Solutionbank C2 Edexcel Modular Mathematics for AS and A-Level

Transcription

1 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise A, Question Find the values of for which f() is an increasing function, given that f() equals: (a) (b) (c) 5 8 (d) (e) + + (f) 5 + (g) + (h) 8 (a) f ( ) = f ( ) = f ( ) > > 0 So > 8 6 i.e. > (b) f ( ) = f ( ) = 6 f ( ) > 0 6 > 0 So > 6 i.e. 6 < < 6 < (c) f ( ) = 5 8 f ( ) = 8 f ( ) > 0 8 > 0 So 8 > (add to both sides) i.e. < 8 < (d) f ( ) = f ( ) =

2 Heinemann Solutionbank: Core Maths C Page of f ( ) > > 0 So 6 ( ) > 0 i.e. 6 ( ) ( ) > 0 By considering the regions < < < > 6( )( ) +ve ve +ve Then < or > (e) f ( ) = + + f ( ) = 6 + f ( ) > > 0 So ( + ) > 0 i.e. ( ) > 0 So R, (f) f ( ) = 5 + f ( ) = 5 + f ( ) > > 0 This is true for all real values of. So R (g) f ( ) = + f ( ) = + f ( ) > 0 + > 0 So ( + ) > 0 As + > 0 for all, > 0 (h) f ( ) = 8 f ( ) = f ( ) > 0 > 0 So ( 6 ) > 0 As > 0 for all, 6 > 0 So > 6 Pearson Education Ltd 008

3 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise A, Question Find the values of for which f() is a decreasing function, given that f() equals: (a) 9 (b) 5 (c) (d) (e) 7 + (f) + 5 (g) + 9 (h) ( + ) (a) f ( ) = 9 f ( ) = 9 f ( ) < 0 9 < 0 So < 9 i.e. <.5 (b) f ( ) = 5 f ( ) = 5 f ( ) < 0 5 < 0 So 5 < i.e. > 5 >.5 (c) f ( ) = f ( ) = f ( ) < 0 < 0 So < i.e. > > (d) f ( ) = f ( ) = 6 6 f ( ) < < 0 So 6 ( ) < 0 i.e. 6 ( ) ( + ) < 0 By considering the regions <, < <, > determine < <

4 Heinemann Solutionbank: Core Maths C Page of (e) f ( ) = 7 + f ( ) = 7 + f ( ) < < 0 So < 7 i.e. < 9 < < (f) f = + f = f < 0 < 0 So < 5 Multiply both sides by : < 5 5 < < 5 (g) f = + 9 f = 9 5 f < 0 < 0 So 9 < 0 > 0 or the function is not defined So 0 < < 9 5 (h) f ( ) = + f ( ) = + 6 f ( ) < < 0 So ( + ) < 0 Consider the regions <, < < 0 and > 0 to give < < Pearson Education Ltd 008

5 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise B, Question Find the least value of each of the following functions: (a) f ( ) = + 8 (b) f ( ) = 8 (c) f ( ) = 5 + (a) f ( ) = + 8 f ( ) = Put f ( ) = 0, then = 0, i.e. = 6 f ( 6 ) = = 8 The least value of f() is 8. (b) f ( ) = 8 f ( ) = 8 Put f ( ) = 0, then 8 = 0, i.e. = f ( ) = 8 = 7 The minimum value of f() is 7. (c) f ( ) = 5 + f ( ) = 0 + Put f ( ) = 0, then 0 + = 0, i.e. = or = f = 5 + = = 5 The least value of f() is Pearson Education Ltd 008

6 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise B, Question Find the greatest value of each of the following functions: (a) f ( ) = 0 5 (b) f ( ) = + (c) f ( ) = ( 6 + ) ( ) (a) f ( ) = 0 5 f ( ) = 0 Put f ( ) = 0, then 0 = 0, i.e. = 0 f ( 0 ) = = 0 Maimum value of f() is 0. (b) f ( ) = + f ( ) = Put f ( ) = 0, then = 0, i.e. = f ( ) = + = The greatest value of f() is. (c) f ( ) = ( 6 + ) ( ) = 6 5 f ( ) = 5 Put f ( ) = 0, then 5 = 0, i.e. = f = = The maimum value of f() is. Pearson Education Ltd 008

7 Heinemann Solutionbank: Core Maths C Page of 5 Solutionbank C Eercise B, Question Find the coordinates of the points where the gradient is zero on the curves with the given equations. Establish whether these points are maimum points, minimum points or points of infleion, by considering the second derivative in each case. (a) y = + 6 (b) y = 9 + (c) y = + (d) y = ( ) (e) y = + (f) y = + 5 (g) y = (h) y = 6 (i) y = (a) y = + 6 = Put = 0 Then = 0 8 = 6 = When =, y = + 6 = = So, is a point of zero gradient d y = 8 >

8 Heinemann Solutionbank: Core Maths C Page of 5 So, is a minimum point (b) y = 9 + = Put = 0 Then = 0 = When =, y = 9 + = 9 So, 9 is a point with zero gradient d y = < 0 So, 9 is a maimum point (c) y = + = 9 Put = 0 Then = 0 ( + ) ( ) = 0 = or = When =, y = + = When =, y = + = 0 So, and (, 0) are points of zero gradient d y = When =, = < 0 So, is a maimum point When =, = 6 = > 0 So (, 0) is a minimum point (d) y = ( ) = d y = 8 d y 5 7

9 Heinemann Solutionbank: Core Maths C Page of 5 Put = 0 Then 8 = 0 ( + ) ( ) = 0 = or When =, y = = When =, y = = 8 So, and (, 8 ) are points with zero gradient d y = When =, = 0 < 0 So, is a maimum point When =, = + 0 > 0 So (, 8 ) is a minimum point (e) y = + = + d y d y 7 = Put = 0 Then = 0 = = ± When =, y = + = When =, y = + = So (, ) and (, ) are points with zero gradient d y = d y When =, = > 0 So (, ) is a minimum point When =, = < 0 So (, ) is a maimum point 5 (f) y = + = + 5 d y = 5 Put = 0

10 Heinemann Solutionbank: Core Maths C Page of 5 Then 5 = 0 = 5 = 7 = When =, y = + = 7 So (, 7) is a point of zero gradient d y = + 08 d y 5 When =, = 6 > 0 So (, 7) is a minimum point (g) y = = = Put = 0 Then = 0 = = = 9 9 When =, y = \ = 9 So, is a point with zero gradient d y = When =, = = = > 0 9 d y So, is a minimum point (h) y = 6 = 6 = Put = 0 Then = 0

11 Heinemann Solutionbank: Core Maths C Page 5 of 5 = Multiply both sides by : = = When =, y = = So (, ) is a point with zero gradient d y = + When =, = + > 0 So (, ) is a minimum point (i) y = = Put = 0 Then = 0 ( 6 ) = 0 = 0 or = ± 6 When = 0, y = 0 When = ± 6, y = 6 So ( 0, 0 ), ( 6, 6 ) and ( 6, 6 ) are points with zero gradient d y = When = 0, = < 0 So (0, 0) is a maimum point When = 6, = 8 > 0 d y d y d y So ( 6, 6 ) and ( 6, 6 ) are minimum points Pearson Education Ltd 008

12 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise B, Question Sketch the curves with equations given in question parts (a), (b), (c) and (d) labelling any stationary values.

13 Heinemann Solutionbank: Core Maths C Page of

14 Heinemann Solutionbank: Core Maths C Page of. Pearson Education Ltd 008

15 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise B, Question 5 By considering the gradient on either side of the stationary point on the curve y = +, show that this point is a point of infleion. Sketch the curve y = +. y = + = 6 + Put = 0 Then 6 + = 0 ( + ) = 0 ( ) = 0 = when =, y = So (, ) is a point with zero gradient. Consider points near to (, ) and find the gradient at these points. The gradient on either side of (, ) is positive. This is not a turning point it is a point of infleion. Pearson Education Ltd 008

16 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise B, Question 6 Find the maimum value and hence the range of values for the function f ( ) = 7. f ( ) = 7 f ( ) = 8 Put f ( ) = 0 Then 8 = 0 So = 0 f ( 0 ) = 7 So (0, 7) is a point of zero gradient f ( ) = f ( 0 ) = 0 not conclusive Find gradient on either side of (0, 7): There is a maimum turning point at (0, 7). So the maimum value of f ( ) is 7 and range of values is f ( ) 7. Pearson Education Ltd 008

17 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise C, Question A rectangular garden is fenced on three sides, and the house forms the fourth side of the rectangle. Given that the total length of the fence is 80 m show that the area, A, of the garden is given by the formula A = y ( 80 y ), where y is the distance from the house to the end of the garden. Given that the area is a maimum for this length of fence, find the dimensions of the enclosed garden, and the area which is enclosed. Let the width of the garden be m. Then + y = 80 So = 80 y * Area A = y So A = y ( 80 y ) A = 80y y da Put = 80 y da = 0 for maimum area Then 80 y = 0 So y = 0 Substitute in * to give = 0. So area = 0 m 0 m = 800 m Pearson Education Ltd 008

18 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise C, Question A closed cylinder has total surface area equal to 600π. Show that the volume, V cm, of this cylinder is given by the formula V = 00πr πr, where r cm is the radius of the cylinder. Find the maimum volume of such a cylinder. Total surface area = πrh + πr So πrh + πr = 600π rh = 00 r Volume =πr h =πr ( rh ) =πr ( 00 r ) So V = 00πr πr dv For maimum volume = 0 dr dv dr = 00π πr dv Put = 0 dr Then 00π πr = 0 So r = 00 r = 0 Substitute r = 0 into V to give V = 00π 0 π 0 = 000π Maimum volume = 000π cm Pearson Education Ltd 008

19 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise C, Question A sector of a circle has area 00 cm. Show that the perimeter of this sector is given by the formula P = r +, r > \ 00 π. Find the minimum value for the perimeter of such a sector. 00 r Let angle MON =θ radians. Then perimeter P = r + rθ and area A = r θ But area is 00 cm so r θ = 00 rθ = 00 r Substitute into to give P = r + Since area of circle > area of sector πr > 00 So r > \ 00 r 00 π

20 Heinemann Solutionbank: Core Maths C Page of For the minimum perimeter = 0 dp dr = dp Put = 0 dr 00 r 00 Then = 0 r So r = 0 Substitute into to give P = = 0 Minimum perimeter = 0 cm dp dr Pearson Education Ltd 008

21 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise C, Question A shape consists of a rectangular base with a semicircular top, as shown. Given that the perimeter of the shape is 0 cm, show that its area, A cm, is given by the formula A = 0r r πr where r cm is the radius of the semicircle. Find the maimum value for this area. Let the rectangle have dimensions r by cm. Then perimeter of figure is ( r + +πr ) cm But perimeter is 0 cm so r + +πr = 0 0 πr r = * Area = r + πr (rectangle + semicircle) So A = r 0 πr r + πr (substituting from *) A = 0r r πr To find maimum value, put = 0: 0 r πr = 0 r = 0 +π da dr Substitute into epression for A:

22 Heinemann Solutionbank: Core Maths C Page of A = 0 π A = + π A = A = 600 +π 600 +π 600 +π π A = cm +π +π 800 +π 0 +π 600 ( +π) 0 +π 0 +π Pearson Education Ltd 008

23 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise C, Question 5 The shape shown is a wire frame in the form of a large rectangle split by parallel lengths of wire into smaller equal-sized rectangles. Given that the total length of wire used to complete the whole frame is 5 mm, show that the area of the whole shape is A mm, where A = , where mm is the width of one of the smaller rectangles. Find the maimum area which can be enclosed in this way. Total length of wire is ( 8 + y ) But length = 5 mm so 8 + y = 5 y = Total area A mm is given by A = y 6 Substitute into to give A = A = 96 * 7 mm da For maimum area, put = 0: da 6 = 96 7 da when = 0, = = Substitute = into * to give A = 76 Maimum area = 76 mm d A 6 7 (Check: = < 0 maimum) Pearson Education Ltd 008

24 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise D, Question Given that: y = + > 0 8 (a) Find the value of and the value of y when = 0. (b) Show that the value of y which you found in (a) is a minimum. Given that y = + > 0 8 (a) = Put = 0: 8 = = = 8 Substitute = into y = + to give y = 8 + = 0 So = and y = 0 when = 0 8 d y (b) = + 96 d y 8 96 When =, = + = > 0 minimum Pearson Education Ltd 008

25 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise D, Question A curve has equation y = Determine, by calculation, the coordinates of the stationary points of the curve C. y = = When = = 0 ( 7 ) ( ) = 0 7 = or = When =, y = When =, y = So, and (, ) are stationary points (where the gradient is zero) 5 7 Pearson Education Ltd 008

26 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise D, Question The function f, defined for R, > 0, is such that: f = + (a) Find the value of f ( ) at =. (b) Given that f ( ) = 0, find f ( ). (c) Prove that f is an increasing function. f = + > 0 (a) f = At =, f = 7 (b) f = + c f = 0 + c = 0 c = So f = (c) For an increasing function, f ( ) > 0 + > 0 > 0 This is true for all, ecept = [where f ( ) = 0 ]. So the function is an increasing function. Pearson Education Ltd 008

27 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise D, Question A curve has equation y = Find the coordinates of its maimum turning point. y = = + 9 Put = 0 Then + 9 = 0 ( + ) = 0 ( ) ( ) = 0 = or = d y = 6 When =, d y = 6 < 0 maimum point When =, d y = + 6 > 0 minimum point So the maimum point is where =. Substitute = into y = Then y = = So (, ) is the maimum turning point. Pearson Education Ltd 008

28 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise D, Question 5 A wire is bent into the plane shape ABCDEA as shown. Shape ABDE is a rectangle and BCD is a semicircle with diameter BD. The area of the region enclosed by the wire is R m, AE = metres, AB = ED = y metres. The total length of the wire is m. (a) Find an epression for y in terms of. (b) Prove that R = 8 8 π Given that can vary, using calculus and showing your working, (c) find the maimum value of R. (You do not have to prove that the value you obtain is a maimum.) (a) The total length of wire is y + + m As total length is m so y + + = y = + (b) Area R = y + π Substitute from to give π R = + π π π 8 π

29 Heinemann Solutionbank: Core Maths C Page of R = 8 8 π +π R = 8 π 8 (c) For maimum R, = 0 R = dr So = π 8 π dr dr Put = 0 to obtain = + π So = +π Substitute into to give R = 8 ( +π) R = ( +π) 6 +π + 8π 6 π +π π +π R = ( +π) 6 + π +π R = R = ( +π) ( +π) +π Pearson Education Ltd 008

30 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise D, Question 6 The fied point A has coordinates (8, 6, 5) and the variable point P has coordinates (t, t, t). (a) Show that AP = 6t t + 5. (b) Hence find the value of t for which the distance AP is least. (c) Determine this least distance. (a) From Pythagoras AP = ( 8 t ) + ( 6 t ) + ( 5 t ) AP = 6 6t + t t + t + 5 0t + t AP = 6t t + 5 * (b) AP is least when AP is least. d ( AP ) dt d ( AP ) = t Put = 0, then t = dt (c) Substitute t = into * to obtain AP = = 0 So AP = \ 0 Pearson Education Ltd 008

31 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise D, Question 7 A cylindrical biscuit tin has a close-fitting lid which overlaps the tin by cm, as shown. The radii of the tin and the lid are both cm. The tin and the lid are made from a thin sheet of metal of area 80πcm and there is no wastage. The volume of the tin is V cm. (a) Show that V =π(0 ). Given that can vary: (b) Use differentiation to find the positive value of for which V is stationary. (c) Prove that this value of gives a maimum value of V. (d) Find this maimum value of V. (e) Determine the percentage of the sheet metal used in the lid when V is a maimum. (a) Let the height of the tin be h cm. The area of the curved surface of the tin = πh cm The area of the base of the tin =π cm The area of the curved surface of the lid = π cm The area of the top of the lid =π cm Total area of sheet metal is 80π cm So π + π + πh = 80π Rearrange to give h = 0 The volume, V, of the tin is given by V =π h

32 Heinemann Solutionbank: Core Maths C Page of π ( 0 ) So V = =π 0 dv (b) =π 0 When V is stationary = 0 So 0 = 0 ( 0 ) ( + ) = 0 0 = or dv But is positive so = 0 is the required value. d V (c) =π 6 0 When =, =π 0 < 0 d V So V has a maimum value. (d) Substitute = 0 into the epression given in part (a): V = 00π 7 (e) The metal used in the lid = π +π with = 0 i.e. A lid = 60π 9 Total area = 80π 60π So percentage used in the lid = 80π 00 = %. 9 9 Pearson Education Ltd 008

33 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise D, Question 8 The diagram shows an open tank for storing water, ABCDEF. The sides ABFE and CDEF are rectangles. The triangular ends ADE and BCF are isosceles, and AED = BFC = 90. The ends ADE and BCF are vertical and EF is horizontal. Given that AD = metres: (a) show that the area of triangle ADE is m. Given also that the capacity of the container is 000 m and that the total area of the two triangular and two rectangular sides of the container is S m : (b) Show that S = +. Given that can vary: 6000 (c) Use calculus to find the minimum value of S. (d) Justify that the value of S you have found is a minimum. (a) Let the equal sides of ADE be a metres.

34 Heinemann Solutionbank: Core Maths C Page of Then a + a = (Pythagoras' Theorem) So a = a = Area of ADE = base height = a a = (b) Area of two triangular sides is = Let the length AB = CD = y metres Area of two rectangular sides is ay = ay = \ y Then S = + \ y * But capacity of storage tank is y so y = y = Substitute this into equation * to give S = (c) ds ds = Put = Then = 0 = = 6000 = 0 or 8.8 Substitute into epression for S to give S = = 00 d S (d) = When = 0, d S = > 0 minimum value Pearson Education Ltd 008

35 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise D, Question 9 The diagram shows part of the curve with equation y = f ( ), where: f 00, > 0 The curve cuts the -ais at the points A and C. The point B is the maimum point of the curve. (a) Find f ( ). 50 (b) Use your answer to part (a) to calculate the coordinates of B. (a) f = f = 50 (b) At the maimum point, B, f ( ) = 0. So 50 = 0 50 = 50 = = 5 = 5 at point B As y = f ( ), y = f ( 5 ) at point B. So y = 5. The coordinates of B are ( 5, 5 ). Pearson Education Ltd 008

36 Heinemann Solutionbank: Core Maths C Page of

37 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise D, Question 0 The diagram shows the part of the curve with equation y = 5 for which y 0. The point P(, y) lies on the curve and O is the origin. (a) Show that OP = + 5. Taking f + 5 : (b) Find the values of for which f ( ) = 0. (c) Hence, or otherwise, find the minimum distance from O to the curve, showing that your answer is a minimum. (a) P has coordinates, 5. So OP = ( 0 ) = = + 5 (b) Given f = + 5 f ( ) = 8 When f ( ) = 0, 8 = 0 ( 8 ) = 0 = 0 or = 8

38 Heinemann Solutionbank: Core Maths C Page of = 0 or = ± (c) Substitute = 8 into f ( ) : OP = = 9 So OP = when = ± f ( ) = 8 = 6 > 0 when = 8 minimum value for OP and hence OP. So minimum distance from O to the curve is. Pearson Education Ltd 008

39 Heinemann Solutionbank: Core Maths C Page of Solutionbank C Eercise D, Question The diagram shows part of the curve with equation y = The curve touches the -ais at A and crosses the -ais at C. The points A and B are stationary points on the curve. (a) Show that C has coordinates (, 0). (b) Using calculus and showing all your working, find the coordinates of A and B. (a) y = Let y = 0, then = 0 ( ) ( + + ) = 0 ( ) ( + ) = 0 = or = when y = 0 The curve touches the -ais at = ( A ) and cuts the ais at = ( C ). C has coordinates (, 0 ) (b) = 5 + Put = 0, then 5 + = 0 ( 5 ) ( + ) = 0 5 = or = When =, y = = 9 5 So, 9 is the point B. When =, y =

40 Heinemann Solutionbank: Core Maths C Page of So (, 0 ) is the point A. Pearson Education Ltd 008