AP Precalculus -3.11 Secant, Cosecant, and Cotangent- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -3.11 Secant, Cosecant, and Cotangent- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -3.11 Secant, Cosecant, and Cotangent- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The function $ f $ is defined by $ f(x) = \sec\left(\frac{1}{2}\left(x – \frac{\pi}{2}\right)\right) $. Which of the following describes the domain of $ f $?

(A) The domain is the set of all real numbers $ x $, except when $ x = \frac{\pi}{2} + n\pi k $, where $ k $ is any integer.
(B) The domain is the set of all real numbers $ x $, except when $ x = \frac{\pi}{2} + 2n\pi k $, where $ k $ is any integer.
(C) The domain is the set of all real numbers $ x $, except when $ x = \pi + 2n\pi k $, where $ k $ is any integer.
(D) The domain is the set of all real numbers $ x $, except when $ x = \frac{3\pi}{2} + 2n\pi k $, where $ k $ is any integer.

▶️ Answer/Explanation

Detailed solution

$ f(x) = \sec\left(\frac12\left(x – \frac{\pi}{2}\right)\right) $ is undefined where $ \cos\left(\frac12\left(x – \frac{\pi}{2}\right)\right) = 0 $.
Let $ u = \frac12\left(x – \frac{\pi}{2}\right) $. Cosine is zero when $ u = \frac{\pi}{2} + n\pi $.
Thus $ \frac12\left(x – \frac{\pi}{2}\right) = \frac{\pi}{2} + n\pi $
Multiply by 2: $ x – \frac{\pi}{2} = \pi + 2n\pi $
So $ x = \frac{\pi}{2} + \pi + 2n\pi = \frac{3\pi}{2} + 2n\pi $.
That’s $ x = \frac{3\pi}{2} + 2\pi k $ (letting $ k = n $).
✅ Answer: (D)

Question

In the $xy$-plane, the graph of which of the following functions has a vertical asymptote at $x=\frac{3\pi}{4}$?

(A) $f(x)=cot~x$
(B) $f(x)=cot(x-\frac{\pi}{2})$
(C) $f(x)=cot(x-\frac{\pi}{4})$
(D) $f(x)=cot(x+\frac{\pi}{4})$

▶️ Answer/Explanation

Detailed solution

1. Condition for Cotangent Asymptotes:
The function $y = \cot(u)$ has vertical asymptotes where $u = k\pi$ (where $k$ is an integer), because $\sin(u) = 0$.

2. Test the Options at $x = \frac{3\pi}{4}$:
We need the argument of the cotangent function to be a multiple of $\pi$.

(A) $x = \frac{3\pi}{4}$ (Not a multiple of $\pi$)
(B) $x – \frac{\pi}{2} = \frac{3\pi}{4} – \frac{2\pi}{4} = \frac{\pi}{4}$ (Not a multiple of $\pi$)
(C) $x – \frac{\pi}{4} = \frac{3\pi}{4} – \frac{\pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2}$ (Cotangent is 0 here, not undefined)
(D) $x + \frac{\pi}{4} = \frac{3\pi}{4} + \frac{\pi}{4} = \frac{4\pi}{4} = \pi$

Since $\pi$ is an integer multiple of $\pi$, the function in (D) is undefined and has a vertical asymptote.

✅ Answer: (D)

Question

The range of the function $f(x) = -2 \sec(3x + 2) + 1$ can be defined as what?

a. $(-\infty, 1) \cup (1, \infty)$

b. $(-\infty, -3) \cup (1, \infty)$

c. $(-\infty, -1) \cup (3, \infty)$

d. $(-\infty, 0) \cup (2, \infty)$

▶️ Answer/Explanation

Detailed solution

The parent function $\sec(\theta)$ has a range of $(-\infty, -1] \cup [1, \infty)$.
The multiplier $-2$ stretches the graph vertically, changing the range to $(-\infty, -2] \cup [2, \infty)$.
The constant $+1$ shifts the entire range up by $1$ unit.
Lower bound: $-2 + 1 = -1$.
Upper bound: $2 + 1 = 3$.
The final range is $(-\infty, -1] \cup [3, \infty)$.
Note: While the options use parentheses, option c represents the correct interval values.
Correct Option: c

Question

Consider the functions $g$ and $h$ given by $g(\theta) = \sec \theta$ and $h(\theta) = \csc \theta$. Which of the following statements is true for both $g$ and $h$?

(A) The domain is all real numbers.
(B) The range is $(-\infty, -1] \cup [1, \infty)$.
(C) The graphs of the functions have vertical asymptotes at $\theta = \pi n$, where $n$ is an integer.
(D) The graphs of the functions have vertical asymptotes at $\theta = \frac{\pi}{2} + \pi n$, where $n$ is an integer.

▶️ Answer/Explanation

Detailed solution

Since $g(\theta) = \frac{1}{\cos \theta}$ and $h(\theta) = \frac{1}{\sin \theta}$, both functions have restricted domains where their denominators equal zero.
The range of $\cos \theta$ and $\sin \theta$ is $[-1, 1]$.
Taking the reciprocal of values in $[-1, 1]$ results in values $\ge 1$ or $\le -1$.
Therefore, the range for both functions is $(-\infty, -1] \cup [1, \infty)$.
Vertical asymptotes for $g(\theta)$ occur at $\theta = \frac{\pi}{2} + \pi n$ where $\cos \theta = 0$.
Vertical asymptotes for $h(\theta)$ occur at $\theta = \pi n$ where $\sin \theta = 0$.
Thus, statement (B) is the only property shared by both functions.

Question

Let $h(x) = 2 \csc(4x)$. Which of the following gives the range of $h$?

(A) $(-\infty, -1] \cup [1, \infty)$

(B) $(-\infty, -2] \cup [2, \infty)$

(D) $\left( -\infty, -\frac{1}{4} \right] \cup \left[ \frac{1}{4}, \infty \right)$

▶️ Answer/Explanation

Detailed solution

The parent function $f(\theta) = \csc(\theta)$ has a range of $(-\infty, -1] \cup [1, \infty)$.
The horizontal compression by the factor in $4x$ affects the period, but does not change the range.
The vertical stretch by a factor of $2$ multiplies the output values of the cosecant function.
Multiplying the boundaries of the parent range by $2$ gives $2 \times 1 = 2$ and $2 \times -1 = -2$.
Therefore, the minimum positive value is $2$ and the maximum negative value is $-2$.
The resulting range of $h(x)$ is $(-\infty, -2] \cup [2, \infty)$.
This corresponds to option (B).

AP Precalculus -3.11 Secant, Cosecant, and Cotangent- MCQ Exam Style Questions - Effective Fall 2023