Take this after Lesson 25, in one 3-hour sitting with a 5-minute break between sections. This is the dress rehearsal: full length, full timing, real rules.
| Section | Items | Time | Calculator |
|---|---|---|---|
| I, Part A | Questions 1–28 | 80 min | 🚫 NONE |
| I, Part B | Questions 29–40 | 40 min | 📱 graphing calculator |
| II, Part A | FRQ 1 & FRQ 2 | 30 min | 📱 graphing calculator |
| II, Part B | FRQ 3 & FRQ 4 | 30 min | 🚫 NONE |
1. (A) f(3) = 15, f(0) = 0 → 15/3 = 5. [L1]
2. (B) Positive, shrinking rate → increasing, concave down. [L1]
3. (C) Inflection = concavity change = rate of change switches trend. [L3]
4. (D) Multiplicity 2 at 3 (touch), 1 at −1 (cross). [L4]
5. (A) 2i forces −2i; remaining two zeros are 1 and one more real (a lone non-real can't exist) → 2 real zeros with multiplicity. [L4]
lim x→∞ (5x³ − x⁷) =6. (B) Leading term −x⁷ → −∞. [L5]
7. (C) Bottom-heavy (1 < 3) → y = 0. [L5]
8. (A) At 4: num mult 2 ≥ den mult 1 → hole. At −1: den mult 2 > num mult 1 → VA. [L6]
9. (A) x² − 4 = (x + 3)(x − 3) + 5 → quotient x − 3. [L7]
10. (B) p(−2) = −8 + 2 + 1 = −5 (remainder theorem). [L7]
11. (D) a = −1 then d = +2: reflect over x-axis, then up 2. (B) computes −(f + 2). [L8]
12. (A) [1 − 4, 3 − 4] = [−3, −1]. [L8]
13. (B) One max, rise-then-fall → concave-down quadratic. [L9]
14. (A) 7 + 6(−3) = −11. [L10]
15. (C) Exponential dominance: g exceeds f eventually, always. [L10]
16. (D) (27^(1/3))² = 3² = 9. [L11]
17. (B) Shift up 1: y = 1. [L11]
18. (A) g(2) = 7; f(7) = 13. (B) is g(f(2)). [L13]
19. (D) Strictly increasing everywhere → one-to-one: x³ + 2. [L14]
20. (B) 2⁻³ = 1/8 → −3. [L15]
21. (A) 3 log x − log y. (B) wrongly applies 3 to y as well. [L15]
22. (C) 2x = ln 5 → x = (ln 5)/2. [L16]
23. (D) x² + 3x − 10 = 0 → x = 2 or −5; log(−5) undefined → {2}. [L16]
24. (C) Change of base, input on top: ln 12/ln 5. [L15]
25. (B) 240·π/180 = 4π/3. [L17]
26. (C) QII, reference π/6, cosine negative: −√3/2. [L17]
27. (B) Midline 2 ± amplitude 3: [−1, 5]. [L18]
28. (D) T = 2π/(1/3) = 6π. [L19]
29. (A) Graphical root: f(2.69) ≈ −0.007 → zero ≈ 2.690. [L3/L4]
30. (B) Vertex t = 30/9.8 ≈ 3.061; h ≈ 47.9. (A) reports the input. [L2/L9]
31. (A) e^(0.04t) = 5/3 → t = ln(5/3)/0.04 ≈ 0.51083/0.04 ≈ 12.771. [L16]
32. (D) Patternless-and-small beats patterned: exponential more appropriate. [L12]
33. (C) 1.062⁹ ≈ 1.7184 → 850·1.7184 ≈ $1,461. [L12]
34. (B) Intersection of y = ln x and y = 3 − x: x ≈ 2.208 (ln 2.208 ≈ 0.792 ≈ 3 − 2.208 ✓). [L16]
35. (A) sin(π(3)/6) = sin(π/2) = 1 → 7 + 3 = 10. [L19]
36. (C) cos θ = 0.4: θ₁ = arccos 0.4 ≈ 1.159; twin 2π − 1.159 ≈ 5.124. (B) uses π + θ₁. [L21]
37. (B) cos(πt/40) = −10/18 → πt/40 = arccos(−5/9) ≈ 2.160 → t ≈ 40(2.160)/π ≈ 27.5 s. [L19/L21]
38. (D) n = 5 odd → 5 petals. [L23]
39. (A) (r(π/2) − r(0))/(π/2) = (2 − 4)/(π/2) = −4/π ≈ −1.273. [L24]
40. (B) Reference arcsin 0.72 ≈ 0.8038; QII: π − 0.8038 ≈ 2.338. [L20]
Weekly ticket sales for a theater production, in hundreds:
| week t | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| S (hundreds) | 12 | 20 | 24 | 24 | 20 |
(a) (i) Find the average rate of change of S over [1, 4]. (ii) Interpret it in context with units.
(b) (i) Explain why a quadratic model is appropriate, using rates of change over equal-length intervals. (ii) Find a quadratic model for S(t) (regression or algebra — the data cooperate).
(c) (i) According to your model, in which week do sales fall to zero? (ii) The producers want to predict sales in week 8. What does the model say, and why should they not trust it?
In a northern city, the number of daylight hours varies sinusoidally through the year. The maximum, 15.2 hours, occurs on day 172; the following minimum, 9.2 hours, occurs on day 355. (Use a period of 366 days.)
(a) (i) State the midline and amplitude of the daylight function. (ii) Verify that the given max and min days are consistent with the stated period.
(b) Write a function D(t) for the hours of daylight on day t.
(c) (i) Find the daylight hours on day 264, to two decimal places. (ii) On day 264, is daylight increasing or decreasing, and at a relatively fast or slow rate? Justify using the function's structure.
(a) Solve exactly: 9^(x−1) = 27^(x/2).
(b) Solve exactly, rejecting any extraneous solutions: 2 ln x − ln(x + 4) = 0.
(c) Solve on [0, 2π): sin(2θ) = sin θ. (Use the double-angle formula and factor; do not divide by sin θ.)
Let f(x) = (3x² − 12)/(x² − 2x − 8).
(a) (i) Write f in factored form and state its domain. (ii) Classify the graph's behavior at x = −2 and x = 4, justifying with multiplicities, and give the coordinates of any hole.
(b) (i) Describe the end behavior of f using limit notation and identify the horizontal asymptote. (ii) Determine whether the graph of f ever crosses its horizontal asymptote. Justify algebraically.
(c) Write one-sided limit statements describing the behavior of f on each side of its vertical asymptote, with a sign analysis to justify each.
1. (A) f(3) = 15, f(0) = 0 → 15/3 = 5. [L1] 2. (B) Positive, shrinking rate → increasing, concave down. [L1] 3. (C) Inflection = concavity change = rate of change switches trend. [L3] 4. (D) Multiplicity 2 at 3 (touch), 1 at −1 (cross). [L4] 5. (A) 2i forces −2i; remaining two zeros are 1 and one more real (a lone non-real can't exist) → 2 real zeros with multiplicity. [L4] 6. (B) Leading term −x⁷ → −∞. [L5] 7. (C) Bottom-heavy (1 < 3) → y = 0. [L5] 8. (A) At 4: num mult 2 ≥ den mult 1 → hole. At −1: den mult 2 > num mult 1 → VA. [L6] 9. (A) x² − 4 = (x + 3)(x − 3) + 5 → quotient x − 3. [L7] 10. (B) p(−2) = −8 + 2 + 1 = −5 (remainder theorem). [L7] 11. (D) a = −1 then d = +2: reflect over x-axis, then up 2. (B) computes −(f + 2). [L8] 12. (A) [1 − 4, 3 − 4] = [−3, −1]. [L8] 13. (B) One max, rise-then-fall → concave-down quadratic. [L9] 14. (A) 7 + 6(−3) = −11. [L10] 15. (C) Exponential dominance: g exceeds f eventually, always. [L10] 16. (D) (27^(1/3))² = 3² = 9. [L11] 17. (B) Shift up 1: y = 1. [L11] 18. (A) g(2) = 7; f(7) = 13. (B) is g(f(2)). [L13] 19. (D) Strictly increasing everywhere → one-to-one: x³ + 2. [L14] 20. (B) 2⁻³ = 1/8 → −3. [L15] 21. (A) 3 log x − log y. (B) wrongly applies 3 to y as well. [L15] 22. (C) 2x = ln 5 → x = (ln 5)/2. [L16] 23. (D) x² + 3x − 10 = 0 → x = 2 or −5; log(−5) undefined → {2}. [L16] 24. (C) Change of base, input on top: ln 12/ln 5. [L15] 25. (B) 240·π/180 = 4π/3. [L17] 26. (C) QII, reference π/6, cosine negative: −√3/2. [L17] 27. (B) Midline 2 ± amplitude 3: [−1, 5]. [L18] 28. (D) T = 2π/(1/3) = 6π. [L19]
29. (A) Graphical root: f(2.69) ≈ −0.007 → zero ≈ 2.690. [L3/L4] 30. (B) Vertex t = 30/9.8 ≈ 3.061; h ≈ 47.9. (A) reports the input. [L2/L9] 31. (A) e^(0.04t) = 5/3 → t = ln(5/3)/0.04 ≈ 0.51083/0.04 ≈ 12.771. [L16] 32. (D) Patternless-and-small beats patterned: exponential more appropriate. [L12] 33. (C) 1.062⁹ ≈ 1.7184 → 850·1.7184 ≈ $1,461. [L12] 34. (B) Intersection of y = ln x and y = 3 − x: x ≈ 2.208 (ln 2.208 ≈ 0.792 ≈ 3 − 2.208 ✓). [L16] 35. (A) sin(π(3)/6) = sin(π/2) = 1 → 7 + 3 = 10. [L19] 36. (C) cos θ = 0.4: θ₁ = arccos 0.4 ≈ 1.159; twin 2π − 1.159 ≈ 5.124. (B) uses π + θ₁. [L21] 37. (B) cos(πt/40) = −10/18 → πt/40 = arccos(−5/9) ≈ 2.160 → t ≈ 40(2.160)/π ≈ 27.5 s. [L19/L21] 38. (D) n = 5 odd → 5 petals. [L23] 39. (A) (r(π/2) − r(0))/(π/2) = (2 − 4)/(π/2) = −4/π ≈ −1.273. [L24] 40. (B) Reference arcsin 0.72 ≈ 0.8038; QII: π − 0.8038 ≈ 2.338. [L20]
(a) (i) [1] (24 − 12)/(4 − 1) = 4 hundred tickets per week. (ii) [1] From week 1 to week 4, weekly sales grew by an average of about 400 tickets per week. (b) (i) [1] Rates over 1-week intervals: 8, 4, 0, −4 — decreasing by a constant 4 each week; a rate changing at a constant rate is the quadratic signature. (ii) [1] Δ² = −4 → a = −4/2 = −2; fitting: S(t) = −2t² + 14t (check: S(3) = −18 + 42 = 24 ✓, S(5) = −50 + 70 = 20 ✓; regression returns the same). (c) (i) [1] −2t² + 14t = 0 → 2t(7 − t) = 0 → t = 7: week 7. (ii) [1] S(8) = −128 + 112 = −16 — negative sales, impossible. The model breaks down past its zero at t = 7, and week 8 lies outside the observed weeks 1–5; the prediction should not be used.
(a) (i) [1] Midline (15.2 + 9.2)/2 = 12.2 h; amplitude (15.2 − 9.2)/2 = 3 h. (ii) [1] Max to min should be half a period: 355 − 172 = 183 = 366/2 ✓. (b) [2] b = 2π/366 = π/183; cosine peaking at t = 172: D(t) = 12.2 + 3 cos(π(t − 172)/183). [1 pt for a, d and structure; 1 pt for b and the shift. Any equivalent sine form earns full credit.] Check: D(172) = 15.2 ✓; D(355) = 12.2 + 3cos(π) = 9.2 ✓. (c) (i) [1] D(264) = 12.2 + 3 cos(π(92)/183) ≈ 12.2 + 3(−0.009) ≈ 12.17 h. (ii) [1] Day 264 is between the max (172) and min (355), so daylight is decreasing; D(264) ≈ 12.17 is essentially at the midline (12.2), and a sinusoid changes fastest at its midline crossings — so daylight is shortening at close to its maximum rate (roughly 2π·3/366 ≈ 0.05 h/day near the crossing).
(a) [2] Base 3: 3^(2(x−1)) = 3^(3x/2) [1] → 2x − 2 = 3x/2 → 4x − 4 = 3x → x = 4 [1]. (Check: 9³ = 729 = 27² ✓) (b) [2] ln(x²/(x + 4)) = 0 → x² = x + 4 → x² − x − 4 = 0 → x = (1 ± √17)/2 [1]. Need x > 0 (for ln x): reject (1 − √17)/2 < 0. x = (1 + √17)/2 [1]. (c) [2] 2 sin θ cos θ − sin θ = 0 → sin θ(2 cos θ − 1) = 0 [1] → sin θ = 0: θ = 0, π; cos θ = 1/2: θ = π/3, 5π/3. {0, π/3, π, 5π/3} [1]. (Dividing by sin θ would have lost 0 and π.)
(a) (i) [1] f(x) = 3(x − 2)(x + 2)/((x − 4)(x + 2)); domain x ≠ −2, 4. (ii) [1] At x = −2: multiplicities 1 = 1 → hole; simplified form 3(x − 2)/(x − 4) at −2: 3(−4)/(−6) = 2 → hole at (−2, 2). At x = 4: denominator only → vertical asymptote x = 4.
(b) (i) [1] Equal degrees, leading ratio 3x²/x² = 3: lim x→±∞ f(x) = 3; horizontal asymptote y = 3. (ii) [1] Crossing requires f(x) = 3: 3x² − 12 = 3(x² − 2x − 8) → −12 = −6x − 24 → x = −2 — but x = −2 is excluded from the domain (the hole). The graph never crosses y = 3; the only algebraic candidate is the missing point.
(c) [2] Using 3(x − 2)/(x − 4) near x = 4: numerator ≈ 3(2) = 6 > 0. As x → 4⁻, (x − 4) → 0⁻: lim x→4⁻ f(x) = −∞; as x → 4⁺, (x − 4) → 0⁺: lim x→4⁺ f(x) = +∞. [1 pt for correct limit statements, 1 pt for the sign justification]
Estimated AP score bands (estimates only — real cut points vary by year and are set after scaling):
| Composite | Estimated AP score |
|---|---|
| ≥ 0.70 | 5 |
| 0.60–0.69 | 4 |
| 0.47–0.59 | 3 |
| 0.35–0.46 | 2 |
| < 0.35 | 1 |
Your running multiple-choice score appears in the bar below. Self-score the free-response section with the rubrics in the answer key, then use the diagnostic table to target review.