misswelton.weebly.com€¦ · Web viewA large number of candidates, either by graphical (mostly) or algebraic or via use of a GDC solver, were able to readily obtain . Most candidates

2.5 HL

1a. [2 marks]

Markscheme

(M1)

Note: Award M1 for a valid attempt at remainder theorem or polynomial division.

= −12 A1

remainder = −12

[2 marks]

Examiners report

[N/A]

1b. [1 mark]

Markscheme

= 0 A1

remainder = 0

[1 mark]

Examiners report

[N/A]

1c. [4 marks]

Markscheme

(is a zero) A1

Note: Can be seen anywhere.

EITHER

factorise to get (M1)A1

1

(for ) (or equivalent statement) R1

Note: Award R1 if correct two complex roots are given.

OR

A1

attempting to show M1

eg discriminant = 36 – 96 < 0, completing the square

no turning points R1

THEN

only one real zero (as the curve is continuous) AG

[4 marks]

Examiners report

[N/A]

1d. [2 marks]

Markscheme

new graph is (M1)

stretch parallel to the -axis (with invariant), scale factor 0.5 A1

Note: Accept “horizontal” instead of “parallel to the -axis”.

[2 marks]

Examiners report

[N/A]

1e. [6 marks]

Markscheme

M1A1

2

Note: Allow factorials in the denominator for A1.

A1

Note: Accept any correct cubic equation without factorials and .

EITHER

(M1)

(A1)

OR

(M1)(A1)

THEN

= 1.5 A1

[6 marks]

Examiners report

[N/A]

2a. [7 marks]

Markscheme

METHOD 1

attempt to find roots or factors (M1)

roots are A1A1

Note: Award A1 for each pair of roots or factors, real and complex.

attempt to form quadratic M1

A1

(A1)

3

A1

METHOD 2

attempt to find roots or factors (M1)

real roots are (or real factors , ) A1

attempt to form quadratic M1

A1

equate coefficients of M1

A1

solve to give A1

[7 marks]

Examiners report

[N/A]

2b. [6 marks]

Markscheme

4

shape A1

-axis intercepts at (−3, 0), (1, 0) and -axis intercept at (0, −51) A1A1

minimum points at (−1.62, −118) and (3.72, 19.7) A1A1

maximum point at (2.40, 26.9) A1

Note: Coordinates may be seen on the graph or elsewhere.

Note: Accept −3, 1 and −51 marked on the axes.

[6 marks]

Examiners report

[N/A]

2c. [2 marks]

Markscheme

from graph, 19.7 ≤ ≤ 26.9 A1A1

Note: Award A1 for correct endpoints and A1 for correct inequalities.5

[2 marks]

Examiners report

[N/A]

3a. [2 marks]

Markscheme

attempt to substitute and set equal to zero, or use of long / synthetic division (M1)

A1

[2 marks]

Examiners report

[N/A]

3b. [1 mark]

Markscheme

0 A1

[1 mark]

Examiners report

[N/A]

3c. [3 marks]

Markscheme

EITHER

attempt to solve (M1)

6

OR

(M1)

comparing coefficients gives = 5, = 2

THEN

= 105 A1

= 50 A1

[3 marks]

Examiners report

[N/A]

4. [5 marks]

Markscheme

attempt to substitute x = −1 or x = 2 or to divide polynomials (M1)

1 − p − q + 5 = 7, 16 + 8p + 2q + 5 = 1 or equivalent A1A1

attempt to solve their two equations M1

p = −3, q = 2 A1

[5 marks]

Examiners report

[N/A]

5. [5 marks]

7

Markscheme

METHOD 1

substitute each of = 1,2 and 3 into the quartic and equate to zero (M1)

or equivalent (A2)

Note: Award A2 for all three equations correct, A1 for two correct.

attempting to solve the system of equations (M1)

= −7, = 17, = −17 A1

Note: Only award M1 when some numerical values are found when solving algebraically or using GDC.

METHOD 2

attempt to find fourth factor (M1)

A1

attempt to expand M1

( = −7, = 17, = −17) A2

Note: Award A2 for all three values correct, A1 for two correct.

Note: Accept long / synthetic division.

[5 marks]

Examiners report

[N/A]

6a. [3 marks]

Markscheme

8

(M1)

A1

A1

[3 marks]

Examiners report

[N/A]

6b. [3 marks]

Markscheme

equate coefficients of : (M1)

(A1)

A1

Note: Allow part (b) marks if any of this work is seen in part (a).

Note: Allow equivalent methods (eg, synthetic division) for the M marks in each part.

[3 marks]

Examiners report

[N/A]

7a. [2 marks]

9

Markscheme

A1

A1

[2 marks]

Examiners report

[N/A]

7b. [2 marks]

Markscheme

M1A1

is a factor of AG

Note: Accept use of division to show remainder is zero.

[2 marks]

Examiners report

[N/A]

7c. [5 marks]

Markscheme

METHOD 1

(M1)

by inspection A1

(M1)(A1)

10

A1

METHOD 2

, are two roots of the quadratic

(A1)

from part (a) (M1)

A1

(M1)

A1

Note: Award FT if following through from their sum .

METHOD 3

(M1)A1

Note: This may have been seen in part (b).

(M1)

A1A1

[5 marks]

Examiners report

[N/A]

7d. [3 marks]

Markscheme

11

M1

M1

A1

(or )

Notes: Award the second M1 for an attempt to use the quadratic formula or to complete the square.

Do not award FT from (c).

[3 marks]

Examiners report

[N/A]

7e. [3 marks]

Markscheme

M1A1

for concave up R1AG

[3 marks]

Examiners report

[N/A]

7f. [3 marks]

Markscheme

12

-intercept at A1

-intercept at A1

stationary point of inflexion at with correct curvature either side A1

[3 marks]

Examiners report

[N/A]

8a. [4 marks]

Markscheme

M1A1

A1

A1

[4 marks]

Examiners report

[N/A]

8b. [3 marks]

13

Markscheme

attempt to equate coefficients (M1)

(A1)

A1

Note: Accept any equivalent valid method.

[3 marks]

Examiners report

[N/A]

8c. [3 marks]

Markscheme

14

A1 for correct shape (ie with correct number of max/min points)

A1 for correct and intercepts

A1 for correct maximum and minimum points

[3 marks]

Examiners report

[N/A]

8d. [3 marks]

Markscheme

A1

A1A1

Note: Award A1 for correct end points and A1 for correct inequalities.

15

Note: If the candidate has misdrawn the graph and omitted the first minimum point, the maximum

mark that may be awarded is A1FTA0A0 for seen.

[3 marks]

Examiners report

[N/A]

9. [5 marks]

Markscheme

M1A1

EITHER

as is real M1A1

OR

M1

(A1)

THEN

hence smallest possible value for is A1

[5 marks]

Examiners report

For quite a difficult question, there were many good solutions for this, including many different

methods. It was disturbing to see how many students did not seem to be aware of the remainder

theorem, instead choosing to divide the polynomial.

16

10. [6 marks]

Markscheme

using to obtain M1A1

(M1)(A1)

Note: Award M1 for a cubic graph with correct shape and A1 for clearly showing that the above cubic

crosses the horizontal axis at only.

A1

EITHER

showing that has no real (two complex) solutions for R1

OR

showing that has one real (and two complex) solutions for R1

Note: Award R1 for solutions that make specific reference to an appropriate graph.

[6 marks]

Examiners report

A large number of candidates, either by graphical (mostly) or algebraic or via use of a GDC solver, were

able to readily obtain . Most candidates who were awarded full marks however, made specific

reference to an appropriate graph. Only a small percentage of candidates used the discriminant to

justify that only one value of satisfied the required condition. A number of candidates erroneously

obtained or equivalent rather than .

11. [5 marks]

Markscheme17

M1A1A1

Note: Award M1 for substitution of 2 or −1 and equating to remainder, A1 for each correct equation.

attempt to solve simultaneously M1

A1

[5 marks]

Examiners report

[N/A]

12. [4 marks]

Markscheme

METHOD 1

substituting

(A1)

equating real or imaginary parts (M1)

A1

A1

METHOD 2

other root is (A1)

considering either the sum or product of roots or multiplying factors (M1)

(sum of roots) so A1

(product of roots) A1

18

[4 marks]

Examiners report

[N/A]

13. [5 marks]

Markscheme

M1

M1

Note: In each case award the M marks if correct substitution attempted and right-hand side correct.

attempt to solve simultaneously M1

A1

A1

[5 marks]

Examiners report

Many candidates scored full marks on what was thought to be an easy first question. However, a

number of candidates wrote down two correct equations but proceeded to make algebraic errors and

thus found incorrect values for p and q. A small number also attempted to answer this question using

long division, but fully correct answers using this technique were rarely seen.

14a. [2 marks]

Markscheme

A1A1

Note: A1 for two correct parameters, A2 for all three correct.

19

[2 marks]

Examiners report

This question covered many syllabus areas, completing the square, transformations of graphs, range,

integration by substitution and compound angle formulae. There were many good solutions to parts (a)

– (e).

14b. [3 marks]

Markscheme

translation (allow “0.5 to the right”) A1

stretch parallel to y-axis, scale factor 4 (allow vertical stretch or similar) A1

translation (allow “4 up”) A1

Note: All transformations must state magnitude and direction.

Note: First two transformations can be in either order.

It could be a stretch followed by a single translation of . If the vertical translation is before the

stretch it is .

[3 marks]

Examiners report



– (e) but the following points caused some difficulties.

(b) Exam technique would have helped those candidates who could not get part (a) correct as any

solution of the form given in the question could have led to full marks in part (b). Several candidates

20

obtained expressions which were not of this form in (a) and so were unable to receive any marks in (b)

Many missed the fact that if a vertical translation is performed before the vertical stretch it has a

different magnitude to if it is done afterwards. Though on this occasion the markscheme was fairly

flexible in the words it allowed to be used by candidates to describe the transformations it would be

less risky to use the correct expressions.

14c. [2 marks]

Markscheme

general shape (including asymptote and single maximum in first quadrant), A1

intercept or maximum shown A1

[2 marks]

Examiners report

21




(c) Generally the sketches were poor. The general rule for all sketch questions should be that any

asymptotes or intercepts should be clearly labelled. Sketches do not need to be done on graph paper,

but a ruler should be used, particularly when asymptotes are involved.

14d. [2 marks]

Markscheme

A1A1

Note: A1 for , A1 for .

[2 marks]

Examiners report



– (e).

14e. [3 marks]

Markscheme

let A1

A1

A1

AG

Note: If following through an incorrect answer to part (a), do not award final A1 mark.

22

[3 marks]

Examiners report




(e) and (f) were well done up to the final part of (f), in which candidates did not realise they needed to

use the compound angle formula.

14f. [7 marks]

Markscheme

A1

Note: A1 for correct change of limits. Award also if they do not change limits but go back to x values

when substituting the limit (even if there is an error in the integral).

(M1)

A1

let the integral = I

M1

(M1)A1

A1AG

[7 marks]

Examiners report

23




(e) and (f) were well done up to the final part of (f), in which candidates did not realise they needed to

use the compound angle formula.

15a. [9 marks]

Markscheme

(i) A1A1A1

Note: Accept modulus and argument given separately, or the use of exponential (Euler) form.

Note: Accept arguments given in rational degrees, except where exponential form is used.

(ii) the points lie on a circle of radius 2 centre the origin A1

differences are all A1

points equally spaced triangle is equilateral R1AG

Note: Accept an approach based on a clearly marked diagram.

(iii) M1

A1

A1AG

[9 marks]

Examiners report

24

(i) A disappointingly large number of candidates were unable to give the correct arguments for the

three complex numbers. Such errors undermined their efforts to tackle parts (ii) and (iii).

15b. [9 marks]

Markscheme

(i) attempt to obtain seven solutions in modulus argument form M1

A1

(ii) w has argument and 1 + w has argument ,

then M1

A1

A1

Note: Accept alternative approaches.

(iii) since roots occur in conjugate pairs, (R1)

has a quadratic factor A1

AG

other quadratic factors are A1

and A1

[9 marks]

25

Examiners report

Many candidates were successful in part (i), but failed to capitalise on that – in particular, few used the

fact that roots of come in complex conjugate pairs.

16. [7 marks]

Markscheme

other root is 2 – i (A1)

a quadratic factor is therefore (M1)

A1

x + 1 is a factor A1

is a factor A1

(M1)

A1

[7 marks]

Examiners report

Whilst most candidates knew that another root was , much fewer were able to find the quadratic

factor. Surprisingly few candidates knew that must be a repeated factor and less surprisingly

many did not recognise that the whole expression needed to be multiplied by .

17. [6 marks]

Markscheme

let

let

26

A1

A1

M1

(M1)

A1A1

[6 marks]

Examiners report

Candidates who used the remainder theorem usually went on to find the two possible values of k. Some

candidates, however, attempted to find the remainders using long division. While this is a valid method,

the algebra involved proved to be too difficult for most of these candidates.

18a. [2 marks]

Markscheme

(i)

A1

(ii) equating real and imaginary parts M1

AG

AG

[2 marks]

Examiners report

27

Since (a) was a ‘show that’ question, it was essential for candidates to give a convincing explanation of

how the quoted results were obtained. Many candidates just wrote

This was not given full credit since it simply repeated what was given in the question. Candidates were

expected to make it clear that they were equating real and imaginary parts. In (b), candidates who

attempted to use de Moivre’s Theorem to find the square roots were given no credit since the question

stated ‘hence’.

18b. [5 marks]

Markscheme

substituting M1

EITHER

A1

A1

and (A1)

OR

A1

A1

28

and (A1)

Note: Accept solution by inspection if completely correct.

THEN

the square roots are and A1

[5 marks]

Examiners report






stated ‘hence’.

18c. [3 marks]

Markscheme

EITHER

consider

A1

29

A1

A1

AG

OR

A1

A1

A1

AG

[3 marks]

Examiners report






stated ‘hence’.

18d. [2 marks]

30

Markscheme

and A1A1

[2 marks]

Examiners report






stated ‘hence’.

18e. [2 marks]

Markscheme

the graph crosses the x-axis twice, indicating two real roots R1

since the quartic equation has four roots and only two are real, the other two roots must be complex

R1

[2 marks]

Examiners report

In (a), the explanations were often unconvincing. Candidates were expected to make it clear that the

two intersections with the x-axis gave two real roots and, since the polynomial was a quartic and

therefore had four zeros, the other two roots must be complex. Candidates who made vague statements

such as ‘the graph shows two real roots’ were not given full credit. In (b), most candidates stated the

31

values of a and b correctly but algebraic errors often led to incorrect values for the other parameters.

Candidates who failed to solve (b) correctly were unable to solve (c), (d) and (e) correctly although

follow through was used where possible.

18f. [5 marks]

Markscheme

A1A1

A1

Since the curve passes through ,

M1

A1

Hence

[5 marks]

Examiners report








18g. [2 marks]

Markscheme

(M1)

32

A1

[2 marks]

Examiners report








18h. [2 marks]

Markscheme

33

A1A1

Note: Accept points or vectors on complex plane.

Award A1 for two real roots and A1 for two complex roots.

[2 marks]

Examiners report34








18i. [6 marks]

Markscheme

real roots are and A1A1

considering

A1

finding using M1

A1

A1

Note: Accept arguments in the range .

Accept answers in degrees.

[6 marks]

Examiners report





35




19a. [2 marks]

Markscheme

using the factor theorem z +1 is a factor (M1)

A1

[2 marks]

Examiners report

In part a) the factorisation was, on the whole, well done.

19b. [9 marks]

Markscheme

(i) METHOD 1

(M1)

solving M1

A1

therefore one cube root of −1 is AG

METHOD 2

M1A1

A1

= −1 AG

METHOD 3

36

M1A1

A1

(ii) METHOD 1

as is a root of then M1R1

AG

Note: Award M1 for the use of in any way.

Award R1 for a correct reasoned approach.

METHOD 2

M1

A1

(iii) METHOD 1

(M1)

A1

(M1)

A1

METHOD 2

M1A1

37

Note: Award M1 for attempt at binomial expansion.

use of any previous result e.g. M1

= 1 A1

Note: As the question uses the word ‘hence’, other methods that do not use previous results are

awarded no marks.

[9 marks]

Examiners report

Part (b) was done well by most although using a substitution method rather than the result above. This

used much m retime than was necessary but was successful. A number of candidates did not use the

previous results in part (iii) and so seemed to not understand the use of the ‘hence’.

20a. [3 marks]

Markscheme

M1

or sketch or GDC (M1)

A1

[3 marks]

Examiners report

Candidates found this question surprisingly challenging. The most straightforward approach was use of

the Remainder Theorem but a significant number of candidates seemed unaware of this technique. This

lack of knowledge led many candidates to attempt an algebraically laborious use of long division. In (b)

a number of candidates did not seem to appreciate the significance of the word unique and hence found

it difficult to provide sufficient detail to make a meaningful argument. However, most candidates did

recognize that they needed a technological approach when attempting (b).

38

20b. [2 marks]

Markscheme

substituting into

A1

EITHER

graph showing unique solution which is indicated (must include max and min) R1

OR

convincing argument that only one of the solutions is real R1

(−1.74, 0.868 ±1.12i)

[5 marks]

Examiners report

Candidates found this question surprisingly challenging. The most straightforward approach was use of

the Remainder Theorem but a significant number of candidates seemed unaware of this technique. This

lack of knowledge led many candidates to attempt an algebraically laborious use of long division. In (b)

a number of candidates did not seem to appreciate the significance of the word unique and hence found

it difficult to provide sufficient detail to make a meaningful argument. However, most candidates did

recognize that they needed a technological approach when attempting (b).

21. [5 marks]

Markscheme

using the factor theorem or long division (M1)

(A1)

(A1)

(A1)

(A1) (N3)

Note: Award M1A0A0A1A1 for using as the third factor, without justification that the leading

coefficient is 1.

39

[5 marks]

Examiners report

Most candidates attempted this question and it was the best done question on the paper with many

fully correct answers. It was good to see a range of approaches used (mainly factor theorem or long

division). A number of candidates assumed was the missing factor without justification.

22. [5 marks]

Markscheme

(a) (A1)

(A1)

M1

A1 N4

(b) b is any real number A1

[5 marks]

Examiners report

Many candidates answered part (a) successfully. For part (b), some candidates did not consider that the

entire set of real numbers was asked for.

23. [5 marks]

Markscheme

M1A1

A1

M1

A1

40

Notes: The first M1 is for one substitution and the consequent equations.

Accept expressions for and that are not simplified.

[5 marks]

Examiners report

Most candidates were able to access this question although the number who used either synthetic

division or long division was surprising as this often lead to difficulty and errors. The most common

error was in applying the factor of 7 to the wrong side of the equation. It was also disappointing the

number of students who made simple algebraic errors late in the question.

24. [6 marks]

Markscheme

M1

A1

M1

A1

A1A1 N0

[6 marks]

Examiners report

Most candidates made a meaningful attempt at this question. Weaker candidates often made arithmetic

errors and a few candidates tried using long division, which also often resulted in arithmetic errors.

Overall there were many fully correct solutions.

25. [6 marks]

Markscheme

METHOD 1

As (x +1) is a factor of P(x), then P(−1) = 0 (M1)

41

(or equivalent) A1

As (x − 2) is a factor of P(x), then P(2) = 0 (M1)

(or equivalent) A1

Attempting to solve for a and b M1

a = −2 and b = −1 A1 N1

[6 marks]

METHOD 2

By inspection third factor must be x −1. (M1)A1

(M1)A1

Equating coefficients a = −2, b = −1 (M1)A1 N1

[6 marks]

METHOD 3

Considering or equivalent (M1)

A1A1

Recognising that (M1)

Attempting to solve for a and b M1

a = −2 and b = −1 A1 N1

[6 marks]

Examiners report

Most candidates successfully answered this question. The majority used the factor theorem, but a few

employed polynomial division or a method based on inspection to determine the third linear factor.

26. [6 marks]

Markscheme

42

M1

A1

M1

A1

solving, A1A1

[6 marks]

Examiners report

[N/A]

Printed for International School of Monza

© International Baccalaureate Organization 2019

International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

43

Documents

misswelton.weebly.com€¦ · Web viewA large number of candidates, either by graphical (mostly) or algebraic or via use of a GDC solver, were able to readily obtain . Most candidates