IEB mock Paper II (third paper)

(a) Answer: EC = 14 units

1. CE/EA = CD/DB (line ∥ one side of Δ; ED ∥ AB)
2. EC/7 = 8/4
3. EC = 14

(b) Answer: EG = 8.4 units

1. EG/GC = EF/FD = 3/2 (line ∥ one side of Δ; FG ∥ DC)
2. EG = 3/5 × EC = 3/5 × 14
3. EG = 8.4

(c) Answer: FG = 4.8 units

1. In ΔEFG and ΔEDC:
2. $\hat{\mathrm{E}}$ = $\hat{\mathrm{E}}$ (common)
3. E$\hat{\mathrm{F}}$G = C$\hat{\mathrm{D}}$E (corresp ∠s equal; FG ∥ DC)
4. ∴ ΔEFG ||| ΔEDC (∠∠∠)
5. FG/DC = EF/ED = 3/5 (∥∥∥ Δs)
6. FG = 3/5 × 8 = 4.8

(a) Answer: M(−5; −1); r = 5

1. (x + 5)^{^2} + (y + 1)^{^2} = −1 + 25 + 1
2. (x + 5)^{^2} + (y + 1)^{^2} = 25
3. M(−5; −1) and r = √(25) = 5

(b) Answer: See the working below.

1. (−8)^{^2} + (−5)^{^2} + 10(−8) + 2(−5) + 1 = 0
2. so P lies on the circle

(c) Answer: y = −(3/4)x − 11

1. m_{_MP} = (−5 − (−1))/(−8 − (−5)) = 4/3
2. the tangent is perpendicular to the radius: m = −3/4
3. −5 = (−3/4)(−8) + c, so c = −11
4. y = −(3/4)x − 11

(d) Answer: QT = 5 units

1. QM^{^2} = (0 − (−5))^{^2} + (4 − (−1))^{^2} = 50
2. MT ⊥ QT (radius ⊥ tangent), so QT^{^2} = QM^{^2} − r^{^2}
3. QT^{^2} = 50 − 25 = 25
4. QT = 5

(a) Answer: M(0; 0)

1. the diagonals of a parallelogram bisect each other: M is the midpoint of AC
2. M = ((2 + −2)/2; (2 + −2)/2) = (0; 0)

(b) Answer: D(−2; 2)

1. M is also the midpoint of BD
2. D = (2(0) − 2; 2(0) − (−2)) = (−2; 2)

(c) Answer: AB = 4 units

1. AB = √((2 − 2)^{^2} + (2 − (−2))^{^2}) = √(16)
2. = 4

(d) Answer: Yes: ABCD is a rectangle

1. one of AB and BC is vertical and the other horizontal
2. so AB ⊥ BC and the parallelogram has a right angle

(e) Answer: 8 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. = 8

(a) Answer: A$\hat{\mathrm{D}}$B = 27°

1. A$\hat{\mathrm{D}}$B = 27° (∠s in the same seg)

(b) Answer: D$\hat{\mathrm{C}}$A = 94°

1. D$\hat{\mathrm{C}}$A = 94° (∠s in the same seg)

(c) Answer: D$\hat{\mathrm{A}}$B = 59°

1. D$\hat{\mathrm{A}}$B = 59° (sum ∠s of Δ)

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

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

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

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

(f) Answer: A$\hat{\mathrm{O}}$M = 56°

1. A$\hat{\mathrm{O}}$M = 56° (sum ∠s of Δ)

(a) Answer: cos θ

1. cos(90° + θ) = −sin θ
2. sin(90° + θ) = cos θ
3. sin(−θ) = −sin θ
4. = cos θ

(b) Answer: 1/2

1. sin 19° cos 11° + cos 19° sin 11° = sin(19° + 11°)
2. = sin 30° = 1/2

(c) Answer: √2/2

1. 2cos^{^2} 22.5° − 1 = cos 45°
2. = √2/2

(d) Answer: θ = −60° + k·180°, k ∈ ℤ

1. tan θ = −√3
2. θ = −60° + k·180°, k ∈ ℤ

(a) Answer: 26.42

1. Σx = 317 and n = 12
2. mean = 317/12 = 26.42

(b) Answer: σ = 6.02

1. σ² = Σ(x − 26.42)²/12 = 36.24
2. σ = 6.02

(c) Answer: 10

1. between 20.40 and 32.44
2. 10 values

(d) Answer: 18; 24; 24.5; 27.5; 43

1. minimum 18; Q1 24; median 24.5; Q3 27.5; maximum 43

(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: Yes: 43 > 32.75

1. IQR = 27.5 − 24 = 3.5
2. Q3 + 1.5 × IQR = 27.5 + 5.25 = 32.75
3. 43 > 32.75

(g) Answer: The mean increases by 10; 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: O$\hat{\mathrm{B}}$A = 45°

1. OA = OB (radii)
2. O$\hat{\mathrm{B}}$A = 45° (∠s opp equal sides)

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

1. A$\hat{\mathrm{O}}$B = 90° (sum ∠s of Δ)

(c) Answer: A$\hat{\mathrm{P}}$B = 45°

1. A$\hat{\mathrm{P}}$B = 45° (∠ at centre = 2 × ∠ at circumference)

(d) Answer: A$\hat{\mathrm{C}}$B = 48°

1. A$\hat{\mathrm{C}}$B = 48° (sum ∠s of Δ)

(e) Answer: D$\hat{\mathrm{C}}$A = 132°

1. D$\hat{\mathrm{C}}$A = 132° (∠s on a str line)

(a) Answer: y = 340.988 − 18.973x

1. A = 340.988
2. B = −18.973

(b) Answer: r = −1.00

1. r = −1.00

(c) Answer: a strong negative correlation

1. r = −1.00: r < 0 and |r| is at least 0.8

(d) Answer: y = 18.45

1. y = 340.988 − 18.973(17)
2. y = 18.45

(e) Answer: No: x lies outside the data (extrapolation)

1. the data runs from x = 1 to x = 11; x = 17 lies outside it

(a) Answer: 180°

1. g(x) = −tan x repeats every 180°

(b) Answer: x = 90° and x = 270°

1. tan x is undefined where cos x = 0: x = 90° + k·180°
2. In [0°; 360°]: x = 90° and x = 270°

(c) Answer: See the sketch below.

1. f(x) = cos 2x + 1
2. f: x-intercepts (90°; 0), (270°; 0)
3. f: y-intercept (0°; 2)
4. f: turning points (90°; 0), (180°; 2), (270°; 0)
5. f: end points (0°; 2), (360°; 2)
6. g(x) = −tan x
7. g: x-intercepts (0°; 0), (180°; 0), (360°; 0)
8. g: y-intercept (0°; 0)
9. g: asymptotes x = 90°, x = 270°
10. g: end points (0°; 0), (360°; 0)

(d) Answer: 90° < x < 180° or 270° < x < 360°

1. f and g have the same sign, and g(x) ≠ 0, read from the sketch
2. 90° < x < 180° or 270° < x < 360°

(a) Answer: m_{_AB} = −2

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

(b) Answer: θ = 116.6°

1. tan θ = −2
2. θ = 180° − 63.4°
3. θ = 116.6°

(c) Answer: y = (1/2)x − 6

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

(d) Answer: k = 5

1. m_{_AD} = m_{_AB}
2. (k − (−1))/(−4 − (−1)) = −2
3. k = 5

(e) Answer: B$\hat{\mathrm{A}}$C = 132.5°

1. the inclination of AB is 116.6°
2. m_{_AC} = −2/7, so the inclination of AC is 164.1°
3. the lines make 47.5°; B$\hat{\mathrm{A}}$C is its supplement: 180° − 47.5°
4. B$\hat{\mathrm{A}}$C = 132.5°

(a) Answer: x = −90° + k·360° or x = −30° + k·360° or x = 210° + k·360°, k ∈ ℤ

1. 2sin^{^2} x + 3sin x + 1 = 0
2. (2sin x + 1)(sin x + 1) = 0
3. sin x = −1/2 or sin x = −1
4. x = −90° + k·360° or x = −30° + k·360° or x = 210° + k·360°, k ∈ ℤ

(b) 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

(c) Answer: x = 210° or x = 330°

1. sin x = −1/2
2. x = −30° + k·360° or x = 210° + k·360°
3. In [0°; 360°]: x = 210° or x = 330°

(a) Answer: See the proof below.

1. Given: ΔABC and ΔDEF with $\hat{\mathrm{A}}$ = $\hat{\mathrm{D}}$, $\hat{\mathrm{B}}$ = $\hat{\mathrm{E}}$ and $\hat{\mathrm{C}}$ = $\hat{\mathrm{F}}$.
2. To prove: AB/DE = AC/DF = BC/EF
3. Construction: Mark G on AB so that AG = DE, and H on AC so that AH = DF. Join GH.
4. In ΔAGH and ΔDEF:
5. AG = DE (construction)
6. $\hat{\mathrm{A}}$ = $\hat{\mathrm{D}}$ (given)
7. AH = DF (construction)
8. ∴ ΔAGH ≡ ΔDEF (SAS)
9. ∴ A$\hat{\mathrm{G}}$H = $\hat{\mathrm{E}}$ = $\hat{\mathrm{B}}$ (≡ Δs)
10. ∴ GH ∥ BC (corresp ∠s equal)
11. ∴ AB/AG = AC/AH (line ∥ one side of Δ)
12. ∴ AB/DE = AC/DF
13. Similarly, cutting from B, AB/DE = BC/EF

(b) Answer: See the proof below.

1. In ΔABD and ΔCAD:
2. A$\hat{\mathrm{D}}$B = A$\hat{\mathrm{D}}$C = 90° (given)
3. $\hat{\mathrm{B}}$ + B$\hat{\mathrm{A}}$D = 90° (sum ∠s of Δ)
4. B$\hat{\mathrm{A}}$D + C$\hat{\mathrm{A}}$D = 90° (given)
5. ∴ $\hat{\mathrm{B}}$ = C$\hat{\mathrm{A}}$D
6. ∴ ΔABD ||| ΔCAD (∠∠∠)

(c) Answer: See the proof below.

1. BD/AD = AD/DC (∥∥∥ Δs)
2. ∴ AD² = BD · DC

(d) Answer: AD = 6 units

1. AD² = BD · DC = 12 × 3 = 36 (proved above)
2. AD = 6

(e) Answer: AB = 6√5 units

1. AB² = AD² + BD² = 36 + 144 = 180 (Pythagoras)
2. AB = √(180) = 6√5