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2025 AP AP Precalculus Practice Paper with Answers

Thinka May 2025 AP-Style Mock — AP Precalculus

24 marks60 mins2025

An original Thinka practice paper modelled on the structure and difficulty of the May 2025 AP AP Precalculus paper. Not affiliated with or reproduced from AP.

Section II Part A (Graphing Calculator Required)

Answer 2 free-response questions in 30 minutes. Graphing calculator required. Decimal approximations must be correct to three decimal places unless specified otherwise. Show all supporting work.

2 Question · 12 marks

Question 1 · Free Response: Function Concepts (Table & Analytical)

6 marks

A marine biologist measures the dissolved oxygen concentration in a coastal estuary following an environmental restoration treatment. The dissolved oxygen concentration, in milligrams per liter ($\text{mg/L}$), is modeled by a function $D(t)$, where $t$ is the time, in days, since the treatment began for $0 \le t \le 12$. Selected values of $D(t)$ are shown in the table below.

$$\begin{array}{|c|c|c|c|c|c|}
\hline
t \text{ (days)} & 0 & 3 & 6 & 9 & 12 \\
\hline
D(t) \text{ (mg/L)} & 2.10 & 4.35 & 6.00 & 7.05 & 7.50 \\
\hline
\end{array}$$

(A)
(i) Use the given data to find the average rate of change of dissolved oxygen concentration, in milligrams per liter per day, from $t = 3$ to $t = 9$ days. Express your answer as a decimal approximation. Show the computations that lead to your answer.
(ii) Interpret the meaning of your answer from part (i) in the context of the problem.

(B)
(i) Based on the values in the table, which of the following function types best models $D(t)$: linear, quadratic, or exponential? Give a reason for your answer based on the rate of change of the data over consecutive equal-length input intervals.
(ii) A quadratic model $M(t) = at^2 + bt + c$ is used to model the data. Use the data at $t = 0$, $t = 6$, and $t = 12$ to write three equations that can be used to find the values for constants $a$, $b$, and $c$.

(C)
In a connected deeper reservoir, the dissolved oxygen concentration is modeled by the function

$$R(t) = \frac{10.5t + 18}{1.2t + 8}$$

for $t \ge 0$, where $R(t)$ is measured in milligrams per liter and $t$ is measured in days.

(i) Find all values of $t$, as decimal approximations, for which $R(t) = 7.2$, or indicate that there are no such values. Show the work that leads to your answer.
(ii) Determine the end behavior of $R(t)$ as $t$ increases without bound. Express your answer using the mathematical notation of a limit.

Show answer & marking scheme

Worked solution

### Part (A)
(i) The average rate of change from $t = 3$ to $t = 9$ is:
$$\frac{D(9) - D(3)}{9 - 3} = \frac{7.05 - 4.35}{6} = \frac{2.70}{6} = 0.45 \text{ mg/L per day}$$

(ii) From day $t = 3$ to day $t = 9$, the dissolved oxygen concentration in the estuary increases at an average rate of $0.45\text{ mg/L}$ per day.

---

### Part (B)
(i) Calculate the average rates of change over consecutive 3-day intervals:
- On $[0, 3]$: $\frac{4.35 - 2.10}{3} = \frac{2.25}{3} = 0.75$
- On $[3, 6]$: $\frac{6.00 - 4.35}{3} = \frac{1.65}{3} = 0.55$
- On $[6, 9]$: $\frac{7.05 - 6.00}{3} = \frac{1.05}{3} = 0.35$
- On $[9, 12]$: $\frac{7.50 - 7.05}{3} = \frac{0.45}{3} = 0.15$

The differences between consecutive average rates of change are:
- $0.55 - 0.75 = -0.20$
- $0.35 - 0.55 = -0.20$
- $0.15 - 0.35 = -0.20$

Because the average rates of change change at a constant rate over consecutive equal-length intervals, a quadratic function best models the data.

(ii) Substitute $t = 0, 6, 12$ and their corresponding output values into $M(t) = at^2 + bt + c$:
1. $a(0)^2 + b(0) + c = 2.10$ (or $c = 2.10$)
2. $a(6)^2 + b(6) + c = 6.00$ (or $36a + 6b + c = 6.00$)
3. $a(12)^2 + b(12) + c = 7.50$ (or $144a + 12b + c = 7.50$)

---

### Part (C)
(i) Set $R(t) = 7.2$ and solve for $t$:
$$\frac{10.5t + 18}{1.2t + 8} = 7.2$$
$$10.5t + 18 = 7.2(1.2t + 8)$$
$$10.5t + 18 = 8.64t + 57.6$$
$$1.86t = 39.6$$
$$t = \frac{39.6}{1.86} \approx 21.290$$

(ii) The end behavior as $t \to \infty$ is given by the horizontal asymptote:
$$\lim_{t \to \infty} R(t) = \lim_{t \to \infty} \frac{10.5t + 18}{1.2t + 8} = \frac{10.5}{1.2} = 8.75$$

Marking scheme

### Point Breakdown

- Point A1 (Part A(i)): 1 point for the correct average rate of change with valid supporting computation $\frac{7.05 - 4.35}{9 - 3} = 0.45$.
- Point A2 (Part A(ii)): 1 point for a correct interpretation in context, including units (mg/L per day) and referencing the time interval $t = 3$ to $t = 9$ days.
- Point B1 (Part B(i)): 1 point for selecting quadratic with a valid justification citing constant second differences or a linear rate of change with supporting values.
- Point B2 (Part B(ii)): 1 point for correctly writing a system of three equations in terms of $a, b,$ and $c$.
- Point C1 (Part C(i)): 1 point for the correct value $t \approx 21.290$ (or $21.291$ rounded) with setup/equation shown.
- Point C2 (Part C(ii)): 1 point for the correct limit statement: $\lim_{t \to \infty} R(t) = 8.75$ (or $\frac{35}{4}$).

### General & Scoring Notes
- Precision: Numerical values must be correct to three decimal places unless exact values are given.
- Supporting Work: Answers with no supporting work receive 0 points where required.
- Part B(ii): Equations may be left in unsimplified form (e.g., $a(6)^2 + b(6) + c = 6.00$).

Question 2 · Free Response: Modeling a Non-Periodic Context

6 marks

The concentration of a chemical compound in an industrial runoff sample is measured over time as a neutralizing agent is added. The table gives the concentration of the compound, $ C(t) $, in milligrams per liter (mg/L), at time $ t $ hours after the neutralizing agent is introduced.

$$\begin{array}{|c|c|c|c|c|}
\hline
t \text{ (hours)} & 0 & 2 & 4 & 6 \\
\hline
C(t) \text{ (mg/L)} & 120 & 84 & 58.8 & 41.16 \\
\hline
\end{array}$$

(A)
(i) Use the given data to find the average rate of change of the concentration of the compound from $ t = 0 $ to $ t = 4 $ hours. Express your answer as a decimal approximation. Show the computations that lead to your answer.
(ii) Interpret the meaning of your answer from part (i) in the context of the problem, including appropriate units.

(B)
(i) Based on the table, which of the following function types best models $ C(t) $: linear, quadratic, exponential, or logarithmic? Give a reason for your answer based on the relationship between the change in the output values of $ C $ and the change in the input values of $ t $. Refer to the values in the table in your reasoning.
(ii) An exponential model $ M(t) = a b^t $ is used to model the concentration of the compound at time $ t $ hours. Use the given data to write two equations that can be used to find the values for constants $ a $ and $ b $, and determine the values of $ a $ and $ b $.

(C)
(i) Use the model $ M(t) $ from part (B) to find the time $ t $, in hours, at which the concentration of the compound is $ 15 \text{ mg/L} $. Express your answer as a decimal approximation. Show the work that leads to your answer.
(ii) Determine the end behavior of $ M(t) $ as $ t $ increases without bound. Express your answer using the mathematical notation of a limit.

Show answer & marking scheme

Worked solution

(A)(i)
The average rate of change from $ t = 0 $ to $ t = 4 $ is given by:
$$\frac{C(4) - C(0)}{4 - 0} = \frac{58.8 - 120}{4 - 0} = \frac{-61.2}{4} = -15.3 \text{ mg/L per hour}$$

(A)(ii)
From $ t = 0 $ to $ t = 4 $ hours, the concentration of the chemical compound decreases by an average of $ 15.3 \text{ mg/L} $ each hour (or the average rate of change is $ -15.3 \text{ mg/L per hour} $).

(B)(i)
An exponential function best models $ C(t) $.
Reason: Over equal increments of time ($ \Delta t = 2 $), the ratio of consecutive output values is constant:
$$\frac{84}{120} = 0.7, \quad \frac{58.8}{84} = 0.7, \quad \frac{41.16}{58.8} = 0.7$$
Because successive output values change by a constant factor for equal intervals of input values, the function is exponential.

(B)(ii)
Using the data points $ (0, 120) $ and $ (2, 84) $:
1. $ 120 = a \cdot b^0 $
2. $ 84 = a \cdot b^2 $

From the first equation, $ a = 120 $.
Substituting into the second equation:
$$84 = 120 b^2 \implies b^2 = \frac{84}{120} = 0.7 \implies b = \sqrt{0.7} \approx 0.837$$
Thus, $ a = 120 $ and $ b = \sqrt{0.7} \approx 0.837 $.

(C)(i)
To find the time $ t $ when $ M(t) = 15 $:
$$120(\sqrt{0.7})^t = 15 \quad \text{or} \quad 120(0.7)^{t/2} = 15$$
$$(0.7)^{t/2} = \frac{15}{120} = 0.125$$
$$\frac{t}{2} \ln(0.7) = \ln(0.125)$$
$$t = 2 \cdot \frac{\ln(0.125)}{\ln(0.7)} \approx 11.660 \text{ hours}$$

(C)(ii)
As $ t $ increases without bound:
$$\lim_{t \to \infty} M(t) = 0$$

Marking scheme

### Scoring Guidelines (6 Points Total)

- Point A1 (1 mark): Average rate of change
- Earned for computing $ \frac{58.8 - 120}{4 - 0} = -15.3 $ with correct work shown.

- Point A2 (1 mark): Interpretation with units
- Earned for correctly interpreting the rate of change in context, referencing the interval $ [0, 4] $, the direction of change (decrease), and using correct units (mg/L per hour).

- Point B1 (1 mark): Function type and justification
- Earned for selecting exponential AND providing a correct justification demonstrating equal ratios (e.g., $ \frac{84}{120} = \frac{58.8}{84} = 0.7 $) over equal intervals of $ \Delta t = 2 $.

- Point B2 (1 mark): System of equations and constants
- Earned for writing two valid equations (e.g., $ 120 = a b^0 $ and $ 84 = a b^2 $) AND finding $ a = 120 $ and $ b = \sqrt{0.7} \approx 0.837 $.

- Point C1 (1 mark): Model evaluation / solving for $ t $
- Earned for setting up $ M(t) = 15 $ and correctly solving to obtain $ t \approx 11.660 $ (or $ 11.666 $ if using rounded $ b = 0.837 $).

- Point C2 (1 mark): End behavior limit notation
- Earned for correctly writing $ \lim_{t \to \infty} M(t) = 0 $.

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Section II Part B (No Calculator Allowed)

Answer 2 free-response questions in 30 minutes. No calculator allowed. Exact values required unless specified otherwise. Show all steps and supporting work.

2 Question · 12 marks

Question 1 · Free Response: Modeling a Periodic Context

6 marks

A robotic arm on an automated manufacturing track oscillates back and forth along a straight horizontal rail. Let $ x(t) $ represent the position of the robotic arm, in centimeters from the left end of the rail, at time $ t $ seconds after the machine is turned on.

At time $ t = 3 $ seconds, the robotic arm is at its minimum distance of $ 18 \text{ cm} $ from the left end. The arm moves to the right and reaches its maximum distance of $ 82 \text{ cm} $ from the left end for the first time at $ t = 11 $ seconds. The position of the arm is modeled by a sinusoidal function $ x(t) $.

(A)
(i) Find the period and amplitude of the function $ x(t) $. Show the computations that lead to your answers.
(ii) Write an equation for the midline of the graph of $ y = x(t) $.

(B) The function $ x $ can be written in the form $ x(t) = a \cos(b(t - c)) + d $, where $ a > 0 $. Find the values of the constants $ a, b, c, $ and $ d $.

(C)
(i) Find the exact value of $ x(7) $. Show the work that leads to your answer.
(ii) In the context of the problem, determine whether the robotic arm is moving to the left or to the right at time $ t = 7 $ seconds. Give a reason for your answer based on the behavior of the model.

Show answer & marking scheme

Worked solution

(A)(i)
The time elapsed between the minimum position at $ t = 3 $ and the consecutive maximum position at $ t = 11 $ represents half of one full period:
\[ \frac{1}{2}\text{Period} = 11 - 3 = 8 \implies \text{Period} = 2 \times 8 = 16 \text{ seconds} \]

The amplitude is half the difference between the maximum and minimum values:
\[ \text{Amplitude} = \frac{82 - 18}{2} = \frac{64}{2} = 32 \text{ cm} \]

(A)(ii)
The midline is the average of the maximum and minimum values:
\[ y = \frac{82 + 18}{2} = \frac{100}{2} = 50 \]
Equation of the midline: $ y = 50 $.

(B)
From the properties of the sinusoidal function $ x(t) = a \cos(b(t - c)) + d $ with $ a > 0 $:
- Amplitude: $ a = 32 $
- Period $ P = \frac{2\pi}{b} = 16 \implies b = \frac{2\pi}{16} = \frac{\pi}{8} $
- Midline / Vertical shift: $ d = 50 $
- Since $ a > 0 $, a standard cosine function achieves a maximum at $ t = c $. A maximum occurs at $ t = 11 $, so $ c = 11 $ (or any value $ 11 + 16k $ for integer $ k $, such as $ c = -5 $ or $ c = 27 $).

Thus, $ a = 32 $, $ b = \frac{\pi}{8} $, $ c = 11 $, and $ d = 50 $.

(C)(i)
Using the model $ x(t) = 32 \cos\left(\frac{\pi}{8}(t - 11)\right) + 50 $:
\[ x(7) = 32 \cos\left(\frac{\pi}{8}(7 - 11)\right) + 50 = 32 \cos\left(\frac{-4\pi}{8}\right) + 50 = 32 \cos\left(-\frac{\pi}{2}\right) + 50 \]
Since $ \cos\left(-\frac{\pi}{2}\right) = 0 $:
\[ x(7) = 32(0) + 50 = 50 \text{ cm} \]

(C)(ii)
The robotic arm is at its minimum position ($ 18 \text{ cm} $) at $ t = 3 $ and reaches its maximum position ($ 82 \text{ cm} $) at $ t = 11 $. Since $ 3 < 7 < 11 $ and $ x(t) $ is strictly increasing on the interval $ (3, 11) $, the distance from the left end is increasing at $ t = 7 $. Therefore, the robotic arm is moving to the right.

Marking scheme

Part A (2 points)
- Point A1 (1 mark): Period = $ 16 $ and Amplitude = $ 32 $ with supporting work.
- Point A2 (1 mark): Equation for midline: $ y = 50 $ (accept $ x = 50 $ or stating the midline is $ 50 $).

Part B (2 points)
- Point B1 (1 mark): Values for $ a = 32 $, $ b = \frac{\pi}{8} $, and $ d = 50 $.
- Point B2 (1 mark): Value for $ c = 11 $ (or equivalent valid phase shift consistent with $ a > 0 $, e.g., $ c = -5 $).

Part C (2 points)
- Point C1 (1 mark): $ x(7) = 50 $ with supporting work.
- Point C2 (1 mark): Direction ("moving to the right") with valid justification based on $ x(t) $ increasing on the interval $ (3, 11) $ (or equivalent explanation showing the arm is traveling from minimum to maximum).

Question 2 · Free Response: Symbolic Manipulations

6 marks

Directions:
- Unless otherwise specified, the domain of a function $f$ is assumed to be the set of all real numbers $x$ for which $f(x)$ is a real number.
- Angle measures for trigonometric functions are given in radians.
- Solutions to equations must be real numbers. Determine the exact value of any expression that can be written without trigonometric or logarithmic functions.

---

(A) The functions $f$ and $g$ are given by
\[f(x) = 2\sin^2(x) - 3\sin(x) + 1\]
\[g(x) = \tan\left(\arccos\left(-\frac{\sqrt{3}}{2}\right)\right)\]
(i) Find all values of $x$ in the interval $[0, 2\pi)$ that satisfy $f(x) = 0$.
(ii) Find the exact value of $g(x)$.

(B) The function $h$ is given by
\[h(x) = 5\cos^2(x) + 2\sin^2(x) - 3\]
(i) Rewrite $h(x)$ as an expression in which $\cos(x)$ appears exactly once and no other trigonometric functions are involved.
(ii) Find all values of $x$ in the interval $[0, 2\pi)$ that satisfy $h(x) = \frac{5}{4}$.

(C) The function $k$ is given by
\[k(x) = \log_3(4x - 5) - \log_3(x - 2)\]
(i) Rewrite $k(x)$ as a single logarithmic expression of the form $\log_3(A(x))$.
(ii) Find all input values in the domain of $k$ that yield an output value of $2$.

Show answer & marking scheme

Worked solution

(A)(i) Set $f(x) = 0$:
\[2\sin^2(x) - 3\sin(x) + 1 = 0\]
Factoring gives:
\[(2\sin(x) - 1)(\sin(x) - 1) = 0\]
Thus, $\sin(x) = \frac{1}{2}$ or $\sin(x) = 1$.
On the interval $[0, 2\pi)$:
- $\sin(x) = \frac{1}{2} \implies x = \frac{\pi}{6}, \frac{5\pi}{6}$
- $\sin(x) = 1 \implies x = \frac{\pi}{2}$
Therefore, the solutions are $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}$.

(A)(ii) Evaluate $\arccos\left(-\frac{\sqrt{3}}{2}\right)$:
The principal range of $\arccos(u)$ is $[0, \pi]$. Since $\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$, we have $\arccos\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6}$.
Then,
\[g(x) = \tan\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{3} \text{ (or } -\frac{1}{\sqrt{3}}\text{)}\]

(B)(i) Using the Pythagorean identity $\sin^2(x) = 1 - \cos^2(x)$:
\[h(x) = 5\cos^2(x) + 2(1 - \cos^2(x)) - 3 = 5\cos^2(x) + 2 - 2\cos^2(x) - 3 = 3\cos^2(x) - 1\]

(B)(ii) Set $h(x) = \frac{5}{4}$:
\[3\cos^2(x) - 1 = \frac{5}{4} \implies 3\cos^2(x) = \frac{9}{4} \implies \cos^2(x) = \frac{3}{4} \implies \cos(x) = \pm\frac{\sqrt{3}}{2}\]
On the interval $[0, 2\pi)$:
- $\cos(x) = \frac{\sqrt{3}}{2} \implies x = \frac{\pi}{6}, \frac{11\pi}{6}$
- $\cos(x) = -\frac{\sqrt{3}}{2} \implies x = \frac{5\pi}{6}, \frac{7\pi}{6}$
Therefore, the solutions are $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$.

(C)(i) Using the logarithm quotient property $\log_b(M) - \log_b(N) = \log_b\left(\frac{M}{N}\right)$:
\[k(x) = \log_3\left(\frac{4x - 5}{x - 2}\right)\]

(C)(ii) Set $k(x) = 2$:
\[\log_3\left(\frac{4x - 5}{x - 2}\right) = 2 \implies \frac{4x - 5}{x - 2} = 3^2 = 9\]
\[4x - 5 = 9(x - 2) \implies 4x - 5 = 9x - 18 \implies 5x = 13 \implies x = \frac{13}{5}\]
Check domain: For $x = \frac{13}{5} = 2.6$, $4(2.6) - 5 = 5.4 > 0$ and $2.6 - 2 = 0.6 > 0$, so $x = \frac{13}{5}$ is in the domain of $k$.

Marking scheme

### Scoring Guidelines (6 Points Total)

- Point A1 (1 mark): Correct values of $x$ for part (A)(i): $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}$.
- Point A2 (1 mark): Correct exact value for part (A)(ii): $-\frac{\sqrt{3}}{3}$ or $-\frac{1}{\sqrt{3}}$.
- Point B1 (1 mark): Correct rewritten expression for $h(x)$: $3\cos^2(x) - 1$ or $3(\cos(x))^2 - 1$.
- Point B2 (1 mark): Correct values of $x$ for part (B)(ii): $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$.
- Point C1 (1 mark): Correct single logarithmic expression for $k(x)$: $\log_3\left(\frac{4x - 5}{x - 2}\right)$.
- Point C2 (1 mark): Correct value $x = \frac{13}{5}$ with valid supporting algebraic steps.

### Scoring Notes:
- In part (A)(i), all three solutions must be present and no extraneous solutions in $[0, 2\pi)$ to earn Point A1.
- In part (A)(ii), unsimplified expressions such as $\tan(5\pi/6)$ earn 0 points. The value must be purely numerical and exact.
- In part (B)(i), $\cos(x)$ must appear only once; expressions containing $\sin(x)$ receive 0 points.
- In part (B)(ii), all four solutions must be listed without extraneous values.
- In part (C)(ii), student must show steps solving $\frac{4x - 5}{x - 2} = 9$ or $3^2$.

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