Week 1 — Drills

Exam format. Show all work. Core items are for everyone; stretch items for whoever finishes early. Solutions at the end, worked the way an examiner wants to see them.

A. Linear equations and systems

D1 (core). Arno Bikes has fixed monthly costs of €60,000 and a variable cost of €400 per bike. Bikes sell at €1,000. (a) Write the cost function C(q) and the revenue function R(q). (b) Find the break-even quantity. (c) The landlord raises the rent so that fixed costs become €72,000. Find the new break-even quantity and state in one sentence what happened to the cost line in the (q, €) plane.

D2 (core). In the Tuscan e-bike market, demand is Q_d = 1200 − 2p and supply is Q_s = −300 + 3p (p in €, Q in bikes per month). (a) Find the equilibrium price and quantity. (b) The region introduces a tax of €50 per bike paid by sellers, so that sellers now respond to the price they keep, p − 50. Write the new supply function and find the new equilibrium. (c) How much of the €50 tax is paid by buyers, in the form of a higher price?

D3 (stretch). A workshop assembles two models. Model A takes 2 hours, model B takes 3 hours; 1,200 hours are available per month and the plan is 500 bikes in total. Set up and solve the system giving the number of each model.

B. Quadratics

D3b (core — the syllabus asks for these and they are quick). Solve, and write each answer as an interval: (a) 3q − 12 > 0 (b) 5 − 2q ≥ 1 (c) 400q + 60,000 ≤ 1,000q (d) What does (c) say about Arno Bikes, in one clause?

D4 (core). A firm’s profit function is π(q) = −2q² + 600q − 40,000. (a) Find the quantities at which profit is zero. (b) For which quantities is the firm profitable? (c) Find the vertex of the parabola and interpret it.

D5 (core). Solve the inequalities: (a) x² − 7x + 10 ≤ 0 (b) 2x² + 3x − 2 > 0

D6 (stretch — the final-exam pattern). Factor numerator and denominator, then simplify: (x − 1)² / (x − x²). Evaluate the simplified expression at x = 1.1 and at x = 0.9 and comment on the sign.

C. Exponentials and logarithms

D7 (core). (a) Simplify: ln(e³), log₂ 32, ln(a²b / c). (b) Solve for x: 3ˣ = 81; 10ˣ = 500 (three decimals).

D8 (core). Arno Bikes’ revenue is €1.2M and grows at 12% a year. (a) Write revenue after t years. (b) How many years to double? (c) How many years to reach €3M?

D9 (stretch). (a) Solve 5e^{0.03t} = 8 for t. (b) A competitor doubled revenue in 5 years. What constant annual growth rate does that imply?

D. Absolute values and intervals

D10 (core). (a) Battery cells must be 500 Wh ± 15 Wh. Write the condition with an absolute value and as an interval. Is a 483 Wh cell acceptable? (b) Solve |x − 3| > 2.

D11 (core — notation for the exam). Write as intervals: (a) x ≥ −3; (b) x > −3; (c) −3 < x ≤ 5. Then state the domain, as an interval, of (d) f(x) = √(x + 3) and (e) g(x) = ln(x − 1).

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Solutions

D1. (a) C(q) = 60,000 + 400q; R(q) = 1,000q. (b) R = C → 1,000q = 60,000 + 400q → 600q = 60,000 → q = 100 bikes. (Contribution per bike €600; 60,000 / 600 = 100.) (c) 1,000q = 72,000 + 400q → q = 120. The cost line shifted up in parallel by €12,000 (same slope, higher intercept); the crossing with the revenue line moved right.

D2. (a) 1200 − 2p = −300 + 3p → 1500 = 5p → p = €300; Q = 1200 − 600 = 600. (b) Q_s = −300 + 3(p − 50) = −450 + 3p. Then 1200 − 2p = −450 + 3p → 1650 = 5p → p = €330, Q = 540. (c) Buyers pay €330 instead of €300: €30 of the €50 is passed to buyers; sellers keep 330 − 50 = 280, i.e. bear €20.

D3. Let x = model A, y = model B. 2x + 3y = 1200, x + y = 500. From the second, x = 500 − y; substitute: 1000 − 2y + 3y = 1200 → y = 200, x = 300. 300 of A, 200 of B.

D3b. (a) 3q > 12 → q > 4 → (4, +∞). (b) −2q ≥ −4; dividing by a negative flips the sign → q ≤ 2 → (−∞, 2], or [0, 2] for a quantity. (c) 60,000 ≤ 600q → q ≥ 100 → [100, +∞). (d) It is the break-even condition written as an inequality: at 100 bikes or more the firm is not losing money. The one that costs marks is (b): multiplying or dividing an inequality by a negative number reverses it.

D4. (a) Divide by −2: q² − 300q + 20,000 = 0. Discriminant 300² − 4·20,000 = 90,000 − 80,000 = 10,000, √ = 100. q = (300 ± 100)/2 → q = 100 and q = 200. (b) Downward parabola (a = −2 < 0) is positive between its roots: 100 < q < 200. (c) Vertex at q = −b/(2a) = −600/(−4) = 150; π(150) = −2·22,500 + 90,000 − 40,000 = **€5,000**. Maximum profit is €5,000 at 150 units; the window is narrow, the business is fragile.

D5. (a) Roots of x² − 7x + 10: (x − 2)(x − 5) → 2 and 5. Upward parabola ≤ 0 between roots: 2 ≤ x ≤ 5, i.e. [2, 5]. (b) Roots: x = (−3 ± √(9 + 16))/4 = (−3 ± 5)/4 → x = 1/2 and x = −2. Upward parabola > 0 outside the roots: x < −2 or x > 1/2.

D6. x − x² = x(1 − x) = −x(x − 1). So (x − 1)² / (−x(x − 1)) = −(x − 1)/x for x ≠ 0, 1. At x = 1.1: −0.1/1.1 ≈ −0.091. At x = 0.9: −(−0.1)/0.9 ≈ +0.111. The expression changes sign across x = 1 — approaching 1 from the right it is a small negative number, from the left a small positive number. (This is the algebra behind the one-sided limits the final exam asks for: factor, cancel, then look at the sign of what is left. The limit from the right is 0⁻.)

D7. (a) ln(e³) = 3; log₂ 32 = 5 (2⁵ = 32); ln(a²b/c) = 2 ln a + ln b − ln c. (b) 3ˣ = 81 = 3⁴ → x = 4. 10ˣ = 500 → x = log₁₀ 500 = **2.699**.

D8. (a) R(t) = 1.2 · 1.12ᵗ (€M). (b) 1.12ᵗ = 2 → t ln 1.12 = ln 2 → t = 0.6931 / 0.1133 = **6.1 years**. (c) 1.12ᵗ = 2.5 → t = ln 2.5 / ln 1.12 = 0.9163 / 0.1133 = **8.1 years**.

D9. (a) e^{0.03t} = 1.6 → 0.03t = ln 1.6 = 0.4700 → t ≈ 15.7. (b) (1 + g)⁵ = 2 → 1 + g = 2^{1/5} = e^{(ln 2)/5} = e^{0.1386} = 1.1487 → g ≈ 14.9% a year.

D10. (a) |c − 500| ≤ 15 ↔︎ 485 ≤ c ≤ 515, i.e. [485, 515]. 483 < 485: not acceptable. (b) |x − 3| > 2 ↔︎ x − 3 > 2 or x − 3 < −2 ↔︎ x > 5 or x < 1.

D11. (a) [−3, +∞); (b) (−3, +∞); (c) (−3, 5]. (d) Need x + 3 ≥ 0 → [−3, +∞). (e) Need x − 1 > 0 → (1, +∞). (Square bracket when the endpoint is included, round when it is not or when it is ±∞. The resit asks exactly this.)