HL Week 1 Revision – Polynomials Mark Scheme

1. [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]

2. [6 marks]

Markscheme

METHOD 1

     (M1)(A1)

Note: Award M1A0 if only one equation obtained.

     A1

     (M1)

       A1A1

 

METHOD 2

     (M1)(A1)

     A1

     (M1)

1

        A1A1

Note: Exact answers only.

[6 marks]

 

3. [5 marks]

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]2

4a. [3 marks]

Markscheme

    (M1)

    A1

    A1

[3 marks]

4b. [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]

5. [5 marks]

Markscheme

METHOD 1

3

let roots be and     (M1)

sum of roots     M1

    A1

EITHER

product of roots     M1

OR

    M1

THEN

    A1

 

METHOD 2

    (M1)

    M1A1

    (M1)

    A1

[5 marks]

6a. [2 marks]

Markscheme

4

     A1

    A1

 

[2 marks]

6b. [2 marks]

Markscheme

     M1A1

is a factor of      AG

 

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

 

[2 marks]

6c. [5 marks]

Markscheme

METHOD 1

     (M1)

by inspection      A1

     (M1)(A1)

     A1

METHOD 2

, are two roots of the quadratic

     (A1)

from part (a)      (M1)

5

     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]

6d. [3 marks]

Markscheme

     M1

     M1

     A1

(or )

 

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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]

6e. [3 marks]

Markscheme

     M1A1

for concave up     R1AG

 

[3 marks]

6f. [3 marks]

Markscheme

-intercept at      A1

-intercept at      A1

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stationary point of inflexion at with correct curvature either side     A1

[3 marks]

7a. [4 marks]

Markscheme

     M1A1

     A1

     A1

[4 marks]

7b. [3 marks]

Markscheme

attempt to equate coefficients     (M1)

     (A1)

     A1

 

Note:     Accept any equivalent valid method.

 

[3 marks]

7c. [3 marks]

Markscheme

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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]

7d. [3 marks]

Markscheme

     A1

     A1A1

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

 

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]

8. [6 marks]

Markscheme

   A1

   A1

   (M1)

   A1

attempt to solve quadratic     (M1)

   A1

[6 marks]

9. [5 marks]

Markscheme

    M1A1

EITHER

10

as is real     M1A1

OR

    M1

    (A1)

THEN

hence smallest possible value for is     A1

[5 marks]

10a. [4 marks]

Markscheme

the sum of the roots of the polynomial     (A1)

    M1A1

 

Note:     The formula for the sum of a geometric sequence must be equated to a value for the M1 to be

awarded.

 

    A1

[4 marks]

10b. [2 marks]

Markscheme

    M1

11

    A1

[2 marks]

Total [6 marks]

11a. [3 marks]

Markscheme

(i)         A1

(ii)         A1A1

 

Note:     First A1 is for the negative sign.

[3 marks]

11b. [2 marks]

Markscheme

METHOD 1

if satisfies then

so the roots of are each  less than the roots of     (R1)

so sum of roots is     A1

METHOD 2

    (M1)

so sum of roots is     A1

[2 marks]

12

Tofal [5 marks]

12a. [3 marks]

Markscheme

(i)-(iii) given the three roots , we have

    M1

    A1

    A1

comparing coefficients:

    AG

    AG

    AG

[3 marks]

12b. [5 marks]

Markscheme

METHOD 1

(i)     Given

And

Let the three roots be

So     M1

or

Attempt to solve simultaneous equations:     M1

    A1

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    AG

(ii)    

    (A1)

Therefore     A1

METHOD 2

(i)     let the three roots be     M1

adding roots     M1

to give     A1

    AG

(ii)     is a root, so     M1

    A1

METHOD 3

(i)     let the three roots be     M1

adding roots     M1

to give     A1

    AG

(ii)         M1

    A1

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[5 marks]

12c. [6 marks]

Markscheme

METHOD 1

Given

And

Let the three roots be .

So     M1

or

Attempt to solve simultaneous equations:     M1

    A1

    (A1)(A1)

Therefore     A1

METHOD 2

let the three roots be     M1

attempt at substitution of and and into equations from (a)     M1

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    A1

    A1

therefore     A1

therefore     A1

[6 marks]

Total [14 marks]

13a. [2 marks]

Markscheme

using the formulae for the sum and product of roots:

(i)         A1

(ii)         A1

 

Note:     Award A0A0 if the above results are obtained by solving the original equation (except for the

purpose of checking).

[2 marks]

13b. [4 marks]

Markscheme

METHOD 1

required quadratic is of the form     (M1)

    A1

16

    M1

    A1

 

Note:     Accept the use of exact roots

 

METHOD 2

 

replacing with     M1

    (A1)

and     A1A1

 

Note:     Award A1A0 for ie, if and are not explicitly stated.

[4 marks]

Total [6 marks]

14. [6 marks]

Markscheme

using to obtain     M1A1

    (M1)(A1)

 

17

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]

15. [5 marks]

Markscheme

    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]

16. [6 marks]

Markscheme18

    (M1)(A1)

    (M1)(A1)

    M1

    A1

 

Note:     Award M1A0M1A0M1A1 if answer of 50 is found using and .

 

[6 marks]

17. [6 marks]

Markscheme

(a) using the formulae for the sum and product of roots:

    A1

    A1

    M1

    A1

 

Note:     Award M0 for attempt to solve quadratic equation.

 

[4 marks]

 

(b)         M1

19

    A1

 

Note:     Final answer must be an equation. Accept alternative correct forms.

 

[2 marks]

 

Total [6 marks]

18. [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

[4 marks]

19. [5 marks]

20

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]

20a. [2 marks]

Markscheme

    A1A1

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

 

[2 marks]

20b. [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.

21

 

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]

20c. [2 marks]

Markscheme

22

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

intercept or maximum shown     A1

[2 marks]

20d. [2 marks]

Markscheme

    A1A1

Note: A1 for , A1 for .

 

[2 marks]

20e. [3 marks]

Markscheme

let     A1

    A1

    A1

    AG

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

 

[3 marks]

20f. [7 marks]

Markscheme

    A1

23

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]

21a. [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.

 

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(iii)         M1

    A1

    A1AG

[9 marks]

21b. [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

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other quadratic factors are     A1

and     A1

[9 marks]

22. [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]

23a. [4 marks]

Markscheme

    M1A1

Note: Award M1A1 if expression seen within quadratic formula.

 

EITHER

    M1

always positive, therefore the equation always has two distinct real roots     R126

(and cannot be always negative as )

OR

sketch of or not crossing the x-axis     M1

always positive, therefore the equation always has two distinct real roots     R1

OR

write as     M1

always positive, therefore the equation always has two distinct real roots     R1

[4 marks]

23b. [3 marks]

Markscheme

closest together when is least     (M1)

minimum value occurs when k = 1.5     (M1)A1

[3 marks]

24. [4 marks]

Markscheme

    M1A1

    (A1)

which is positive for all k     R1

Note: Accept analytical, graphical or other correct methods. In all cases only award R1 if a reason is

given in words or graphically. Award M1A1A0R1 if mistakes are made in the simplification but the

argument given is correct.

 

[4 marks]

25. [6 marks]

27

Markscheme

 

let

let

    A1

    A1

    M1

    (M1)

    A1A1

[6 marks]

 

26a. [2 marks]

Markscheme

(i)    

    A1

(ii)     equating real and imaginary parts     M1

    AG

    AG

[2 marks]

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26b. [5 marks]

Markscheme

substituting     M1

EITHER

    A1

    A1

and     (A1)

OR

    A1

    A1

and     (A1) 

Note: Accept solution by inspection if completely correct.

 

THEN

the square roots are and     A1

[5 marks]

26c. [3 marks]

29

Markscheme

EITHER

consider

    A1

    A1

    A1

    AG

OR

    A1

    A1

    A1

    AG

[3 marks]

26d. [2 marks]

Markscheme

and     A1A1

[2 marks]

26e. [2 marks]

30

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]

26f. [5 marks]

Markscheme

    A1A1

    A1

Since the curve passes through ,

    M1

    A1

Hence

[5 marks]

26g. [2 marks]

Markscheme

    (M1)

    A1

[2 marks]

31

26h. [2 marks]

Markscheme

 

A1A1

 

Note: Accept points or vectors on complex plane.

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

 

32

[2 marks]

26i. [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]

27. [5 marks]

Markscheme

    M1A1

For use of discriminant or completing the square      (M1)

      (A1)

 Note: Accept trial and error, sketches of parabolas with vertex (2,0) or use of second derivative.

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    A1

[5 marks] 

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