misswelton.weebly.com€¦ · Web viewFew candidates were successful in part (b) with many unable...

1.1 SL

1a. [5 marks]

Markscheme

correct working (A1)

eg ,

(seen anywhere) (A1)

correct working (ignore additional values) (A1)

eg ,

= 2, 10 A1A1 N1N1

[5 marks]

Examiners report

[N/A]

1b. [4 marks]

Markscheme


eg , ,

valid approach (M1)

eg , , = common difference

= −6, = 8 (accept ) A1A1 N2N2

[4 marks]

Examiners report

[N/A]

1

1c. [4 marks]

Markscheme

valid approach (M1)

eg first intersection at ,

correct working A1

eg , ,

P(154, ) (accept and ) A1A1 N3

[4 marks]

Examiners report

[N/A]

1d. [4 marks]

Markscheme

valid attempt to find upper boundary (M1)

eg half way between and , , 154 + 4, , at least two values of new sequence {6,

14, ...}

upper boundary at (seen anywhere) (A1)

correct integral expression (accept missing ) A1A1 N4

eg , ,

Note: Award A1 for two correct limits and A1 for correct integrand. The A1 for correct integrand may

be awarded independently of all the other marks.

[4 marks]

Examiners report

[N/A]

2

2a. [3 marks]

Markscheme

attempt to add corresponding terms (M1)

eg

correct value for (A1)

eg 324

4, 36, 324 (accept 4 + 36 + 324) A1 N3

[3 marks]

Examiners report

[N/A]

2b. [2 marks]

Markscheme

valid approach (M1)

eg ,

(accept ; may be incorrect) A1 N2

[2 marks]

Examiners report

[N/A]

2c. [2 marks]

Markscheme

recognition that 225 terms of consists of 113 non-zero terms (M1)

eg , , 113

3

(accept ; may be incorrect) A1 N2

[2 marks]

Examiners report

[N/A]

3a. [4 marks]

Markscheme

attempt to find (M1)

eg 1.4 − 1.3 , ,

(may be seen in expression for ) (A1)

correct equation (A1)

eg ,

A1 N3

[4 marks]

Examiners report

[N/A]

3b. [2 marks]

Markscheme

correct substitution (A1)

eg , ,

A1 N2

[2 marks]

Examiners report

4

[N/A]

3c. [5 marks]

Markscheme

recognizing need to find the sequence of multiples of 3 (seen anywhere) (M1)

eg first term is (= 1.5) (accept notation ) ,

(= 0.3) , 100 terms (accept ), last term is 31.2

(accept notation ) , (accept )

correct working for sum of sequence where n is a multiple of 3 A2

, , 1635

valid approach (seen anywhere) (M1)

eg , , (their sum for )

correct working (seen anywhere) A1

eg , 4875 − 1635

AG N0

[5 marks]

Examiners report

[N/A]

3d. [5 marks]

Markscheme


eg dividing consecutive terms

5

correct value of (seen anywhere, including in formula)

eg , 0.707106… ,

correct working (accept equation) (A1)

eg

correct working A1

METHOD 1 (analytical)

eg , , 948.974

METHOD 2 (using table, must find both values)

eg when , AND when ,

A1 N2

[5 marks]

Examiners report

[N/A]

4a. [2 marks]

Markscheme


eg −5 + (8 − 1)(3)

u = 16 A1 N2

6

[2 marks]

Examiners report

[N/A]

4b. [2 marks]

Markscheme


eg −5 + (8 − 1)(3)

u = 16 A1 N2

[2 marks]

Examiners report

[N/A]

4c. [4 marks]

Markscheme

correct substitution into u formula (A1)

eg −5 + 3(n − 1), 3n − 8


eg −5 + 3(n − 1) = 67, 3n − 8 = 67, 3(n − 1) = 72


eg 3n = 75, n − 1 = 24

n = 25 A1 N3

[4 marks]

7

Examiners report

[N/A]

5. [6 marks]

Markscheme

correct substitution into formula for infinite geometric series (A1)

eg

correct substitution into formula for (seen anywhere) (A1)

eg

attempt to express in terms of (or vice-versa) (M1)

eg , , ,


eg , , (0.4, 19.95), (0.6, 13.3),

, A1A1 N3

[6 marks]

Examiners report

[N/A]

6a. [2 marks]

Markscheme

valid approach (M1)

eg

8

A1 N2

[2 marks]

Examiners report

[N/A]

6b. [3 marks]

Markscheme

recognizing that sinθ is bounded (M1)

eg 0 ≤ sin θ ≤ 1, −1 ≤ sinθ ≤ 1, −1 < sinθ < 1

0 < r ≤ A2 N3

Note: If working shown, award M1A1 for correct values with incorrect inequality sign(s).

If no working shown, award N1 for correct values with incorrect inequality sign(s).

[3 marks]

Examiners report

[N/A]

6c. [4 marks]

Markscheme

correct substitution into formula for infinite sum A1

eg

evidence of choosing an appropriate rule for cos 2θ (seen anywhere) (M1)

eg cos 2 θ = 1 − 2 sin θ

correct substitution of identity/working (seen anywhere) (A1)

eg

9

correct working that clearly leads to the given answer A1

eg

AG N0

[4 marks]

Examiners report

[N/A]

6d. [6 marks]

Markscheme

METHOD 1 (using differentiation)

recognizing (seen anywhere) (M1)

finding any correct expression for (A1)

eg


eg sin 2θ = 0

any correct value for sin(0) (seen anywhere) (A1)

eg 0, , … , sketch of sine curve with x-intercept(s) marked both correct values for 2θ (ignore

additional values) (A1)

2θ = , 3 (accept values in degrees)

both correct answers A1 N4

Note: Award A0 if either or both correct answers are given in degrees.

Award A0 if additional values are given.

10

METHOD 2 (using denominator)

recognizing when S is greatest (M1)

eg 2 + cos 2θ is a minimum, 1−r is smallest


eg minimum value of 2 + cos 2θ is 1, minimum r =


eg

EITHER (using cos 2θ)

any correct value for cos(−1) (seen anywhere) (A1)

eg , 3 , … (accept values in degrees), sketch of cosine curve with x-intercept(s) marked

both correct values for 2θ (ignore additional values) (A1)

2θ = , 3 (accept values in degrees)

OR (using sinθ)

sin =θ ±1 (A1)

sin(1) = (accept values in degrees) (seen anywhere) A1

THEN

both correct answers A1 N4

Note: Award A0 if either or both correct answers are given in degrees.

Award A0 if additional values are given.

[6 marks]

Examiners report

[N/A]

7a. [2 marks]

11

Markscheme

valid approach involving addition or subtraction M1

eg

correct application of log law A1

eg

AG N0

[2 marks]

Examiners report

[N/A]

7b. [6 marks]

Markscheme

METHOD 1 (finding and d)

recognizing (seen anywhere) (A1)

attempt to find or d using (M1)

eg , , correct value of or d

= 2, d = 3 (seen anywhere) (A1)(A1)


eg

= 610 A1 N2

METHOD 2 (expressing S in terms of c)


12

correct expression for S in terms of c (A1)

eg

(seen anywhere) (A1)(A1)


eg

= 610 A1 N2

METHOD 3 (expressing S in terms of c)


correct expression for S in terms of c (A1)

eg

correct application of log law (A1)

eg

correct application of definition of log (A1)

eg


eg

= 610 A1 N2

[6 marks]

Examiners report

[N/A]

13

8a. [4 marks]

Markscheme

valid approach to find maxima (M1)

eg one correct value of x, sketch of f

any two correct consecutive values of x (A1)(A1)

eg x = 1, x = 5

a = 4 A1 N3

[4 marks]

Examiners report

[N/A]

8b. [4 marks]

Markscheme

recognizing the sequence x x x x is arithmetic (M1)

eg d = 4

correct expression for sum (A1)

eg

valid attempt to solve for n (M1)

eg graph, 2n − n − 861 = 0

n = 21 A1 N2

[4 marks]

Examiners report

[N/A]

9a. [2 marks]

14

Markscheme

correct substitution into infinite sum (A1)

eg

r = 0.98 (exact) A1 N2

[2 marks]

Examiners report

[N/A]

9b. [2 marks]

Markscheme


29.8473

29.8 A1 N2

[2 marks]

Examiners report

[N/A]

9c. [3 marks]

Markscheme

attempt to set up inequality (accept equation) (M1)

eg

correct inequality for n (accept equation) or crossover values (A1)

eg n > 83.5234, n = 83.5234, S = 162.606 and S = 163.354

n = 84 A1 N1

[3 marks]

15

Examiners report

[N/A]

10a. [5 marks]

Markscheme

infinite sum of segments is 2 (seen anywhere) (A1)

eg

recognizing GP (M1)

eg ratio is

correct substitution into formula (may be seen in equation) A1

eg


eg

correct working leading to answer A1

eg

AG N0

[5 marks]

Examiners report

[N/A]

10b. [9 marks]

Markscheme

recognizing infinite geometric series with squares (M1)

eg

16

correct substitution into (must substitute into formula) (A2)

eg


eg

(seen anywhere) A1

valid approach with segments and CD (may be seen earlier) (M1)

eg

correct expression for in terms of (may be seen earlier) (A1)

eg

substituting their value of into their formula for (M1)

eg

A1 N3

[9 marks]

Examiners report

[N/A]

11a. [2 marks]

Markscheme

subtracting terms (M1)

eg

A1 N2

17

[2 marks]

Examiners report

[N/A]

11b. [2 marks]

Markscheme

correct substitution into formula (A1)

eg

A1 N2

[2 marks]

Examiners report

[N/A]

11c. [2 marks]

Markscheme

correct substitution into formula for sum (A1)

eg

A1 N2

[2 marks]

Examiners report

[N/A]

12a. [3 marks]

Markscheme

correct use A1

eg

18

valid approach to find (M1)

eg

A1 N2

[3 marks]

Examiners report

[N/A]

12b. [5 marks]

Markscheme

recognizing a sum (finite or infinite) (M1)

eg

valid approach (seen anywhere) (M1)

eg recognizing GP is the same as part (a), using their value from part (a),

correct substitution into infinite sum (only if is a constant and less than 1) A1

eg


eg

A1 N3

[5 marks]

Examiners report

[N/A]

13a. [2 marks]

Markscheme19

attempt to subtract terms (M1)

eg

A1 N2

[2 marks]

Examiners report

[N/A]

13b. [2 marks]

Markscheme

correct approach (A1)

eg

A1 N2

[2 marks]

Examiners report

[N/A]

13c. [2 marks]

Markscheme

correct substitution into sum (A1)

eg

A1 N2

[2 marks]

Examiners report

[N/A]

14. [6 marks]

20

Markscheme


eg

correct expression for (A1)

eg

EITHER (solving inequality)

valid approach (accept equation) (M1)

eg

valid approach to find M1

eg , sketch

correct value

eg (A1)

(must be an integer) A1 N2

OR (table of values)

valid approach (M1)

eg , one correct crossover value

both crossover values, and A2


OR (sketch of functions)

valid approach M1

eg sketch of appropriate functions

valid approach (M1)

eg finding intersections or roots (depending on function sketched)

21

correct value

eg (A1)


[6 marks]

Examiners report

[N/A]

15a. [2 marks]

Markscheme

evidence of dividing terms (in any order) (M1)

eg

A1 N2

[2 marks]

Examiners report

[N/A]

15b. [2 marks]

Markscheme


eg

correct working A1

eg

AG N0

[2 marks]

22

Examiners report

[N/A]

15c. [4 marks]

Markscheme

evidence of subtracting two terms (in any order) (M1)

eg

correct application of the properties of logs (A1)

eg


eg

A1 N3

[4 marks]

Examiners report

[N/A]

15d. [2 marks]

Markscheme

correct substitution into the formula for the sum of an arithmetic sequence (A1)

eg

correct working A1

eg

AG N0

[2 marks]

23

Examiners report

[N/A]

15e. [3 marks]

Markscheme


eg


eg

(accept ) A1 N2

[3 marks]

Examiners report

[N/A]

16a. [2 marks]

Markscheme

attempt to substitute into formula for mean (M1)

eg

mean A1 N2

[2 marks]

Examiners report

[N/A]

16b. [2 marks]

Markscheme

24

(i) mean A1 N1

(ii) A1 N1

[2 marks]

Examiners report

[N/A]

16c. [6 marks]

Markscheme

(i) valid approach (M1)

eg 95%, 5% of 27


eg

median A1 N2

(ii) METHOD 1

variance (seen anywhere) (A1)

valid attempt to find new standard deviation (M1)

eg

variance A1 N2

METHOD 2

variance (seen anywhere) (A1)

valid attempt to find new variance (M1)

eg

new variance A1 N2

[6 marks]25

Examiners report

[N/A]

16d. [6 marks]

Markscheme

(i) both correct frequencies (A1)

eg 80, 150

subtracting their frequencies in either order (M1)

eg

70 (students) A1 N2

(ii) evidence of a valid approach (M1)

eg 10% of 200, 90%


eg , 180 students

A1 N3

[6 marks]

Examiners report

[N/A]

17a. [3 marks]

Markscheme

recognizing that it is an arithmetic sequence (M1)

eg

correct equation A1

eg

26

correct working (do not accept substituting ) A1

eg

AG N0

[3 marks]

Examiners report

Most candidates recognized that the series was arithmetic but many worked backwards using

rather than creating and solving an equation of their own to produce the given answer.

17b. [3 marks]

Markscheme

recognition of sum (M1)

eg

correct working for AP (A1)

eg

A1 N2

[3 marks]

Examiners report

Almost all students answered (b) correctly.

18. [6 marks]

Markscheme

METHOD 1

valid approach (M1)

eg

correct equation in terms of only A1

27

eg


eg

valid attempt to solve their quadratic equation (M1)

eg factorizing, formula, completing the square

evidence of correct working (A1)

eg

A1 N4

METHOD 2 (finding r first)

valid approach (M1)

eg

correct equation in terms of only A1

eg

evidence of correct working (A1)

eg

A1

substituting their values of to find (M1)

eg

A1 N4

[6 marks]

Examiners report

28

Nearly all candidates attempted to set up an expression, or pair of expressions, for the common ratio of

the geometric sequence. When done correctly, these expressions led to a quadratic equation which was

solved correctly by many candidates.

19. [6 marks]

Markscheme

correct equation to find (A1)

eg

(seen anywhere) (A1)

correct equation to find A1

eg

(A1)

(M1)

1280 A1 N4

[6 marks]

Examiners report

The majority of candidates did well on this question although identifying the common ratio was not

always as easily done and some candidates lost marks as a result of using an inappropriate method

such as 8/4. Other candidates correctly guessed the value of or did so by showing that if , then

the ratio of the fourth term to the first term would be 8. It was also disappointing to see some

candidates use an incorrect formula for the sum of the first terms of a geometric series: a common

error seen was , which led to the wrong answer of 22.5.

20a. [3 marks]

Markscheme


eg

29

A1 N2

(ii) correct interpretation R1 N1

eg population is decreasing, growth rate is negative

[3 marks]

Examiners report

Part (a) was generally done well, with many candidates able to find the value of correctly and to

interpret its meaning. Lack of accuracy was occasionally a concern, with some candidates writing their

value of to 2 significant figures or evaluating incorrectly.

20b. [5 marks]

Markscheme

METHOD 1

valid approach (accept an equality, but do not accept 0.74) (M1)

eg

valid approach to solve their inequality (M1)

eg logs, graph

A1

28 years A2 N2

METHOD 2

valid approach which gives both crossover values accurate to at least 2 sf A2

eg

(A1)

28 years A2 N2

[5 marks]

30

Examiners report

Few candidates were successful in part (b) with many unable to set up an inequality or equation which

would allow them to find the condition on . Some were able to find the value of in decades but most

were unable to correctly interpret their inequality in terms of the least number of whole years. While a

solution through analytic methods was readily available, very few students attempted to use their GDC

to solve their initial equation or inequality.

21a. [2 marks]

Markscheme

valid approach (M1)

eg

A1 N2

[2 marks]

Examiners report

Most candidates found this question straightforward and accessible. They could find the correct

difference and substituted correctly into term and sum formula respectively.

21b. [2 marks]

Markscheme

correct substitution into term formula (A1)

eg

A1 N2

[2 marks]

Examiners report



21c. [2 marks]

31

Markscheme

correct substitution into sum formula (A1)

eg

A1 N2

[2 marks]

Examiners report



22. [6 marks]

Markscheme

Note: There are many approaches to this question, and the steps may be done in any order. There are

3 relationships they may need to apply at some stage, for the 3rd, 4th and 5th marks. These are

equating bases eg recognising 9 is

log rules: ,

exponent rule: .

The exception to the FT rule applies here, so that if they demonstrate correct application of the 3

relationships, they may be awarded the A marks, even if they have made a previous error. However all

applications of a relationship need to be correct. Once an error has been made, do not award A1FT for

their final answer, even if it follows from their working.

Please check working and award marks in line with the markscheme.

correct substitution into formula (A1)

eg

set up equation for in any form (seen anywhere) (M1)

32

eg

correct application of relationships (A1)(A1)(A1)

A1 N3

[6 marks]

Examples of application of relationships

Example 1

correct application of exponent rule for logs (A1)

eg

correct application of addition rule for logs (A1)

eg

substituting for 9 or 3 in ln expression in equation (A1)

eg

Example 2

recognising (A1)

eg

one correct application of exponent rule for logs relating to (A1)

eg

another correct application of exponent rule for logs (A1)

eg

Examiners report

[N/A]

23a. [2 marks]

Markscheme33

valid approach (M1)

eg

A1 N2

[2 marks]

Examiners report

[N/A]

23b. [2 marks]

Markscheme

correct substitution into (A1)

eg

(exact), A1 N2

[2 marks]

Examiners report

[N/A]

23c. [3 marks]

Markscheme

METHOD 1 (analytic)

valid approach (M1)

eg

correct inequality (accept equation) (A1)

eg

A1 N1

34

METHOD 2 (table of values)

both crossover values A2

eg

A1 N1

[3 marks]

Total [7 marks]

Examiners report

[N/A]

24a. [2 marks]

Markscheme

recognizing Ann rolls green (M1)

eg

A1 N2

[2 marks]

Examiners report

Some teachers’ comments suggested that the word ‘loses’ in the diagram was misleading. But candidate

scripts did not indicate any adverse effect.

a) Very well answered.

b) i) Probabilities and were typically found correctly. ii) Fewer candidates identified the common

ratio and number of rolls correctly.

Few candidates recognized that this was an infinite geometric sum although some did see that a

geometric progression was involved.

24b. [7 marks]

Markscheme

35

recognize the probability is an infinite sum (M1)

eg Ann wins on her roll or roll or roll…,

recognizing GP (M1)

(seen anywhere) A1

(seen anywhere) A1

correct substitution into infinite sum of GP A1

eg


eg

A1 N1

[7 marks]

Total [15 marks]

Examiners report

Some teachers’ comments suggested that the word ‘loses’ in the diagram was misleading, But candidate

scripts did not indicate any adverse effect.

a) Very well answered.

b) i) Probabilities and were typically found correctly. ii) Fewer candidates identified the common

ratio and number of rolls correctly.

Few candidates recognized that this was an infinite geometric sum although some did see that a

geometric progression was involved.

25a. [1 mark]

Markscheme

36

A1 N1

[1 mark]

Examiners report

In general, candidates showed confidence in this area of the syllabus. Appropriate formulae were

chosen for parts (b) and (c) and many candidates were able to achieve full marks. However, many

candidates found the common difference to be in part (a) by subtracting or believing

that the common difference should always be positive.

25b. [3 marks]

Markscheme

METHOD 1

valid approach (M1)

eg


eg

A1 N2

METHOD 2

attempt to list 3 or more terms in either direction (M1)

eg

correct list of 4 or more terms in correct direction (A1)

eg

A1 N2

[3 marks]

Examiners report



37



25c. [2 marks]

Markscheme

correct expression (A1)

eg

A1 N2

[2 marks]

Total [6 marks]

Examiners report





26. [7 marks]

Markscheme

METHOD 1

recognize that the distance walked each minute is a geometric sequence (M1)

eg , valid use of

recognize that total distance walked is the sum of a geometric sequence (M1)

eg

correct substitution into the sum of a geometric sequence (A1)

eg

38

any correct equation with sum of a geometric sequence (A1)

eg

attempt to solve their equation involving the sum of a GP (M1)

eg graph, algebraic approach

A1

since R1

he will be late AG N0

Note: Do not award the R mark without the preceding A mark.

METHOD 2


eg , valid use of


eg

correct substitution into the sum of a geometric sequence (A1)

eg

attempt to substitute into sum of a geometric sequence (M1)

eg


eg

A1

39

since R1

he will not be there on time AG N0


METHOD 3


eg , valid use of


eg

listing at least 5 correct terms of the GP (M1)

15 correct terms A1

attempt to find the sum of the terms (M1)

eg

A1

since R1

he will not be there on time AG N0


[7 marks]

Examiners report

Many found this question accessible, although the most common approach was to calculate each term

by brute force, which at times contained small errors or inaccuracies that affected the overall sum.

40

Although this was a valid method, it meant an inefficient use of time that could have affected the

performance on other questions.

Those who applied the formula for geometric series were typically more successful and far more

efficient in answering the question.

27a. [2 marks]

Markscheme


eg

A1 N2

[2 marks]

Examiners report

All three parts of this question were very well done by the candidates. The occasional mistakes that

were seen tended to be arithmetic errors which happened after the candidates had substituted

correctly into the formulas given in the formula booklet.

27b. [2 marks]

Markscheme


eg , listing terms

A1 N2

[2 marks]

Examiners report




27c. [2 marks]

Markscheme41


eg , listing terms,

A1 N2

[2 marks]

Total [6 marks]

Examiners report




28a. [5 marks]

Markscheme


eg

A1 N2

(ii) attempt to substitute into formula, with their (M1)

eg


eg

A1 N2

[5 marks]

Examiners report

Most candidates answered part (a) correctly.

28b. [5 marks]

42

Markscheme

(i) attempt to substitute (M1)

eg

A1 N2

(ii) correct substitution of into A1

eg

correct work (A1)

eg finding intersection point,

A1 N2

[5 marks]

Examiners report

A surprising number assumed the second sequence to be geometric as well, and thus part (b) was

confusing for many. It was quite common that students did not clearly show which work was relevant

to part (i) and which to part (ii), thus often losing marks.

28c. [5 marks]

Markscheme

correct expression for (A1)

eg

EITHER

correct substitution into inequality (accept equation) (A1)

eg

valid approach to solve inequality (accept equation) (M1)

43

eg finding point of intersection,


OR

table of values

when A1

when A1


[4 marks]

Total [14 marks]

Examiners report

Few students successfully completed part (c) as tried to solve algebraically instead of graphically.

Those who used the table of values did not always show two sets of values and consequently lost

marks.

29a. [4 marks]

Markscheme

valid method for finding side length (M1)

eg

correct working for area (A1)

eg

1 2 3 8 4 32 16 8

44

A1A1 N2N2

[4 marks]

Examiners report

[N/A]

29b. [4 marks]

Markscheme

METHOD 1

recognize geometric progression for (R1)

eg

(A1)


eg

A1 N3

METHOD 2


eg

(A1)


eg

A1 N3

[4 marks]

45

Examiners report

[N/A]

29c. [7 marks]

Markscheme

METHOD 1

recognize infinite geometric series (R1)

eg

area of first triangle in terms of (A1)

eg

attempt to substitute into sum of infinite geometric series (must have ) (M1)

eg

correct equation A1

eg


eg

valid attempt to solve their quadratic (M1)

eg

A1 N2

METHOD 2

recognizing that there are four sets of infinitely shaded regions with equal area R1

area of original square is (A1)

46

so total shaded area is (A1)

correct equation A1

(A1)

valid attempt to solve their quadratic (M1)

eg

A1 N2

[7 marks]

Examiners report

[N/A]

30a. [2 marks]

Markscheme


eg

A1 N2

[2 marks]

Examiners report

[N/A]

30b. [2 marks]

Markscheme


eg

A1 N2

47

[2 marks]

Examiners report

[N/A]

30c. [3 marks]

Markscheme

correct substitution into sum or term formula (A1)

eg

correct simplification (A1)

eg

A1 N2

[3 marks]

Examiners report

[N/A]

31a. [4 marks]

Markscheme

valid method (M1)

eg

A1A1A1 N4

[4 marks]

Examiners report

[N/A]

31b. [4 marks]

Markscheme

48

correct AP or GP (A1)

eg finding common difference is , common ratio is

valid approach using arithmetic and geometric formulas (M1)

eg and

A1A1 N4

Note: Award A1 for , A1 for .

[4 marks]

Examiners report

[N/A]

32a. [2 marks]

Markscheme

correct expression for A1 N1

eg

[2 marks]

Examiners report

[N/A]

32b. [2 marks]

Markscheme

correct equation A1

eg


49

eg

correct working A1

eg

AG N0

[2 marks]

Examiners report

[N/A]

32c. [3 marks]

Markscheme

valid attempt to solve (M1)

eg

A1A1 N3

[3 marks]

Examiners report

[N/A]

32d. [3 marks]

Markscheme

attempt to substitute any value of to find (M1)

eg

A1A1 N3

[3 marks]

Examiners report

50

[N/A]

32e. [3 marks]

Markscheme

(may be seen in justification) A1

valid reason R1 N0

eg

Notes: Award R1 for only if A1 awarded.

[2 marks]

Examiners report

[N/A]

32f. [3 marks]

Markscheme

finding the first term of the sequence which has (A1)

eg

(may be seen in formula) (A1)

correct substitution of and their into , as long as A1

eg

A1 N3

[4 marks]

51

Examiners report

[N/A]

33a. [1 mark]

Markscheme

A1 N1

[1 mark]

Examiners report

The majority of candidates had little difficulty with this question. If errors were made, they were

normally made out of carelessness. A very few candidates mistakenly used the formulas for geometric

sequences and series.

33b. [4 marks]

Markscheme

(i) correct substitution into term formula (A1)

e.g. ,

A1 N2

(ii) correct substitution into sum formula (A1)

eg ,

A1 N2

[4 marks]

Examiners report




33c. [2 marks]

Markscheme52


eg ,

A1 N2

[2 marks]

Total [7 marks]

Examiners report




34. [6 marks]

Markscheme

correct substitution into sum of a geometric sequence A1

eg ,

correct substitution into sum to infinity A1

eg

attempt to eliminate one variable (M1)

eg substituting

correct equation in one variable (A1)

eg ,

evidence of attempting to solve the equation in a single variable (M1)

eg sketch, setting equation equal to zero,

53

A1 N4

[6 marks]

Examiners report

Many candidates were able to successfully obtain two equations in two variables, but far fewer were

able to correctly solve for the value of . Some candidates had misread errors for either or ,

with some candidates taking the French and Spanish exams mistaking the decimal comma for a

thousands comma. Many candidates who attempted to solve algebraically did not cancel the from

both sides and ended up with a 4 degree equation that they could not solve. Some of these candidates

obtained the extraneous answer of as well. Some candidates used a minimum of algebra to

eliminate the first term and then quickly solved the resulting equation on their GDC.

35a. [2 marks]

Markscheme

valid method (M1)

e.g. subtracting terms, using sequence formula

A1 N2

[2 marks]

Examiners report

Most candidates performed well on this question. A few were confused between the term number and

the value of a term.

35b. [2 marks]

Markscheme


e.g.

28 term is 50.9 (exact) A1 N2

[2 marks]

Examiners report54



35c. [2 marks]

Markscheme

correct substitution into sum formula (A1)

e.g. ,

(exact) [ , ] A1 N2

[2 marks]

Examiners report



36a. [3 marks]

Markscheme

(i) A1 N1

(ii) evidence of valid approach (M1)

e.g. , repeated addition of d from 36

A1 N2

[3 marks]

Examiners report

The majority of candidates were successful with this question. Most had little difficulty with part (a).

36b. [3 marks]

Markscheme

(i) correct substitution into sum formula A1

e.g. ,

55

evidence of simplifying

e.g. A1

AG N0

(ii) A1 N1

[3 marks]

Examiners report

Some candidates were unable to show the required result in part (b), often substituting values for n

rather than working with the formula for the sum of an arithmetic series.

Printed for International School of Monza

© International Baccalaureate Organization 2019

International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

56

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