Quadratic equations. Fluency
Solve: 3x² − 7x − 6 = 0.
(a) x = (7 ± √(49+72))/6 = (7 ± √121)/6 = (7 ± 11)/6. x = 3 or x = −2/3 .
(b) Sum = 3 + (−2/3) = 7/3 . Product = 3 × (−2/3) = −2 . (Check: −b/a = 7/3 ✓, c/a = −2 ✓.)
(c) Δ = 36 − 12k < 0 ⇒ k > 3. Also k ≠ 0 for quadratic. k > 3 .
(d) x-intercepts: (3,0) and (−2/3,0). y-intercept: (0,−6). Vertex: x=7/6, y=3(49/36)−7(7/6)−6=49/12−49/6−6=−121/12≈−10.1 . Parabola opens upward.
Compound interest and depreciation. Fluency
(a) A = 8000(1+0.05/4)12 = 8000(1.0125)12 ≈ 8000 × 1.1608 ≈ $9 286 .
(b) V = 2500 × 0.84 = 2500 × 0.4096 ≈ $1 024 .
(c) 2500 × 0.8t < 500 ⇒ 0.8t < 0.2. t log 0.8 < log 0.2 ⇒ t > log 0.2/log 0.8 ≈ 7.2. After 8 complete years .
(d) (1+r)10 =2 ⇒ 1+r=20.1 ≈1.0718 ⇒ r ≈ 7.18% per annum .
Sine rule application. Fluency
In triangle ABC, angle A = 48°, angle B = 63° and AB = 15 cm.
(a) C = 180°−48°−63° = 69° .
(b) BC/sin A = AB/sin C ⇒ BC = 15 sin 48°/sin 69° ≈ 15(0.743)/0.934 ≈ 11.9 cm .
(c) AC = 15 sin 63°/sin 69° ≈ 15(0.891)/0.934 ≈ 14.3 cm .
(d) Area = ½ × AB × AC × sin A = ½(15)(14.3)sin48° ≈ ½(15)(14.3)(0.743) ≈ 79.7 cm² .
Circle geometry proof. Fluency
O is the centre of a circle. Points A, B and C lie on the circle. Angle AOB = 100° (reflex).
(a) 360° − 100° = 260° . Wait — if reflex AOB = 100° then non-reflex = 360°−100° = 260°... Re-read: if the given 100° is the non-reflex angle, reflex = 260°. Taking AOB = 100° (non-reflex): reflex AOB = 260° .
(b) Angle at circumference on major arc = ½ × non-reflex AOB = ½ × 100° = 50° .
(c) Angle at circumference on minor arc = ½ × reflex AOB = ½ × 260° = 130° .
(d) 50° + 130° = 180° . This illustrates that opposite angles of a cyclic quadrilateral sum to 180° (ACBD is a cyclic quadrilateral).
Line of best fit. Understanding
The line of best fit for a data set is y = −2.5x + 80, where x = age of car (years) and y = resale value ($thousands).
(a) The car loses $2 500 in value per year on average.
(b) y = −2.5(6)+80 = −15+80 = $65 000 .
(c) 0 = −2.5x+80 ⇒ x = 32 years. Not realistic — most cars have a scrap value and aren't driven for 32 years. The model breaks down long-term.
(d) y = −2.5(15)+80 = 42.5 ($42 500). The model predicts more than the asking price. However, x=15 is outside the data range (extrapolation) so the model prediction is unreliable.
Indices and logarithms. Understanding
(a) (4a³b)−2 = 1/(16a&sup6b²). ×8a&sup5b³ = 8a&sup5b³/(16a&sup6b²) = b/(2a) .
(b) log3 81=4, log3 (1/9)=−2. Sum = 2 .
(c) (x+1)log2 = x log3 ⇒ x log2+log2=x log3 ⇒ log2=x(log3−log2) ⇒ x=log2/(log3−log2)=0.3010/0.1761≈1.71 .
(d) log(x²y/(xy²)) = log(x/y) = log x − log y .
Functions and inverses. Understanding
Let f(x) = 2x² − 3 (x ≥ 0) and g(x) = √((x+3)/2).
(a) f(4) = 2(16)−3 = 29 .
(b) f(g(x)) = 2((x+3)/2)−3 = (x+3)−3 = x ✓. g(f(x)) = √((2x²−3+3)/2) = √(x²) = x (for x≥0) ✓.
(c) f: domain x≥0, range f(x)≥−3. f−1 : domain x≥−3, range f−1 (x)≥0.
(d) 2x²−3 = √((x+3)/2). Let u=√((x+3)/2), so f(x)=x means 2x²−3=x ⇒ 2x²−x−3=0 ⇒ (2x−3)(x+1)=0. Since x≥0: x = 3/2 .
Coordinate geometry. Understanding
A circle has equation (x−2)² + (y+1)² = 25. A line has equation y = x + 2.
(a) Centre (2, −1), radius 5 .
(b) Substitute y=x+2: (x−2)²+(x+3)²=25 ⇒ x²−4x+4+x²+6x+9=25 ⇒ 2x²+2x−12=0 ⇒ x²+x−6=0 ⇒ (x+3)(x−2)=0. x=−3,y=−1 and x=2,y=4. Points: (−3,−1) and (2,4) .
(c) M = ((−3+2)/2, (−1+4)/2) = (−1/2, 3/2) .
(d) Gradient of chord = 1. Gradient from centre (2,−1) to midpoint (−1/2, 3/2): m = (3/2+1)/(−1/2−2) = (5/2)/(−5/2) = −1 . Product = 1×−1=−1 ✓ perpendicular.
Optimisation problem. Problem Solving
A farmer has 200 m of fencing. He wants to create three equal rectangular paddocks side by side (sharing internal fences).
(a) 4 widths (two outer + two internal) + 2 lengths: 4x + 2L = 200 ⇒ L = 100 − 2x .
(b) A = L × (total width). Total width = 3x. A(x) = 3x(100−2x) = −6x² + 300x .
(c) x = −300/(2×−6) = 25 m. L = 100−50 = 50 m. A(25) = −6(625)+7500 = −3750+7500 = 3 750 m² . Each paddock: 25 m × 50/3 m (or total dimensions 75 m × 50 m total area 3 750 m²).
(d) x > 0 and L > 0: 100−2x > 0 ⇒ x < 50. Domain: 0 < x < 50 . Vertex x=25 is inside ✓.
Extended mixed problem. Problem Solving
A company sells x hundred units per week. The demand function is p(x) = 80 − 4x dollars per unit. Fixed costs are $400 per week and variable cost is $12 per unit.
(a) R(x) = 100x × (80−4x) = −400x² + 8000x . C(x) = 12(100x)+400 = 1200x + 400 .
(b) P(x) = R−C = −400x²+8000x−1200x−400 = −400x² + 6800x − 400 .
(c) x = −6800/(2×−400) = 8.5 hundred units (850 units).
(d) P(8.5) = −400(72.25)+6800(8.5)−400 = −28900+57800−400 = $28 500 . Price = 80−4(8.5) = $46 per unit .