IEB mock Paper II (second paper)

(a) Answer: sin 68° = 2m

1. sin 68° = 2 sin 34° cos 34° = 2m

(b) Answer: cos 22° = 2m

1. cos 22° = cos(90° − 68°) = sin 68° = 2m

(c) Answer: cos 136° = 1 − 8m^{^2}

1. cos 136° = 1 − 2sin^{^2} 68°
2. = 1 − 2(2m)^{^2} = 1 − 8m^{^2}

(d) Answer: cos 68° = √(1 − 4m^{^2})

1. cos^{^2} 68° = 1 − sin^{^2} 68° = 1 − 4m^{^2}
2. 68° is acute, so cos 68° > 0
3. cos 68° = √(1 − 4m^{^2})

(a) Answer: See the proof below.

1. Given: Circle with centre O. A$\hat{\mathrm{O}}$B is at the centre and A$\hat{\mathrm{P}}$B at the circumference, both on arc AB.
2. To prove: A$\hat{\mathrm{O}}$B = 2 × A$\hat{\mathrm{P}}$B
3. Construction: Join PO and produce it to S.
4. Let A$\hat{\mathrm{P}}$O = x and B$\hat{\mathrm{P}}$O = y
5. OA = OP (radii)
6. ∴ O$\hat{\mathrm{A}}$P = x (∠s opp equal sides)
7. ∴ A$\hat{\mathrm{O}}$S = 2x (ext ∠ of Δ)
8. Similarly B$\hat{\mathrm{O}}$S = 2y
9. ∴ A$\hat{\mathrm{O}}$B = 2x + 2y = 2(x + y)
10. ∴ A$\hat{\mathrm{O}}$B = 2 × A$\hat{\mathrm{P}}$B

(b) Answer: A$\hat{\mathrm{O}}$B = 100°

1. A$\hat{\mathrm{O}}$B = 100° (∠ at centre = 2 × ∠ at circumference)

(c) Answer: O$\hat{\mathrm{A}}$B = 40°

1. OA = OB (radii)
2. O$\hat{\mathrm{A}}$B = O$\hat{\mathrm{B}}$A (∠s opp equal sides)
3. O$\hat{\mathrm{A}}$B = 40° (sum ∠s of Δ)

(d) Answer: O$\hat{\mathrm{B}}$A = 40°

1. O$\hat{\mathrm{B}}$A = 40° (∠s opp equal sides)

(a) Answer: M(4; 4); r = √26

1. (x − 4)^{^2} + (y − 4)^{^2} = −6 + 16 + 16
2. (x − 4)^{^2} + (y − 4)^{^2} = 26
3. M(4; 4) and r = √(26) = √26

(b) Answer: See the working below.

1. (9)^{^2} + (3)^{^2} − 8(9) − 8(3) + 6 = 0
2. so P lies on the circle

(c) Answer: y = 5x − 42

1. m_{_MP} = (3 − (4))/(9 − (4)) = −1/5
2. the tangent is perpendicular to the radius: m = 5
3. 3 = (5)(9) + c, so c = −42
4. y = 5x − 42

(d) Answer: QT = 2√2 units

1. QM^{^2} = (1 − (4))^{^2} + (−1 − (4))^{^2} = 34
2. MT ⊥ QT (radius ⊥ tangent), so QT^{^2} = QM^{^2} − r^{^2}
3. QT^{^2} = 34 − 26 = 8
4. QT = 2√2

(a) Answer: See the proof below.

1. Given: Circle with centre O. AB is a chord. OM ⊥ AB.
2. To prove: AM = MB
3. Construction: Join OA and OB.
4. In ΔOMA and ΔOMB:
5. OA = OB (radii)
6. OM = OM (common)
7. O$\hat{\mathrm{M}}$A = O$\hat{\mathrm{M}}$B = 90° (given)
8. ∴ ΔOMA ≡ ΔOMB (RHS)
9. ∴ AM = MB (≡ Δs)

(b) Answer: O$\hat{\mathrm{M}}$A = 90°

1. O$\hat{\mathrm{M}}$A = 90° (line from centre ⊥ to chord)

(c) Answer: O$\hat{\mathrm{M}}$B = 90°

1. O$\hat{\mathrm{M}}$B = 90° (line from centre ⊥ to chord)

(d) Answer: A$\hat{\mathrm{O}}$M = 80°

1. A$\hat{\mathrm{O}}$M = 80° (sum ∠s of Δ)

(a) Answer: See the working below.

1. A$\hat{\mathrm{C}}$B = 180° − (α + β) (sum ∠s of Δ)
2. AC/(sin β) = d/(sin(180° − (α + β))) (sine rule)
3. sin(180° − (α + β)) = sin(α + β)
4. AC = (d sin β)/(sin(α + β))

(b) Answer: See the working below.

1. In ΔACT, A$\hat{\mathrm{C}}$T = 90°: tan γ = TC/AC
2. TC = AC tan γ = (d sin β tan γ)/(sin(α + β))

(c) Answer: TC = 107.85 m

1. TC = (150 sin 65° tan 38°)/(sin 100°)
2. = 107.85 m

(d) Answer: BT = 138.80 m

1. BC = (150 sin 35°)/(sin 100°) = 87.36 m (sine rule)
2. BT^{^2} = BC^{^2} + TC^{^2} = 87.36^{^2} + 107.85^{^2}
3. BT = 138.80 m

(e) Answer: 50.99°

1. tan T$\hat{\mathrm{B}}$C = TC/BC = 107.85/87.36
2. T$\hat{\mathrm{B}}$C = 50.99°

(a) Answer: 22.33

1. Σx = 268 and n = 12
2. mean = 268/12 = 22.33

(b) Answer: σ = 2.95

1. σ² = Σ(x − 22.33)²/12 = 8.72
2. σ = 2.95

(c) Answer: 7

1. between 19.38 and 25.29
2. 7 values

(d) Answer: 18; 19.5; 22.5; 23.5; 28

1. minimum 18; Q1 19.5; median 22.5; Q3 23.5; maximum 28

(e) Answer: See the plot below.

1. the box from Q1 to Q3 with the median marked; whiskers to the minimum and maximum

(f) Answer: No: 28 ≤ 29.5

1. IQR = 23.5 − 19.5 = 4
2. Q3 + 1.5 × IQR = 23.5 + 6 = 29.5
3. 28 ≤ 29.5

(g) Answer: The mean increases by 3; the standard deviation stays the same.

1. every value, and so their mean, moves by the same amount
2. the distances from the mean do not change

(a) Answer: (11; 25)

1. (11; 25) lies far from the trend of the other points

(b) Answer: y = 95.879 − 3.901x

1. A = 95.879
2. B = −3.901

(c) Answer: r = −1.00

1. r = −1.00

(d) Answer: y decreases by about 3.901

1. B = −3.901 is the change in y for each 1 added to x

(e) Answer: Weaker

1. with the outlier r = −0.92; without it r = −1.00
2. |r| is smaller with the outlier: it weakened the correlation

(a) Answer: M(1; 4)

1. the diagonals of a parallelogram bisect each other: M is the midpoint of AC
2. M = ((4 + −2)/2; (6 + 2)/2) = (1; 4)

(b) Answer: D(−1; 7)

1. M is also the midpoint of BD
2. D = (2(1) − (3); 2(4) − (1)) = (−1; 7)

(c) Answer: AB = √26 units

1. AB = √((4 − (3))^{^2} + (6 − (1))^{^2}) = √(26)
2. = √26

(d) Answer: Yes: ABCD is a rectangle

1. m_{_AB} × m_{_BC} = (5)(−1/5) = −1
2. so AB ⊥ BC and the parallelogram has a right angle

(e) Answer: 13 square units

1. area = (1/2)|x_{_A}(y_{_B} − y_{_C}) + x_{_B}(y_{_C} − y_{_A}) + x_{_C}(y_{_A} − y_{_B})|
2. = 13

(a) Answer: See the proof below.

1. LHS = (1 − (1 − 2sin^{^2} x))/(2 sin x)
2. = (2sin^{^2} x)/(2 sin x)
3. = sin x
4. = RHS

(b) Answer: x = 0° or x = 180° or x = 360°

1. The identity is not valid where sin x = 0.
2. sin x = 0: x = k·180°
3. In [0°; 360°]: x = 0° or x = 180° or x = 360°

(c) Answer: x = −30° or x = 210°

1. sin x = −1/2
2. x = −30° + k·360° or x = 210° + k·360°
3. In [−90°; 270°]: x = −30° or x = 210°

(a) Answer: m_{_AB} = 1

1. m_{_AB} = (8 − (1))/(8 − (1))
2. = 1

(b) Answer: θ = 45.0°

1. tan θ = 1
2. θ = 45.0°

(c) Answer: y = −x + 8

1. m = −1/(1) = −1
2. 6 = (−1)(2) + c, so c = 8
3. y = −x + 8

(d) Answer: k = −4

1. m_{_AD} = m_{_AB}
2. (k − (1))/(−4 − (1)) = 1
3. k = −4

(e) Answer: B$\hat{\mathrm{A}}$C = 33.7°

1. the inclination of AB is 45.0°
2. m_{_AC} = 5, so the inclination of AC is 78.7°
3. B$\hat{\mathrm{A}}$C = 78.7° − 45.0°
4. B$\hat{\mathrm{A}}$C = 33.7°

(a) Answer: B$\hat{\mathrm{A}}$S = 116°

1. B$\hat{\mathrm{A}}$S = 116° (∠s on a str line)

(b) Answer: O$\hat{\mathrm{A}}$S = 90°

1. O$\hat{\mathrm{A}}$S = 90° (tan ⊥ radius)

(c) Answer: O$\hat{\mathrm{A}}$T = 90°

1. O$\hat{\mathrm{A}}$T = 90° (tan ⊥ radius)

(d) Answer: O$\hat{\mathrm{B}}$A = 26°

1. B$\hat{\mathrm{A}}$O = 26°
2. OA = OB (radii)
3. O$\hat{\mathrm{B}}$A = 26° (∠s opp equal sides)

(e) Answer: B$\hat{\mathrm{O}}$A = 128°

1. B$\hat{\mathrm{O}}$A = 128° (sum ∠s of Δ)

(f) Answer: B$\hat{\mathrm{A}}$T = 58°

1. B$\hat{\mathrm{A}}$T = 58° (∠s on a str line)

(g) Answer: A$\hat{\mathrm{C}}$B = 58°

1. A$\hat{\mathrm{C}}$B = 58° (tan chord theorem)

(a) Answer: 180°

1. f(x) = cos 2x repeats every 180°

(b) Answer: See the sketch below.

1. f(x) = cos 2x
2. f: x-intercepts (−45°; 0), (45°; 0), (135°; 0), (225°; 0)
3. f: y-intercept (0°; 1)
4. f: turning points (0°; 1), (90°; −1), (180°; 1)
5. f: end points (−90°; −1), (270°; −1)
6. g(x) = sin(x − 60°)
7. g: x-intercepts (60°; 0), (240°; 0)
8. g: y-intercept (0°; −√3/2)
9. g: turning points (−30°; −1), (150°; 1)
10. g: end points (−90°; −1/2), (270°; −1/2)

(c) Answer: x = −70° or x = 50° or x = 170° or x = 210°

1. cos 2x = sin(x − 60°)
2. sin(x − 60°) = cos(90° − (x − 60°)) = cos(150° − x)
3. 2x = 150° − x + k·360°, so x = 50° + k·120°
4. or 2x = −(150° − x) + k·360°, so x = −150° + k·360°
5. In [−90°; 270°]: x = −70° or x = 50° or x = 170° or x = 210°

(d) Answer: −90° ≤ x ≤ −70° or 50° ≤ x ≤ 170° or 210° ≤ x ≤ 270°

1. f lies on or below g, read from the sketch
2. −90° ≤ x ≤ −70° or 50° ≤ x ≤ 170° or 210° ≤ x ≤ 270°

(e) Answer: −45° < x < 45° or 60° < x < 135° or 225° < x < 240°

1. f and g have opposite signs, and g(x) ≠ 0, read from the sketch
2. −45° < x < 45° or 60° < x < 135° or 225° < x < 240°