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AP Precalculus -3.7 Sinusoidal Modeling- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -3.7 Sinusoidal Modeling- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -3.7 Sinusoidal Modeling- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question 

The table gives ordered pairs for five points from a larger data set. The larger data set can be modeled by a sinusoidal function \(f\) with a period of 4. The maximum values of the data set occur at \(x\)-values that are multiples of 4. Which of the following best defines \(f(x)\) for the larger data set?
(A) \(\cos\left(\frac{\pi}{2}x\right) + 4\)
(B) \(\cos(\pi x) + 4\)
(C) \(2\cos\left(\frac{\pi}{2}x\right) + 4\)
(D) \(2\cos(\pi x) + 4\)
▶️ Answer/Explanation
Detailed solution

From the table:
Max value \(y=5\) at \(x=0,4\) (multiples of 4)
Min value \(y=3\) at \(x=2\)
Midline: \( \frac{5+3}{2} = 4 \)
Amplitude: \( \frac{5-3}{2} = 1 \)
Period = 4 → \( \frac{2\pi}{b} = 4 \) → \( b = \frac{\pi}{2} \)
Using cosine with max at \(x=0\): \( f(x) = 1\cdot \cos\left(\frac{\pi}{2}x\right) + 4 \)
✅ Answer: (A)

Question 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
A table gives temperatures in degrees Fahrenheit in a town on a given day. The sinusoidal function \( F(t) = 8 \cos \left( \frac{\pi}{12} (t + c) \right) + 30 \) models the data, where \( t \) is hours past midnight. Which of the following is true about the value of \( c \)?
(A) The value of \( c \) is \( 2 \) because this accounts for a phase shift that aligns a minimum value of the data set with a minimum value of \( F \).
(B) The value of \( c \) is \( 2 \) because this accounts for a vertical shift that aligns a minimum value of the data set with a minimum value of \( F \).
(C) The value of \( c \) is \( 14 \) because this accounts for a phase shift that aligns a maximum value of the data set with a maximum value of \( F \).
(D) The value of \( c \) is \( 14 \) because this accounts for a vertical shift that aligns a maximum value of the data set with a maximum value of \( F \).
▶️ Answer/Explanation
Detailed solution

Step 1: Identify key features of the data

From the table, the maximum temperature is \(38^\circ\mathrm{F}\) at \(t = 14\),
and the minimum temperature is \(22^\circ\mathrm{F}\) at \(t = 2\).

Step 2: Interpret the sinusoidal model

The model is \(F(t) = 8\cos\left(\dfrac{\pi}{12}(t + c)\right) + 30\).

The amplitude is \(8\), the midline is \(30\), and the period is
\(\dfrac{2\pi}{\pi/12} = 24\) hours, which matches a daily temperature cycle.

Step 3: Use the cosine maximum

A cosine function reaches its maximum when its argument is \(0\).
Thus, the maximum of \(F\) occurs when
\(\dfrac{\pi}{12}(t + c) = 0\), which gives \(t + c = 0\).

Step 4: Align the maximum of the model with the data

The data show a maximum temperature at \(t = 14\).
Substituting gives \(14 + c = 0\), so \(c = -14\).

Since cosine is periodic, a phase shift of \(-14\) is equivalent to \(14\).
Thus, \(c = 14\).

Step 5: Interpret the meaning of \(c\)

The constant \(c\) appears inside the cosine function, so it represents a
phase shift. The value \(c = 14\) aligns a maximum of the model with a
maximum of the data.

\(\boxed{\text{Correct answer: (C)}}\)

Question 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
The vertical motion of the tip of a needle on a sewing machine is periodic with respect to time and can be modeled by a sinusoidal function. The figure shows the location of the tip of the needle in relation to a cloth plate. The tip of the needle moves straight up and down above and below the cloth plate. The table gives a consecutive maximum and minimum displacement of the tip at two times. The function \( d \) models the displacement, in centimeters, of the tip of the needle from the cloth plate at time \( t \), in seconds. Which of the following expressions could define \( d(t) \)?
Time \( t \) (seconds)Displacement (cm)
02.5
1/24–2.5
(A) \( 2.5 \cos\left(\frac{1}{12}t\right) \)
(B) \( 2.5 \cos\left(\frac{\pi}{6}t\right) \)
(C) \( 2.5 \cos(24\pi t) \)
(D) \( 2.5 \cos(48\pi t) \)
▶️ Answer/Explanation
Detailed solution

From the table: at \( t = 0 \), displacement = \( 2.5 \) cm (maximum); at \( t = \frac{1}{24} \) s, displacement = \( -2.5 \) cm (minimum).
Amplitude \( A = \frac{2.5 – (-2.5)}{2} = 2.5 \) cm.
The time from a maximum to the next minimum is half the period, so \( T/2 = \frac{1}{24} \) ⇒ \( T = \frac{1}{12} \) seconds.
For \( d(t) = A\cos(bt) \), the period is \( T = \frac{2\pi}{b} \).
Set \( \frac{2\pi}{b} = \frac{1}{12} \) ⇒ \( b = 24\pi \).
Thus \( d(t) = 2.5\cos(24\pi t) \), which matches choice (C).
✅ Answer: (C)

Question 

In the tidal area of a certain city, a sinusoidal function \( f(x) = a \sin(b(x + c)) + d \), where \( a, b, c, \) and \( d \) are constants, is used to model one cycle of high and low tides. The maximum value of the tide is 8.88 feet, and the minimum value of the tide is 0.54 feet in that cycle. If the values of \( b, c, \) and \( d \) have already been determined to fit the data, which of the following would best define \( f(x) \)?
(A) \( 4.17 \sin(b(x + c)) + d \)
(B) \( 4.44 \sin(b(x + c)) + d \)
(C) \( 4.71 \sin(b(x + c)) + d \)
(D) \( 8.34 \sin(b(x + c)) + d \)
▶️ Answer/Explanation
Detailed solution

Amplitude \( a \) is half the difference between the maximum and minimum:
\( a = \frac{8.88 – 0.54}{2} = \frac{8.34}{2} = 4.17 \).
Thus the function is \( f(x) = 4.17 \sin(b(x + c)) + d \).
✅ Answer: (A)

Question 

The table gives the maximum temperature in degrees Celsius on the first day of each of nine months. The function \( f(\theta) = a \sin(b(\theta + c)) + d \) models these data with period 12 months. Based on the table, which is the best value for \( d \)?
Month \(\theta\)123456789
Temp (°C)6.1-5.5-6.010.017.225.630.632.226.1
(A) \( \frac{\pi}{6} \)
(B) 13
(C) 19
(D) 38
▶️ Answer/Explanation
Detailed solution

The parameter \( d \) represents the vertical shift (midline) of the sinusoidal model, which is the average of the maximum and minimum temperatures in the data.
Maximum temperature = 32.2°C, Minimum temperature = -6.0°C.
Midline \( d = \frac{32.2 + (-6.0)}{2} = \frac{26.2}{2} = 13.1 \approx 13 \).
✅ Answer: (B)

Question 

 
 
 
 
 
 
 
 
 
 
 
The height of point \( P \) is given by \( h(t) = 6 + 6\cos\left(\frac{\pi}{3}t\right) \) inches. If the child rides for 5 minutes (300 seconds), how many times will point \( P \) touch the ground?
(A) 1
(B) 50
(C) 100
(D) 286
▶️ Answer/Explanation
Detailed solution

Ground level corresponds to \( h(t) = 0 \).
The minimum of \( h(t) \) occurs when \( \cos\left(\frac{\pi}{3}t\right) = -1 \), giving \( h_{\min} = 6 + 6(-1) = 0 \).
Thus \( P \) touches ground once per period when \(\cos = -1\).
Period \( T = \frac{2\pi}{\pi/3} = 6 \) seconds.
Number of periods in 300 seconds: \( \frac{300}{6} = 50 \).
So \( P \) touches ground 50 times.
✅ Answer: (B)

Question 

Part of a video game design involves the use of one period of a sinusoidal function as the path that a spaceship will follow across a rectangular video screen. The video screen has a width of 1000 pixels and a height of 600 pixels. The values \( x = 0 \) and \( x = 1000 \) represent the left and right sides of the screen, respectively. The values \( y = 0 \) and \( y = 600 \) represent the bottom and top sides of the screen, respectively.

The path of the spaceship begins on the left side of the screen, \( x = 0 \), and completes one period of a sinusoidal function by ending on the right side of the screen, \( x = 1000 \). During its path, the spaceship reaches its minimum height of \( y = 200 \) before reaching its maximum height of \( y = 500 \). If \( y = f(x) \) models the path of the spaceship, which of the following could define \( f(x) \)?

(A) \(-300 \sin \left( \frac{\pi}{500} x \right) + 350\)
(B) \(-150 \sin \left( \frac{\pi}{500} x \right) + 350\)
(C) \( 150 \sin \left( \frac{\pi}{500} x \right) + 350\)
(D) \( 300 \sin \left( \frac{\pi}{500} x \right) + 350\)
▶️ Answer/Explanation
Detailed solution

Midline: \( \frac{500 + 200}{2} = 350 \).
Amplitude: \( \frac{500 – 200}{2} = 150 \).
Period = 1000 ⇒ \( \frac{2\pi}{b} = 1000 \) ⇒ \( b = \frac{\pi}{500} \).

Since the ship hits the minimum before the maximum, the sine wave is inverted relative to \( \sin x \) (which goes from 0 up to max first).
So we use \( -A \sin(bx) + k \).
Thus \( f(x) = -150 \sin\left( \frac{\pi}{500} x \right) + 350 \).

✅ Answer: (B)

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