AP Precalculus -3.5 Sinusoidal Functions- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -3.5 Sinusoidal Functions- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -3.5 Sinusoidal Functions- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The figure shows the graph of a sinusoidal function $ g $. What are the values of the period and amplitude of $ g $?

(A) The period is 4, and the amplitude is 3.
(B) The period is 8, and the amplitude is 3.
(C) The period is 4, and the amplitude is 6.
(D) The period is 8, and the amplitude is 6.

▶️ Answer/Explanation

Detailed solution

From the graph:
– Maximum $ g_{\text{max}} = 3 $, Minimum $ g_{\text{min}} = -3 $
– Amplitude = $ \frac{g_{\text{max}} – g_{\text{min}}}{2} = \frac{3 – (-3)}{2} = 3 $
– Consecutive maxima occur at $ x = 3 $ and $ x = 11 $, so period = $ 11 – 3 = 8 $
– Consecutive minima occur at $ x = -1 $ and $ x = 7 $, also giving period = $ 7 – (-1) = 8 $
✅ Answer: (B)

Question

The graph of the function $ h $ is given in the $ xy $-plane. If $ h(x) = a \tan(bx) + 10 $, where $ a $ and $ b $ are constants, which of the following is true?

(A) $ a > 0 $ and $ b > 1 $
(B) $ a > 0 $ and $ 0 < b < 1 $
(C) $ a < 0 $ and $ b > 1 $
(D) $ a < 0 $ and $ 0 < b < 1 $

▶️ Answer/Explanation

Detailed solution

From the graph:
– The standard tangent curve increases through its midline; here it also increases (no vertical reflection), so $ a > 0 $.
– The period of $ h $ is shorter than $ \pi $ (the period of $ \tan x $). Since period = $ \frac{\pi}{b} $, we have $ \frac{\pi}{b} < \pi $ → $ b > 1 $.
Thus $ a > 0 $ and $ b > 1 $.
✅ Answer: (A)

Question

The function $ g $ is given by $ g(x) = 3 \csc(\pi(x + 2)) – 1 $. Which of the following describes the range of $ g $?

(A) The range of $ g $ is $[-3, 3]$.
(B) The range of $ g $ is $(-\infty, -2] \cup [4, \infty)$.
(C) The range of $ g $ is $(-\infty, -3] \cup [3, \infty)$.
(D) The range of $ g $ is $(-\infty, -4] \cup [2, \infty)$.

▶️ Answer/Explanation

Detailed solution

$ g(x) = 3\csc(\pi(x+2)) – 1 $.
Cosecant function: $ \csc u $ has range $ (-\infty, -1] \cup [1, \infty) $.
Multiplying by 3 stretches vertically: range becomes $ (-\infty, -3] \cup [3, \infty) $.
Subtracting 1 shifts down by 1: range becomes $ (-\infty, -4] \cup [2, \infty) $.
✅ Answer: (D)

Question

The function $ p $ is given by $ p(\theta) = 3\tan\left(\frac{\pi}{2}(\theta + 1)\right) – 4 $. What is the period of $ p $?

(A) $ \frac{2}{\pi} $
(B) $ \frac{\pi}{2} $
(C) 2
(D) 4

▶️ Answer/Explanation

Detailed solution

For $ y = \tan(k\theta) $, the period is $ \frac{\pi}{|k|} $.
Here $ p(\theta) = 3\tan\left(\frac{\pi}{2}(\theta + 1)\right) – 4 $.
The coefficient of $ \theta $ inside the tangent is $ \frac{\pi}{2} $, so $ k = \frac{\pi}{2} $.
Period $ = \frac{\pi}{k} = \frac{\pi}{\pi/2} = 2 $.
✅ Answer: (C)

Question

The function $ f $ is given by $ f(x) = a\tan(bx) $, where $ a $ and $ b $ are constants. Which of the following is true about the period of $ f $?

(A) Both $ a $ and $ b $ have an impact on the period.
(B) Only $ a $ has an impact on the period.
(C) Only $ b $ has an impact on the period.
(D) Neither $ a $ nor $ b $ has an impact on the period.

▶️ Answer/Explanation

Detailed solution

For $ y = \tan(bx) $, the period is $ \frac{\pi}{|b|} $.
The amplitude-like factor $ a $ affects vertical stretching but not the period.
Thus only $ b $ affects the period.
✅ Answer: (C)

Question

The function $ f $ is defined by $ f(x) = a \sin(b(x + c)) + d $, for constants $ a, b, c, $ and $ d $. In the $ xy $-plane, the points (2, 2) and (4, 4) represent a minimum value and a maximum value, respectively, on the graph of $ f $. What are the values of $ a $ and $ d $?

(A) $ a = 1 $ and $ d = 3 $
(B) $ a = 1 $ and $ d = 2 $
(C) $ a = 2 $ and $ d = 3 $
(D) $ a = 2 $ and $ d = 2 $

▶️ Answer/Explanation

Detailed solution

Midline $ d $ is halfway between max and min:
\[ d = \frac{4 + 2}{2} = 3 \]
Amplitude $ a $ is half the difference:
\[ a = \frac{4 – 2}{2} = 1 \]
Thus $ a = 1 $, $ d = 3 $.
✅ Answer: (A)

Question

The function $ f $ is given by $ f(x) = 2\sin(4x) + \cos(2x) $. Using the period of $ f $, which of the following is the number of complete cycles of the graph of $ f $ in the $ xy $-plane on the interval $ 0 \leq x \leq 1000 $?

(A) 159
(B) 318
(C) 602
(D) 636

▶️ Answer/Explanation

Detailed solution

Find the period of $ f(x) = 2\sin(4x) + \cos(2x) $.
• $ 2\sin(4x) $ has period $ \frac{2\pi}{4} = \frac{\pi}{2} $.
• $ \cos(2x) $ has period $ \frac{2\pi}{2} = \pi $.
The combined function $ f $ is periodic with period equal to the least common multiple of $ \frac{\pi}{2} $ and $ \pi $, which is $ \pi $.
Thus the period of $ f $ is $ \pi $.
Number of complete cycles in $ 0 \leq x \leq 1000 $:
\[ \frac{1000}{\pi} \approx 318.31 \]
So there are 318 complete cycles.
✅ Answer: (B)

Question

Is the function $f(x) = \sin 3\theta – \cos 3\theta$ a sinusoidal function on all its domain?

a. No, since it has both a sine and a cosine, while a sinusoidal function can only have one sine or one cosine function.

b. No, since it has a sine being subtracted by a cosine function, while a sinusoidal function has it being added together.

c. Yes, since the sine and cosine functions have the same input angle, $3\theta$.

d. Yes, since it incorporates trigonometric functions and not inverse trigonometric functions.

▶️ Answer/Explanation

Detailed solution

The correct answer is c.
Any linear combination of a sine and cosine with the same frequency results in a single shifted sinusoid.
Using the identity $A\sin(Bx) + B\cos(Bx) = R\sin(Bx + \phi)$, we can rewrite the function.
In this case, both terms share the same angular frequency (or input angle) of $3\theta$.
The function $f(x) = \sin 3\theta – \cos 3\theta$ can be simplified to $\sqrt{2}\sin(3\theta – 45^\circ)$.
Because it can be expressed as a single sine wave, it is classified as a sinusoidal function.
Options a and b are incorrect because they misunderstand the additive properties of waves.

Question

Which of the following characteristics of $f(x) = -3\sin(3x – 3\pi)$ is true?

i. $f(x)$ is odd
ii. $f(x)$ has an amplitude of $6$
iii. $f(x)$ has a period of $\frac{2\pi}{3}$

a. i and iii only
b. i only
c. iii only
d. i, ii, and iii only

▶️ Answer/Explanation

Detailed solution

The function is $f(x) = -3\sin(3(x – \pi))$, which simplifies using $\sin(\theta – 3\pi) = -\sin(\theta)$ to $f(x) = 3\sin(3x)$.
Checking (i): $f(-x) = 3\sin(-3x) = -3\sin(3x) = -f(x)$, so the function is odd.
Checking (ii): The amplitude is $|a| = |-3| = 3$, so the statement “amplitude of $6$” is false.
Checking (iii): The period is calculated as $T = \frac{2\pi}{|b|} = \frac{2\pi}{3}$, so this statement is true.
Since statements i and iii are correct, the correct option is a.

Question

The figure shows the graph of a sinusoidal function $ f $. Which of the following values is the period of $ f $?

(A) $ 1 $

(B) $ 3 $

(D) $ 2\pi $

▶️ Answer/Explanation

Detailed solution

The period of a sinusoidal function is defined as the horizontal length of one complete cycle of the graph.
To determine this, identify a starting point on the graph, such as the y-intercept at $ \theta = 0 $.
At $ \theta = 0 $, the graph is at the midline $ y=3 $ and is sloping upwards.
Trace the graph along the horizontal axis until this pattern (midline value with upward slope) repeats.
The graph completes a full wave—reaching a maximum, crossing the midline, reaching a minimum, and returning to the midline—at $ \theta = \pi $.
Therefore, the length of the period is $ \pi – 0 = \pi $.
As highlighted by the red brace in the image, the interval from $ 0 $ to $ \pi $ represents one full period.

Correct Option: (C)

Question

The figure shows the graph of a sinusoidal function $ g $. What is the amplitude of $ g $?

(A) 1

(B) 2

(D) 4

▶️ Answer/Explanation

Detailed solution

First, identify the maximum and minimum values of the function $ g(\theta) $ from the graph.
The maximum value (peak) is $ y_{\text{max}} = 3 $.
The minimum value (trough) is $ y_{\text{min}} = -1 $.
The amplitude of a sinusoidal function is half the difference between its maximum and minimum values.
Using the formula: $ \text{Amplitude} = \frac{1}{2}(\text{max} – \text{min}) $.
Substitute the values: $ \text{Amplitude} = \frac{1}{2}(3 – (-1)) = \frac{1}{2}(4) $.
Therefore, $ \text{Amplitude} = 2 $.
The correct option is (B).

Question

The graph of the sinusoidal function $ g $ is shown in the figure above. The function $ g $ can be written as $ g(\theta) = a \sin(\theta) + d $. What are the values of the constants $ a $ and $ d $?

(A) $ a = -2 $ and $ d = -2 $

(B) $ a = -2 $ and $ d = 2 $

(D) $ a = 2 $ and $ d = 2 $

▶️ Answer/Explanation

Detailed solution

From the graph, the maximum y-value is $ 0 $ and the minimum y-value is $ -4 $.
The constant $ d $ is the midline (vertical shift), found by averaging the max and min: $ d = \frac{0 + (-4)}{2} = -2 $.
The constant $ a $ is the amplitude, found by taking half the distance between max and min: $ a = \frac{0 – (-4)}{2} = \frac{4}{2} = 2 $.
Since the graph starts at the midline and goes up (like a standard sine wave), $ a $ is positive.
Therefore, the correct values are $ a = 2 $ and $ d = -2 $.
Correct Option: (C)

Question

The graph of the sinusoidal function $h$ is shown in the figure above. The function $h$ can be written as $h(\theta) = a\cos(b\theta) – 2$. What are the values of the constants $a$ and $b$?

(A) $a = 6$ and $b = \frac{1}{2}$

(B) $a = 12$ and $b = \frac{1}{2}$

(D) $a = 12$ and $b = 2$

▶️ Answer/Explanation

Detailed solution

Identify the maximum and minimum values from the graph: $y_{\text{max}} = 4$ and $y_{\text{min}} = -8$.

Calculate the amplitude $a$ using the formula $a = \frac{\text{max} – \text{min}}{2}$.

Substitute the known values: $a = \frac{4 – (-8)}{2} = \frac{12}{2} = 6$.

Identify the period $T$ of the function from the graph, which shows one full cycle from $0$ to $\pi$, so $T = \pi$.

Use the period formula $b = \frac{2\pi}{T}$ to determine the frequency constant.

Calculate $b$: $b = \frac{2\pi}{\pi} = 2$.

Therefore, the constants are $a = 6$ and $b = 2$.

This corresponds to option (C).

Question

The figure shows the graph of a sinusoidal function $f$. What are the values of the period and amplitude of $f$?

(A) The period is $\pi$ , and the amplitude is $1$ .

(B) The period is $\pi$ , and the amplitude is $2$ .

(D) The period is $\frac{\pi}{2}$ , and the amplitude is $2$ .

▶️ Answer/Explanation

Detailed solution

From the graph, identify the maximum value as $3$ and the minimum value as $-1$.
The amplitude is half the distance between the maximum and minimum values: $\frac{1}{2}(\text{max} – \text{min})$.
Substituting the values, we get: $\text{Amplitude} = \frac{1}{2}(3 – (-1)) = \frac{1}{2}(4) = 2$.
Next, identify the period by finding the length of one complete cycle.
The graph completes one full wave starting from $0$ and ending at $\frac{\pi}{2}$.
Therefore, the period is $\frac{\pi}{2}$.
Matching these results with the options, we find the period is $\frac{\pi}{2}$ and the amplitude is $2$.
Correct Option: (D)

Question

The graph of the sinusoidal function $ k $ is shown in the figure above. The function $ k $ can be written as $ k(\theta) = a \sin(\theta) + d $. What are the values of the constants $ a $ and $ d $?

(A) $ a = 2 $ and $ d = 4 $

(B) $ a = 2 $ and $ d = 6 $

(D) $ a = 6 $ and $ d = 2 $

▶️ Answer/Explanation

Detailed solution

The correct answer is (C).
From the graph, we identify the maximum value as $ y_{\text{max}} = 6 $ and the minimum value as $ y_{\text{min}} = -2 $.
The constant $ d $ represents the midline (vertical shift), which is the average of the maximum and minimum values: $ d = \frac{6 + (-2)}{2} = \frac{4}{2} = 2 $.
The constant $ a $ represents the amplitude, which is half the distance between the maximum and minimum values: $ a = \frac{6 – (-2)}{2} = \frac{8}{2} = 4 $.
Thus, the values are $ a = 4 $ and $ d = 2 $, matching option (C).

Question

The figure shows the graph of a sinusoidal function $g$. What are the values of the period and amplitude of $g$?

(A) The period is $2$, and the amplitude is $2$.

(B) The period is $2$, and the amplitude is $4$.

(D) The period is $4$, and the amplitude is $4$.

▶️ Answer/Explanation

Detailed solution

To find the amplitude, we identify the maximum and minimum values from the graph.
The maximum value is $3$ and the minimum value is $-1$.
Using the formula $\text{Amplitude} = \frac{1}{2}(\text{max} – \text{min})$, we get $\frac{1}{2}(3 – (-1)) = \frac{1}{2}(4) = 2$.
To find the period, we measure the horizontal distance between two consecutive peaks.
The peaks in the graph occur at $x = -2$ and $x = 2$.
The period is the difference: $2 – (-2) = 4$.
Thus, the period is $4$ and the amplitude is $2$, corresponding to option (C).

Question

The graph of the sinusoidal function $ f $ is shown in the figure above. The function can be written as $ f(\theta) = a \sin(b\theta) + 2 $. What are the values of the constants $ a $ and $ b $?

(A) $ a = 4 $ and $ b = 8 $

(B) $ a = 4 $ and $ b = \frac{\pi}{4} $

(D) $ a = 6 $ and $ b = \frac{\pi}{4} $

▶️ Answer/Explanation

Detailed solution

The amplitude $ a $ corresponds to the vertical stretch from the midline.
From the graph, the maximum value is $ 6 $ and the minimum value is $ -2 $.
Calculate $ a = \frac{\text{max} – \text{min}}{2} = \frac{6 – (-2)}{2} = \frac{8}{2} = 4 $.
The period is the horizontal length of one complete cycle, found between two peaks.
The peaks occur at $ x = 2 $ and $ x = 10 $, so the Period $ = 10 – 2 = 8 $.
The formula for the period of a sine function is $ \frac{2\pi}{b} $.
Set $ \frac{2\pi}{b} = 8 $ and solve for $ b $: $ b = \frac{2\pi}{8} = \frac{\pi}{4} $.
Thus, $ a = 4 $ and $ b = \frac{\pi}{4} $, which matches option (B).

Question

The sinusoidal function $ h $ has a minimum at the point $ (2\pi, -1) $. The first maximum after reaching this minimum value occurs at the point $ (5\pi, 7) $. What are the values of the period and amplitude of $ h $?

(A) The period is $ 6\pi $, and the amplitude is $ 4 $.

(B) The period is $ 6\pi $, and the amplitude is $ 8 $.

(D) The period is $ 12\pi $, and the amplitude is $ 8 $.

▶️ Answer/Explanation

Detailed solution

The amplitude is half the difference between the maximum and minimum values:
$ \text{Amplitude} = \frac{1}{2}(\text{max} – \text{min}) = \frac{1}{2}(7 – (-1)) = \frac{1}{2}(8) = 4 $.
The horizontal distance between a minimum and the consecutive maximum represents half of the period:
$ \text{Half Period} = 5\pi – 2\pi = 3\pi $.
To find the full period, we multiply this difference by $ 2 $:
$ \text{Period} = 2 \times 3\pi = 6\pi $.
Therefore, the period is $ 6\pi $ and the amplitude is $ 4 $, which corresponds to option (A).

Question

The sinusoidal function $ k $ has a minimum at the point $ \left(\frac{\pi}{2}, -12\right) $. The first maximum after reaching this minimum value occurs at the point $ (\pi, -2) $. Which of the following gives the period and the midline of $ k $?

(A) The period is $ \frac{\pi}{2} $, and the midline is $ y = -7 $.

(B) The period is $ \frac{\pi}{2} $, and the midline is $ y = 5 $.

(D) The period is $ \pi $, and the midline is $ y = 5 $.

▶️ Answer/Explanation

Detailed solution

To find the midline, we take the average of the maximum and minimum values: $ y = \frac{\text{max} + \text{min}}{2} $.
Substituting the given $y$-values: $ y = \frac{-2 + (-12)}{2} = \frac{-14}{2} = -7 $.
The horizontal distance from a minimum to the next maximum corresponds to half of the period.
Calculate this distance using the $x$-coordinates: $ \pi – \frac{\pi}{2} = \frac{\pi}{2} $.
Since half the period is $ \frac{\pi}{2} $, the full period is $ 2 \times \frac{\pi}{2} = \pi $.
Thus, the function has a period of $ \pi $ and a midline of $ y = -7 $, matching option (C).

Question

The function $ g $ is defined by $ g(x) = a \sin(b(x + c)) + d $, for constants $ a, b, c, $ and $ d $. In the $ xy $-plane, the points $ (3, -1) $ and $ (7, 3) $ represent a minimum value and a maximum value, respectively, on the graph of $ g $. What are the values of $ a $ and $ d $?

(A) $ a = 2 $ and $ d = -1 $

(B) $ a = 2 $ and $ d = 1 $

(D) $ a = 4 $ and $ d = 1 $

▶️ Answer/Explanation

Detailed solution

From the given coordinates, the maximum value is $ y_{\text{max}} = 3 $ and the minimum value is $ y_{\text{min}} = -1 $.
The constant $ d $ represents the midline (vertical shift), which is the average of the maximum and minimum values.
Calculation for $ d $: $ d = \frac{y_{\text{max}} + y_{\text{min}}}{2} = \frac{3 + (-1)}{2} = \frac{2}{2} = 1 $.
The constant $ a $ represents the amplitude, which is half the distance between the maximum and minimum values.
Calculation for $ a $: $ a = \frac{y_{\text{max}} – y_{\text{min}}}{2} = \frac{3 – (-1)}{2} = \frac{4}{2} = 2 $.
Therefore, the values are $ a = 2 $ and $ d = 1 $.
This corresponds to option (B).

Question

The function $ h $ is defined by $ h(x) = a \sin(b(x+c)) + d $, for constants $ a, b, c, $ and $ d $. In the $ xy $-plane, the points $ (2, 4) $ and $ (4, 10) $ represent a minimum value and a maximum value, respectively, on the graph of $ h $. What is the value of $ b $?

(A) $ b = 2 $

(B) $ b = 4 $

(D) $ b = \pi $

▶️ Answer/Explanation

Detailed solution

The horizontal distance between a consecutive minimum and maximum represents half of the period.
Given the minimum is at $ x = 2 $ and the maximum is at $ x = 4 $, the half-period is $ 4 – 2 = 2 $.
Thus, the full period of the function is $ 2 \times 2 = 4 $.
The formula relating the period to the coefficient $ b $ is $ \text{Period} = \frac{2\pi}{b} $.
Setting up the equation: $ 4 = \frac{2\pi}{b} $.
Solving for $ b $, we get $ b = \frac{2\pi}{4} $.
Simplifying the fraction results in $ b = \frac{\pi}{2} $.

The correct option is (C).

Question

The function $f$ is defined by $f(x) = a \cos(b(x+c)) + d$, for constants $a, b, c,$ and $d$. In the $xy$-plane, the points $(\pi, 6)$ and $(2\pi, 2)$ represent a maximum value and a minimum value, respectively, on the graph of $f$. What are the values of $b$ and $d$?

(A) $b = 2\pi$ and $d = 2$
(B) $b = 1$ and $d = 2$
(C) $b = 2\pi$ and $d = 4$
(D) $b = 1$ and $d = 4$

▶️ Answer/Explanation

Detailed solution

The vertical shift $d$ (midline) is the average of the maximum and minimum values.
$d = \frac{\text{max} + \text{min}}{2} = \frac{6 + 2}{2} = \frac{8}{2} = 4$.
The distance from a maximum at $x = \pi$ to the next minimum at $x = 2\pi$ is half the period.
Half-period $= 2\pi – \pi = \pi$, so the full period is $2\pi$.
The coefficient $b$ is calculated using the period formula: $b = \frac{2\pi}{\text{Period}}$.
Substituting the period: $b = \frac{2\pi}{2\pi} = 1$.
Therefore, $b = 1$ and $d = 4$.

Question

The function $ k $ is defined by $ k(x) = a \cos(b(x+c)) + d $, for constants $ a, b, c, $ and $ d $. In the $ xy $-plane, the points $ \left( \frac{\pi}{4}, 10 \right) $ and $ \left( \frac{3\pi}{4}, 40 \right) $ represent a minimum value and a maximum value, respectively, on the graph of $ k $.

What are the period and amplitude of the function $ k $?

(A) The period is $ \frac{\pi}{2} $ and the amplitude is $ 15 $.

(B) The period is $ \frac{\pi}{2} $ and the amplitude is $ 30 $.

(D) The period is $ \pi $ and the amplitude is $ 30 $.

▶️ Answer/Explanation

Detailed solution

The amplitude is half the difference between the maximum value $ 40 $ and the minimum value $ 10 $.

Calculating the amplitude: $ \text{Amplitude} = \frac{1}{2}(\text{max} – \text{min}) = \frac{1}{2}(40 – 10) = \frac{30}{2} = 15 $.

The horizontal distance between the minimum at $ x = \frac{\pi}{4} $ and the maximum at $ x = \frac{3\pi}{4} $ represents half of the period.

Calculating half the period: $ \frac{3\pi}{4} – \frac{\pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2} $.

Since $ \frac{\text{Period}}{2} = \frac{\pi}{2} $, multiplying by $ 2 $ gives the full period: $ \text{Period} = \pi $.

Therefore, the function has a period of $ \pi $ and an amplitude of $ 15 $, which corresponds to option (C).

Question

At a coastal city, the height of the tide, in feet (ft), is modeled by the function (h), defined by (h(t) = 6.3\cos\left(\frac{\pi}{6}t\right)) for (0 \le t \le 12) hours. Based on the model, which of the following is true?

(A) The maximum height of the tide is 13.8 ft
(B) The maximum height of the tide occurs at (t = 6) hours
(C) The minimum height of the tide is 1 ft
(D) The minimum height of the tide occurs at (t = 12) hours

▶️ Answer/Explanation

Detailed solution

The function (h(t) = A\cos(Bt) + K) models the tide, where (A=6.3) and (B=\frac{\pi}{6}).
The period is $\frac{2\pi}{B} = \frac{2\pi}{\pi/6} = 12$ hours.
A positive cosine function starts at its maximum at (t=0) and reaches the maximum again at (t=12).
The minimum value of a positive cosine function occurs halfway through the period at (t=6).
This incorrect phase behavior eliminates option (B) (which claims max at (t=6)) and option (D) (which claims min at (t=12)).
For option (A) to be true, the maximum height (K + 6.3) must equal (13.8), implying a vertical shift (K = 7.5).
Using (K=7.5), the minimum height would be (7.5 – 6.3 = 1.2), which makes option (C) incorrect (as it states 1 ft).
Thus, assuming the implied vertical shift, option (A) is the correct statement.

Correct Option: (A)

Question

The function $H$ models a periodic phenomenon. The maximum value of $H$ is 8, which occurs at time $t = 0$ hours. The period of $H$ is 12 hours. For $0 < t < 24$ hours, which of the following could be an expression for $H(t)$?

(A) $3.5 \sin\left(\frac{\pi}{12}t\right) + 4.5$

(B) $3.5 \sin\left(\frac{\pi}{6}t\right) + 4.5$

(D) $3.5 \cos\left(\frac{\pi}{6}t\right) + 4.5$

▶️ Answer/Explanation

Detailed solution

The correct answer is (D).

Since the maximum value occurs at $t=0$, the function should involve cosine, as $\cos(0)=1$ (maximum) while $\sin(0)=0$ (midline).
This eliminates options (A) and (B), leaving only the cosine functions.
The period is given as 12 hours. The formula for period is $P = \frac{2\pi}{B}$, where $B$ is the coefficient of $t$.
We solve for $B$: $12 = \frac{2\pi}{B} \Rightarrow B = \frac{2\pi}{12} = \frac{\pi}{6}$.
Option (C) has a coefficient of $\frac{\pi}{12}$, which results in an incorrect period of 24.
Option (D) has the correct coefficient $\frac{\pi}{6}$.
Checking the maximum value for (D): $3.5\cos(0) + 4.5 = 3.5(1) + 4.5 = 8$, which is correct.

Question

A sinusoidal function model $ y = f(x) $ is constructed from the data in the table and results in $ f(x) = a\sin(b(x + c)) + d $, where $ a $, $ b $, $ c $, and $ d $ are constants. Of the following, which is the best approximation for the value of $ b $?

(A) 0.7
(B) 1
(C) 6
(D) 9

▶️ Answer/Explanation

Detailed solution

The constant $ b $ determines the frequency and is related to the period $ P $ by the formula $ b = \frac{2\pi}{P} $.
To find the period, we identify the distance between two consecutive peaks (maximum values) in the data.
The first peak occurs at $ x = 4.5 $ with $ y = 7.02 $, and the next peak is at $ x = 13.5 $ with $ y = 7.04 $.
Calculating the difference gives the period: $ P = 13.5 – 4.5 = 9 $.
Now, we substitute $ P = 9 $ into the formula for $ b $: $ b = \frac{2\pi}{9} $.
Approximating $ \pi \approx 3.14 $, we get $ b \approx \frac{6.28}{9} \approx 0.698 $.
Rounding to the nearest option, the value is approximately $ 0.7 $.

Question

A circular clock face on a wall has hours marked with integers from $1$ to $12$ and the hour hand that rotates as time passes. The hour hand points to the number $12$ at midnight. The height above the floor, in inches, of the tip of the hour hand at $t$ hours after midnight is modeled by the periodic function $y$. Values of $y(t)$ at selected values of $t$ are given in the table. For $t \geq 0$, which of the following could be an expression for $y(t)$?

$t$ hours	$0$	$3$	$6$	$9$	$12$
$y(t)$ inches	$78$	$70$	$62$	$70$	$78$

(A) $-8 \sin\left(\frac{\pi}{12}(t-3)\right) + 70$

(B) $-8 \sin\left(\frac{\pi}{6}(t-3)\right) + 70$

(D) $8 \cos\left(\frac{\pi}{6}(t-3)\right) + 70$

▶️ Answer/Explanation

Detailed solution

The maximum height is $78$ and the minimum is $62$, so the vertical shift (midline) is $\frac{78 + 62}{2} = 70$ and amplitude is $78 – 70 = 8$.
The function repeats its values from $t=0$ to $t=12$, so the period is $12$.
The coefficient $B$ is calculated as $\frac{2\pi}{\text{Period}} = \frac{2\pi}{12} = \frac{\pi}{6}$. This eliminates options (A) and (C).
We test the point $t=0$ (where $y=78$) on the remaining options to find the correct expression.
For option (D): $8 \cos\left(\frac{\pi}{6}(0-3)\right) + 70 = 8 \cos\left(-\frac{\pi}{2}\right) + 70 = 0 + 70 = 70$ (Incorrect).
For option (B): $-8 \sin\left(\frac{\pi}{6}(0-3)\right) + 70 = -8 \sin\left(-\frac{\pi}{2}\right) + 70 = -8(-1) + 70 = 78$ (Correct).
Therefore, the correct expression is (B).

\(t\) hours	\(0\)	\(3\)	\(6\)	\(9\)	\(12\)
\(y(t)\) inches	\(78\)	\(70\)	\(62\)	\(70\)	\(78\)

AP Precalculus -3.5 Sinusoidal Functions- MCQ Exam Style Questions - Effective Fall 2023