This closing lesson converts 14 lessons of content into exam-day execution, then stress-tests the whole course with a review set that mirrors the official topic distribution.
| Section | Questions | Time | Pace | Calculator |
|---|---|---|---|---|
| 1 | 25 | 50 min | 2.0 min/q | Online graphing calculator (non-CAS TI-84 Plus CE built into the exam software) — needed for only some questions |
| 2 | 23 | 40 min | ~1.7 min/q | None |
Approximately 48 questions total, including some unscored pretest items you can't identify — so treat every question as scored. Content follows the six official weights: Algebraic Expressions, Equations and Inequalities 20%; Functions: Concept, Properties and Operations 15%; Representations of Functions: Symbolic, Graphical and Tabular 30%; Analytic Geometry 10%; Trigonometry and Its Applications 15%; Functions as Models 10%. One asterisk on those numbers: trig permeates the other categories, so expect trig content in roughly a third of the questions, not 15%.
Scoring: your raw score is simply the number of correct answers — there is no penalty for wrong answers. The typical credit-granting threshold is a scaled score of 50. Practically: never leave anything blank, and remember you don't need anywhere near 100% — a comfortable majority of 48 questions clears 50 on the scale. That changes pacing psychology: banking 40 solid answers beats agonizing over 48.
The embedded calculator is a graphing tool, not a co-author. It earns its keep on exactly these tasks:
Everything else — identities, factoring, special angles, transformations — is faster by hand, and Section 2 requires it by hand. Practice both modes; College Board offers a free 30-day trial of the same calculator, and walking in without having driven it is a self-inflicted handicap.
Radian-mode discipline. The single costliest settings error: evaluating trig in the wrong angle mode. A quick self-test when you start Section 1 — ask for sin(30). If the answer is −0.988, the calculator is in radian mode (it computed the sine of 30 radians); if 0.5, degree mode. Match the mode to the problem's units: models like h(t) = 25 − 20cos(πt/30) are built in radians; a problem stated in degrees (tan 35°) needs degree mode or a manual conversion. Know how to check and switch in one motion.
Multiple choice, five options A–E. Elimination is worth real points: wrong options are built from known misconceptions (sign flips, swapped formulas, off-by-one-step arithmetic), so knocking out two or three is usually fast. With no wrong-answer penalty, an educated guess among the survivors is always correct strategy — and so is a blind guess in the final seconds. Blank earns nothing; guessed earns 20% at worst.
Numeric entry. Some items say "Type your answer into the box below," often with a rounding instruction ("round to the nearest tenth"). Rules of engagement: carry full precision through the computation and round only at the end; obey the stated rounding exactly (8.1, not 8.11 or 8); re-read the question to confirm what quantity goes in the box (years, not the final balance).
Pacing. Two passes per section. First pass: answer everything answerable in under 90 seconds, flag the rest. Second pass: flagged items with the remaining time. Never park four minutes on one question in a 2-minutes-per-item budget — every question carries equal weight, and question 24 may be a gift. In the last minute, fill every remaining blank with a guess.
| Weight | Topic | Core moves (lesson) |
|---|---|---|
| 20% | Algebraic expressions, equations, inequalities | exponent/radical rules, factoring, quadratics, absolute-value and rational inequalities, log equations with extraneous-root checks |
| 15% | Functions: concept, properties, operations | domain/range, composition, inverses, even/odd, one-to-one |
| 30% | Representations: symbolic ↔ graphical ↔ tabular | transformations, matching graphs to formulas, reading tables, zeros/intercepts/asymptotes/end behavior |
| 10% | Analytic geometry | circles (complete the square), parabolas, ellipses, hyperbolas — recognizing each from its equation |
| 15% | Trigonometry | unit circle, identities, equations on an interval, triangle laws |
| 10% | Functions as models | exponential growth/decay, interest, sinusoidal models, parameter interpretation |
The 12 questions below follow those proportions (2–3 algebra, 2 functions, 3 representations, 1 analytic geometry, 2 trig-weighted, 1 model — with the first two testing exam craft itself). Time the set at 24 minutes to rehearse the 2-minute pace.
Q1. D. In radian mode the calculator reads 30 as 30 radians; 30 radians ≈ 4.775 full turns lands at an angle whose sine is about −0.988. The tool did exactly what it was told — the mode mismatched the problem's units. A) the calculator is working correctly; a surprising trig value should trigger a mode check, not a reset. B) 30° is a first-quadrant angle with sine +0.5; there is no "fourth quadrant" reading of it. C) cos(30°) ≈ 0.866 — swapping functions doesn't explain a negative output. E) sine is positive for all acute angles. Fix: sin(30) is the built-in mode test — −0.988 means radians, 0.5 means degrees; run it before the first trig question.
Q2. A. CLEP raw scores count correct answers only; a wrong answer and a blank both contribute zero, so guessing has positive expected value (at worst a 1-in-5 chance of a point) and zero downside. B) describes a formula-scored exam with a guessing penalty — CLEP has none. C) and D) each surrender a free 20% chance on the abandoned question for no benefit. E) unanswered questions are not excluded; they simply earn no point, exactly like wrong answers. Fix: on the CLEP exam, never leave a blank — in the final minute, fill every remaining answer.
Q3. D. |2x − 3| < 5 means −5 < 2x − 3 < 5; adding 3 gives −2 < 2x < 8, so −1 < x < 4. A) keeps only the right-hand inequality. B) keeps only the left-hand one. C) solves the "greater than" version — |expr| > 5 splits outward into two rays. E) divides before adding 3, mangling the order of operations on the compound inequality. Fix: |expr| < k traps the expression between −k and k (one connected interval); |expr| > k splits outside (two rays).
Q4. E. Combine: log₂[x(x − 2)] = 3 → x(x − 2) = 2³ = 8 → x² − 2x − 8 = 0 → (x − 4)(x + 2) = 0 → x = 4 or x = −2. But x = −2 makes both original logs undefined — extraneous. Only x = 4 survives (check: log₂4 + log₂2 = 2 + 1 = 3 ✓). A) keeps the extraneous root, skipping the domain check. B) keeps only the extraneous root. C) solves x(x − 2) = 3·... treating the 3 as a factor rather than an exponent, or stops at the log argument's zero. D) reports 2³ = 8 itself — the product of the arguments, not x. Fix: after solving any log equation, substitute every candidate into the original equation and discard any that make a log argument ≤ 0.
Q5. B. Inside-out: g(3) = 9, then f(9) = 2(9) − 1 = 17. A) computes the reversed composition g(f(3)) = g(5) = 25 — order matters in composition. C) computes 2·(g(3)·... doubling 17 plus a slip, or f(g(3)) with g(3) mistaken as 18. D) evaluates f(3)·... = 2(3) − 1 = 5 then doubles and adds 1 → 11 comes from f(f(3)). E) stops at g(3) = 9, never applying f. Fix: evaluate compositions inside-out, and read f(g(x)) as "g first, then f."
Q6. C. Set y = (x + 3)/2, swap and solve: x = (y + 3)/2 → 2x = y + 3 → y = 2x − 3. Check: f(2x − 3) = (2x − 3 + 3)/2 = x ✓. A) is the reciprocal 1/f(x), confusing "inverse function" with "multiplicative inverse." B) undoes the operations in the wrong order (subtract 3, then divide). D) has the right shape but the wrong sign — it fails the composition check: f(2x + 3) = x + 3. E) applies the original operations in reverse direction but not reverse order. Fix: to invert, undo operations in reverse order (last applied, first undone), then verify f(f⁻¹(x)) = x.
Q7. C. An upward parabola with vertex (h, k) = (2, −1) is y = (x − 2)² − 1; it checks against every plotted feature: zeros at 1 and 3 ✓, y-intercept (0−2)² − 1 = 3 ✓. A) shifts left 2 (vertex at (−2, −1)) — the sign-of-h misread. B) puts the vertex at (2, +1), above the x-axis, so the graph would never cross it. D) opens downward with maximum −1, entirely below the x-axis. E) has vertex (−2, 1) and no x-intercepts. Fix: vertex form y = a(x − h)² + k — the h keeps its sign inside the parentheses reversed; confirm a candidate equation against two features of the graph (vertex plus one intercept).
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| g(x) | 3 | 5 | 1 | 2 | 4 |
The value of g(g(1)) is
Q8. A. Inside-out from the table: g(1) = 3, then g(3) = 1. B) reads g(4) — the value in the wrong column after a first-step slip. C) stops at g(1) = 3 without applying g again. D) computes g(g(2)) = g(5) = 4, starting from the wrong input. E) reads g(2) = 5 for the first step, an adjacent-column error. Fix: table compositions are two separate look-ups — write the intermediate value down, then look it up as a fresh input.
Q9. B. Factor: x³ − 4x = x(x² − 4) = x(x − 2)(x + 2); zeros at x = −2, 0, 2, and all three are x-intercepts. A) drops the negative root — even though odd functions pair their nonzero roots symmetrically. C) keeps a single root, as if the factors repeated. D) forgets the factored-out x, losing the intercept at the origin. E) reads the coefficient 4 as a root without taking its square root. Fix: factor completely — GCF first (the x is an intercept too), then difference of squares; every distinct real zero is an x-intercept.
Q10. E. Equal coefficients on x² and y² (both 1) with no xy term is the circle signature; completing the square confirms it's real: (x − 2)² + (y + 3)² = 12 + 4 + 9 = 25, a circle of radius 5 centered at (2, −3). A) has a minus between the squared terms — a hyperbola. B) has unequal coefficients (1 and 4) — an ellipse. C) has only one squared variable — a parabola. D) is xy = constant — a rotated hyperbola. Fix: circle = both squares present, same sign, equal coefficients; complete the square to verify the right side ends positive.
Q11. B. Let u = 2x; as x runs over [−π, π], u runs over [−2π, 2π]. cos u = 0 at u = ±π/2 and ±3π/2 — four values in the window, each giving one x (x = ±π/4, ±3π/4). A) counts only the base period's two solutions, forgetting the doubled window. C) miscounts by treating an endpoint as a solution boundary (cos(±2π) = 1 ≠ 0, so nothing is gained or lost at the ends). D) stretches the window by 3 instead of 2. E) doubles the correct count, stretching twice. Fix: for f(bx) = k on an interval, multiply the interval by b, count solutions there (cosine hits each level twice per 2π), and map one-to-one back to x.
Q12. A. A = Pe^(rt) = 2000e^(0.06·5) = 2000e^0.3 ≈ 2000(1.34986) ≈ $2,699.72 → $2,700 to the nearest dollar. B) is simple interest: 2000(1 + 0.06·5) = 2,600 — no compounding at all. C) is annual compounding, 2000(1.06)⁵ ≈ 2,676 — the wrong compounding scheme for "continuously." D) applies one year of growth (2000 × 1.06), ignoring the 5. E) adds a single 3% half-step, misreading rt = 0.3 as 3%... of one year. Fix: "compounded continuously" always means A = Pe^(rt) — compute rt first, exponentiate once, and round only at the end.
Q1. D. In radian mode the calculator reads 30 as 30 radians; 30 radians ≈ 4.775 full turns lands at an angle whose sine is about −0.988. The tool did exactly what it was told — the mode mismatched the problem's units. A) the calculator is working correctly; a surprising trig value should trigger a mode check, not a reset. B) 30° is a first-quadrant angle with sine +0.5; there is no "fourth quadrant" reading of it. C) cos(30°) ≈ 0.866 — swapping functions doesn't explain a negative output. E) sine is positive for all acute angles. Fix: sin(30) is the built-in mode test — −0.988 means radians, 0.5 means degrees; run it before the first trig question.
Q2. A. CLEP raw scores count correct answers only; a wrong answer and a blank both contribute zero, so guessing has positive expected value (at worst a 1-in-5 chance of a point) and zero downside. B) describes a formula-scored exam with a guessing penalty — CLEP has none. C) and D) each surrender a free 20% chance on the abandoned question for no benefit. E) unanswered questions are not excluded; they simply earn no point, exactly like wrong answers. Fix: on the CLEP exam, never leave a blank — in the final minute, fill every remaining answer.
Q3. D. |2x − 3| < 5 means −5 < 2x − 3 < 5; adding 3 gives −2 < 2x < 8, so −1 < x < 4. A) keeps only the right-hand inequality. B) keeps only the left-hand one. C) solves the "greater than" version — |expr| > 5 splits outward into two rays. E) divides before adding 3, mangling the order of operations on the compound inequality. Fix: |expr| < k traps the expression between −k and k (one connected interval); |expr| > k splits outside (two rays).
Q4. E. Combine: log₂[x(x − 2)] = 3 → x(x − 2) = 2³ = 8 → x² − 2x − 8 = 0 → (x − 4)(x + 2) = 0 → x = 4 or x = −2. But x = −2 makes both original logs undefined — extraneous. Only x = 4 survives (check: log₂4 + log₂2 = 2 + 1 = 3 ✓). A) keeps the extraneous root, skipping the domain check. B) keeps only the extraneous root. C) solves x(x − 2) = 3·... treating the 3 as a factor rather than an exponent, or stops at the log argument's zero. D) reports 2³ = 8 itself — the product of the arguments, not x. Fix: after solving any log equation, substitute every candidate into the original equation and discard any that make a log argument ≤ 0.
Q5. B. Inside-out: g(3) = 9, then f(9) = 2(9) − 1 = 17. A) computes the reversed composition g(f(3)) = g(5) = 25 — order matters in composition. C) computes 2·(g(3)·... doubling 17 plus a slip, or f(g(3)) with g(3) mistaken as 18. D) evaluates f(3)·... = 2(3) − 1 = 5 then doubles and adds 1 → 11 comes from f(f(3)). E) stops at g(3) = 9, never applying f. Fix: evaluate compositions inside-out, and read f(g(x)) as "g first, then f."
Q6. C. Set y = (x + 3)/2, swap and solve: x = (y + 3)/2 → 2x = y + 3 → y = 2x − 3. Check: f(2x − 3) = (2x − 3 + 3)/2 = x ✓. A) is the reciprocal 1/f(x), confusing "inverse function" with "multiplicative inverse." B) undoes the operations in the wrong order (subtract 3, then divide). D) has the right shape but the wrong sign — it fails the composition check: f(2x + 3) = x + 3. E) applies the original operations in reverse direction but not reverse order. Fix: to invert, undo operations in reverse order (last applied, first undone), then verify f(f⁻¹(x)) = x.
Q7. C. An upward parabola with vertex (h, k) = (2, −1) is y = (x − 2)² − 1; it checks against every plotted feature: zeros at 1 and 3 ✓, y-intercept (0−2)² − 1 = 3 ✓. A) shifts left 2 (vertex at (−2, −1)) — the sign-of-h misread. B) puts the vertex at (2, +1), above the x-axis, so the graph would never cross it. D) opens downward with maximum −1, entirely below the x-axis. E) has vertex (−2, 1) and no x-intercepts. Fix: vertex form y = a(x − h)² + k — the h keeps its sign inside the parentheses reversed; confirm a candidate equation against two features of the graph (vertex plus one intercept).
Q8. A. Inside-out from the table: g(1) = 3, then g(3) = 1. B) reads g(4) — the value in the wrong column after a first-step slip. C) stops at g(1) = 3 without applying g again. D) computes g(g(2)) = g(5) = 4, starting from the wrong input. E) reads g(2) = 5 for the first step, an adjacent-column error. Fix: table compositions are two separate look-ups — write the intermediate value down, then look it up as a fresh input.
Q9. B. Factor: x³ − 4x = x(x² − 4) = x(x − 2)(x + 2); zeros at x = −2, 0, 2, and all three are x-intercepts. A) drops the negative root — even though odd functions pair their nonzero roots symmetrically. C) keeps a single root, as if the factors repeated. D) forgets the factored-out x, losing the intercept at the origin. E) reads the coefficient 4 as a root without taking its square root. Fix: factor completely — GCF first (the x is an intercept too), then difference of squares; every distinct real zero is an x-intercept.
Q10. E. Equal coefficients on x² and y² (both 1) with no xy term is the circle signature; completing the square confirms it's real: (x − 2)² + (y + 3)² = 12 + 4 + 9 = 25, a circle of radius 5 centered at (2, −3). A) has a minus between the squared terms — a hyperbola. B) has unequal coefficients (1 and 4) — an ellipse. C) has only one squared variable — a parabola. D) is xy = constant — a rotated hyperbola. Fix: circle = both squares present, same sign, equal coefficients; complete the square to verify the right side ends positive.
Q11. B. Let u = 2x; as x runs over [−π, π], u runs over [−2π, 2π]. cos u = 0 at u = ±π/2 and ±3π/2 — four values in the window, each giving one x (x = ±π/4, ±3π/4). A) counts only the base period's two solutions, forgetting the doubled window. C) miscounts by treating an endpoint as a solution boundary (cos(±2π) = 1 ≠ 0, so nothing is gained or lost at the ends). D) stretches the window by 3 instead of 2. E) doubles the correct count, stretching twice. Fix: for f(bx) = k on an interval, multiply the interval by b, count solutions there (cosine hits each level twice per 2π), and map one-to-one back to x.
Q12. A. A = Pe^(rt) = 2000e^(0.06·5) = 2000e^0.3 ≈ 2000(1.34986) ≈ $2,699.72 → $2,700 to the nearest dollar. B) is simple interest: 2000(1 + 0.06·5) = 2,600 — no compounding at all. C) is annual compounding, 2000(1.06)⁵ ≈ 2,676 — the wrong compounding scheme for "continuously." D) applies one year of growth (2000 × 1.06), ignoring the 5. E) adds a single 3% half-step, misreading rt = 0.3 as 3%... of one year. Fix: "compounded continuously" always means A = Pe^(rt) — compute rt first, exponentiate once, and round only at the end.