Work across topics, choose a method and justify the result. This form contains 42 multiple-choice questions and four free-response questions, followed by optional extended practice.
This is original Learn by Exploring practice, not a College Board exam, an endorsed product or a prediction of an AP score. Its question counts and section times follow the announced May 2027 format. Difficulty and scoring have not been calibrated against an official examination.
Write your choices and mathematical work on paper or in your own document, recording revision PC-1.1 with the item ID. This page does not collect or save responses. Keep solutions closed until you have recorded an attempt. You can use sections independently, take breaks or use appropriate accommodations; there is no enforced timer or navigation lock.
For a timed rehearsal, use 65 minutes without a calculator, then 40 minutes with a graphing calculator for multiple choice. Use 35 minutes with a graphing calculator for free responses 1–2, then 35 minutes without one for free responses 3–4. Use a permitted device you can operate independently. Public solutions make this exposed practice, not secure assessment evidence.
Choose one answer per multiple-choice question. For free response, show the mathematical expression, conditions and reasoning, not only a decimal. Use radians unless stated otherwise. Carry unrounded values through calculations. Unless a question specifies otherwise, report numerical answers to three decimal places; the solution explains any acceptable numerical tolerance.
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29 questions · suggested time 65 minutes. Record one letter per question.
A function has the values in the table. What is its average rate of change on [1,5]?
| x | f(x) |
|---|---|
| 1 | 7 |
| 2 | 11 |
| 5 | 17 |
Answer C. The output change is 17−7=10 and the input change is 5−1=4. The rate is 10/4=5/2. Averaging the two neighboring secant rates without weighting their unequal widths would not compute this rate.
Review focus: Calculate and compare average rates from a table. Unit 1: Polynomial and rational functions. Item pc1-nc-01.
For f(x)=(x²−5x+6)/(x−2), which statement is correct?
Answer A. Factor the numerator as (x−2)(x−3). For x≠2, f(x)=x−3, but the original denominator still excludes 2. The missing point would have height 2−3=−1; cancellation does not restore that input.
Review focus: Relate rational expressions to exclusions and holes. Unit 1: Polynomial and rational functions. Item pc1-nc-02.
Let p(x)=(x+1)²(x−3)³. Which describes its zeros?
Answer D. The even power at −1 does not change sign, and the remaining factor is nonzero there. The odd power at 3 changes sign while the other factor stays positive. Thus only the zero at 3 is a crossing.
Review focus: Interpret polynomial multiplicities. Unit 1: Polynomial and rational functions. Item pc1-nc-03.
What is the remainder when 2x³−x²+4 is divided by x+2?
Answer B. The remainder is p(−2)=2(−8)−4+4=−16. A divisor x+2 corresponds to input −2, not 2.
Review focus: Apply the remainder theorem. Unit 1: Polynomial and rational functions. Item pc1-nc-04.
As x increases without bound, (3x²−2)/(2x²+7) approaches which value?
Answer B. Divide numerator and denominator by x². The ratio becomes (3−2/x²)/(2+7/x²), which approaches 3/2 because both reciprocal-square terms approach zero.
Review focus: Interpret rational end behavior. Unit 1: Polynomial and rational functions. Item pc1-nc-05.
The table gives the COMPLETE function f on its four-input domain. What is f⁻¹(7)?
| x | f(x) |
|---|---|
| −3 | 4 |
| 0 | −2 |
| 2 | 7 |
| 5 | 1 |
Answer D. An inverse exchanges the input and output roles. The output 7 comes from input 2, so f⁻¹(7)=2. All four outputs are distinct, making the inverse a function on those outputs. No interpolation between the four inputs is stipulated.
Review focus: Recover an inverse value from a one-to-one finite table. Unit 1: Polynomial and rational functions. Item pc1-nc-06.
The two tables list the COMPLETE domains of f and g. What is the domain of f∘g?
| Input of f | Output of f | Input of g | Output of g |
|---|---|---|---|
| −2 | 5 | 0 | −2 |
| 1 | 4 | 2 | 3 |
| 3 | −1 | 4 | 5 |
| 6 | 1 |
Answer A. A valid input must first belong to g and then produce an output in the domain {−2,1,3} of f. Inputs 0,2,6 produce −2,3,1 respectively. Input 4 produces 5, where f is undefined. Thus the domain is {0,2,6}; the output values of f are not its admissible inputs.
Review focus: Determine a composition domain from complete tables. Unit 1: Polynomial and rational functions. Item pc1-nc-07.
The supplied graph defines f exactly by straight segments. When the input is restricted to 0≤x≤4, what is the greatest value of f(x)?
Graph description: Exact straight segments join (−1,5),(1,1),(3,4),(5,0). The question restricts the input to [0,4]. f (solid): straight segments through (-1, 5), (1, 1), (3, 4), (5, 0). Axes use the labeled scales; no values outside the specified domain are supplied.
Answer C. Within [0,4], the relevant endpoint and corner heights are f(0)=3,f(1)=1,f(3)=4,f(4)=2. Each straight segment has values between its endpoint heights, so the greatest allowed value is 4 at x=3. The height 5 at x=−1 is outside the restricted domain.
Review focus: Read a maximum on a restricted supplied graph. Unit 1: Polynomial and rational functions. Item pc1-nc-08.
A polynomial q matches measured values at x=0,1,2. Which expression must match those same values regardless of q?
Answer D. The added product vanishes at all three inputs. The other modifications fail at least one input. Thus three values do not uniquely identify an unrestricted polynomial degree.
Review focus: Distinguish model evidence from unique identification. Unit 1: Polynomial and rational functions. Item pc1-nc-09.
The supplied graph joins its marked points by straight segments. If h(x)=f(x+1)−2, what is h(1)?
Graph description: The complete graph of f on [−1,3], formed by two straight segments. f (solid): straight segments through (-1, 1), (1, 3), (3, 1). Axes use the labeled scales; no values outside the specified domain are supplied.
Answer A. h(1)=f(2)−2. On the segment from (1,3) to (3,1), the value at x=2 is 2. Therefore h(1)=0; the horizontal shift changes the input before the vertical shift is applied.
Review focus: Read and transform a supplied graph. Unit 1: Polynomial and rational functions. Item pc1-nc-10.
An exponential model f(x)=A b^x has f(0)=5 and f(2)=45, with b>0. What is f(3)?
Answer C. A=5 and 5b²=45, so b²=9. The positive-base condition gives b=3. Thus f(3)=5·27=135.
Review focus: Recover an exponential model from values. Unit 2: Exponential and logarithmic functions. Item pc1-nc-11.
What is the real solution of log₂(x−1)+log₂(x+1)=3?
Answer B. Both logarithms require x>1. Combining them gives log₂(x²−1)=3, hence x²−1=8 and x=±3. Only 3 satisfies the original domain.
Review focus: Solve a logarithmic equation with its domain. Unit 2: Exponential and logarithmic functions. Item pc1-nc-12.
Which is the inverse of f(x)=4+3·2^x?
Answer A. From y=4+3·2^x, obtain (y−4)/3=2^x and x=log₂((y−4)/3). The exponential output is positive, so the original range and inverse domain are y>4.
Review focus: Invert an exponential transformation. Unit 2: Exponential and logarithmic functions. Item pc1-nc-13.
A sequence begins 3,6,12,24,…, with each term twice the preceding one. What is the sum of its first five terms?
Answer D. The first five terms are 3,6,12,24,48, whose sum is 93. Equivalently 3(2⁵−1)/(2−1)=93. The fifth term alone is not the sum.
Review focus: Sum a finite geometric sequence. Unit 2: Exponential and logarithmic functions. Item pc1-nc-14.
The table records the number of items produced by a machine since it started. What is the average production rate from 0 to 3 hours, in items per minute?
| Elapsed time (hours) | Items produced |
|---|---|
| 0 | 0 |
| 1 | 45 |
| 3 | 150 |
Answer B. The endpoint increase is 150 items over 3 hours, or 180 minutes. The average is 150/180=5/6 item per minute. The table does not establish a constant rate within the interval. Fifty items per hour must be divided by 60 to convert to items per minute.
Review focus: Convert an average rate from a supplied time table. Unit 2: Exponential and logarithmic functions. Item pc1-nc-15.
Positive values satisfy log₁₀ y=2−x. Which original-scale equation follows?
Answer C. Exponentiating gives y=10^(2−x)=10²·10^(−x)=100(1/10)^x. Equal increments of x multiply y by 1/10 rather than subtracting a constant.
Review focus: Translate a semilog equation to original units. Unit 2: Exponential and logarithmic functions. Item pc1-nc-16.
For f(x)=7−4(1/2)^x, which statement describes all real x?
Answer C. As x increases, (1/2)^x decreases; multiplying by −4 reverses the direction, making f increase. The exponential remains strictly positive, so f<7. It becomes arbitrarily large as x tends to negative infinity, making f arbitrarily negative. The value 7 is approached but never attained.
Review focus: Interpret decay with a shifted baseline. Unit 2: Exponential and logarithmic functions. Item pc1-nc-17.
If log₃ a=2 and log₃ b=−1, with a,b positive, what is log₃(a²/b)?
Answer A. The power and quotient rules give 2log₃ a−log₃ b=4−(−1)=5. Directly a=9,b=1/3, so a²/b=243=3⁵.
Review focus: Evaluate logarithms through exponent relationships. Unit 2: Exponential and logarithmic functions. Item pc1-nc-18.
A model has the form y=k+A b^x. The exact table below is consistent with which parameters?
| x | y |
|---|---|
| 0 | 5 |
| 1 | 8 |
| 2 | 14 |
| 3 | 26 |
Answer B. Subtracting baseline 2 gives 3,6,12,24, a geometric sequence with factor 2. Thus y=2+3·2^x. The raw values do not have a constant ratio; their successive differences 3,6,12 reveal the shifted structure.
Review focus: Identify shifted exponential structure. Unit 2: Exponential and logarithmic functions. Item pc1-nc-19.
A monitor reports only at integer times n=0,1,2,3,4,5. Its complete readings appear below. What is the FIRST observation time with a reading of at least 20?
| n | Reading |
|---|---|
| 0 | 12 |
| 1 | 18 |
| 2 | 17 |
| 3 | 23 |
| 4 | 21 |
| 5 | 26 |
Answer D. The readings at 0,1,2 are all below 20, while the reading at 3 is 23. Therefore 3 is the first observation meeting the threshold. Interpolating a crossing between observations would answer a different question and would require an additional model.
Review focus: Find the first threshold in a complete discrete table. Unit 2: Exponential and logarithmic functions. Item pc1-nc-20.
What is sin(7π/6)?
Answer A. The angle is π+π/6, in the third quadrant, where sine is negative. Its reference-angle sine is 1/2, so the value is −1/2.
Review focus: Evaluate exact trigonometric values. Unit 3: Trigonometric and polar functions. Item pc1-nc-21.
For y=3−2cos(πx/3), what are the period and range?
Answer C. The cosine argument changes by 2π when x changes by 6. Since cosine ranges from −1 to 1, multiplying by −2 and adding 3 gives [1,5]. The negative coefficient reflects the graph without making the amplitude negative.
Review focus: Interpret a transformed sinusoidal period and range. Unit 3: Trigonometric and polar functions. Item pc1-nc-22.
What is arcsin(sin(5π/6))?
Answer D. sin(5π/6)=1/2. The inverse sine returns the angle in [−π/2,π/2] with that sine, namely π/6. Inverse sine is not an unrestricted cancellation of every angle.
Review focus: Use the principal inverse-trigonometric branch. Unit 3: Trigonometric and polar functions. Item pc1-nc-23.
Which is the complete set of excluded inputs for y=tan(2x−π/4)? In the options, k is any integer.
Answer B. Tangent is undefined when its argument is π/2+kπ. Solving 2x−π/4=π/2+kπ gives 2x=3π/4+kπ and x=3π/8+kπ/2.
Review focus: Determine tangent exclusions after transformation. Unit 3: Trigonometric and polar functions. Item pc1-nc-24.
What is the range of y=2−3sec x on its real domain?
Answer C. Secant has range (−∞,−1] union [1,∞). Values sec x≥1 give y≤−1; values sec x≤−1 give y≥5. Both boundary values are attained. The entire interval (−1,5), not only 2, is excluded.
Review focus: Transform a reciprocal-trigonometric range. Unit 3: Trigonometric and polar functions. Item pc1-nc-25.
What are all solutions of 2sin²x−1=0 on 0≤x<2π?
Answer B. The equation gives sin²x=1/2, so sin x=±√2/2. The reference angle π/4 occurs in all four quadrants. Squaring does not restrict sine to its positive value.
Review focus: Solve a trigonometric equation over a full cycle. Unit 3: Trigonometric and polar functions. Item pc1-nc-26.
For the polar curve r=2+3sinθ, what Cartesian point corresponds to θ=3π/2?
Answer A. At θ=3π/2, r=2−3=−1. Then x=r cosθ=0 and y=r sinθ=(−1)(−1)=1. Negative radius reverses the nominal downward ray; its distance is 1.
Review focus: Convert a signed polar point. Unit 3: Trigonometric and polar functions. Item pc1-nc-27.
Which interval traces the entire left petal of r=3cos(2θ), from origin to tip and back?
Answer D. Radius vanishes at 3π/4 and 5π/4 and is positive between them. At the middle angle π, r=3 places the tip at (−3,0). The positive-radius ray sweeps the complete left petal between the two origin visits.
Review focus: Choose a complete polar petal restriction. Unit 3: Trigonometric and polar functions. Item pc1-nc-28.
The functions f(x)=4+cos(πx/2) and g(x)=4+cos(3πx/2) agree at every integer x. Which input gives different values and can distinguish these exact candidates?
Answer A. At 1/2 the cosine arguments are π/4 and 3π/4, giving outputs 4+√2/2 and 4−√2/2. For integer n, cos(3πn/2)=cos(2πn−πn/2)=cos(πn/2), so the other inputs cannot distinguish them. This is a sampling ambiguity, not equality of the functions.
Review focus: Distinguish periodic models from their sampled values. Unit 3: Trigonometric and polar functions. Item pc1-nc-29.
13 questions · suggested time 40 minutes. Record one letter per question.
Graph f(x)=x³−2x²−4x+1 using an initial window −2≤x≤4 and −12≤y≤12, then refine its zeros. Which list contains ALL zeros in the domain [0,4], rounded to three decimals?
Answer B. Numerical zero searches give −1.3913823806, 0.2271344422 and 3.1642479385. The first lies outside the stated domain. The two permitted zeros round to 0.227 and 3.164. A cubic has at most three distinct real zeros, and the three located sign changes account for all of them. A graph window is a search aid, not permission to include roots outside the question domain.
Review focus: Find all polynomial zeros in a restricted interval with graphing technology. Unit 1: Polynomial and rational functions. Item pc1-c-01.
A model compares R(x)=(2x+1)/(x−1) with L(x)=0.4x+1, only for 1<x≤8. Use a graph window 1≤x≤8, 0≤y≤15 and refine the intersection. What is its x-coordinate, rounded to three decimals?
Answer C. Multiplication by x−1 is valid only away from x=1. The equation becomes 0.4x²−1.4x−2=0, or 2x²−7x−10=0. Its roots are (7±√129)/4. Only the positive root 4.5894541729 belongs to (1,8]. The vertical asymptote x=1 is not an intersection, and a plotted segment across that discontinuity must not be mistaken for one.
Review focus: Find a rational-function intersection while preserving domain exclusions. Unit 1: Polynomial and rational functions. Item pc1-c-02.
Squares of side x centimeters are cut from the corners of an 8-by-6 centimeter sheet, and its sides are folded to make an open box. For 0<x<3, V(x)=x(8−2x)(6−2x). Graph V using 0≤x≤3, 0≤V≤30 and refine the maximum. What is the largest volume, to three decimals, rather than the input that produces it?
Answer D. The physical domain requires both base dimensions positive. Graphing and refining the sole interior maximum gives x≈1.1314829082 and V≈24.2584197277 cm³. The endpoint extensions V(0)=V(3)=0 do not give positive-volume boxes. The requested output is 24.258 cm³, not the cut length. Substitution of the retained x value checks the volume. This numerical maximum can also be checked algebraically without calculus: with a=(7−√13)/3, V(a)−V(x)=4(x−a)²(7−2a−x), which is nonnegative for 0≤x≤3.
Review focus: Use technology to identify an extremum of a contextual polynomial on its physical domain. Unit 1: Polynomial and rational functions. Item pc1-c-03.
For the synthetic table, fit y=a+bx+cx²+dx³ by unweighted least squares in the ORIGINAL y units, using all six rows. Retain the coefficients. What is the fitted prediction at x=2.5, to three decimals? The model family is assumed for this question; fitting does not establish its truth.
| x | y |
|---|---|
| 0 | 2 |
| 1 | 4 |
| 2 | 11 |
| 3 | 28 |
| 4 | 62 |
| 5 | 116 |
Answer A. The fitted coefficients in ascending powers are a=43/21≈2.0476190476, b=67/63≈1.0634920635, c=−2/21≈−0.0952380952 and d=8/9≈0.8888888889. Substituting 2.5 gives exactly 18 for the least-squares solution, rounding to 18.000. The value 19.500 is linear interpolation between the neighboring observations, a different procedure. More observations than coefficients make this regression rather than an instruction to interpolate every row; a fresh observation would be needed to test its predictive usefulness.
Review focus: Perform cubic least-squares regression and make a conditional interpolated prediction. Unit 1: Polynomial and rational functions. Item pc1-c-04.
A stipulated model gives a quantity Q(n)=80(1.07)ⁿ, observed only at integer days n≥0. What is the FIRST observation day on which the model predicts Q(n)≥150? Solve with logarithms or a numerical table and check the neighboring days.
Answer C. The continuous crossing is n=ln(150/80)/ln(1.07)≈9.2908840718. Because the positive base exceeds 1, Q increases with n. The first eligible integer is 10, not the rounded crossing 9. Day 9 is below 150 and day 10 is above 150. This is a conditional prediction under the given growth model, not evidence that a real quantity follows the model indefinitely.
Review focus: Convert an exponential threshold into the first allowable discrete observation time. Unit 2: Exponential and logarithmic functions. Item pc1-c-05.
Fit y=a+b log₂x by unweighted least squares in y to the supplied data. Here x is a positive dimensionless ratio to a reference amount. Using the retained fitted coefficients, what positive x predicts y=10, to three decimals? If your calculator fits a+b ln x, convert the slope or use that fitted equation consistently.
| x | y |
|---|---|
| 1 | 3.2 |
| 2 | 5.9 |
| 4 | 9.1 |
| 8 | 11.8 |
| 16 | 15.2 |
Answer B. Regressing y on log₂x=0,1,2,3,4 gives y=3.06+2.99log₂x. Thus x=2^((10−3.06)/2.99)≈4.9970277649. The equivalent natural-log slope is 2.99/ln2, not 2.99. The inverse prediction lies within the training x-range, but its precision as a real prediction is not established by the displayed numerical decimals. This is inverse prediction from a fixed fitted equation, not a new regression with x and y exchanged.
Review focus: Fit a logarithmic response model and invert the retained fitted equation. Unit 2: Exponential and logarithmic functions. Item pc1-c-06.
Let f(t)=12−5(0.6)^(t−1), with real domain. Find f⁻¹(5), rounded to three decimals. Use the entire transformed equation, including its horizontal and vertical shifts.
Answer A. Set 5=12−5(0.6)^(t−1), giving (0.6)^(t−1)=7/5. Hence t=1+ln(1.4)/ln(0.6)≈0.3413168389. The function is increasing because multiplying the decreasing positive exponential by −5 reverses order; its range is (−∞,12), so the output 5 is admissible. Substituting the computed t back into f checks the inversion. Dropping the horizontal shift would instead give −0.659.
Review focus: Invert a shifted exponential while respecting the output range. Unit 2: Exponential and logarithmic functions. Item pc1-c-07.
Two candidate cost models are E(t)=1.6ᵗ and L(t)=t+2 for 0≤t≤6. Graph them using 0≤t≤6 and 0≤y≤18, then refine their intersection. Which t gives the crossover, to three decimals?
Answer D. Solving 1.6ᵗ=t+2 numerically gives t≈3.7049305673. The sign of E−L is negative at 3 and positive at 4, so refine that bracket instead of relying on screen pixels. There is only one nonnegative crossing: E/(t+2) decreases only initially and then increases, or equivalently exponential strict convexity and the negative difference at 0 exclude two positive crossings. The numerical answer compares the stipulated formulas on [0,6]; it does not establish that either cost law is empirically correct.
Review focus: Refine an exponential-linear crossover on a stated model domain. Unit 2: Exponential and logarithmic functions. Item pc1-c-08.
For these synthetic readings, the candidate period is FIXED at 6 input units. Fit y=c+a cos(πx/3)+b sin(πx/3) by unweighted least squares in y to all seven rows. Use radians and retain all coefficients. What is the fixed-fit prediction at x=0.5, rounded to three decimals? Do not search for a new period. To use a calculator without constrained sinusoidal regression, solve the supplied least-squares coefficient equations 7c+a=60, c+4a=23.65 and 3b=√3/20, then evaluate the model.
| x | y |
|---|---|
| 0 | 12.1 |
| 1 | 9.8 |
| 2 | 6.2 |
| 3 | 4.1 |
| 4 | 6.0 |
| 5 | 9.9 |
| 6 | 11.9 |
Answer A. The supplied equations come from minimizing squared residuals for this fixed frequency; they are not a free-period sinusoidal regression command. Eliminate a to obtain 27c=216.35, then recover a and b. The coefficients are c≈8.0129629630, a≈3.9092592593 and b≈0.0288675135. Evaluate c+a cos(π/6)+b sin(π/6)≈11.4129145482. This gives 11.413. The input 0.5 is not added as a new observation, and the query does not refit the model. Period 6 is a supplied candidate: this least-squares result does not certify a globally best period or validate future predictions. Linear interpolation of the two nearest observations would give 10.950 instead.
Review focus: Fit a sinusoidal model with a fixed candidate period and predict without refitting. Unit 3: Trigonometric and polar functions. Item pc1-c-09.
How many solutions does sinθ=0.35 have on 0≤θ≤7, with angles in radians? Use a numerical graph or inverse sine to locate candidates, and retain the periodic branches.
Answer C. The first two solutions are α=arcsin(0.35)≈0.3575711036 and π−α≈2.7840215499. Add integer multiples of 2π to each branch. The next α+2π≈6.6407564108 still lies in[0,7], while π−α+2π≈9.0672068571 does not. Negative shifts give negative angles. Exactly three solutions remain. A principal inverse-sine output alone finds only one branch and does not enumerate a periodic solution set.
Review focus: Recover all periodic equation solutions before applying an interval restriction. Unit 3: Trigonometric and polar functions. Item pc1-c-10.
On the restricted domain −π/12<x<5π/12, let T(x)=2−3tan(2x−π/3). What x satisfies T(x)=−2.5, rounded to three decimals? Use radians.
Answer B. The specified interval maps under 2x−π/3 onto(−π/2,π/2), one continuous strictly increasing tangent branch. The equation requires tan(2x−π/3)=1.5. Therefore x=π/6+(1/2)arctan(1.5)≈1.0149956372, which is inside the interval. Other periodic tangent solutions lie outside this branch. The endpoints are poles and cannot be included.
Review focus: Solve a transformed tangent equation on one pole-free branch. Unit 3: Trigonometric and polar functions. Item pc1-c-11.
A polar curve has r=1+2.5cosθ. On 0≤θ≤π, at what angle does the point have distance1 from the origin AND negative signed radius? Round the angle in radians to three decimals.
Answer B. Distance is |r|. The extra negative-radius condition requires r=−1, not r=1. Thus cosθ=−2/2.5=−0.8 and θ=arccos(−0.8)≈2.4980915448. Cosine is strictly decreasing on[0,π], so this angle is unique. Choosing r=1 would give π/2 but violates the sign requirement. The Cartesian point is(0.8,−0.6), whose distance is 1; its direction is opposite the nominal angle ray.
Review focus: Solve a polar distance condition using absolute signed radius. Unit 3: Trigonometric and polar functions. Item pc1-c-12.
A stipulated height model is H(t)=4+2.5cos(π(t−1)/6) meters for 0≤t≤12 hours. For how many total hours in this interval does H(t)>5? Round to three decimals; include both pieces if the interval cuts a high-height portion.
Answer C. The period is 12 hours. The inequality is cos(π(t−1)/6)>0.4. With α=arccos(0.4), the half-width of each high portion is w=6α/π≈2.2140607174 hours. On[0,12], the two pieces are[0,1+w) and(13−w,12], of lengths1+w andw−1. Total duration2w=12α/π≈4.4281214348 hours. Strict versus non-strict threshold endpoints do not change duration. Retain both pieces; counting only the first gives approximately3.214 hours.
Review focus: Measure the total duration above a periodic threshold in a contextual interval. Unit 3: Trigonometric and polar functions. Item pc1-c-13.
2 questions · suggested time 35 minutes. Each question has 6 practice-rubric points.
CALCULATOR PERMITTED. PC-1.1 Questions 1–2 share a 35-minute section. Let F(x)=ln(x+1)+0.2x² with domain [0,4]. The supplied graph defines G exactly by straight segments through (0,1), (1,3), (2,2), (4,6), with domain [0,4]. T is a discrete function defined only at the five inputs in the table; do not interpolate T. All coordinates are dimensionless. Show equations and lookups. Report numerical answers to three decimal places, retaining extra digits in intermediate work. Each rubric entry is worth one point only when all its named components are present.
| x | T(x) |
|---|---|
| 0 | 2 |
| 1 | 0 |
| 2 | 4 |
| 3 | 1 |
| 4 | 3 |
Graph description: Exact continuous piecewise-linear graph G with vertices (0,1),(1,3),(2,2),(4,6). It rises, then falls between inputs 1 and 2, then rises again. Each listed vertex is included. G: exact joined segments (solid): straight segments through (0, 1), (1, 3), (2, 2), (4, 6). Axes use the labeled scales; no values outside the specified domain are supplied.
Find G(T(3)) and T(G(1)). Show both inner and outer lookups for each composition.
Find F⁻¹(T(4)). Give the equation solved, a bracket for its solution, and a reason that this inverse value is unique without using derivatives.
Determine the complete domain of G∘F. Show the outer-function restriction, the numerical boundary equation and a bracket, and explain endpoint inclusion.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
T(3)=1 and G(1)=3, so G(T(3))=3. Conversely G(1)=3 and T(3)=1, so T(G(1))=1. Composition order matters.
T(4)=3, so solve ln(x+1)+0.2x²=3. F(2)≈1.898612 and F(3)≈3.186294 bracket 3. Both ln(x+1) and x² increase on [0,4], with ln strictly increasing; F is continuous and strictly increasing. Thus the unique answer is x≈2.869548818337, or 2.870. The inverse is not a reciprocal.
Require x∈[0,4] and 0≤F(x)≤4. F(0)=0 and F is continuous and strictly increasing. F(3)≈3.186294 and F(4)≈4.809438 bracket 4. The unique solution of ln(x+1)+0.2x²=4 is b≈3.528222328269. Therefore the domain is [0,b], approximately [0,3.528]. Both endpoints are included because G accepts 0 and 4. Strict increase excludes every larger x in the original domain.
Review focus: Function Concepts: table and graph composition, numerical inverse and composite domain. Unit 1: Polynomial and rational functions. Item pc1-fr-01.
CALCULATOR PERMITTED. Questions 1–2 share a 35-minute section. A synthetic cooling model has temperature C(t)=k+Aq^(t/2) degrees Celsius on the proposed domain 0≤t≤8 hours, where A>0 and 0<q<1. The unknown k is a limiting temperature and q is a multiplicative factor per two hours. Treat the table as exact calibration data for constructing a candidate model, not independent validation. Report numerical answers to three decimal places and keep unrounded intermediate values. Each one-point rubric criterion requires all its named components.
| t (hours) | C(t) (°C) |
|---|---|
| 0 | 74 |
| 2 | 42 |
| 4 | 26 |
Write a system using all three observations and solve for k, A and q. Explain how the consecutive changes determine q, and state the resulting model.
Use your model to find the average rate of temperature change from t=1 to t=5. Include units and interpret the sign and what this average does not assert.
Predict the first time at which C(t)=17°C. Show a calculator equation or equivalent logarithmic setup. Classify the prediction relative to the calibration interval and proposed domain, and propose a new check before relying on it.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
The system is k+A=74, k+Aq=42, k+Aq²=26. Subtraction gives A(q−1)=−32 and Aq(q−1)=−16; their ratio gives q=1/2. Then A=64 and k=10. Thus C(t)=10+64(1/2)^(t/2). The excess above 10°C halves every two hours.
C(1)=10+32√2≈55.254834 and C(5)=10+8√2≈21.313708. The average is [C(5)−C(1)]/(5−1)=−6√2≈−8.485281°C per hour. Temperature falls by an average of about 8.485°C per hour over this interval; this does not assert a constant or instantaneous rate.
Solve 10+64(1/2)^(t/2)=17, so t=2ln(7/64)/ln(1/2)≈6.385290155885 hours. Strict decrease makes this crossing unique and first. It lies within the proposed domain [0,8] but outside the calibration interval [0,4], so it is extrapolation. Compare an unused measurement near this time with the fixed prediction using a stated measurement allowance before revising the model. Exact calibration agreement does not establish accuracy there.
Review focus: Nonperiodic modeling: parameter system, contextual average rate and extrapolation. Unit 2: Exponential and logarithmic functions. Item pc1-fr-02.
2 questions · suggested time 35 minutes. Each question has 6 practice-rubric points.
NO CALCULATOR. PC-1.1 Questions 3–4 share a 35-minute section. The supplied graph is a stipulated sinusoidal level H(t), in meters, on 2≤t≤26 hours and shows two complete cycles. For an equivalent numerical description, a minimum of 2 meters occurs at t=2, a maximum of 8 meters at t=8, and the next minimum at t=14. A, B, C, D and E are consecutive quarter-cycle landmarks in the first cycle: initial minimum, rising midline crossing, maximum, falling midline crossing, next minimum. Give exact results and use no derivatives. Each one-point rubric criterion requires all its named components.
The supplied smooth curve is stipulated to be sinusoidal. A, B, C, D and E are consecutive quarter-cycle landmarks: minimum, rising midline crossing, maximum, falling midline crossing, then minimum. The graph shows two complete cycles, with the second repeating the first. The horizontal axis begins at 2 hours; it does not show time zero. Axis scales supply the numerical readings. The dashed line marks the midline, while filled circles identify the five labeled points.
Give the coordinates of the five labeled points A, B, C, D and E.
Determine the midline, amplitude and period, with units. Construct a cosine formula for H(t) and explain the phase choice.
On 5<t<8, state whether height increases or decreases, whether equal-width average rates increase or decrease, and whether the graph is concave up or down. Justify the whole-interval behavior from the sinusoidal graph or formula; do not equate positive height with a positive or increasing rate.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
The period is 14−2=12 hours, so quarter-cycle spacing is 3 hours and the midline is (2+8)/2=5 meters. A=(2,2), B=(5,5), C=(8,8), D=(11,5), E=(14,2).
Midline 5 meters, amplitude 3 meters, period 12 hours. One formula is H(t)=5+3cos(π(t−8)/6); the cosine maximum is placed at t=8 and repeats at t=20. Equivalently H(t)=5−3cos(π(t−2)/6), which places a minimum at t=2.
Height increases while positive equal-width average rates decrease; the graph is concave down. Write v=π(t−8)/6, which increases from −π/2 to 0. Cosine increases there, but successive chords flatten as the maximum is approached. For a formula justification of concavity, for endpoints m−h and m+h within (5,8), H(m)−[H(m−h)+H(m+h)]/2=3cos(π(m−8)/6)[1−cos(πh/6)]≥0. Midpoint subdivision and continuity give the above-chord property throughout this interval and decreasing equal-width chord rates. For illustration, rates on [5,13/2] and [13/2,8] are √2 and 2−√2 meters/hour; these two samples alone would not prove the whole-interval claim.
Review focus: Periodic modeling: two-cycle graph, parameter recovery and rate/concavity reasoning. Unit 3: Trigonometric and polar functions. Item pc1-fr-03.
NO CALCULATOR. Questions 3–4 share a 35-minute section. Work over the real numbers. Solve the two independent equations and rewrite the expression with its original domain retained. Give exact answers and justify transformations that might introduce or remove solutions. Each one-point rubric criterion requires all its named components.
Solve 3^(2x−1)=5^(x+1) over the real numbers.
Solve cos(2u)=sin u for 0≤u≤2π.
Rewrite E(z)=ln((z−2)²/(z+1))+ln(z+1) first as a single logarithm and then as twice a logarithm involving an absolute value. Determine and retain the original real domain, explaining why cancellation must not enlarge it.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
Both sides are positive for every real x, so taking natural logarithms is reversible. (2x−1)ln3=(x+1)ln5, hence x(2ln3−ln5)=ln3+ln5 and x=ln15/ln(9/5). The denominator is nonzero because 9/5≠1, so this is the unique solution.
Use cos(2u)=1−2sin²u. Then 2sin²u+sin u−1=0, or (2sin u−1)(sin u+1)=0. Thus sin u=1/2 or −1, giving u=π/6, 5π/6, 3π/2. These exhaust the prescribed interval and satisfy the original equation.
The second logarithm requires z+1>0. On that set, the first argument is positive exactly when z≠2. Thus the original domain is z>−1 with z≠2. On this domain the product rule gives E(z)=ln((z−2)²)=2ln|z−2|. The absolute value is needed for valid inputs with z−2<0. The simplified expression alone would also admit z<−1, but those values are excluded by the original logarithms and must remain excluded.
Review focus: Symbolic manipulation: exponential and trigonometric equations, equivalent logarithmic expressions. Unit 2: Exponential and logarithmic functions. Item pc1-fr-04.
These four earlier tasks extend the reasoning practice. They are outside the timed form, its 24-point free-response total and the printed question paper. Keep their item IDs with any responses. Their worked rubrics support feedback rather than an additional examination score.
CALCULATOR PERMITTED. Let F(x)=ln(x+1)+0.2x² with domain [0,4]. The supplied graph defines G exactly by straight segments joining (0,1),(1,3),(2,2),(4,6), with domain [0,4]. The segments are a stipulated function, not an interpolation claim about unknown measurements. All coordinates are dimensionless. Show numerical equations and the graph intervals you use; report numerical answers to three decimal places while retaining extra digits in intermediate work. No calculus is needed.
Graph description: Exact continuous piecewise-linear graph G with vertices (0,1),(1,3),(2,2),(4,6). It rises, then falls between inputs 1 and 2, then rises again. Each listed vertex is included. G: exact joined segments (solid): straight segments through (0, 1), (1, 3), (2, 2), (4, 6). Axes use the labeled scales; no values outside the specified domain are supplied.
Find G(F(2)), showing how you choose the relevant segment of G.
Find the complete domain of the composite G∘F. State the equation used for any numerical boundary, give a bracket for it, and justify that no other boundary is missing without using derivatives.
Find F⁻¹(G(1)). Show the numerical equation and explain why this inverse value exists and is unique on the stated domain.
Find the average rate of change of G on [1,4]. Does its sign establish that G increases throughout that interval? Justify from the supplied graph.
F(2)=ln3+0.8≈1.898612289 lies between 1 and 2. On that segment G(u)=4−u. Thus G(F(2))=3.2−ln3≈2.101387711, or 2.101. Use the computed inner value as the input to G, not the original input 2.
The composite requires x∈[0,4] and F(x)∈[0,4]. On [0,4], ln(x+1) and x² are increasing, with the logarithm strictly increasing, so F is continuous and strictly increasing. F(0)=0, F(3)≈3.186294361 and F(4)≈4.809437912. The upper boundary is the unique solution of ln(x+1)+0.2x²=4 in (3,4), approximately 3.528222328. Therefore the complete domain is [0,3.528222328…], with both endpoints included. A graphing or equation-solver calculation should use this bracket and the stated real domain.
G(1)=3, so solve ln(x+1)+0.2x²=3. Since F(2)≈1.898612289<3<F(3)≈3.186294361 and F is continuous and strictly increasing, the unique inverse input lies in (2,3). The solution is approximately 2.869548818, or 2.870. The symbol F⁻¹ denotes an inverse function, not 1/F.
The average is [G(4)−G(1)]/(4−1)=(6−3)/3=1 output unit per input unit. Its positive sign describes the net endpoint change. G decreases from 3 to 2 over [1,2], so it is not increasing throughout [1,4].
Item pc1-extension-01.
CALCULATOR PERMITTED. A synthetic instrument study records the calibration pairs in the table. The input x is a positive dimensionless ratio to a fixed reference input; response R is measured in sensor units. Fit the family R(x)=a+b ln x by least squares in the original response R, not in ln R. Use all five calibration rows. Keep unrounded coefficients for later calculations, identify your regression variables/settings, and report requested numerical values to three decimal places. A separately supplied check must not be included in the regression.
| Input ratio x | Response R (sensor units) |
|---|---|
| 1 | 1.4 |
| 2 | 3.7 |
| 4 | 6.5 |
| 8 | 8.8 |
| 16 | 11.6 |
Execute the regression and report a and b. Calculate the original-response residuals, defined as observed minus predicted, and their sum of squares. State clearly which quantity was transformed for the fit.
Use the fitted rule to find the input ratio that predicts response 10. Show the inverse equation. Is this an interpolation or extrapolation relative to the calibration input range, and does the calculation establish that a physical instrument will give exactly 10 there?
Before refitting anything, predict the response at x=√8. A fresh measurement there is 5.4 with a stated allowance of ±0.1 sensor units. Calculate its residual, assess compatibility with that allowance, and state how using this check to revise the model changes its evidence status.
Fit R against u=ln x, retaining original R. Equivalently the base-two transformed inputs are 0,1,2,3,4, with mean response 6.4, slope 2.55 and intercept 1.3. Thus a=1.300 and b=2.55/ln2≈3.678872354. In input order, predictions are 1.3,3.85,6.4,8.95,11.5; residuals are 0.10,−0.15,0.10,−0.15,0.10, and SSE=0.075 squared sensor units. A base-two equation R=1.3+2.55 log₂x is equivalent, but b in the requested natural-log form is approximately 3.679.
Solve 10=a+b ln x, so x=exp((10−a)/b)=2^((10−1.3)/2.55)≈10.642496466, or 10.642. It lies between 1 and 16, so it is interpolation in input location, even though no observation was taken at precisely that ratio. It is a prediction of the fitted model, not a verified physical output; parameter uncertainty, measurement variability and family adequacy remain. The inverse uses x>0 and b≠0.
Since log₂√8=1.5, the fixed model predicts 1.3+2.55(1.5)=5.125. Fresh residual=5.4−5.125=0.275 sensor units. The permitted interval [5.3,5.5] excludes 5.125, so the frozen prediction fails this specified allowance-based check. This does not by itself identify the cause or prove every logarithmic model unsuitable. If the new reading is used to revise coefficients or select a model, it becomes development/fitting evidence; a later independent check needs another unused observation.
Item pc1-extension-02.
NO CALCULATOR. A synthetic periodic level H(t), in meters, is stipulated to be sinusoidal for 0≤t≤24 hours. The table gives exact consecutive quarter-cycle landmarks. The listed minimum and maximum values are the global extremes of the stipulated sinusoid, not merely the smallest/largest of a sparse unknown dataset. Give exact expressions and complete intervals; no derivatives are needed.
| Time t (hours) | Level H(t) (meters) |
|---|---|
| 2 | 2 |
| 5 | 5 |
| 8 | 8 |
| 11 | 5 |
| 14 | 2 |
| 17 | 5 |
| 20 | 8 |
Determine the midline, amplitude and period, then write one cosine equation for H. Construct a complete cycle on [2,14] with its exact landmarks and direction between them.
Find every time in [0,24] when H(t)≥13/2. Justify completeness using the periodic model and indicate whether the threshold endpoints are included.
Calculate the average rates of change on [6,7] and [7,8]. Use exact values to explain how the level can increase over both intervals while its average rate becomes smaller. Do not call these finite averages instantaneous rates.
The midline is (8+2)/2=5 m, amplitude (8−2)/2=3 m and period 14−2=12 hours (also 20−8). One rule is H(t)=5+3cos(π(t−8)/6), equivalently 5−3cos(π(t−2)/6). A complete cycle has points (2,2),(5,5),(8,8),(11,5),(14,2), rising from t=2 to 8 and falling from 8 to 14. The argument coefficient π/6 has units radians per hour.
The inequality becomes cos(π(t−8)/6)≥1/2. Thus the phase lies in [−π/3+2kπ,π/3+2kπ], giving t∈[6+12k,10+12k]. Intersecting with [0,24] leaves [6,10] union [18,22]. Both ends of both intervals are included because equality is allowed. Other translates lie outside the time domain.
H(6)=13/2, H(7)=5+3√3/2 and H(8)=8. The first one-hour rate is (3√3−3)/2 m/h; the second is (6−3√3)/2 m/h. Both are positive, and the first exceeds the second because 6√3>9 (equivalently √3>3/2). The level rises while the successive average increases become smaller as it approaches its maximum. These are rates over specified finite intervals, not a derivative value.
Item pc1-extension-03.
NO CALCULATOR. Work over the real numbers and show every relevant domain restriction before cancellation, inversion or comparison. Exact answers are required. The parts concern different functions; restrictions from one part must not be silently transferred to another.
Let R(x)=(x²−x−6)/(x²−9). Simplify R while retaining its original domain, then solve R(x)=5/6. Explain any algebraic candidate that must be rejected.
Let f(x)=1−2·3^(x+1), restricted to x≥−1. Derive f⁻¹, state its exact domain and range, and verify both compositions on their proper domains.
Solve log_(1/2)((x−1)/(x+2))≥1. Give the original logarithm domain, account for the decreasing base and return the complete solution set.
Factor R=[(x−3)(x+2)]/[(x−3)(x+3)]. Its original domain excludes x=−3 and x=3. On that domain R=(x+2)/(x+3), but the canceled input x=3 remains excluded. Solving 6(x+2)=5(x+3) gives x=3, which is invalid. Hence the original equation has no real solution; the simplified expression by itself would have filled the missing point.
The function decreases from −1 at x=−1 without a lower bound, so its range is (−∞,−1]. Solving y=1−2·3^(x+1) gives f⁻¹(y)=−1+log₃((1−y)/2). Its domain is y≤−1 and its range is x≥−1; merely requiring y<1 would ignore the original restriction. For x≥−1, f⁻¹(f(x))=−1+log₃(3^(x+1))=x. For y≤−1, f(f⁻¹(y))=1−2((1−y)/2)=y, with a positive log argument.
The argument must be positive and its denominator nonzero, giving x<−2 or x>1. Because base 1/2 is decreasing, the inequality is equivalent to 0<(x−1)/(x+2)≤1/2. Subtracting 1/2 gives (x−4)/[2(x+2)]≤0, whose solution away from its pole is −2<x≤4. Intersecting with the original log domain yields 1<x≤4. At x=4 the logarithm is exactly 1; x=1 is excluded because its argument is zero. No point in x<−2 satisfies the upper-ratio bound.
Item pc1-extension-04.
Each free-response rubric entry is worth one point; a bundled criterion earns that point only when all its named components are present. Accept equivalent exact expressions. For calculator results, accept values within 0.001 of the stated reference unless an exact value is requested; retain unrounded intermediate values. This tolerance is a convention for this practice form, not a universal official scoring rule.
Award one point for each correct multiple-choice answer and zero for an incorrect or omitted answer. For free response, use the stated criteria rather than rewarding a matching final number alone. Accept equivalent valid mathematics. Unless a criterion explicitly depends on a correct earlier value, award later method points for mathematically consistent carry-forward work; do not award a correctness point for a wrong value. Record the specific error rather than subtracting the same point twice.
Keep three raw totals separate: noncalculator multiple choice out of 29, calculator multiple choice out of 13, and free response out of 24. An optional practice-only weighted percentage is 43.75×(MC-A/29)+18.75×(MC-B/13)+37.5×(FR/24). The two multiple-choice sections have different weights per item; do not combine them as 62.5×(MC/42). This is not an AP 1–5 conversion or a passing threshold. An independently reviewed explanation is stronger evidence than self-marking after reading a solution.
For each missed or uncertain item, record its ID, your original reasoning, the error type, a corrected argument and a later unfamiliar retry. Distinguish a method-selection gap from algebra, graph reading, missing justification, calculator operation or premature rounding. Repeating this exposed form does not create a fresh independent test.
Each multiple-choice item has one primary unit label. Questions can use several ideas; this table is not proof that every learning objective is tested.
| Primary unit | Questions |
|---|---|
| 1. Polynomial and rational functions | 14 |
| 2. Exponential and logarithmic functions | 14 |
| 3. Trigonometric and polar functions | 14 |
| Function type | Questions |
|---|---|
| General functions through non-analytical representations | 7 |
| Polynomial and rational functions | 9 |
| Exponential and logarithmic functions | 12 |
| Trigonometric and polar functions | 14 |
Repair gaps with rates and function evidence, polynomial models, exponential and logarithmic models, semilog representations, periodic and reciprocal graphs, and polar curves. Use the ordinary lesson response and revision tools for further practice; this examination page keeps no response history.
The motion, matrix and transition extensions support broader precalculus study but are not included in this examined-content form. Additional forms, finer objective alignment and independent learner review remain in development.
Format checked 25 October 2026 against the official Precalculus exam overview; six-point free-response allocation follows the Fall 2026 Course and Exam Description, printed page 126; unit organization follows the official course overview. The actual examination combines digital question presentation with handwritten free responses. Use official Bluebook practice and the official graphing-calculator guidance to practice the actual device workflow.
