IEB mock Paper II

A complete 150-mark practice paper: 11 questions in 56 parts, 31 topics, Section A 70 marks and Section B 80. Every question was generated for this page and every answer was independently verified.

The split between the sections, and how much of the paper each topic is worth, were measured from 415 real IEB questions — not guessed. None of those questions appears here.

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The paper to print, and its memorandum as a separate file so you can sit it without the answers in front of you. In English and Afrikaans.

Questions (English) 13 pages, 0.3 MB
Memorandum (English) 12 pages, 0.3 MB
Vrae (Afrikaans) 13 pages, 0.3 MB
Memorandum (Afrikaans) 12 pages, 0.3 MB

The paper

Section A[70 marks]

QUESTION 1

The scatter plot shows the number of hours (x) a learner studied and the learner's test mark out of 50 (y): (1; 9); (2; 13); (3; 12); (4; 0); (5; 19); (6; 23); (10; 33); (11; 35); (12; 40).

(a)
Identify the outlier in the data.
(1)
(b)
The outlier is removed. Determine the equation of the least squares regression line for the remaining data, in the form y = A + Bx, with A and B correct to three decimal places.
(3)
(c)
Determine the correlation coefficient r of the remaining data, correct to two decimal places.
(1)
(d)
By how much does y change, on average, for each increase of 1 in x? Use your equation.
(2)
(e)
Did the outlier make the correlation stronger or weaker? Explain.
(2)

[9]

QUESTION 2

In the diagram:

O is the centre of the circle.
PA and PB are tangents to the circle at A and B respectively.
C is a point on the circle such that AC ∥ PB.
A $\hat{C}$ B = x

State all relevant reasons with your statements.

(a)
Express P $\hat{B}$ A in terms of x.
(2)
(b)
Express A $\hat{P}$ B in terms of x.
(3)
(c)
Is ΔABC an isosceles triangle? Explain.
(4)
(d)
Prove that PAOB is a cyclic quadrilateral.
(3)
(e)
If x = 38°, determine the size of A $\hat{O}$ B.
(2)

[14]

QUESTION 3

(a)
Simplify fully to a single trigonometric ratio: (cos(180° + θ) · cos(90° + θ))/sin(360° − θ)
(6)
(b)
Determine, without the use of a calculator, the value of cos 113° cos 37° − sin 113° sin 37°.
(3)
(c)
Determine, without the use of a calculator, the value of cos^{^2} 75° − sin^{^2} 75°.
(2)
(d)
Determine the general solution of √3 sin θ = cos θ.
(3)

[14]

QUESTION 4

In the diagram:

The straight line through A(2; −9) and B(7; 6) is drawn.
P(−4; 3) is a point that does not lie on AB.

(a)
Determine the equation of the line AB.
(2)
(b)
The line AB cuts the x-axis at C and the y-axis at D. Calculate the area of ΔOCD, where O is the origin.
(3)
(c)
F is the foot of the perpendicular from P to AB. Determine the equation of PF.
(3)
(d)
Determine the coordinates of F.
(3)
(e)
Determine the shortest distance from P to the line AB, in simplest surd form.
(2)
(f)
Calculate the area of ΔPAB.
(2)

[15]

QUESTION 5

Given: f(x) = sin x + 1 and g(x) = cos 2x for x ∈ [−90°; 270°].

(a)
Determine the general solution of f(x) = g(x).
(5)
(b)
Hence write down the values of x ∈ [−90°; 270°] for which f(x) = g(x).
(2)
(c)
Sketch the graphs of f and g on the set of axes provided. Label all intercepts with the axes, turning points and end points.
(6)
(d)
Use your graph to determine the values of x ∈ [−90°; 270°] for which f(x)·g(x) ≥ 0.
(3)
(e)
Write down the range of h(x) = 2g(x) − 1.
(2)

[18]

Section B[80 marks]

QUESTION 6

In the diagram:

Circle P has centre P(3; 4) and passes through the origin.
Circle Q has the equation (x + 5)^{^2} + (y + 2)^{^2} = 25.

(a)
Determine the equation of circle P.
(2)
(b)
Calculate the distance PQ between the centres.
(2)
(c)
Show that the circles touch externally.
(2)
(d)
Determine the coordinates of T, the point where the circles touch.
(3)
(e)
Determine the equation of the common tangent to the circles at T.
(3)

[12]

QUESTION 7

In the diagram:

ABCD is a cyclic quadrilateral in the circle with centre O.

State all relevant reasons with your statements.

(a)
Use the diagram to prove the theorem that states that the opposite angles of a cyclic quadrilateral are supplementary.
(5)
In the diagram below, ABCD is a cyclic quadrilateral. AB is produced to E and DC is produced to F so that EF ∥ BC.
(b)
Prove that A $\hat{E}$ F = A $\hat{B}$ C.
(1)
(c)
Hence prove that ADFE is a cyclic quadrilateral.
(4)
(d)
If A $\hat{D}$ C = 79°, determine the size of A $\hat{E}$ F.
(2)

[12]

QUESTION 8

In the diagram:

The circle with centre M passes through the origin O.
The circle cuts the x-axis at A(−6; 0) and the y-axis at B(0; −10).

(a)
Explain why AB is a diameter of the circle.
(2)
(b)
Determine the coordinates of the centre M and the radius of the circle.
(3)
(c)
Show that the equation of the circle can be written as x^{^2} + y^{^2} + 6x + 10y = 0.
(2)
(d)
Determine the equation of the tangent to the circle at O.
(3)
(e)
The line y = −x cuts the circle at O and at C. Determine the coordinates of C.
(3)

[13]

QUESTION 9

In the diagram:

AB is a diameter of the circle with centre O.
The tangent to the circle at C meets AB produced at D.
${\hat{C}}_{3}$ = x

State all relevant reasons with your statements.

(a)
Express C $\hat{A}$ B in terms of x.
(2)
(b)
Express ${\hat{C}}_{2}$ in terms of x.
(3)
(c)
Prove that ΔDCB ||| ΔDAC.
(3)
(d)
Hence prove that DC^{^2} = DB · DA.
(2)
(e)
If DB = 2 units and DC = 4 units, calculate the radius of the circle.
(3)

[13]

QUESTION 10

The ogive (cumulative frequency curve) below shows the time in minutes that 60 people waited at a clinic.

(a)
How many data values does the ogive represent?
(1)
(b)
Estimate the median.
(2)
(c)
Estimate the interquartile range.
(3)
(d)
How many of the data values lie between 20 and 40?
(2)
(e)
The lowest value is 1 and the highest is 47. Draw a box-and-whisker plot of the data.
(3)
(f)
What percentage of the data values are 10 or more?
(2)
(g)
Is the data skewed? Use the quartiles to explain.
(2)

[15]

QUESTION 11

In the diagram (answers correct to two decimal places):

A, B, C and D lie on the circle.
AB = 5 cm, AD = 9 cm and BC = 6 cm.
B $\hat{A}$ D = 55°.

(a)
Calculate the length of BD.
(3)
(b)
Write down the size of B $\hat{C}$ D. Give a reason.
(2)
(c)
Calculate the size of B $\hat{D}$ C.
(3)
(d)
Calculate the length of CD.
(3)
(e)
Calculate the area of ABCD.
(4)

[15]

Memorandum

Every answer below was confirmed by an independent verifier at the moment the question was made.

QUESTION 1

(a) Answer: (4; 0)

(4; 0) lies far from the trend of the other points

(b) Answer: y = 5.892 + 2.737x

A = 5.892
B = 2.737

(c) Answer: r = 0.99

(d) Answer: y increases by about 2.737

B = 2.737 is the change in y for each 1 added to x

(e) Answer: Weaker

with the outlier r = 0.91; without it r = 0.99
|r| is smaller with the outlier: it weakened the correlation

Scatter plots

QUESTION 2

(a) Answer: P $\hat{B}$ A = x

P $\hat{B}$ A = A $\hat{C}$ B = x (tan chord theorem)

(b) Answer: A $\hat{P}$ B = 180° − 2x

PA = PB (tans from common pt)
P $\hat{A}$ B = P $\hat{B}$ A = x (∠s opp equal sides)
∴ A $\hat{P}$ B = 180° − 2x (sum ∠s of Δ)

(c) Answer: Yes: BA = BC, so ΔABC is isosceles

C $\hat{A}$ B = A $\hat{B}$ P = x (alt ∠s; AC ∥ PB)
A $\hat{C}$ B = x (given)
∴ C $\hat{A}$ B = A $\hat{C}$ B
∴ BA = BC (sides opp equal ∠s)

(d) Answer: PAOB is a cyclic quadrilateral

O $\hat{A}$ P = 90° (tan ⊥ radius)
O $\hat{B}$ P = 90° (tan ⊥ radius)
∴ O $\hat{A}$ P + O $\hat{B}$ P = 180°
∴ PAOB is a cyclic quad (opp ∠s of quad supp)

(e) Answer: A $\hat{O}$ B = 76°

A $\hat{O}$ B = 2 × A $\hat{C}$ B = 76° (∠ at centre = 2 × ∠ at circumference)

Tangents, and tangents from a point outside

QUESTION 3

(a) Answer: −cos θ

cos(180° + θ) = −cos θ
cos(90° + θ) = −sin θ
sin(360° − θ) = −sin θ
= −cos θ

(b) Answer: −√3/2

cos 113° cos 37° − sin 113° sin 37° = cos(113° + 37°)
= cos 150° = −√3/2

(c) Answer: −√3/2

cos^{^2} 75° − sin^{^2} 75° = cos 150°
= −√3/2

(d) Answer: θ = 30° + k·180°, k ∈ ℤ

tan θ = 1/√3

Identities, reduction and compound angles

QUESTION 4

(a) Answer: y = 3x − 15

m = (6 − (−9))/(7 − 2) = 3
−9 = (3)(2) + c, so c = −15

(b) Answer: 75/2 square units

C(5; 0) and D(0; −15)
area = ½ × 5 × 15 = 75/2

(c) Answer: y = −(1/3)x + 5/3

PF ⊥ AB: m_{_PF} = −1/(3) = −1/3
3 = (−1/3)(−4) + c, so c = 5/3

(d) Answer: F(5; 0)

3x − 15 = −(1/3)x + 5/3
x = 5, y = 0

(e) Answer: PF = 3√10 units

PF^{^2} = (−4 − 5)^{^2} + (3 − 0)^{^2} = 90
PF = 3√10

(f) Answer: 75 square units

AB = 5√10 and the height is PF = 3√10
area = ½ × AB × PF = 75

The equation of a straight line

QUESTION 5

(a) Answer: x = k·180° or x = −30° + k·360° or x = 210° + k·360°

sin x + 1 = cos 2x
1 + sin x = 1 − 2sin^{^2} x
2sin^{^2} x + sin x = 0
sin x(2 sin x + 1) = 0
sin x = 0: x = k·180°
or sin x = −1/2: x = −30° + k·360° or x = 210° + k·360°

(b) Answer: x = −30° or x = 0° or x = 180° or x = 210°

k = …, −1, 0, 1, … in x = k·180° or x = −30° + k·360° or x = 210° + k·360°
In [−90°; 270°]: x = −30° or x = 0° or x = 180° or x = 210°

(c) Answer: See the sketch below.

f(x) = sin x + 1
f: x-intercepts (−90°; 0), (270°; 0)
f: y-intercept (0°; 1)
f: turning points (90°; 2)
f: end points (−90°; 0), (270°; 0)
g(x) = cos 2x
g: x-intercepts (−45°; 0), (45°; 0), (135°; 0), (225°; 0)
g: y-intercept (0°; 1)
g: turning points (0°; 1), (90°; −1), (180°; 1)
g: end points (−90°; −1), (270°; −1)

(d) Answer: x = −90° or −45° ≤ x ≤ 45° or 135° ≤ x ≤ 225° or x = 270°

f and g have the same sign, or one of them is 0, read from the sketch

(e) Answer: −3 ≤ y ≤ 1

−1 ≤ g(x) ≤ 1
−3 ≤ 2g(x) − 1 ≤ 1

Trigonometric graphs

QUESTION 6

(a) Answer: (x − 3)^{^2} + (y − 4)^{^2} = 25

it passes through O: r = PO = √((3)^{^2} + (4)^{^2}) = 5

(b) Answer: PQ = 10 units

PQ = √((−5 − 3)^{^2} + (−2 − 4)^{^2})
= 10

(c) Answer: See the working below.

r_{_P} + r_{_Q} = 5 + 5 = 10 = PQ
∴ the circles touch externally

(d) Answer: T(−1; 1)

T lies on PQ, 5 units from P: 5/10 of the way to Q
(3 + (5/10)(−8); 4 + (5/10)(−6)) = (−1; 1)

(e) Answer: y = −(4/3)x − 1/3

m_{_PQ} = 3/4, so the tangent has m = −4/3
1 = (−4/3)(−1) + c, so c = −1/3

The equation of a circle

QUESTION 7

(a) Answer: See the proof below.

Required to prove: B $\hat{A}$ D + B $\hat{C}$ D = 180°
Construction: Join OB and OD.
${\hat{O}}_{1}$ = 2 × B $\hat{A}$ D (∠ at centre = 2 × ∠ at circumference)
reflex B $\hat{O}$ D = 2 × B $\hat{C}$ D (∠ at centre = 2 × ∠ at circumference)
${\hat{O}}_{1}$ + reflex B $\hat{O}$ D = 360° (∠s round a pt)
∴ 2 × B $\hat{A}$ D + 2 × B $\hat{C}$ D = 360°
∴ B $\hat{A}$ D + B $\hat{C}$ D = 180°

(b) Answer: A $\hat{E}$ F = A $\hat{B}$ C

A $\hat{E}$ F = A $\hat{B}$ C (corresp ∠s; BC ∥ EF)

(c) Answer: ADFE is a cyclic quadrilateral

A $\hat{B}$ C + A $\hat{D}$ C = 180° (opp ∠s of cyclic quad)
A $\hat{E}$ F = A $\hat{B}$ C (proved)
∴ A $\hat{E}$ F + A $\hat{D}$ F = 180°
∴ ADFE is a cyclic quad (opp ∠s of quad supp)

(d) Answer: A $\hat{E}$ F = 101°

A $\hat{B}$ C = 180° − A $\hat{D}$ C = 101° (opp ∠s of cyclic quad)
∴ A $\hat{E}$ F = 101° (proved)

Proving the theorem itself, not using it

QUESTION 8

(a) Answer: See the explanation below.

A $\hat{O}$ B = 90°: the axes are perpendicular
AB subtends a right angle at O, a point on the circle, so AB is a diameter (converse ∠ in semi-circle)

(b) Answer: M(−3; −5); r = √34

M is the midpoint of AB: ((−6 + 0)/2; (0 + (−10))/2) = (−3; −5)
r = MO = √((−3)^{^2} + (−5)^{^2}) = √34

(c) Answer: See the working below.

(x + 3)^{^2} + (y + 5)^{^2} = 34
x^{^2} + 6x + 9 + y^{^2} + 10y + 25 = 34
x^{^2} + y^{^2} + 6x + 10y = 0

(d) Answer: y = −(3/5)x

m_{_MO} = (−5 − 0)/(−3 − 0) = 5/3
the tangent is perpendicular to the radius MO: m = −3/5
it passes through O: y = −(3/5)x

(e) Answer: C(2; −2)

substitute y = −x: x^{^2} + (−x)^{^2} + 6x − 10x = 0
2x(x − 2) = 0
x = 0 (the point O) or x = 2

The midpoint of a line segment

QUESTION 9

(a) Answer: C $\hat{A}$ B = x

C $\hat{A}$ B = ${\hat{C}}_{3}$ = x (tan chord theorem)

(b) Answer: ${\hat{C}}_{2}$ = 90° − x

A $\hat{C}$ B = 90° (∠ in semi-circle)
${\hat{C}}_{1}$ = C $\hat{A}$ O = x (∠s opp equal sides)
∴ ${\hat{C}}_{2}$ = 90° − x

(c) Answer: ΔDCB ||| ΔDAC

In ΔDCB and ΔDAC:
C $\hat{D}$ B = A $\hat{D}$ C (common)
${\hat{C}}_{3}$ = C $\hat{A}$ B = x (proved)
∴ ΔDCB ||| ΔDAC (∠∠∠)

(d) Answer: DC^{^2} = DB · DA

DC/DA = DB/DC (||| Δs)
∴ DC^{^2} = DB · DA

(e) Answer: r = 3 units

4^{^2} = 2(2 + 2r) (proved)
16 = 4 + 4r
∴ r = 3 units

The angle in a semi-circle

QUESTION 10

(a) Answer: 60

the ogive ends at a cumulative frequency of 60

(b) Answer: about 17.9

the median is at a cumulative frequency of 60/2 = 30
read across to the ogive and down: about 17.9

(c) Answer: about 18.0

Q1 at 15: about 13.5; Q3 at 45: about 31.5
IQR = 31.5 − 13.5 = 18.0

(d) Answer: 19

read the ogive at 20 and at 40: 37 and 56
56 − 37 = 19

(e) Answer: See the box-and-whisker plot below.

minimum 1, Q1 ≈ 13.5, median ≈ 17.9, Q3 ≈ 31.5, maximum 47

(f) Answer: 95%

3 values lie below 10, so 57 values are 10 or more
57/60 × 100 = 95%

(g) Answer: Skewed to the right (positively)

Q3 − median ≈ 13.6 and median − Q1 ≈ 4.4
the upper half of the box is longer

Cumulative frequency curves

QUESTION 11

(a) Answer: BD = 7.37 cm

BD^{^2} = AB^{^2} + AD^{^2} − 2(AB)(AD) cos B $\hat{A}$ D (cosine rule)
= 5^{^2} + 9^{^2} − 2(5)(9) cos 55°

(b) Answer: B $\hat{C}$ D = 125° (opp ∠s of cyclic quad)

B $\hat{C}$ D = 180° − 55° = 125° (opp ∠s of cyclic quad)

(c) Answer: B $\hat{D}$ C = 41.80°

(sin B $\hat{D}$ C)/BC = (sin B $\hat{C}$ D)/BD (sine rule)
sin B $\hat{D}$ C = (6 sin 125°)/7.37

(d) Answer: CD = 2.06 cm

C $\hat{B}$ D = 180° − 125° − 41.80° = 13.20° (sum ∠s of Δ)
CD/(sin C $\hat{B}$ D) = BD/(sin B $\hat{C}$ D) (sine rule)
CD = (7.37 sin 13.20°)/(sin 125°) = 2.06 cm

(e) Answer: area ABCD = 23.48 cm²

area ΔABD = (1/2)(5)(9) sin 55° (area rule)
area ΔBCD = (1/2)(6)(7.37) sin 13.20° (area rule)

Sine, cosine and area rules (2D)