee-0b227ae55a1a-4_545_784_420_628}

1. 1. Given that \(y = x + 1 + \frac { 4 } { x ^ { 2 } }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. The curve \(C\) has a minimum point \(M\). Find the coordinates of \(M\).
   3. Find an equation of the normal to \(C\) at the point ( 1,6 ).
2. 1. Find \(\int \left( x + 1 + \frac { 4 } { x ^ { 2 } } \right) \mathrm { d } x\).
   2. Hence find the area of the region bounded by the curve \(C\), the lines \(x = 1\) and \(x = 4\) and the \(x\)-axis.

9

1. Given that \(y = x ^ { - 2 } \ln x\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 - 2 \ln x } { x ^ { 3 } }\).
2. Using integration by parts, find \(\int x ^ { - 2 } \ln x \mathrm {~d} x\).
3. The sketch shows the graph of \(y = x ^ { - 2 } \ln x\). \includegraphics[max width=\textwidth, alt={}, center]{9aac4ee4-2435-4315-a87d-fe9fa8e15665-007_593_1034_696_543}
   1. Using the answer to part (a), find, in terms of e, the \(x\)-coordinate of the stationary point \(A\).
   2. The region \(R\) is bounded by the curve, the \(x\)-axis and the line \(x = 5\). Using your answer to part (b), show that the area of \(R\) is $$\frac { 1 } { 5 } ( 4 - \ln 5 )$$