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).

Question

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 assumed to be in radians.
Solutions to equations must be real numbers. Determine the exact value of any expression that can be obtained without a calculator. For example, $ \log_2 8 $, $ \cos(\frac{\pi}{2}) $, and $ \sin^{-1}(1) $ can be evaluated without a calculator.
Unless otherwise specified, combine terms using algebraic methods and rules for exponents and logarithms, where applicable. For example, $ 2x + 3x $, $ 5^2 \cdot 5^3 $, $ \frac{x^5}{x^2} $, and $ \ln 3 + \ln 5 $ should be rewritten in equivalent forms.
For each part of the question, show the work that leads to your answers.

Part A
The functions $ g $ and $ h $ are given by

$ g(x) = \log_5(4x – 2) $
$ h(x) = \sin^{-1}(8x) $

(i) Solve $ g(x) = 3 $ for values of $ x $ in the domain of $ g $.

(ii) Solve $ h(x) = \frac{\pi}{4} $ for values of $ x $ in the domain of $ h $.

Part B
The functions $ j $ and $ k $ are given by

$ j(x) = (\sec x)(\cot x) $
$ k(x) = \frac{(16^{3x}) \cdot 4^x}{2} $

(i) Rewrite $ j(x) $ as an expression involving $ \sin x $ and no other trigonometric functions.

(ii) Rewrite $ k(x) $ as an expression of the form $ 4^{(ax+b)} $, where $ a $ and $ b $ are constants.

Part C
The function $ m $ is given by

$ m(x) = \sqrt{3}\tan(x + \frac{\pi}{2}) $.

Find all values in the domain of $ m $ that yield an output value of 1.

▶️ Answer/Explanation

Detailed solution

Part A
(i) Solve $ g(x) = 3 $
Start with the given equation:
$ \log_5(4x – 2) = 3 $
Convert the logarithmic equation to exponential form ($ y = \log_b x \iff x = b^y $):
$ 4x – 2 = 5^3 $
Evaluate the exponent:
$ 4x – 2 = 125 $
Add 2 to both sides:
$ 4x = 127 $
Divide by 4:
$ x = \frac{127}{4} $

(ii) Solve $ h(x) = \frac{\pi}{4} $
Start with the given equation:
$ \sin^{-1}(8x) = \frac{\pi}{4} $
Take the sine of both sides to isolate the argument:
$ 8x = \sin\left(\frac{\pi}{4}\right) $
Substitute the exact value of $ \sin\left(\frac{\pi}{4}\right) $:
$ 8x = \frac{\sqrt{2}}{2} $
Divide by 8:
$ x = \frac{\sqrt{2}}{16} $

Part B
(i) Rewrite $ j(x) $
Start with the function definition:
$ j(x) = (\sec x)(\cot x) $
Substitute the reciprocal and quotient identities ($ \sec x = \frac{1}{\cos x} $ and $ \cot x = \frac{\cos x}{\sin x} $):
$ j(x) = \left(\frac{1}{\cos x}\right) \left(\frac{\cos x}{\sin x}\right) $
Cancel the $ \cos x $ terms:
$ j(x) = \frac{1}{\sin x} $

(ii) Rewrite $ k(x) $
Start with the function definition:
$ k(x) = \frac{(16^{3x}) \cdot 4^x}{2} $
To write in the form $ 4^{(ax+b)} $, convert bases 16 and 2 to base 4.
Since $ 16 = 4^2 $ and $ 2 = \sqrt{4} = 4^{1/2} = 4^{0.5} $:
$ k(x) = \frac{(4^2)^{3x} \cdot 4^x}{4^{0.5}} $
Apply the power of a power rule ($ (a^m)^n = a^{mn} $):
$ k(x) = \frac{4^{6x} \cdot 4^x}{4^{0.5}} $
Apply the product rule for exponents ($ a^m \cdot a^n = a^{m+n} $) in the numerator:
$ k(x) = \frac{4^{6x + x}}{4^{0.5}} = \frac{4^{7x}}{4^{0.5}} $
Apply the quotient rule for exponents ($ \frac{a^m}{a^n} = a^{m-n} $):
$ k(x) = 4^{7x – 0.5} $ (or $ 4^{7x – \frac{1}{2}} $)
Thus, $ a = 7 $ and $ b = -0.5 $.

Part C
Find values where $ m(x) = 1 $
Set the function equal to 1:
$ \sqrt{3}\tan\left(x + \frac{\pi}{2}\right) = 1 $
Isolate the tangent function by dividing by $ \sqrt{3} $:
$ \tan\left(x + \frac{\pi}{2}\right) = \frac{1}{\sqrt{3}} $
Determine the reference angle. We know that $ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} $.
Set up the general solution for tangent ($ \theta = \text{ref} + n\pi $):
$ x + \frac{\pi}{2} = \frac{\pi}{6} + n\pi $, where $ n $ is any integer.
Solve for $ x $ by subtracting $ \frac{\pi}{2} $ from both sides:
$ x = \frac{\pi}{6} – \frac{\pi}{2} + n\pi $
Find a common denominator (6) to combine fractions:
$ x = \frac{\pi}{6} – \frac{3\pi}{6} + n\pi $
$ x = -\frac{2\pi}{6} + n\pi $
Simplify the fraction:
$ x = -\frac{\pi}{3} + n\pi $

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