AP Precalculus -1.14 Function Models: Construction and Application- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.14 Function Models: Construction and Application- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.14 Function Models: Construction and Application- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

In 2016, the cost to mail a package was $2.54 for up to 3 ounces, plus an additional cost of $0.20 for each additional ounce or portion of an ounce less than a full ounce. A portion of the graph of this relationship is given with cost, in dollars, as a function of ounces. Which of the following describes the restrictions on the range for such a function?

(A) The range is positive real numbers.
(B) The range is positive integers.
(C) The range is values of the form $ 2.54 + 0.2x $, where $ x $ is a nonnegative real number.
(D) The range is values of the form $ 2.54 + 0.2x $, where $ x $ is a nonnegative integer.

▶️ Answer/Explanation

Detailed solution

Cost structure: $2.54 for up to 3 ounces, then $0.20 for each additional ounce or portion of an ounce less than a full ounce. This means weight exceeding 3 ounces is rounded up to the next integer ounce for pricing beyond 3 ounces.

Let $ w $ = weight in ounces, $ w > 3 $. Additional ounces counted = $ \lceil w – 3 \rceil $ (ceiling function). Let $ x = \lceil w – 3 \rceil $, a nonnegative integer.

Cost = $ 2.54 + 0.20x $, $ x $ ∈ {0, 1, 2, …}. Thus the range is a set of discrete values of that form, not all real numbers, because $ x $ is integer.

✅ Answer: (D)

Question

Jordan’s cell phone plan includes 5 gigabytes (GB) of data per month and has a monthly cost of $79.95. If Jordan uses more than 5 GB of data within the month, there is a charge of $10 per additional gigabyte of data used. Function $ C $ is used to model Jordan’s monthly cell phone bill, where $ d $ is the number of gigabytes of data used and $ C(d) $ is the cost in dollars. Which of the following defines function $ C $?

(A) $ C(d) = \begin{cases} 79.95 & \text{for } 0 \leq d \leq 5 \\ 79.95 + 10(d – 5) & \text{for } d > 5 \end{cases} $
(B) $ C(d) = \begin{cases} 79.95 & \text{for } 0 \leq d \leq 5 \\ 79.95 + 10d & \text{for } d > 5 \end{cases} $
(C) $ C(d) = \begin{cases} 79.95 & \text{for } 0 \leq d \leq 5 \\ 79.95d + 10 & \text{for } d > 5 \end{cases} $
(D) $ C(d) = 79.95 + 10d $

▶️ Answer/Explanation

Detailed solution

For $ 0 \leq d \leq 5 $, cost = base price $79.95.
For $ d > 5 $, extra data = $ d – 5 $ GB, charged at $10 per extra GB, so cost = $ 79.95 + 10(d – 5) $.
The piecewise function matches this.
✅ Answer: (A)

Question

The function $g$ is given by $g(x)=2\cos(\pi x)+1$. Which of the following is the graph of $g$ for $0\le x\le4$?

▶️ Answer/Explanation

Detailed solution

1. Analyze Parameters:
Amplitude = 2, Midline = 1. Max = 3, Min = -1.
Period = $2\pi / \pi = 2$.

2. Check Key Points:
At $x=0$, $g(0) = 2\cos(0)+1 = 3$.
At $x=1$ (half period), $g(1) = 2\cos(\pi)+1 = -1$.
At $x=2$ (full period), $g(2) = 3$.

3. Match Graph:
Graph (A) starts at 3, drops to -1 at $x=1$, and returns to 3 at $x=2$.

✅ Answer: (A)

Question

The figure shows a square sheet of cardboard on the left and an open box on the right. The square sheet has side length $12$ inches (in) and is used to make the open box by removing from each of the four corners a square of side length $x$ inches, and folding up the sides. For what value of $x$ does the box have the maximum possible volume?

Note: The volume of a rectangular solid with length $l$, width $w$, and height $h$ is $V = lwh$.

(A) $2$

(B) $3$

(D) $6$

▶️ Answer/Explanation

Detailed solution

The side of the box base is $l = w = 12 – 2x$ and the height is $h = x$.
The volume function is $V(x) = x(12 – 2x)^2 = 4x^3 – 48x^2 + 144x$.
To find the maximum, find the derivative: $V'(x) = 12x^2 – 96x + 144$.
Set the derivative to zero: $12(x^2 – 8x + 12) = 0$.
Factor the quadratic: $12(x – 6)(x – 2) = 0$.
Critical values are $x = 2$ and $x = 6$.
Since $x = 6$ results in a volume of $0$, the maximum occurs at $x = 2$.
The correct option is (A).

Question

A builder is trying to find the maximum area that can be bounded by a rectangular region on a certain plot of land. The builder determined four different areas based on the length $x$ of one side of the rectangle, for four different rectangles. The table gives values of the one side of the rectangle, in meters (m), and the corresponding area, in square meters ($\text{m}^2$). After noticing that as $x$ increases, the area values increase and then decrease, the builder used a quadratic regression to produce a function $A$ that models the area based on the constraints. Of the following, what is the maximum value of $A(x)$, to the nearest integer, based on the model?

A) $343$
B) $347$
C) $354$
D) $362$

▶️ Answer/Explanation

Detailed solution

The quadratic regression for the points $(16, 332), (18, 343), (25, 325), (28, 291)$ is $A(x) = -0.932x^2 + 35.85x + -2.48$.
The $x$-coordinate of the vertex (maximum) is found using $x = \frac{-b}{2a} = \frac{-35.85}{2(-0.932)} \approx 19.23$.
Substituting $x \approx 19.23$ back into the model: $A(19.23) = -0.932(19.23)^2 + 35.85(19.23) – 2.48$.
Calculating the value gives $A(19.23) \approx 342.22 + 689.39 – 2.48 \approx 342.3$.
Alternatively, checking a simplified model $A(x) = x(L-x)$, where $P=2L \approx 77$, yields a peak near $19.25$.
Since $18$ and $20$ are symmetric around the peak and $A(18)=343$, the maximum must be slightly higher than $343$.
The regression results confirm the nearest integer maximum value is approximately $347$.
Therefore, the correct choice is B) $347$.

Question

The function $L$ is defined by $L(t) = ae^{-0.0149t} + b$, where $a$ and $b$ are positive constants. If $L(0) = 15$ and $\lim_{t \to \infty} L(t) = 1$, for what value of $t$ is $L(t) = 6$?

(A) $56.866$

(B) $61.496$

(D) $73.732$

▶️ Answer/Explanation

Detailed solution

As $t \to \infty$, $e^{-0.0149t} \to 0$, so $\lim_{t \to \infty} L(t) = b$, which means $b = 1$.
Using $L(0) = 15$, we get $a e^0 + 1 = 15$, which simplifies to $a + 1 = 15$, so $a = 14$.
The function is $L(t) = 14e^{-0.0149t} + 1$.
To find $t$ when $L(t) = 6$, set $14e^{-0.0149t} + 1 = 6$.
Subtract 1 and divide by 14 to get $e^{-0.0149t} = \frac{5}{14}$.
Take the natural log: $-0.0149t = \ln\left(\frac{5}{14}\right)$.
Solve for $t = \frac{\ln(5/14)}{-0.0149} \approx 69.102$.
The correct option is (C).

Question

The figure shows a conical tank with water pouring into it. At $t = 0$, water begins to pour into the empty tank. The rate that water pours into the tank varies in such a way that the depth of the water inside the tank increases at a constant rate with respect to time.

Which of the following graphs could depict this situation, where the depth of the water inside the tank, in feet, is the independent variable and the volume of water inside the tank, in cubic feet, is the dependent variable?

▶️ Answer/Explanation

Detailed solution

The correct option is (B).

The volume $V$ of a cone is defined by the formula $V = \frac{1}{3}\pi r^2 h$.
Since the tank is conical, the radius $r$ is directly proportional to the depth $h$; we can write $r = k \cdot h$.
Substituting $r = kh$ into the volume formula gives $V = \frac{1}{3}\pi (kh)^2 h = \frac{1}{3}\pi k^2 h^3$.
This simplifies to $V = C \cdot h^3$, where $C$ is a positive constant.
This equation indicates that Volume is a cubic function of Depth.
A cubic function $y = x^3$ (for $x > 0$) starts at the origin and increases with an increasing slope (concave up).
Graphs (C) and (D) are incorrect because volume must increase as depth increases.
Graph (A) is concave down, whereas Graph (B) is concave up, matching the cubic relationship.

Question

Jordan’s cell phone plan includes $5$ gigabytes (GB) of data per month and has a monthly cost of $\$79.95$. If Jordan uses more than $5$ GB of data within the month, there is a charge of $\$10$ per additional gigabyte of data used. Function $C$ is used to model Jordan’s monthly cell phone bill, where $d$ is the number of gigabytes of data used and $C(d)$ is the cost in dollars. Which of the following defines function $C$?

(A) $C(d) = \begin{cases} 79.95 & \text{for } 0 \le d \le 5 \\ 79.95 + 10(d – 5) & \text{for } d > 5 \end{cases}$

(B) $C(d) = \begin{cases} 79.95 & \text{for } 0 \le d \le 5 \\ 79.95 + 10d & \text{for } d > 5 \end{cases}$

▶️ Answer/Explanation

Detailed solution

The correct option is (A).
For data usage $0 \le d \le 5$, the cost is a constant flat rate of $\$79.95$.
For data usage $d > 5$, the “additional” data used is calculated as $(d – 5)$.
The extra charge is the rate of $\$10$ multiplied by these additional gigabytes: $10(d – 5)$.
The total cost for $d > 5$ is the base cost plus the extra charge: $79.95 + 10(d – 5)$.
This matches the piecewise structure shown in Option (A).

Question (Calc Allowed)

Boyle’s Law states that the pressure of a gas is inversely proportional to the volume of the gas at a constant temperature. The table gives the volume $V$, in milliliters (mL), of a gas for selected pressures $P$. Which of the following gives a model for the volume of the gas as a function of pressure? (Note: the units for pressure are not included.)

(A) $V(P) = 11.458P$
(B) $V(P) = 1650P$
(C) $V(P) = \frac{11.458}{P}$
(D) $V(P) = \frac{1650}{P}$

▶️ Answer/Explanation

Detailed solution

Boyle’s Law implies an inverse relationship: $P \cdot V = k$, where $k$ is a constant.
Using the first data point from the table: $P = 137.500$ and $V = 12$.
Calculate the constant: $k = 137.500 \times 12 = 1650$.
Verify with another point: $103.125 \times 16 = 1650$.
Rearranging the equation $P \cdot V = 1650$ to solve for $V$ gives $V(P) = \frac{1650}{P}$.
Therefore, the correct model matching the data is option (D).

Question (Calc Allowed)

A ball is thrown through an open window to the ground below. The height of the ball, in meters, at time $t$ seconds after it is thrown can be modeled by the function $h$, given by $h(t) = -4.9t^2 + 4.4t + 15.24$. Which of the following describes the height of the ball above the ground?

(A) The ball begins at its maximum height of $15.240$ meters. The height of the ball decreases until it reaches the ground $1.820$ seconds after it leaves the window.
(B) The ball begins at its maximum height of $15.240$ meters. The height of the ball decreases until it reaches the ground $2.269$ seconds after it leaves the window.
(C) After leaving the window, the height of the ball increases to its maximum height of $16.228$ meters. Then the height of the ball decreases until it reaches the ground $1.820$ seconds after reaching its maximum height.
(D) After leaving the window, the height of the ball increases to its maximum height of $16.228$ meters. Then the height of the ball decreases until it reaches the ground $2.269$ seconds after reaching its maximum height.

▶️ Answer/Explanation

Detailed solution

The initial velocity is $v_0 = 4.4$ m/s (positive), so the ball first moves upward.
The time to reach maximum height is $t = \frac{-b}{2a} = \frac{-4.4}{2(-4.9)} \approx 0.449$ seconds.
The maximum height is $h(0.449) = -4.9(0.449)^2 + 4.4(0.449) + 15.24 \approx 16.228$ meters.
To find when it hits the ground, set $h(t) = 0$ and solve $-4.9t^2 + 4.4t + 15.24 = 0$ using the quadratic formula.
The total time $t \approx 2.269$ seconds is the time from $t=0$ until impact.
The time from maximum height to the ground is $2.269 – 0.449 = 1.820$ seconds.
Therefore, Option (C) correctly describes the upward motion to $16.228$ m and the subsequent $1.820$ s fall.

Question (Calc Allowed)

An X-ray machine is used to eliminate germs in certain food processes. The intensity $I$, in millirads per hour, of X-rays produced by the machine is inversely proportional to the square of the distance $d$, in meters, from the machine. For a certain machine, the intensity is $26.5$ millirads per hour at a distance of $4$ meters. Based on this information and using the same machine, what is the intensity, in millirads per hour, at a distance of $3.3$ meters?

(A) $18.037$
(B) $29.176$
(C) $32.121$
(D) $38.935$

▶️ Answer/Explanation

Detailed solution

The relationship is given as $I \propto \frac{1}{d^2}$, which implies $I \cdot d^2 = k$.
Using the initial values, calculate the constant: $k = 26.5 \cdot 4^2 = 26.5 \cdot 16 = 424$.
Set up the equation for the new distance: $I_{new} \cdot (3.3)^2 = 424$.
Calculate the new distance squared: $(3.3)^2 = 10.89$.
Solve for the new intensity: $I_{new} = \frac{424}{10.89}$.
Perform the division: $I_{new} \approx 38.9348$.
The result rounds to $38.935$, which matches option (D).

Question

Let $f$ be an increasing function defined for $x > 0$. The table gives values for $f(x)$ at selected values of $x$. The function $g$ is given by $g(x) = 0.25x^3 – 9.5x^2 + 110x – 399$.

Part A

(i) The function $h$ is defined by $h(x) = (g \circ f)(x) = g(f(x))$. Find the value of $h(8)$ as a decimal approximation, or indicate that it is not defined.

(ii) Find the value of $f^{-1}(20)$, or indicate that it is not defined.

Part B

(i) Find all values of $x$, as decimal approximations, for which $g(x) = -45$, or indicate there are no such values.

(ii) Determine the end behavior of $g$ as $x$ increases without bound. Express your answer using the mathematical notation of a limit.

Part C

(i) Use the table of values of $f(x)$ to determine if $f$ is best modeled by a linear, quadratic, exponential, or logarithmic function.

(ii) Give a reason for your answer based on the relationship between the change in the output values of $f$ and the change in the input values of $f$.

▶️ Answer/Explanation

Detailed solution

Part A

(i) From the table, we find $f(8) = 15$.

Substitute $15$ into the function $g(x)$: $h(8) = g(15) = 0.25(15)^3 – 9.5(15)^2 + 110(15) – 399$.

$h(8) = 0.25(3375) – 9.5(225) + 1650 – 399$.

$h(8) = 843.75 – 2137.5 + 1650 – 399$.

$h(8) = -42.75$.

(ii) To find $f^{-1}(20)$, we look for the $x$ value where $f(x) = 20$.

From the table, $f(16) = 20$.

Therefore, $f^{-1}(20) = 16$.

Part B

(i) We solve the equation $0.25x^3 – 9.5x^2 + 110x – 399 = -45$.

Set the equation to zero: $0.25x^3 – 9.5x^2 + 110x – 354 = 0$.

Using numerical methods or a graphing calculator, the real solutions are approximately:

$x \approx 5.242$

$x \approx 12.188$

$x \approx 20.570$

(ii) The end behavior of a polynomial is determined by its leading term, $0.25x^3$.

Since the leading coefficient is positive and the degree is odd, as $x \to \infty$, $g(x) \to \infty$.

The limit notation is $\lim_{x \to \infty} g(x) = \infty$.

Part C

(i) The function $f$ is best modeled by a logarithmic function.

(ii) In a logarithmic model, constant changes in the output values correspond to proportional changes in the input values.

As the output $f(x)$ increases by a constant $5$ ($0, 5, 10, 15 \dots$),

The input $x$ values are multiplied by a constant factor of $2$ ($1, 2, 4, 8 \dots$).

This constant ratio of inputs for constant additions of outputs is the hallmark of logarithmic growth.

Question

A graphing calculator is required for these questions.

Let $f$ be an increasing function defined for $x \ge 0$. The table gives values of $f(x)$ at selected values of $x$. The function $g$ is given by $g(x) = \frac{x^3 – 14x – 27}{x+2}$.

(A) (i) The function $h$ is defined by $h(x) = (g \circ f)(x) = g(f(x))$. Find the value of $h(5)$ as a decimal approximation, or indicate that it is not defined.

(ii) Find the value of $f^{-1}(4)$, or indicate that it is not defined.

(B) (i) Find all values of $x$, as decimal approximations, for which $g(x) = 3$, or indicate there are no such values.

(ii) Determine the end behavior of $g$ as $x$ decreases without bound. Express your answer using the mathematical notation of a limit.

(C) (i) Use the table of values of $f(x)$ to determine if $f$ is best modeled by a linear, quadratic, exponential, or logarithmic function.

(ii) Give a reason for your answer based on the relationship between the change in the output values of $f$ and the change in the input values of $f$.

▶️ Answer/Explanation

Detailed solution

(A) (i)
First, evaluate the inner function $f(5)$ using the table: $f(5) = 34$.
Next, substitute this value into $g(x)$ to find $h(5) = g(34)$.
$g(34) = \frac{34^3 – 14(34) – 27}{34 + 2}$
$g(34) = \frac{39304 – 476 – 27}{36} = \frac{38801}{36}$
$h(5) \approx 1077.806$

(A) (ii)
To find $f^{-1}(4)$, we look for the input value $x$ in the table that produces an output of $4$.
The table shows that $f(3) = 4$.
Therefore, $f^{-1}(4) = 3$.

(B) (i)
Set $g(x) = 3$ and solve for $x$:
$\frac{x^3 – 14x – 27}{x + 2} = 3$
Multiply both sides by $(x+2)$: $x^3 – 14x – 27 = 3(x + 2)$
Simplify: $x^3 – 14x – 27 = 3x + 6$
Rearrange into polynomial form: $x^3 – 17x – 33 = 0$
Using a graphing calculator to find the zero of this polynomial:
$x \approx 4.879$

(B) (ii)
We evaluate the limit as $x \to -\infty$ for $g(x)$.
$\lim_{x \to -\infty} \frac{x^3 – 14x – 27}{x + 2}$
By examining the leading terms, the function behaves like $\frac{x^3}{x} = x^2$ for large absolute values of $x$.
As $x \to -\infty$, $x^2 \to \infty$.
$\lim_{x \to -\infty} g(x) = \infty$

(C) (i)
Calculate the first differences of $f(x)$: $(-5) – (-10) = 5$, $4 – (-5) = 9$, $17 – 4 = 13$, $34 – 17 = 17$.
Calculate the second differences: $9 – 5 = 4$, $13 – 9 = 4$, $17 – 13 = 4$.
Since the second differences are constant, $f$ is best modeled by a quadratic function.

(C) (ii)
The model is quadratic because for equal intervals of the input values $x$ (step size of $1$), the rate of change of the output values increases by a constant amount (constant second difference of $4$).

AP Precalculus -1.14 Function Models: Construction and Application- MCQ Exam Style Questions - Effective Fall 2023