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IB UNIT TEST – Quadratic Functions and Descriptive Statistics

Name_______________________________________

Unless otherwise stated, all answers are to be exact or to 3 significant figures.

Section A – Answer all questions in the spaces provided

1. The graph of the function y = x2 – x – 2 is drawn below.

x

y

A B

C

0

(a) _______________________

(b) _______________________

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2. In the following ordered data, the mean is 6 and the median is 5.

2, b, 3, a, 6, 9, 10, 12

Find each of the following

(a) the value of a; (b) the value of b. (Total 4 marks)

………………………………………………………………………………………………………………….

………………………………………………………………………………………………………………….

…………………………………………………………………………………………………………………

………………………………………………………………………………………………………………….

………………………………………………………………………………………………………………….

(c) _______________________

(d) _______________________

3. The diagram below shows the graph of y = c + kx – x2, where k and c are constants.

Q

y

xO P(5, 0)

(a) Find the values of k and c.

(b) Find the coordinates of Q, the highest point on the graph.

(Total 8 marks)

………………………………………………………………………………………………………………….

………………………………………………………………………………………………………………….

…………………………………………………………………………………………………………………

………………………………………………………………………………………………………………….

………………………………………………………………………………………………………………….

(a) _______________________

(b) _______________________

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4. A survey was conducted of the number of bedrooms in 208 randomly chosen houses. The results are

shown in the following table.

Number of bedrooms 1 2 3 4 5 6

Number of houses 41 60 52 32 15 8

(a) State whether the data is discrete or continuous. (1)

(b) Write down the mean number of bedrooms per house. (2)

(c) Write down the standard deviation of the number of bedrooms per house. (1)

(d) Find how many houses have a number of bedrooms greater than one standard deviation above the

mean. (2)

………………………………………………………………………………………………………………….

………………………………………………………………………………………………………………….

…………………………………………………………………………………………………………………

………………………………………………………………………………………………………………….

………………………………………………………………………………………………………………….

(Total 6 Marks)

(a) _______________________

(b) _______________________

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(3)

(Total 6 marks)

………………………………………………………………………………………………………………….

…………………………………………………………………………………………………………………

………………………………………………………………………………………………………………….

………………………………………………………………………………………………………………….

(a) _______________________

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(b) _______________________

Section B – Answer on the paper provided.

6. The table below shows the number and weight (w) of fish delivered to a local fish market one morning.

weight (kg) frequency cumulative frequency

0.50 ≤ w < 0.70 16 16

0.70 ≤ w < 0.90 37 53

0.90 ≤ w < 1.10 44 c

1.10 ≤ w < 1.30 23 120

1.30 ≤ w < 1.50 10 130

(a) (i) Write down the value of c. (1)

(ii) On graph paper, draw the cumulative frequency curve for this data. Use a scale of 1 cm to

represent 0.1 kg on the horizontal axis and 1 cm to represent 10 units on the vertical axis.

Label the axes clearly. (4)

(iii) Use the graph to show that the median weight of the fish is 0.95 kg. (1)

(b) (i) The zoo buys the heaviest 10 % of fish. How many fish does the zoo buy? (2)

(ii) A pet food company buys all the fish in the lowest quartile. What is the maximum weight of

a fish bought by the company? (3)

(Total 11 marks)

6

7

8

9

10

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14. (a) Put x = 0 to find y = –2 (M1)

Coordinates are (0, –2) (A1) (C2)

Note: Award (M1)(A0) for –2 if working is shown. If not, award

(M0)(A0).

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(b) Factorise fully, y = (x – 2) (x + 1). (A1)(A1)

y = 0 when x = –1, 2. (A1)(A1)

Coordinates are A(–1, 0), B(2, 0). (A1)(A1) (C6)

Note: Award (C2) for each correct x value if no method shown and full

coordinates not given. If the quadratic formula is used correctly award

(M1)(A1)(A1)(A1)(A1)(A1). If the formula is incorrect award only the

last (A1)(A1) as ft. [8]

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15. (a)

(0, 5)

(A3) (C3)

Notes: Award (A1) for point (0,5) indicated.

Award (A2) for correct shape.

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(b) (1.5, 0.5) (A1)(A1) (C2)

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(c) x = 1.5 (A1) (C1) [6]

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16. (a) y = x(5 – x) or y = 5x – x2 or 25 = c + 5k (M1)

c = 0, k = 5 (A1)(A1) (C3)

Note: Award (A1) if no method is indicated but c = 0 or k = 5 is given

alone.

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(b) Vertex at x = 2

5

2

a

b= 2.5 (M1)(A1)

y = 5(2.5) – 2.52 = 6.25 (M1)(A1)

Note: The substitutions must be attempted to receive the method marks.

Q(2.5, 6.25) (A1) (C5)

Notes: Coordinate pair is required for (A1) but Q is not essential. If no

working shown and answer not fully correct, award (G2) for each

correct value and (A1) for coordinate brackets. However, if values are

close but not exactly correct (eg (2.49, 6.25)) award only (G1) for each

less precise value. In this case AP might also apply if number of digits is

inappropriate.

If differentiation is used, award (M1) for correct process, (A1) for x =

2.5, (M1)(A1) or (G2) for 6.25 and (A1) for coordinate brackets. [8]

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17. (a) (i) a = 60 (A1)

b = 48 (A1) 2

(ii)

Labels and scale (A1)

all points correct (A2)(ft)

Note: At most one error (A1)(A0)

Smooth curve (not straight lines) drawn in given domain only (A1)(ft)

Note: Graphs not drawn on graph paper must be drawn

very accurately to receive marks 4

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(iii) 5.2)2(2

)10(

2

a

bx (M1)(A1)

(AG)

Note: (A1) is for correct substitution

OR

104 xdx

dy (A1)

4x 10 = 0, x = 2.5 (M1)

(AG)

Note: (M1) is for setting derivative to 0.

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OR

For correct symmetry argument leading to x = 2.5 (R2)

(AG)

f(2.5) = 2(2.5)2 – 10(2.5)+60=47.5 (M1)

(AG) 3

Notes: If graph is used and lines are drawn to the vertex from x = 2.5 and

y = 47.5, award (M1)(A0)(M1).

If vertex is indicated on graph, with or without coordinates, Award (M1)

only.

If GDC is used and a sketch showing the vertex is given, award (G1)

only.

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(iv) The correct answer is 2.5 x < 8 or [2.5, 8)

Accept x 2.5 or [2.5, ∞) (A2) 2

Notes: Allow the inequalities to be strict or non-strict

Award (A1) for 2.5 and 8 both seen.

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(b) (i) See graph in part (a) (A1)(ft)

Domain

Note: Penalize domain only once in the question.

horizontal straight line, intercept at 80 (A1) 2

Note: If graphs are drawn on separate axes award at most (A0)(A1).

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(ii) From GDC, x = 6.5311…

Point of intersection is (6.53, 80) (A1)(A1) 2

Notes: Award (A1) for each coordinate. Award (A1)(A0) if brackets are

missing. Allow x = 6.53, y = 80

(–1.53,80) receives (A0)(A1).

If both points are given, award appropriate marks for (6.53, 80).

If the x- coordinate is read from the graph, the method must be shown by

a line drawn on the graph or the point of intersection clearly marked.

The answer can be given to 1dp. Award (A1)(ft) for candidates intercept

value 0.1.

If no method shown an answer of (6.5, 80) or 6.5, 80 receives (A0)(A1).

This is not an AP.

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(iii) 80 – 47.5

For subtracting appropriate values or for showing the distance on

the graph (M1)

For 32.5.

Note: Award (A0) for any other answer (A1)(G2) 2

[17]

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18. (a) 220 = 2(W + x) (M1)

Therefore W = 2

2220 x or 110 – x (A1)

(b) Area = x(110 – x) (allow follow through from part (a)) (A1)

(c) Area = 70(110 – 70) = 2800 m2 (allow follow through from part (b)) (A1)

[4]

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19. (a) 5.5 = 10

385937834 a (M1)

55 = 50 + a

5 = a (A1) (C2)

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(b) 3, 3, 3, 4, 5, 5, 7, 8, 8, 9 (M1)

Median = 5 (A1) (C2)

Note: Award (M1) for arranging scores in ascending or descending

order. Follow through with candidate’s a [4]

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20. (a) Mean = 10

60

= 6 (A1) (C1)

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(b) Mode = 2 (A1) (C1)

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(c) 2, 2, 2, 4, 5, 6, 8, 9, 10, 12

Median =

2

65

(M1)

= 5.5 (A1) (C2) [4]

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21. (a) 2

6a = 5 (M1)(A1)

a + 6 = 10 (A1)

a = 4 (A1) (C4)

(b) 8

42 ba = 6 (M1)

42 + a + b = 48 (A1)

a + b = 6

4 + b = 6 (A1)

b = 2 (A1) (C4) [8]

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22. (a) 63 kg (A1) 1

(b) (i) 70.5 kg (G1)

(ii) 14.6 kg (also accept 15.2 kg) (G1) 2

(c) Total weight of 12 students = 846 kg

Total weight of 11 students = 11 × 70 = 770 kg (M1)

Weight of student who left = 846 – 770 = 76 kg (A1) 2 [5]

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23. (a) Modal group =170 h < 180 (A1) 1

(b) Mean = 171 (G2)

Standard deviation = 11.1 (G2) 4

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(c) Median = 171 (1) (A1) 1

(d) Lower quartile = 164.5 (1) (A1)

Inter-quartile range = 177.3 – 164.5 = 12.8 (A1) 2

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(e) number = 52 (2) (A1)

percentage = 200

52 × 100 = 26% (A1) 2

[10]

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24. (a) (i) 25 minutes (2 minutes) (A2)

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(ii) Lower quartile = 18 (±1 minute) (A1)

Upper quartile = 32 (1 minute) (A1)

Interquartile range = 32 – 18 = 14 minutes (±2 minutes) (A1)

OR

Accept [18 to 32] as interval for the interquartile range. (A3) 5

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(b) p = 20 (A1)

q = 30 (A1) 2

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(c)

Midpoint Frequency M × f

2.5 20 50

10 20 200

17.5 20 350

22.5 40 900

30 60 1800

42.5 30 1275

55 10 550

(A1) Total = 200 Total = 5125

(A1)

Mean = 200

5125 = 25.625 (exact) or 25.6 (3 s.f.) (M1)(A1)

Note: Not every step needs to be seen to get the marks.

OR

Mean = 25.625 or 25.6 (using GDC) (G4) 4 [11]

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25. Unit penalty (UP) is applicable where indicated.

(a) 26cm (A1) (C1)

(UP)

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(b) 33 −19 for identifying correct quartiles. (A1)

=14cm. (A1)(ft)

(UP)

(ft) on their quartiles. (C2)

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(c)

50454035302520151050

length (cm)

correct median (A1)(ft)

correct quartiles and box (A1)(ft)

endpoints at 6 and 47, joined to box by straight lines. (A1) (C3) [6]