Solutionbank C2 Edexcel Modular Mathematics for AS and A-Level

Transcription

1 file://c:\users\buba\kaz\ouba\c_rev_a_.html Eercise A, Question Epand and simplify ( ) 5. ( ) 5 = + 5 ( ) + 0 ( ) + 0 ( ) + 5 ( ) + ( ) 5 = Compare ( + ) n with ( ) n. Replace n by 5 and by.

2 file://c:\users\buba\kaz\ouba\c_rev_a_.html Eercise A, Question In the diagram, ABC is an equilateral triangle with side 8 cm. PQ is an arc of a circle centre A, radius 6 cm. Find the perimeter of the shaded region in the diagram. Remember: The length of an arc of a circle is L = rθ. The area of a sector is A = r θ. Draw a diagram. Remember: 60 = π radius Length of arc PQ = rθ π = 6 ( ) = π cm Perimeter of shaded region = π = + π = 8.8 cm

3 file://c:\users\buba\kaz\ouba\c_rev_a_.html Eercise A, Question The sum to infinity of a geometric series is 5. Given that the first term is 5, (a) find the common ratio, (b) find the third term. (a) a r 5 r = 5, a = 5 = 5 Remember: s = a r a = 5 so that 5 =. 5 r, where r <. Here s = 5 and r = r = (b) ar = 5 ( ) = 5 9 = 0 9 Remember: nth term = ar n. Here a = 5, r = and n =, so that ar n = 5 ( ) = 5 ( )

4 file://c:\users\buba\kaz\ouba\c_rev_a_.html Eercise A, Question Sketch the graph of y = sin θ in the interval π θ<π. Remember: 80 =π radians

5 file://c:\users\buba\kaz\ouba\c_rev_a_5.html Eercise A, Question 5 Find the first three terms, in descending powers of b, of the binomial epansion of ( a + b ) 6, giving each term in its simplest form. ( a + b ) 6 = ( a ) ( ) ( a ) 5 6 ( b ) + ( ) ( a ) ( b ) + = 6 a a 5 b + 5 a b + = 6a a 5 b + 60a b + Compare ( a + b ) n with ( a + b ) n. Replace n by 6, a by a and b by b.

6 file://c:\users\buba\kaz\ouba\c_rev_a_6.html Eercise A, Question 6 AB is an arc of a circle centre O. Arc AB = 8 cm and OA = OB = cm. (a) Find, in radians, AOB. (b) Calculate the length of the chord AB, giving your answer to significant figures. Draw a diagram. Let AOB =θ. (a) AOB =θ θ = 8 Use l = rθ. Here l = 8 and r =. so θ = (b) = AB + ( ) ( ) cos ( ) = 6.66 AB = 7.85 cm ( s.f. ) Use the cosine formula c = a + b ab cos C. Here c = AB, a =, b = and C =θ= change your to radians.. Remember to

7 file://c:\users\buba\kaz\ouba\c_rev_a_7.html Eercise A, Question 7 A geometric series has first term and common ratio r. The sum of the first three terms of the series is 7. (a) Show that r + r = 0. (b) Find the two possible values of r. Given that r is positive, (c) find the sum to infinity of the series. (a), r, r, Use ar n to write down epressions for the first terms. Here a = and n =,,. + r + r = 7 r + r = 0 (as required) (b) r + r = 0 ( r ) ( r + ) = 0 r =, r = Factorize r + r. ac =. ( ) + ( + 6 ) = +, so r r + 6r = r ( r ) + ( r ) = ( r ) ( r + ). (c) r = a r = = 8 a Use S =. Here a = and r =, so that a r = r = = 8.

8 file://c:\users\buba\kaz\ouba\c_rev_a_8.html Eercise A, Question 8 (a) Write down the number of cycles of the graph y = sin n in the interval (b) Hence write down the period of the graph y = sin n. (a) n Consider the graphs of y = sin, y = sin, y = sin y = sin has cycle in the interval y = sin has cycles in the interval y = sin has cycles in the interval etc. So y = sin n has n cycles in the internal (b) 60 n π ( or ) n Period = length of cycle. If there are n cycles in the interval 0 60, 60 the length of each cycle will be. n

9 file://c:\users\buba\kaz\ouba\c_rev_a_9.html Eercise A, Question 9 (a) Find the first four terms, in ascending powers of, of the binomial epansion of ( + p ) 7, where p is a non-zero constant. Given that, in this epansion, the coefficients of and are equal, (b) find the value of p, (c) find the coefficient of. (a) ( + p ) 7 n ( n ) = + n + + n ( n ) ( n )!! + 7 ( 6 ) = + 7 ( p ) + ( p ) + 7 ( 6 ) ( 5 )!! ( p ) + = + 7p + p + 5p + Compare ( + ) n with ( + p ) n. Replace n by 7 and by p. (b) 7p = p p 0, so 7 = p The coefficients of and are equal, so 7p = p. p = (c) 5p = 5 ( ) = 5 7 The coefficient of is 5p. Here p =, so that 5p = 5 ( ).

10 file://c:\users\buba\kaz\ouba\c_rev_a_0.html Eercise A, Question 0 A sector of a circle of radius 8 cm contains an angle of θ radians. Given that the perimeter of the sector is 0 cm, find the area of the sector. Draw a diagram. Perimeter of sector = 0cm, so arc length = cm. 8θ = θ = 8 Area of sector = ( 8 ) θ = ( 8 ) ( ) = 56 cm 8 Find the value of θ. Use L = rθ. Here L = and r = 8 so that 8θ =. Use A = r θ. Here r = 8 and θ =, so that A = ( 8 ) ( ). 8 8

11 file://c:\users\buba\kaz\ouba\c_rev_a_.html Eercise A, Question A pendulum is set swinging. Its first oscillation is through 0. Each succeeding oscillation is What is the total angle described by the pendulum before it stops? 9 0 of the one before it. 0, 0 ( ), 0 ( ), a r = = ( ) 0 = Write down the first term. Use ar n. Here a = 0, r = 9 0 and n =,,. Use S =. Here a = 0 and r = so that S = a r 9 0

12 file://c:\users\buba\kaz\ouba\c_rev_a_.html Page of Eercise A, Question Write down the eact value (a) sin0, (b) cos0, (c) tan ( 60 ). (a) Draw a diagram for each part. Remember sin 0 = (b) cos 0 = cos 0 = \ 0 is in the fourth quadrant. (c)

13 file://c:\users\buba\kaz\ouba\c_rev_a_.html Page of tan ( 60 ) = tan 60 = \ = \ 60 is in the fourth quadrant.

14 file://c:\users\buba\kaz\ouba\c_rev_a_.html Eercise A, Question (a) Find the first three terms, in ascending powers of, of the binomial epansion of ( a ) 8, where a is a non-zero integer. The first three terms are, and b, where b is a constant. (b) Find the value of a and the value of b. (a) ( a ) n ( n ) 8 = + n + +! = + 8 ( a ) + ( a ) + 8 ( 8 )! = 8a + 8a + Compare ( + ) n with ( a ) n Replace n by 8 and by a. (b) 8a = a = b = 8a = 8 ( ) = 5 so a = and b = 5 Compare coefficients of, so that 8a =.

15 file://c:\users\buba\kaz\ouba\c_rev_a_.html Eercise A, Question In the diagram, A and B are points on the circumference of a circle centre O and radius 5 cm. AOB =θ radians. AB = 6 cm. (a) Find the value of θ. (b) Calculate the length of the minor arc AB. (a) cos θ = ( 5 ) ( 5 ) = 7 5 θ =.87 radians Use the cosine formula cos C = a = 5, b = 5 and c = 6. a + b c ab. Here c =θ, (b) arc AB = 5θ = 5.87 = 6. cm Use C = rθ. Here C = arc AB, r = 5 and θ =.87 radians.

16 file://c:\users\buba\kaz\ouba\c_rev_a_5.html Eercise A, Question 5 The fifth and sith terms of a geometric series are.5 and 6.75 respectively. (a) Find the common ratio. (b) Find the first term. (c) Find the sum of the first 0 terms, giving your answer to decimal places. (a) ar =.5, ar 5 = 6.75 ar = ar.5 r = Find r. Divide ar 5 by ar ar 5 so that = ar5 ar a 6.75 = r and =.5..5 (b) a (.5 ) =.5 a =.5 (.5 ) = 8 9 (c) S 0 = 8 ( (.5 ) 0 ) 9.5 = ( d.p. ) a ( r n ) Use S n =. Here a =, r =.5 and n = 0. r 8 9

17 file://c:\users\buba\kaz\ouba\c_rev_a_6.html Page of Eercise A, Question 6 Given that θ is an acute angle measured in degrees, epress in term of cosθ (a) cos ( 60 + θ ), (b) cos ( θ ), (c) cos ( 80 θ ) (a) Draw a diagram for each part. cos ( 60 + θ ) = cos θ 60 + θ is in the first quadrant. (b) cos ( θ ) = cos θ θ is in the fourth quadrant. (c)

18 file://c:\users\buba\kaz\ouba\c_rev_a_6.html Page of cos ( 80 θ ) = cos θ 80 θ is in the second quadrant.

19 file://c:\users\buba\kaz\ouba\c_rev_a_7.html Eercise A, Question 7 (a) Epand ( ) 0 in ascending powers of up to and including the term in. (b) Use your answer to part (a) to evaluate ( 0.98 ) 0 correct to decimal places. (a) ( ) 0 n ( n ) = + n + + n ( n ) ( n )!! + 0 ( a ) = + 0 ( ) + ( ) + 0 ( 9 ) ( 8 ) 6 ( ) + = Compare ( + ) n with ( ) n. Replace n by 0 and by. (b) ( ( 0.0 ) ) 0 = 0 ( 0.0 ) + 80 ( 0.0 ) 960 ( 0.0 ) ~ 0.87 ( d.p. ) Find the value of = 0.0 = ( 0.0 )

20 file://c:\users\buba\kaz\ouba\c_rev_a_8.html Eercise A, Question 8 In the diagram, AB = 0 cm, AC = cm. CAB = 0.6 radians. BD is an arc of a circle centre A and radius 0 cm. (a) Calculate the length of the arc BD. (b) Calculate the shaded area in the diagram. (a) arc BD = = 6 cm (b) Shaded area = ( 0 ) ( ) sin ( 0.6 ) ( 0 ) ( 0.6 ) = 6.70 cm ( s.f. ) Use L = rθ. Here L = arc BD, r = 0 and θ = 0.6 radians. Use area of triangle = bc sin A and area of sector = r θ. Here b =, c = 0 and A = (θ= ) 0.6 ; r = 0 and θ = 0.6.

21 file://c:\users\buba\kaz\ouba\c_rev_a_9.html Eercise A, Question 9 The value of a gold coin in 000 was 80. The value of the coin increases by 5% per annum. (a) Write down an epression for the value of the coin after n years. (b) Find the year in which the value of the coin eceeds , 80 (.05 ), 80 (.05 ), (a) Value after n years = 80 (.05 ) n (b) 80 (.05 ) n > (.05 ) = (.05 ) 5 = 7. The value of the coin will eceed 60 in 0. Write down the first terms. Use ar n. Here a = 80, r =.05 and n =,,. Substitute values of n. The value of the coin after years is 56.9, and after 5 years is 7.. So the value of the coin will eceed 60 in the 5th year.

22 file://c:\users\buba\kaz\ouba\c_rev_a_0.html Page of Eercise A, Question 0 Given that is an acute angle measured in radians, epress in terms of sin (a) sin ( π ), (b) sin (π+), (c) cos. π (a) Draw a diagram for each part. Remember: π Radians = 80 sin ( π ) = sin π is in the fourth quadrant. (b) sin (π+) = sin π + is in the third quadrant. (c) 80 =π radians, so 90 π = radians.

23 file://c:\users\buba\kaz\ouba\c_rev_a_0.html Page of π cos ( ) = sin ( = ). a c

24 file://c:\users\buba\kaz\ouba\c_rev_a_.html Eercise A, Question Epand and simplify 6 ( ) 6 = ( ) 5 6 ( ) + ( ) ( ) 6 + ( ) ( ) 5 + ( ) 6 ) 6 + ( ) ( ) 6 + ( ) ( ) 5 = ( ) + 5 ( ) + 0 ( + 5 ( ) + 6 ( ) = Compare ( ) n with ( a + b ) n. Replace n by 6, a with and b with. ( ) = = ( ) = = ( ) = =

25 file://c:\users\buba\kaz\ouba\c_rev_a_.html Eercise A, Question A cylindrical log, length m, radius 0 cm, floats with its ais horizontal and with its highest point cm above the water level. Find the volume of the log in the water. Draw a diagram. Let sector angle = φ. 6 cos φ = ( = 0.8 ) 0 Area above water level = r ( φ ) r sin ( φ ) ( 0 ) sin ( φ ) = ( 0 ) ( φ ) = 65.0 cm Area below water level =π(0 ) 65.0 = 9. cm Volume below water level = = 88 cm ( = 0.8 cm ) Use area of segment = r θ r sin θ. Here r = 0 cm and θ = cos ( 0.8 )

26 file://c:\users\buba\kaz\ouba\c_rev_a_.html Eercise A, Question (a) On the same aes, in the interval 0 60, sketch the graphs of y = tan ( 90 ) and y = sin. (b) Hence write down the number of solutions of the equation tan ( 90 ) = sin in the interval (a) y = tan ( 90 ) is a translation of y = tan by + 90 in the -direction. (b) solutions in the interval From the sketch, the graphs of y = tan ( 90 ) and y = sin meet at two points. So there are solutions in the internal 0 60.

27 file://c:\users\buba\kaz\ouba\c_rev_a_.html Eercise A, Question A geometric series has first term and common ratio sum eceeding 00.. Find the greatest number of terms the series can have without its a =, r = S n = ( ( ) n ) a ( r n ) Use S n =. Here a = and r =. r = ( ( ) n ) n ) = ( ( ) Now, ( ( ) n ) < 00 ( ) n < 00 ( ) n < + 00 ( ) n < 9 ( ) 7 = 7.9 ( ) 8 = Substitute values of n. The largest value of n for which ( ) n < 9 / is 7. so n = 7

28 file://c:\users\buba\kaz\ouba\c_rev_a_5.html Eercise A, Question 5 Describe geometrically the transformation which maps the graph of (a) y = tan onto the graph of y = tan ( 5 ), (b) y = sin onto the graph of y = sin, (c) y = cos onto the graph of y = cos, (d) y = sin onto the graph of y = sin. (a) A translation of + 5 in the direction (b) A stretch of scale factor in the y direction (c) A stretch of scale factor in the direction (d) A translation of in the y direction

29 file://c:\users\buba\kaz\ouba\c_rev_a_6.html Eercise A, Question 6 If is so small that terms of and higher can be ignored, and ( ) ( + ) 5 a + b + c, find the values of the constants a, b and c. ( + ) 5 = + n + n ( n )! + ( ) + = + 5 ( ) + = ( ) ( ) ( ) = ( ) ( + ) Compare ( + ) n with ( + ) n. Replace n by 5 and by. Epand ( ) ( ) ignoring terms in, so that ( ) and = ( ) = 0 0 simplify so that = so a =, b = 9, c = 70 Compare with a + b + c + so that a =, b = 9 and c = 70.

30 file://c:\users\buba\kaz\ouba\c_rev_a_7.html Eercise A, Question 7 A chord of a circle, radius 0 cm, divides the circumference in the ratio :. Find the ratio of the areas of the segments into which the circle is divided by the chord. Draw a diagram. Let the minor are l have angle θ and the major are l have angle θ. l : l = : 0θ : 0θ = : θ :θ = : Use l = rθ so that l = 0θ and l = 0θ. so θ = π = Shaded area = ( 0 ) θ ( 0 ) sin θ π = 5π 50 Area of large segment =π(0 ) ( 5π 50 ) = 00π 5π + 50 = 75π + 50 Use area of segment = r θ r sin θ. Here r = 0 and θ =θ = Ratio of small segment to large segment is 5π 50 : 75π + 50 Divide throughout by 5. π : π + or : π + π π

31 file://c:\users\buba\kaz\ouba\c_rev_a_8.html Page of Eercise A, Question 8, and + 8 are the fourth, fifth and sith terms of geometric series. (a) Find the two possible values of and the corresponding values of the common ratio. Given that the sum to infinity of the series eists, (b) find the first term, (c) the sum to infinity of the series. ar = ar = ar 5 = + 8 ar 5 ar = + 8 so = ar ar 5 = r and = r so =. ar ar ar ar ar ( + 8 ) = = 0 ( + 9 ) ( ) = 0 =, = 9 ar r = = ar ar ar5 Clear the fractions. Multiply each side by so that + 8 ar = ( + 8 ) and = 9. When =, r = Find r. Substitute =, then = 9, into =, so that When = 9, r = r = and r = =. 9 ar ar (b) r = ar = a ( ) = a = a Remember S = for r <, so r =. r (c)

32 file://c:\users\buba\kaz\ouba\c_rev_a_8.html Page of a r = = 6

33 file://c:\users\buba\kaz\ouba\c_rev_a_9.html Eercise A, Question 9 (a) Sketch the graph of y =.5 cos ( 60 ) in the interval 0 <60 (b) Write down the coordinates of the points where your graph meets the coordinate aes. (a) (b) When = 0, y =.5 cos ( 60 ) = 0.75 so ( 0, 0.75 ) y =.5 cos ( 60 ) The graph of y =.5 cos ( 60 ) meets the y ais when = 0. Substitute = 0 into y =.5 cos ( 60 ) so that y =.5 cos ( 60 ). cos ( 60 ) = cos 60 = so y =.5 = 0.75 y = 0, The graph of y =.5 cos ( 60 ) meets the -ais when = when y = 0. cos ( 60 ) represents a translation of = 50 cos by + 60 in the -direction cos meets the -ais at 90 and 70, so y =.5 ( cos 60 ) meets the - and = ais at = 50 and = 0. = 0

34 file://c:\users\buba\kaz\ouba\c_rev_a_0.html Eercise A, Question 0 Without using a calculator, solve sin ( 0 \ ) = in the interval sin 60 = \ sin = quadrants. \. sin is negative in the rd and th sin ( 0 ) = \ sin ( 00 ) = \ so 0 = 0 sin 0 \ = but sin ( 0 \ ) = so = 60 0 = 0, i.e. = 60. Similarly and 0 = 00 = 0 sin 00 = but sin ( 0 ) = so 0 = 00, i.e. = 0. \ \