Lesson 20: AP Exam Review — Strategy and PracticeAP

This final lesson (all CED units) is about converting eight units of calculus into marks under time pressure. The AP Calculus AB exam is as much a performance format as a knowledge test: its structure is fixed, its question archetypes repeat, and its scoring rewards specific habits. This lesson sets out the format precisely, gives a working strategy for each section, and then runs three mixed worked examples and a final practice set in exact exam style.

The format (know it cold)

• Section I — Multiple Choice (50% of the score): 45 questions, 105 minutes. Part A: 30 questions, 60 minutes, no calculator. Part B: 15 questions, 45 minutes, graphing calculator permitted.
• Section II — Free Response (50%): 6 questions, 90 minutes. Part A: 2 questions, 30 minutes, calculator permitted. Part B: 4 questions, 60 minutes, no calculator — and you may keep working on Part A answers during Part B, but without the calculator.
• Answers are scored 1–5 as a composite; there is no penalty for wrong multiple-choice answers, so never leave a blank.
• The calculator is for the four sanctioned tasks: graphing, finding zeros, computing derivatives at points, and evaluating definite integrals numerically. It is not a substitute for the setups that earn the points.

Scoring mechanics that change how you write

• FRQ points live in the setup. A correct integral expression with correct limits earns its point even if your arithmetic later slips; a bare number, however correct, earns nothing. Write the mathematics first, evaluate second.
• Justifications must cite the reason, in the language of theorems: "f has a relative maximum at x = 2 because f' changes sign from + to − there"; "by the IVT, since f is continuous on [1, 3] and f(1) < 0 < f(3)". The phrases *because*, *since*, and theorem names are the currency.
• Decimals: give answers correct to three decimal places, and carry unrounded values through intermediate steps (store them in the calculator). Rounding mid-problem is a self-inflicted error.
• Units on every applied FRQ part: rates get "units per time", accumulations get "units", and a missing unit forfeits the point for that part.
• Radian mode. Every trig computation on this exam is in radians. Check the mode before Section I, not after.

Worked Example 1: the rate-in/rate-out archetype (table, calculator-style)

Problem. Water flows into a tank at rate R(t) gal/min, tabulated at t = 0, 3, 6, 9, 12 as 10, 14, 12, 18, 15. Water drains at a constant 8 gal/min. The tank starts with 200 gallons. Estimate the amount in the tank at t = 12, and the average inflow rate over [0, 12].

Inflow by trapezoidal rule (Δt = 3):
  3 × (10/2 + 14 + 12 + 18 + 15/2) = 3 × 56.5 = 169.5 gal
Outflow: 8 × 12 = 96 gal
Tank at t = 12: 200 + 169.5 − 96 = 273.5 gal
Average inflow rate: 169.5/12 ≈ 14.125 gal/min

Solution notes. This is the FRQ 1 archetype almost every year: an accumulation estimate from a table, a competing constant rate, and an average-value question ((1/(b−a))∫ = total/time). Each piece is a separate point; none requires anything beyond Lessons 14–15 executed carefully with units attached.

Worked Example 2: derivative skills, mixed

Problem. (i) For f(x) = ln(x^2 + 1), find the tangent line at x = 1. (ii) If h(x) = x^3 + 2x and g is its inverse, find g'(3).

(i) f(1) = ln 2;  f'(x) = 2x/(x^2 + 1), f'(1) = 1.
    Tangent: y = ln 2 + 1·(x − 1).
(ii) g'(3) = 1/h'(a) where h(a) = 3. h(1) = 1 + 2 = 3, so a = 1.
    h'(x) = 3x^2 + 2, h'(1) = 5.  g'(3) = 1/5.

Solution notes. Two Unit 2–3 staples in thirty seconds each: a chain-rule tangent and the inverse-derivative routine of Lesson 8 (find the input first, then reciprocate the derivative there). Review-week goal: these should feel automatic, because FRQs embed them as single parts of larger questions.

Worked Example 3: particle motion, complete

Problem. A particle moves on a line with velocity v(t) = t^2 − 4t + 3 (m/s) for 0 ≤ t ≤ 4, and position s(0) = 5. Find the displacement, the total distance travelled, the position at t = 4, and when the speed is increasing.

Displacement: ∫ from 0 to 4 of v dt = [t^3/3 − 2t^2 + 3t] from 0 to 4 = 64/3 − 32 + 12 = 4/3 m.
v = (t − 1)(t − 3): negative on (1, 3), positive elsewhere.
Total distance: split at 1 and 3 → 4/3 + 4/3 + 4/3 = 4 m (computed in Lesson 15).
s(4) = 5 + 4/3 = 19/3 ≈ 6.333 m.
Speed: |v| increases when v and a = v' = 2t − 4 have the same sign:
  (0, 1): v > 0, a < 0 → decreasing
  (1, 2): v < 0, a < 0 → increasing
  (2, 3): v < 0, a > 0 → decreasing
  (3, 4): v > 0, a > 0 → increasing

Solution notes. Particle motion is the exam's favourite integrator of skills: one problem touches antiderivatives, sign analysis, and the speed-versus-velocity distinction. The discipline: direction changes where v = 0 and changes sign (t = 1, 3), and speed's behaviour is always a two-sign comparison.

A study plan for the final weeks

• Rebuild, don't reread. Redo one past FRQ per day under time (15 minutes each), marking against the published scoring guidelines — calibration against real rubrics is worth more than fresh content.
• Maintain an error log. Every mistake gets classified: algebra slip, concept gap, justification omitted, units, misread question. Patterns in the log tell you what to drill; most students find justification omissions dominate — cheap points to recover.
• Drill the no-calculator core weekly: derivative rules, standard integrals, the special limits, FTC1 forms. Speed here buys time for the hard parts elsewhere.
• One full mock two weeks out, one week out, and three days out; taper the final two days to light review of this course's Key idea boxes.

Common exam-day mistakes (the final list)

• Rounding intermediate values and compounding the error through later parts.
• Answering a different question: value instead of location, rate instead of amount, displacement instead of distance. Reread the prompt's verb before boxing anything.
• Justification-free extrema and IVT/MVT claims — the single largest pool of lost FRQ points nationally, per the Chief Reader reports.
• Leaving the last FRQ blank-ish while polishing an earlier one. Points are spread evenly; triage ruthlessly.

AP Exam Experience

1. (No calculator.) lim(x→0) (e^x − 1)/x =

• (A) 0
• (B) 1
• (C) e
• (D) The limit does not exist.

Answer: (B). Recognise the difference quotient: this is the derivative of e^x at x = 0, which is e^0 = 1. Students who reach for computation miss that half the review course is pattern recognition of exactly this form (compare Lesson 5, MCQ 2).

2. (No calculator.) ∫ from 1 to 3 of (2x − 1) dx =

• (A) 4
• (B) 5
• (C) 6
• (D) 8

Answer: (C). [x^2 − x] from 1 to 3 = (9 − 3) − (1 − 1) = 6. (Also the trapezoid area under the line: heights 1 and 5 over width 2 → (1 + 5)/2 × 2 = 6 — a useful geometric audit.)

3. If f(x) = ln(x^2 + 1), then f'(1) =

• (A) 1/2
• (B) ln 2
• (C) 2
• (D) 1

Answer: (D). f'(x) = 2x/(x^2 + 1), so f'(1) = 2/2 = 1 (Worked Example 2). (B) is f(1) — the eternal "evaluate the function when asked for the derivative" trap, placed here one last time.

Free-response question. A particle moves along a line with velocity v(t) = t^2 − 4t + 3 metres per second for 0 ≤ t ≤ 4, and its position at t = 0 is s(0) = 5 metres.

(a) Find the displacement of the particle and its total distance travelled over [0, 4]. (b) Find s(4). (c) Find all times t in (0, 4) at which the particle changes direction. Justify. (d) Is the particle's speed increasing or decreasing at t = 0.5? Justify.

Scoring guideline.

• (a) 2 points. 1 point for displacement: ∫ from 0 to 4 of v dt = 4/3 m (with the antiderivative or an equivalent computation shown). 1 point for total distance 4 m, earned only with evidence of splitting at the velocity zeros t = 1 and t = 3 (e.g., ∫|v| expressed as three pieces, or the three signed contributions 4/3, −4/3, 4/3 combined as absolute values).
• (b) 1 point. s(4) = s(0) + displacement = 5 + 4/3 = 19/3 ≈ 6.333 m (net change theorem used explicitly or implicitly).
• (c) 1 point. t = 1 and t = 3, justified by v(t) = (t − 1)(t − 3) changing sign at both (positive to negative at 1, negative to positive at 3). Merely solving v = 0 without the sign statement earns no point.
• (d) 1 point. At t = 0.5: v = 0.25 − 2 + 3 = 1.25 > 0 and a = 2(0.5) − 4 = −3 < 0. Opposite signs, so the speed is decreasing (the speed itself is 1.25 m/s; stating it is welcome but the sign comparison is the point).

Extension: the five FRQ archetypes and their first moves

Nearly every free-response question you will meet is one of five shapes. Recognising the shape in the first thirty seconds tells you the first move for each part.

Time architecture. Section II allows 15 minutes per question on average; the calculator pair (Questions 1–2) usually repays a slower, verification-rich pace, because the machine absorbs the arithmetic. In Part B, if a part (b) needs part (a)'s answer and (a) defeated you, *invent nothing* — state "using the value from (a)" symbolically and continue: readers award downstream points for correct method on a declared value. Finally, reserve the last five minutes of each section to audit the cheap things: radian mode, units present, the exact question asked re-read, no blank multiple-choice bubbles. These audits, nationally, are worth more than any single theorem in this course.