ib questionbank test · web viewhl week 1 revision – polynomials mark scheme 1. [5 marks]...
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![Page 1: IB Questionbank Test · Web viewHL Week 1 Revision – Polynomials Mark Scheme 1. [5 marks] Markscheme attempt to substitute x = −1 or x = 2 or to divide polynomials (M1) 1 −](https://reader035.vdocuments.us/reader035/viewer/2022062318/611743c352dfed5a3b794e92/html5/thumbnails/1.jpg)
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
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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
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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
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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
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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)
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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
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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
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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]
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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
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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)
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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
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(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
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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]
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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.
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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
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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
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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
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(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]
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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]
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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]
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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]
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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.
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[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]
Printed for International School of Monza
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International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®
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