AP criterion-linked repair and changed cases
These twenty public retries, together with the existing STAT-1 inference worksheet, provide a selected-criterion repair destination for every free response in the four current forms. They cover selected criteria in Precalculus, AB, BC and Statistics. The original forms remain unchanged. This is a bounded criterion matrix, not a complete current CED learning-objective/essential-knowledge audit. Unit 4 Precalculus enrichment is not counted as examined material. The existing STAT-1 inference repair worksheet remains available separately.
Every free response has a repair destination
This finite matrix covers the twenty authored free responses, not every CED objective or every rubric criterion. Each row states what its changed case does not independently reassess. The original rubric remains authoritative for the original response. Use a reviewer to decide whether evidence transfers; the page makes no automatic diagnosis.
PC-1.1 · pc1-fr-01
Function Concepts: table and graph composition, numerical inverse and composite domain
Original response and rubric· PC-1.1: domain reasoning
The retry changes analytic composite domains. It does not independently reassess the original discrete-table/graph composition or numerical inverse criteria.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Gives both lookups and the value 3 for G(T(3)).
- Gives both lookups and the value 1 for T(G(1)).
Part b
- Uses T(4)=3, the correct equation and a valid bracket such as (2,3).
- Obtains 2.870 within 0.001 and justifies existence and uniqueness by continuity and strict increase.
Part c
- States both restrictions and solves the correct boundary equation using a valid bracket such as (3,4).
- Gives the complete closed interval with upper endpoint within 0.001 and explains both endpoint inclusion and completeness.
PC-1.1 · pc1-fr-02
Nonperiodic modeling: parameter system, contextual average rate and extrapolation
Original response and rubric· PC-1.1: unknown-offset model and validation
The retry supplies exact sequential measurements, not the original calculator system. Calculator procedure and the original model-specific predictions still require the original response.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Writes all three equations and obtains q=1/2 from the ratio of consecutive changes.
- Obtains k=10 and A=64 and gives the complete model with the two-hour interpretation.
Part b
- Evaluates the appropriate endpoints and divides their difference by four hours.
- Obtains −8.485 within 0.001 with °C/hour units and a correct finite-interval interpretation.
Part c
- Gives the equation and time 6.385 within 0.001 and explains uniqueness through strict decrease.
- Distinguishes extrapolation from the proposed domain and supplies an independent contextual check without claiming guaranteed accuracy.
PC-1.1 · pc1-fr-03
Periodic modeling: two-cycle graph, parameter recovery and rate/concavity reasoning
Original response and rubric· PC-1.1: graph recovery
The retry targets parameter/phase recovery. Original average-rate and concavity comparisons are not separately reassessed; held-out validation is an extension.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Correctly gives all three coordinates A, B and C.
- Correctly gives both coordinates D and E.
Part b
- Gives all three parameters with appropriate height/time units.
- Gives a full correct sinusoidal formula and explains its phase using a landmark.
Part c
- Correctly distinguishes increasing height from decreasing positive equal-width average rates.
- Identifies concave down and justifies the whole-interval chord/rate behavior from the stipulated sinusoid, not only height sign or two sampled rates.
PC-1.1 · pc1-fr-04
Symbolic manipulation: exponential and trigonometric equations, equivalent logarithmic expressions
Original response and rubric· PC-1.1: symbolic equations and original domains
The retry changes exponential, trigonometric and logarithmic cases. Review original bundled criteria separately; this is not a score conversion.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Obtains the correct linear equation in x by taking logarithms.
- Gives ln15/ln(9/5) and explains uniqueness and the validity of taking logarithms.
Part b
- Uses a valid identity and obtains both sine values 1/2 and −1.
- Gives exactly π/6, 5π/6 and 3π/2 on the stated interval.
Part c
- Determines the original domain z>−1, z≠2 and obtains the single logarithm by a valid product rule.
- Obtains 2ln|z−2| and explains both the absolute value and the retained original restriction.
AB-1 · ab-fr-01
Numerical accumulation, rates and global optimization
Original response and rubric· AB-1: net-rate maximum and units· AB-1: changed accumulation setup
Incoming-versus-net distinctions and second-derivative interpretation are not all recomputed in the graph retry. Moving bounds are an extension, not an original rubric requirement.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Sets up the integral of E over [0,6], without subtracting departures or adding the initial attendance.
- Evaluates to 132.999 people within tolerance, with an appropriate interpretation.
Part b
- Solves E=L on the stated interval, obtaining r≈3.854 hours.
- Uses strict decrease and the endpoint signs of N, or another valid complete sign/endpoint argument, to establish the absolute maximum.
- Sets up initial attendance plus the net-rate integral to r and obtains approximately 159.973 people; accept correct evaluation using a carried root.
Part c
- Uses net change divided by six, or equivalently the integral of E−L divided by six.
- Obtains approximately 5.166 people/hour and distinguishes net change from incoming flow; allow a correctly carried part (a) entry total.
Part d
- Differentiates both rate functions and obtains −5exp(−1/2)−3, approximately −6.033.
- Gives people/hour² and interprets the negative sign as a decreasing attendance-change rate, not necessarily decreasing attendance.
AB-1 · ab-fr-02
Numerical motion, an integral equation and theorem justification
Original response and rubric· AB-1: distance, arrival and speed
The changed polynomial has an elementary integral. Original non-elementary numerical quadrature and retained calculator output still need review.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Justifies positivity and sets up ∫₀³v, or directly sets up ∫₀³|v| with its correct evaluation.
- Obtains approximately 9.774 meters.
Part b
- Sets up accumulated displacement equal to 6, retaining the initial position.
- Gives a valid sign-changing bracket and explains uniqueness through continuous, strictly increasing position.
- Obtains T≈1.704 seconds; accept correct numerical solution of a meaningful carried accumulation equation.
Part c
- Differentiates velocity correctly and establishes a(1)>0.
- Combines the positive velocity and acceleration signs to conclude increasing speed.
Part d
- Divides displacement by three to obtain approximately 3.258 meters/second; accept a correct carried part (a) value.
- States the applicable continuity/differentiability hypotheses and the mean value theorem conclusion at an interior time.
AB-1 · ab-fr-03
Supplied integrand graph, signed area and FTC
Original response and rubric· AB-1: signed integrand and concavity
The changed graph explicitly rechecks signed areas, endpoint extrema, concavity, derivative existence and composite FTC.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Uses the canceling areas +2 and −2 and obtains H(4)=1.
- Uses the negative triangle of area −1 and obtains H(5)=0, or follows a carried H(4) consistently.
Part b
- Uses the sign of g to identify increasing/decreasing intervals and interior candidates 2 and 5.
- Checks both endpoints together with the interior candidates.
- States maximum 3 at 2 and minimum 0 at 5; accept consistently carried accumulation values with correct comparisons.
Part c
- Gives concave down on (0,4), concave up on (4,6), and the inflection input 4.
- Explains nonexistence of H″(4) using unequal one-sided slopes of g without denying existence of H′(4).
Part d
- Applies FTC and chain rule to obtain 2xg(x²).
- Reads g(3)=−1 and obtains −2√3.
AB-1 · ab-fr-04
Implicit derivatives, tangent geometry and related rates
Original response and rubric· AB-1: implicit tangent and radial motion
The changed case targets first derivatives, related rates and horizontal-tangent regularity. It does not separately reassess second-derivative/concavity part b.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Implicitly differentiates, retaining both terms from xy, and obtains the derivative formula.
- Evaluates the slope and gives the tangent line through (2,3).
Part b
- Obtains a valid second-derivative relation, including the 2(y′)² term.
- Substitutes consistently to obtain −57/256; accept correct arithmetic from a carried first derivative.
- Uses the negative continuous second derivative to conclude local concavity down.
Part c
- Obtains dy/dt=−21/8 and differentiates the distance relation, or gives an equivalent valid combined rate equation.
- Evaluates to −15/(8√13) meters/second with a consistent sign; allow a meaningful carried tangent slope.
Part d
- Solves the numerator-zero condition together with the curve and gives both points.
- Checks the nonzero denominator at those points, rather than treating numerator zero alone as sufficient.
AB-1 · ab-fr-05
Separation, local behavior and tangent-line approximation
Original response and rubric· AB-1: separation and tangent error
The changed case rechecks separation, initial condition, initial local behavior, tangent approximation and interval-wide concavity.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Computes y′(0)=0 and differentiates along the solution to obtain y″=1+y+ty′.
- Obtains y″(0)=2 and explains why the horizontal initial tangent is not evidence of a constant solution.
Part b
- Separates variables and integrates to ln|1+y|=t²/2+C.
- Uses the initial condition and the appropriate positive branch to determine the constant.
- Obtains y=2exp(t²/2)−1 and retains a solution valid for the requested interval.
Part c
- Builds the tangent line using both y(1) and the slope supplied by the differential equation; accept a consistently carried part (b) value.
- Evaluates the line at 9/10 to obtain (9/5)√e−1.
Part d
- Establishes y″=(1+y)(1+t²)>0 throughout the relevant interval, not merely at one endpoint.
- Uses concavity up to conclude that the tangent-line approximation is an underestimate.
AB-1 · ab-fr-06
Area, cross sections, washers and the mean value theorem
Original response and rubric· AB-1: widths, radii and integral mean value
The simpler triangular base retains section dimensions, changed-axis radii and mean-value reasoning. It does not independently calibrate polynomial integration difficulty.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Justifies nonnegative height and sets up upper minus lower over [0,2].
- Evaluates the area as 4/3.
Part b
- Builds the cross-section area from the two specified dimensions and integrates with respect to x.
- Obtains volume 8/3.
Part c
- States the hypotheses and uses the endpoint average to invoke the mean value theorem.
- Solves for both values 1±1/√3 and checks that both lie in the open interval.
Part d
- Identifies outer radius 1+2x−x² and inner radius 1, both measured from the rotation axis.
- Sets up π times the integral of outer-radius squared minus inner-radius squared over [0,2].
- Evaluates to 56π/15; allow consistent carried polynomial arithmetic without repeating an earlier arithmetic deduction.
BC-1 · bc-fr-01
Planar motion, numerical length and changing speed
Original response and rubric· BC-1: vector speed is a norm
The changed case retains vector accumulation, norm length, path slope and speed derivative; elementary exact integrals do not replace original numerical quadrature evidence.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Sets up both component integrals and adds the corresponding initial coordinates.
- Obtains position approximately (5.703,6.990) meters.
Part b
- Sets up the integral of the Euclidean speed over [0,3].
- Evaluates the distance as approximately 8.557 meters.
- Distinguishes both proposed alternatives from traveled length using the component norm and changing direction.
Part c
- Forms the ratio of the vertical and horizontal velocity components at t=1.
- Checks that x′(1) is nonzero and states the resulting slope, rather than a speed or an acceleration ratio.
Part d
- Differentiates both velocity components and establishes a negative dot product at t=1.
- Uses positive speed and the sign of v·a to conclude decreasing speed.
BC-1 · bc-fr-02
Accumulation with exponential weighting, FTC and numerical differential equations
Original response and rubric· BC-1: weighted accumulation and Euler
The changed constant-input case retains weighted accumulation, FTC, unique crossing and Euler comparison; original non-elementary quadrature remains a separate criterion.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Applies the product rule and FTC to obtain R−Q/4.
- Interprets Q/4 as a removal rate and gives consistent units.
Part b
- Sets up the weighted integral and retains both the initial 10 and the outside exponential factor.
- Obtains approximately 15.911 milligrams.
Part c
- States the equation Q(t)=15 and supplies a valid bracket such as (2,3) using endpoint amounts.
- Obtains approximately 2.218 minutes.
- Justifies uniqueness, for example with the strictly increasing weighted difference W; merely exhibiting a plotted crossing is insufficient.
Part d
- Uses the starting amount and derivative to obtain the one-step Euler estimate 13.5 mg.
- Computes Q(1)≈12.979 mg and correctly identifies this estimate as an overestimate.
BC-1 · bc-fr-03
Polar derivatives, geometric region selection and arc length
Original response and rubric· BC-1: signed radius and unique region
The retry transfers geometric region/traversal selection; signed-radius handling and double traversal are extensions beyond the original positive-radius curve. Original tangent derivatives, regularity and arc-length algebra are not all separately reassessed.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Computes the point and both coordinate derivatives using product rules.
- Divides y′ by the nonzero x′ to obtain −3√3.
Part b
- Identifies 0≤θ≤π and inner/outer radii 2 and 2+sinθ.
- Sets up half the integral of the difference of squared radii.
- Evaluates the area as 4+π/4.
Part c
- Sets up the parametric speed integral from the two Cartesian coordinate derivatives.
- Shows the cancellation or equivalent algebra giving √(5+4sinθ), with bounds 0 and π.
Part d
- Finds both angles and their Cartesian points without duplicating the repeated condition.
- Checks x′≠0 at each and concludes both tangents are horizontal.
BC-1 · bc-fr-04
Power-series endpoints, Taylor construction and two remainder arguments
Original response and rubric· BC-1: approximation versus guarantee
The retry targets the separate Taylor guarantee. Original power-series endpoint classification, geometric-series identification and alternating-series error reasoning remain on the original form.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Uses a valid test to obtain radius 3 and the open interval (−1,5).
- Tests x=−1 separately and identifies divergence.
- Tests x=5 separately and identifies conditional convergence, giving (−1,5].
Part b
- Differentiates and sums the geometric series to obtain 1/(x+1).
- Uses A(2)=0 to obtain ln((x+1)/3), not an undetermined integration constant.
Part c
- Evaluates the three-term polynomial to obtain 25/162.
- Justifies decreasing terms tending to zero and obtains the bound 1/5184.
Part d
- Derives or correctly states the fourth derivative and establishes the interval-wide bound 2/27.
- Applies the degree-three remainder formula to obtain 1/5184 and explains the need for interval control.
BC-1 · bc-fr-05
Logistic separation, Euler steps, inflection and improper accumulation
Original response and rubric· BC-1: logistic branch and improper deficit
The changed logistic case rechecks updated Euler slopes, initial-condition separation, sign-changing concavity and limiting improper accumulation.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Uses the initial slope 8/3 to obtain the first numerical value 10/3.
- Uses the updated numerical value for the second slope and obtains 130/27.
Part b
- Separates and integrates with the correct logarithmic factors and signs.
- Uses N(0)=2 to determine the constant or ratio.
- Obtains N=6/(1+2e^(−2t)) on the required branch.
Part c
- Uses the sign of N″ to establish a concavity change at N=3, not just a zero of N″.
- Solves for t=(ln2)/2.
Part d
- Writes a finite-upper-bound integral and obtains the correct primitive or equivalent substitution.
- Takes the limit to obtain 3ln3, rather than substituting infinity as an ordinary endpoint.
BC-1 · bc-fr-06
Integration by parts, partial fractions and integral mean value
Original response and rubric· BC-1: rational tails and integration by parts
The changed integrals recheck parts/division, partial fractions, improper limits and mean-value existence versus uniqueness.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Chooses valid integration-by-parts terms and obtains the boundary term and remaining integral.
- Uses x²/(1+x)=x−1+1/(1+x) and integrates the remainder correctly.
- Combines the terms to obtain 1/4.
Part b
- Finds the correct partial-fraction decomposition and primitive.
- Evaluates the endpoints to obtain ln(4/3).
Part c
- Expresses the improper integral as a limit of a finite endpoint expression without subtracting infinities.
- Evaluates the limit and states convergence to ln(3/2).
Part d
- Uses continuity and the integral average (or an antiderivative MVT) to justify an interior c.
- Establishes strict decrease to justify uniqueness, rather than claiming that the existence theorem alone supplies it.
STAT-1 · stat-fr-01
Formulate investigative questions and design ethical collection
Original response and rubric· STAT-1: design and causal scope
The changed design case targets assignment, experimental unit and causal/generalization scope. Original investigative-question wording, proportional stratified invitations, blocked assignment details and ethical-protection criteria remain separate.
Inspect every original criterion, including criteria not selected by the retry
Part a
- States a measurable, variable survey question with a defined borrower population and response variable.
- States a distinct comparative causal question identifying the reminder treatments, participant group and measurable return outcome.
Part b
- Allocates 30 invitations to North and 90 to South.
- Describes random selection of distinct borrower IDs within each branch rather than convenience selection or choosing whole branches.
- Provides a defensible nonresponse plan and explicitly limits representativeness claims despite the randomized invitations.
Part c
- Gives a genuinely randomized assignment within each branch with 12/12 and 36/36 treatment allocations.
- Defines the standard-wording comparison, holds other reminder procedures fixed, and specifies one prespecified binary outcome per participant.
- Explains how branch blocking addresses a plausible source of outcome variation without confusing blocks with randomly selected treatment groups.
Part d
- States an actionable consent, participant-protection or confidentiality provision tied to this study.
- Distinguishes randomized-treatment causal evidence within the study from unsupported automatic generalization to all borrowers after consent/nonresponse.
STAT-1 · stat-fr-02
Analyze regression summaries, predictions and residuals in context
Original response and rubric· STAT-1: residuals and extrapolation
The changed regression case targets residual sign and extrapolation. Original full least-squares/r² calculation and paired-data interpretation remain separate.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Describes a negative, approximately linear association in the observed batches.
- Identifies the batch as the paired observational unit and explains why independent column reordering changes the relationship.
Part b
- Obtains slope −1.25 and intercept 20.5 with a valid fitted-line calculation, such as the centered sums.
- Interprets the slope as a 1.25-minute decrease in predicted time per additional 1 m³/min of airflow, without a causal assertion.
- Calculates r²≈0.947 (accept within 0.002) and interprets the proportion of observed y variation accounted for by the fitted line.
Part c
- Predicts 11.75 minutes at airflow 7 m³/min.
- Uses observed minus predicted to obtain batch D’s residual +1.5 minutes.
- Explains the positive residual as longer observed drying time than predicted, with the correct context and units.
Part d
- Calculates −2 minutes and rejects this extrapolated physical prediction using the observed airflow range and nonnegative duration.
- Explains why the observational design and possible confounding prevent a causal claim, regardless of r².
STAT-1 · stat-fr-03
Choose and execute independent-mean inference with calculated references
Original response and rubric· Existing four-case inference worksheet
The existing four-case inference worksheet targets design, Welch uncertainty, directional tail decisions and scope. It is public practice, not automatic diagnosis or mastery.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Defines the oriented population mean difference and states the equality null and greater alternative.
- Justifies independent groups and checks both 1% sampling fractions, with the supplied normal-shape assumption.
Part b
- Uses separate variance contributions and Welch df, obtaining SE≈1.76937 and df≈57.8306 (accept within 0.01 and 0.1 respectively).
- Calculates the 95% critical t≈2.00184 and interval approximately [0.758,7.842] minutes (accept endpoints within 0.02).
- Interprets the interval for the population mean difference with appropriate confidence meaning, rather than individual durations.
Part c
- Obtains the signed statistic t≈2.43024 (accept within 0.01) using a zero null difference.
- Computes the upper t-tail p≈0.0091064 (accept within 0.0001), retaining enough precision for the decision.
- Rejects at α=0.01 and states evidence for a larger A population mean without assigning probability to the null.
Part d
- Identifies lack of randomized workflow assignment and a plausible confounding source, distinguishing sampling from causal assignment.
- Distinguishes the two-sided 95% interval from the one-sided 1% test and refers to the computed tail for the requested decision.
STAT-1 · stat-fr-04
Combine categorical inference, conditional distributions and binomial prediction
Original response and rubric· STAT-1: categories, tails and dependence
The changed table and binomial model recheck conditional denominators, homogeneity, expected counts, tail direction, prediction, dependence and nonrejection.
Inspect every original criterion, including criteria not selected by the retry
Part a
- Calculates the two conditional text percentages, about 46.67% and 26.67%, using denominators 90.
- Identifies homogeneity and states equality of the complete population preference distributions as the null.
Part b
- Checks independent-unit/sampling-fraction conditions and obtains expected counts [33,32,25] in each row, all at least 5.
- Calculates χ²≈7.90909 (accept within 0.01) and df=2, excluding the row totals from the category count.
- Computes the upper-tail p≈0.0191674 (accept within 0.0002), without doubling it.
- Rejects homogeneity and states a population-distribution difference with an omnibus or causal limitation.
Part c
- Uses Binomial(12,0.35) and an inclusive upper-event sum or complement through 4.
- Obtains approximately 0.416655 (accept within 0.0001), with correct inclusion of five.
Part d
- Explains that shared-outage dependence invalidates the binomial calculation despite unchanged marginal success probabilities.
- Explains that nonrejection does not establish identical or practically equivalent population preference distributions.
PC-1.1: domain reasoning
Inspect the original form or item · Repair using the teaching lesson
Selected criteria to inspect in the original response
- Identify both function domains.
- Require the inner output to lie in the outer domain.
- Filter algebraic solutions before interpreting them.
Attempt the changed case
Let F(u)=√(u−1) with domain [1, 10], and G(x)=x²−2 with domain [−3, 3]. Find the exact domain and range of F∘G. Explain why requiring only x∈[−3, 3] is insufficient. Then replace G by H(x)=2−x on [−3, 3] and repeat.
Original criterion / error evidence / repair explanation: ____________________
First changed-case response, tools and exposure record:
Reviewer feedback and revision: ____________________
Review changed-case reasoning and rubric
For G, 1≤x²−2≤10 requires 3≤x²≤12. Intersecting with [−3, 3] gives [−3,−√3]∪[√3, 3]. The range is [0,√6]. For H, 1≤2−x≤10 means −8≤x≤1, hence domain [−3, 1] and output √(1−x) has range [0, 2]. The outer domain upper bound also matters in general even though x²≤9 makes it redundant in this first case.
- Writes both inequalities and intersects with inner domain.
- Gives both branches including endpoints for G.
- Obtains both ranges and explains why the changed inner function changes the domain.
PC-1.1: graph recovery
Inspect the original form or item · Repair using the teaching lesson
Parameter and phase recovery map to pc1-fr-03 part b. The held-out validation and aliasing questions below are additional transfer work, not criteria scored in the original item.
Selected criteria to inspect in the original response
- Recover midline/amplitude/period from a supplied graph.
- Select a phase consistent with a marked extremum.
- Give the recovered height and time parameters with appropriate units, as required in original part b.
Attempt the changed case
A sinusoid has consecutive maxima (1, 7),(9, 7) and the intervening minimum (5, 1). Construct a cosine model and predict at t=3. A separate measured value there is 5.2. Explain what may be concluded if measurement tolerance is ±0.1 and the model is asserted exact. What if the two maxima were not known to be consecutive?
Original criterion / error evidence / repair explanation: ____________________
First changed-case response, tools and exposure record:
Reviewer feedback and revision: ____________________
Review changed-case reasoning and rubric
Midline 4, amplitude 3, period 8; y=4+3 cos (π(t−1)/4), prediction 4 at 3. Observed 5.2 differs by 1.2, exceeding tolerance, so the exact sinusoidal model plus the stated tolerance is inconsistent with this validation point. This does not identify which premise failed. Without consecutiveness, shorter periods 8/k can fit both maxima; the marked minimum imposes additional restrictions but does not alone prove that 8 is the fundamental period.
- Recovers phase and period from consecutive extrema.
- Computes held-out prediction and compares with tolerance.
- States model/measurement uncertainty and avoids uniqueness without consecutiveness.
AB-1: net-rate maximum and units
Inspect the original form or item · Repair using the teaching lesson
Selected criteria to inspect in the original response
- Separate incoming amount, net change and initial quantity.
- Use sign changes and endpoints for a global maximum.
- Interpret first and second derivatives with units.
Attempt the changed case
A tank starts with 40 L. Its net rate N(t), in L/min, is the exact piecewise-linear graph joining (0, 2),(2, 6),(5,−3),(7, 1). For 0≤t≤7, find the amount at 5, all candidate absolute extrema and the absolute maximum amount. Explain why N(5)<0 does not mean the amount is negative. Draw the graph and shade signed areas.
Original criterion / error evidence / repair explanation: ____________________
First changed-case response, tools and exposure record:
Reviewer feedback and revision: ____________________
Review changed-case reasoning and rubric
Area 0–2=8. On 2–5, N=12−3 t, with zero 4. Area 2–4=6 and 4–5=−1.5. Thus P(5)=52.5 L. On 5–7, N=2 t−13 with zero 6.5. Candidates t=0, 4, 6.5, 7: amounts 40, 54, 50.25, 50.5 L. Absolute maximum 54 at 4. Negative N means amount decreasing, not negative; the initial quantity and accumulated area determine P.
- Computes signed rather than unsigned integrals and includes initial 40.
- Finds rate zeros 4 and 6.5 plus endpoints, and evaluates quantities.
- Uses L versus L/min and explains sign correctly.
AB-1: changed accumulation setup
Inspect the original form or item · Repair using the teaching lesson
The original ab-fr-01 tests fixed-bound incoming/net accumulation and average change. Moving endpoints and their chain-rule factors below are a transfer extension, not a claim about the original rubric.
Selected criteria to inspect in the original response
- Choose the correct accumulated quantity.
- Set up incoming and net accumulation with the correct bounds.
- Separate average value from total change.
Attempt the changed case
Let R(s)=3+s² L/min for s≥0 and A(t)=∫ from t to 2 t R(s)ds for t≥0. Compute A(1), A′(1), and the average value of R over [1, 2]. Repair the claim A′(t)=R(2 t)−R(t). Explain why A is not the inventory of a tank that started at time 0.
Original criterion / error evidence / repair explanation: ____________________
First changed-case response, tools and exposure record:
Reviewer feedback and revision: ____________________
Review changed-case reasoning and rubric
A(1)=3+7/3=16/3 L. Leibniz/chain rule gives A′(t)=2 R(2 t)−R(t)=3+7 t², hence 10 L/min at 1. Average R on [1, 2]=(16/3)/(2−1)=16/3 L/min. The moving upper endpoint contributes factor 2. A measures quantity over a moving window [t, 2 t], not cumulative inventory from a fixed start; no initial tank amount is supplied.
- Uses both boundary speeds.
- Computes all quantities and distinct units.
- Explains moving versus fixed-start accumulation.
BC-1: signed radius and unique region
Inspect the original form or item · Repair using the teaching lesson
The original bc-fr-03 uses an everywhere-positive radius and tests region selection and radial order. Signed radii and double traversal in this retry are additional transfer work, not original rubric criteria; its tangent and arc-length criteria remain on the original form.
Selected criteria to inspect in the original response
- Identify the actual geometric region and a single traversal.
- Identify the inner and outer radii in the requested geometric region.
- Justify integral bounds rather than using a memorized full-period range.
Attempt the changed case
For r=2 cosθ, identify the Cartesian curve, find an interval tracing it once, and compute its enclosed area. A learner integrates ½r² over [0, 2π] and gets 2π. Explain the error and evaluate the point at θ=2π/3.
Original criterion / error evidence / repair explanation: ____________________
First changed-case response, tools and exposure record:
Reviewer feedback and revision: ____________________
Review changed-case reasoning and rubric
Multiplying r=2 cosθ by r gives x²+y²=2 x, or (x−1)²+y²=1. The interval [−π/2,π/2] traces the circle once. Area=½∫4 cos²θ dθ=π. The full 2π interval retraces the same circle twice. At 2π/3, r=−1, so (x, y)=(1/2,−√3/2); the negative radius moves opposite the displayed angle.
- Derives the translated circle and labels center/radius.
- Chooses and justifies a single traversal for areaπ.
- Computes the negative-radius point and identifies double counting.
BC-1: approximation versus guarantee
Inspect the original form or item · Repair using the teaching lesson
The original bc-fr-04 constructs an approximation from a power series and derives an interval-wide derivative bound. Recovering coefficients from supplied derivative values is a transfer variation here, not the original data format. Endpoint convergence, geometric-series identification and the separate alternating-series bound remain on the original form.
Selected criteria to inspect in the original response
- Construct the specified degree-three approximation and use a valid remainder argument.
- State a valid error theorem and interval hypothesis.
- Keep estimated error distinct from a guaranteed bound.
Attempt the changed case
A C⁴ function has f(0)=2, f′(0)=−1, f″(0)=6, f‴(0)=12, and |f⁽⁴⁾(x)|≤24 for 0≤x≤0.2. Approximate f(0.2) with the cubic Taylor polynomial and give a guaranteed interval. A classmate reports that a numerical fourth derivative near 0 is 0, so the error is zero. Explain why that does not follow.
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P₃(x)=2−x+3 x²+2 x³; P₃(0.2)=1.936. Lagrange error≤24(0.2)^4/4! = 0.0016, giving [1.9344, 1.9376]. A derivative at one point, particularly an approximation, is not a bound on the whole interval or evidence that the fourth derivative vanishes throughout; zero remainder requires additional structure.
- Uses factorials correctly to construct the cubic.
- Applies the fourth-derivative bound on the whole interval and computes 0.0016.
- Explains why local numerical evidence does not certify exactness.
STAT-1: design and causal scope
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Selected criteria to inspect in the original response
- Identify observational units and selection versus assignment.
- Match causal and population conclusions to the actual design.
- Explain why volunteer participation and nonresponse limit population generalization.
Attempt the changed case
Eighty volunteers are randomly assigned equally to two study techniques; their later test scores are compared. Explain what random assignment supports, what voluntary recruitment limits, and what would change if volunteers chose their own technique. Give a plausible confounder in the latter design and a changed design supporting broader population inference.
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Random assignment supports a causal comparison for these volunteers under adherence, no interference and valid measurement assumptions. Voluntary recruitment does not yield a probability sample of all students, so population generalization needs extra justification. Self-selection permits confounding, such as motivation affecting both technique choice and score. A probability sample of the intended student population followed by randomized assignment supports both design-based population inference and causal comparison, subject to nonresponse and implementation limitations.
- Separates recruitment from assignment.
- States a real confounding pathway rather than merely saying biased.
- Proposes sampling plus randomization and retains assumptions.
STAT-1: residuals and extrapolation
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Selected criteria to inspect in the original response
- Compute residual as observed minus predicted.
- Check physical and observed predictor domains.
- Do not infer causation from observational regression.
Attempt the changed case
In observational runs with airflow x between 2 and 10, fitted drying time is ŷ=20−1.5 x minutes. A run at x=8 takes 9 minutes. Compute its residual. Evaluate a proposed prediction at x=16 and the claim that raising airflow causes a 1.5-minute reduction per unit. Design one useful next diagnostic.
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At 8, prediction 8 and residual 9−8=+1 minute. At 16 the line predicts−4 minutes, physically impossible and outside observed range; reject the extrapolation. The slope describes the fitted association, not a causal effect, because composition/initial moisture may confound it. Inspect residuals against airflow and moisture or collect held-out runs within the supported range; randomized controlled airflow at feasible levels would be needed for a causal design.
- Computes residual +1 with units and convention.
- Identifies both extrapolation and impossible duration.
- Gives a specific causal limitation and a relevant diagnostic/design.
PC-1.1: unknown-offset model and validation
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This retry changes calibration and average-rate cases from the original item. It checks extrapolation beyond observed times; the original threshold-time uniqueness and proposed model-domain interpretation still need separate review. The held-out tolerance test is additional transfer work, not a replacement for those original criteria. It does not certify the whole response or current CED coverage.
Selected criteria to inspect in the original response
- Construct a nonperiodic model from supplied data.
- Compute and interpret a finite-interval average rate.
- Distinguish model extrapolation from established physical behavior.
Attempt the changed case
A synthetic cooling model is T(t)=A+Bq^t for t≥0, with 0<q<1. Measurements at t=0,1,2 minutes are 90,66,54 degrees. Determine A,B,q from the three equations. Find the average rate over [0,2], predict T(4), explain whether that prediction extrapolates the observed time range, and assess a separate observation T(4)=47 with tolerance ±0.2 when the model is asserted exact.
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Differences are −24 and −12, so q=1/2. B(q−1)=−24 gives B=48 and A=42. Average rate (54−90)/2=−18 degrees/minute. T(4)=42+48/16=45. The observed 47 differs by 2>0.2, contradicting the exact-model-plus-tolerance claim. It does not identify which model or measurement assumption failed, and the inferred limiting 42 is conditional on continuing the model. The prediction at 4 minutes extrapolates beyond the calibration times 0–2; fitting those three points gives no empirical guarantee at 4.
- Derives q from consecutive differences, then determines both offset and amplitude.
- Uses the full interval and units for average change.
- Compares held-out prediction with tolerance without declaring empirical validation.
PC-1.1: symbolic equations and original domains
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This changed case targets the listed original criteria only. Other original rubric criteria remain visible on the linked form; completing this retry does not certify the whole response or current CED coverage.
Selected criteria to inspect in the original response
- Solve exponential and trigonometric equations with all requested solutions.
- Retain the domain of the original logarithmic expression.
Attempt the changed case
Solve 4^(x−1)=8 exactly. Solve 2sin t=1 for 0≤t<2π. Rewrite ln(x−1)+ln(x+2) as one logarithm while retaining the original real domain; then solve ln(x−1)+ln(x+2)=ln4, explaining which algebraic root is rejected.
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Writing powers of 2 gives 2(x−1)=3, so x=5/2. The angles are π/6 and 5π/6. The original logs require x>1; on that domain their sum is ln((x−1)(x+2)). The equation gives x²+x−6=0, roots 2 and −3; only 2 satisfies x>1. Although the product is positive at −3, the original separate logarithms are undefined there.
- Uses common bases or valid logarithms to solve the exponential equation.
- Finds both angles within the stated half-open interval.
- States x>1 before combining logs and rejects the extraneous root.
AB-1: distance, arrival and speed
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This changed case targets the listed original criteria only. Other original rubric criteria remain visible on the linked form; completing this retry does not certify the whole response or current CED coverage.
Selected criteria to inspect in the original response
- Retain initial position in an accumulation equation.
- Justify unique arrival with monotonic position.
- Use velocity and acceleration together to classify speed.
Attempt the changed case
A particle has v(t)=2+t² m/s for 0≤t≤2 seconds and s(0)=−1 m. Find total distance and final position. Show that it reaches position 2 exactly once and give the time. Is its speed increasing at t=1? State hypotheses ensuring some interior instantaneous velocity equals its average velocity.
Original criterion / error evidence / repair explanation: ____________________
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Positive velocity gives distance ∫₀²(2+t²)dt=20/3 m, final position 17/3 m. Arrival requires 2t+t³/3=3. The left side is continuous and strictly increasing; it is 0 at 0 and 20/3 at 2, so one solution. t=1 is not a root (7/3); numerical root is approximately 1.206960 seconds. A calculator should solve the retained equation, not set v=2 m/s. Acceleration 2t is positive at 1, as is velocity 3 m/s, so speed increases. s is continuous on [0,2] and differentiable inside; MVT gives v(c)=(s(2)−s(0))/2=10/3 m/s at some interior c.
- Integrates speed and distinguishes distance from final position.
- Uses continuity and strict increase for arrival existence and uniqueness; solves the displacement equation.
- Combines velocity/acceleration signs and states the MVT hypotheses.
AB-1: signed integrand and concavity
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This changed case targets the listed original criteria only. Other original rubric criteria remain visible on the linked form; completing this retry does not certify the whole response or current CED coverage.
Selected criteria to inspect in the original response
- Use signed areas and initial value.
- Check extrema and endpoints.
- Distinguish a first derivative from a nonexistent second derivative at a corner.
Attempt the changed case
Let g be the exact joined segments through (0,3),(3,0),(5,−2),(7,2), and H(x)=2+∫₀ˣg. Draw g. Compute H(5), all absolute extrema on [0,7], and concavity intervals of H. Does H″(5) exist? For K(x)=H(x²), calculate K′(2).
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g=3−x on [0,5], then 2x−12 on [5,7]. H(5)=2+9/2−2=9/2. H increases on (0,3), decreases on (3,6), and increases on (6,7). Candidate values H(0)=2,H(3)=13/2,H(6)=7/2,H(7)=9/2. Maximum 13/2 at 3, minimum 2 at 0. H is concave down on (0,5), up on (5,7). H″(5) does not exist because g has slopes −1 and 2; H′(5)=−2 does exist. K′(2)=4g(4)=−4.
- Computes signed areas and keeps initial 2.
- Checks endpoints as well as zeros of g; minimum need not be the negative-to-positive zero.
- Uses slopes of g for concavity and applies FTC plus chain rule.
AB-1: implicit tangent and radial motion
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This changed case targets the listed original criteria only. Other original rubric criteria remain visible on the linked form; completing this retry does not certify the whole response or current CED coverage.
Selected criteria to inspect in the original response
- Retain both product-rule terms in implicit differentiation.
- Link a curve derivative to time rates.
- Check denominator when identifying horizontal tangents.
Attempt the changed case
On x²+xy+y²=7, with coordinates in meters, use the branch through (1,2). Find y′ and the tangent line there. If dx/dt=2 m/s at that point, find dy/dt and the rate of distance from the origin. Find every horizontal-tangent point and verify its regularity.
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y′=−(2x+y)/(x+2y); at (1,2), slope −4/5 and tangent y−2=−4(x−1)/5. dy/dt=−8/5 m/s. With ρ²=x²+y², ρ′=(x x′+y y′)/ρ=(2−16/5)/√5=−6/(5√5) m/s. Horizontal tangents require y=−2x and 3x²=7: (√(7/3),−2√(7/3)) and its negative. The denominator −3x is nonzero at both.
- Differentiates xy correctly and uses the given point.
- Uses time chain rule and differentiates the radial distance, with units and sign.
- Solves numerator zero jointly with the curve and checks nonzero denominator.
AB-1: separation and tangent error
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This changed case targets the listed original criteria only. Other original rubric criteria remain visible on the linked form; completing this retry does not certify the whole response or current CED coverage.
Selected criteria to inspect in the original response
- Separate and integrate with the initial condition.
- Construct a tangent approximation using the differential equation.
- Justify error direction on the intervening interval.
Attempt the changed case
Solve y′=2t(1+y), y(0)=2 on [0,1]. Explain why the horizontal initial tangent does not imply a constant solution. Approximate y(0.9) by the tangent at 1 and justify whether the approximation is too high or too low.
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Integration gives ln|1+y|=t²+C; y(0)=2 chooses 1+y=3e^(t²), hence y=3e^(t²)−1. The excluded constant solution −1 does not meet the initial condition. y′(0)=0 but y″=2(1+y)+2ty′=(2+4t²)(1+y)>0, so the solution is not constant. At 1, y=3e−1 and slope 6e. L(0.9)=3e−1−0.6e=12e/5−1. Positive second derivative throughout [0.9,1] makes the graph strictly convex, so the tangent approximation is an underestimate.
- Integrates with absolute value/valid positive branch and determines the constant.
- Uses y(1) and y′(1) in the line.
- Proves convexity over the interval rather than checking only one point.
AB-1: widths, radii and integral mean value
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This changed case targets the listed original criteria only. Other original rubric criteria remain visible on the linked form; completing this retry does not certify the whole response or current CED coverage.
Selected criteria to inspect in the original response
- Build cross-section area from actual dimensions.
- Measure washer radii from the rotation axis.
- Apply the mean value theorem with its hypotheses.
Attempt the changed case
The region 0≤x≤1,0≤y≤x is a triangular base. Find its area, the volume with rectangular sections perpendicular to x having height 2−x, and its volume when rotated about y=−2. For F(x)=∫₀ˣt dt, justify and find the interior c satisfying F′(c)=[F(1)−F(0)]/(1−0).
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Area ∫₀¹x dx=1/2. Section area x(2−x), giving volume 1−1/3=2/3. Washer radii are 2+x and 2; volume π∫₀¹[(2+x)²−4]dx=π(2+1/3)=7π/3. F is continuous on [0,1] and differentiable within; MVT guarantees an interior c. F′(c)=c=1/2, which lies inside. Area units are squared, volume units cubed.
- Uses specified dimensions for section area, not its square by habit.
- Identifies both radii before integrating.
- States MVT hypotheses and checks the solved interior point.
BC-1: vector speed is a norm
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This changed case targets the listed original criteria only. Other original rubric criteria remain visible on the linked form; completing this retry does not certify the whole response or current CED coverage.
Selected criteria to inspect in the original response
- Accumulate both position components from their initial coordinates.
- Distinguish Euclidean speed from component sum or endpoint distance.
- Use v·a to classify speed change.
Attempt the changed case
A particle starts at (2,−1) with v(t)=(3,4t) for 0≤t≤1 seconds; positions are in meters and velocities in m/s. Find position at 1 and tangent slope there. Set up traveled distance; compare its value with endpoint distance without treating them as equal. Classify speed change at t=1.
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Position is (2+3t,−1+2t²), so (5,1) m at 1 second. Slope y′/x′=4/3 because x′=3≠0. Length=∫₀¹√(9+16t²)dt=5/2+(9/8)ln3≈3.736 m; endpoint distance √(3²+2²)=√13≈3.606 m. The velocity direction changes, so the path is longer than its chord; 3+4t is not speed. Acceleration (0,4), v·a=16 at 1, and speed 5>0, so speed derivative 16/5 m/s²>0.
- Adds the separate initial coordinates and checks nonzero x velocity.
- Integrates the Euclidean norm and distinguishes chord from arc.
- Uses the dot product and positive speed for the conclusion.
BC-1: weighted accumulation and Euler
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This changed case targets the listed original criteria only. Other original rubric criteria remain visible on the linked form; completing this retry does not certify the whole response or current CED coverage.
Selected criteria to inspect in the original response
- Differentiate an integrating-factor formula by product rule and FTC.
- Retain initial amount and decay factor.
- Use current numerical state in an Euler step.
Attempt the changed case
Let Q(t)=e^(−t/2)[4+∫₀ᵗ6e^(u/2)du] mg for t≥0, with t in minutes. Derive its differential equation and interpret the removal term. Simplify Q, find when it first reaches 8 mg and prove uniqueness. Use one Euler step of size 1 from 0 to estimate Q(1), then compare with the exact value.
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Product rule and FTC give Q′=6−Q/2 mg/min, with Q/2 the removal rate and coefficient 1/2 min⁻¹. Integration gives Q=12−8e^(−t/2), hence Q(0)=4. Reaching 8 requires e^(−t/2)=1/2, t=2ln2. Q′=4e^(−t/2)>0, so this crossing is unique. Euler slope at 0 is 6−2=4, estimate 8 mg. Exact Q(1)=12−8e^(−1/2)≈7.14775 mg, so the one-step estimate is high by ≈0.85225 mg; this is a conclusion for this case, not all Euler problems.
- Shows both differentiation rules and physical units.
- Retains the initial term and proves crossing uniqueness.
- Computes the starting-state Euler step and quantifies its error.
BC-1: logistic branch and improper deficit
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This changed case targets the listed original criteria only. Other original rubric criteria remain visible on the linked form; completing this retry does not certify the whole response or current CED coverage.
Selected criteria to inspect in the original response
- Update Euler slopes at numerical states.
- Solve logistic separation with the initial branch.
- Prove a concavity change and express an improper integral using finite upper bounds before taking a limit.
Attempt the changed case
Use dimensionless time and population in this changed case. For N′=N(4−N)/2, N(0)=1, take two Euler steps of size 1/2 to estimate N(1). Solve the IVP, find the time of inflection, and evaluate ∫₀∞(4−N(t))dt by a limit.
Original criterion / error evidence / repair explanation: ____________________
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Initial slope 3/2 gives N₁=7/4. Updated slope (7/4)(9/4)/2=63/32, giving N₂=7/4+63/64=175/64. Separation gives (1/2)ln[N/(4−N)]=t+C and initial ratio 1/3, so N=4/(1+3e^(−2t)), always between 0 and 4. N″=(2−N)N′ changes from positive to negative at N=2, giving t=(ln3)/2. Deficit=12e^(−2t)/(1+3e^(−2t)); integral to T is 2ln4−2ln(1+3e^(−2T)), tending to 2ln4.
- Uses the first numerical update in the second slope.
- Retains the correct logistic branch and initial constant.
- Shows the second derivative changes sign and treats infinity through a limit.
BC-1: rational tails and integration by parts
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This changed case targets the listed original criteria only. Other original rubric criteria remain visible on the linked form; completing this retry does not certify the whole response or current CED coverage.
Selected criteria to inspect in the original response
- Use integration by parts and polynomial division.
- Use partial fractions with a finite endpoint before an improper limit.
- Separate mean-value existence from uniqueness.
Attempt the changed case
Evaluate ∫₀¹x ln(2+x)dx by parts. For f(x)=1/[(x+2)(x+3)], find ∫₀¹f and ∫₁∞f. Justify that exactly one c∈(0,1) has f(c)=∫₀¹f.
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Choose u=ln(2+x),dv=x dx. The remaining rational term has x²/(x+2)=x−2+4/(x+2). The integral is (1/2)ln3−(1/2)[1/2−2+4ln(3/2)]=3/4−(3/2)ln3+2ln2. Partial fractions f=1/(x+2)−1/(x+3); integral from 0 to 1=ln(9/8). Integral from 1 to T=ln((T+2)/(T+3))−ln(3/4), tending to ln(4/3). Continuity gives an integral-mean-value point; f′=−(2x+5)/[(x+2)²(x+3)²]<0 gives strict decrease, interior attainment and uniqueness. Do not subtract two divergent logs separately.
- Includes the boundary term and correct polynomial division.
- Combines logarithms before taking the improper limit.
- Uses continuity for existence and strict decrease for uniqueness.
STAT-1: categories, tails and dependence
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This changed case targets the listed original criteria only. Other original rubric criteria remain visible on the linked form; completing this retry does not certify the whole response or current CED coverage.
Selected criteria to inspect in the original response
- Use row-conditional denominators and the full homogeneity null.
- Compute expected counts, chi-square degrees of freedom and an upper-tail probability.
- Distinguish a predictive binomial tail from a test p-value.
Attempt the changed case
Independent random samples of 60 different people from each of two populations over 600 people give counts A=(30,20,10), B=(20,20,20) in three exclusive preference categories; no person appears in both samples. Compare first-category percentages and test homogeneity at 0.05 with no continuity correction. Separately, for six independent Bernoulli trials with known p=1/2, find P(X≥4). Explain what fails if trials share an outage and why nonrejection does not prove identical preferences.
Original criterion / error evidence / repair explanation: ____________________
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Conditional first-category percentages are 50% and 100/3%≈33.33%. Homogeneity null: all three population category probabilities agree. Expected counts in each row are (25,20,15), all ≥5; independent units, random sampling and fractions below 10% support the design. χ²=2[25/25+0+25/15]=16/3, df=2; upper-tail p=e^(−8/3)≈0.069483>0.05, so fail to reject. This is insufficient evidence of differing distributions, not proof of equality. Independently X~Binomial(6,1/2), P(X≥4)=(15+6+1)/64=11/32. Shared outage destroys independence; common marginal probability does not preserve this binomial tail. The predictive probability is not the homogeneity p-value.
- Uses row totals and states the full distributional null.
- Checks design/count conditions and computes the undoubled upper tail with df 2.
- Includes4 in the event, distinguishes prediction from inference and rejects equality-by-nonrejection.
