AP Precalculus -3.14 Polar Function Graphs- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -3.14 Polar Function Graphs- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -3.14 Polar Function Graphs- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

Which of the following is the graph of the polar function $ r = f(\theta) $, where $ f(\theta) = 2\cos(2\theta) $, in the polar coordinate system for $ -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} $?

▶️ Answer/Explanation

Detailed solution

The equation $ r = 2\cos(2\theta) $ represents a rose curve with 4 petals (since the coefficient of $\theta$ is even, the number of petals is $2n = 4$). Over the domain $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, the curve completes two full petals (one in the right half-plane and one in the upper/lower quadrants depending on $\theta$), and the graph shown in option (B) matches the correctly restricted portion of the 4-petaled rose.
✅ Answer: (B)

Question

The function $ f $ is given by $ f(x) = 3 \cos(\pi x) $, where $ 0 \leq x \leq 1 $. Which of the following gives $ f^{-1}(x) $?

(A) $ 3 \cos^{-1}(\pi x) $, $-3 \leq x \leq 3$
(B) $ \frac{1}{3 \cos(\pi x)} $, $-3 \leq x \leq 3$
(C) $ \pi \cos^{-1}\left( \frac{x}{3} \right) $, $-3 \leq x \leq 3$
(D) $ \frac{1}{\pi} \cos^{-1}\left( \frac{x}{3} \right) $, $-3 \leq x \leq 3$

▶️ Answer/Explanation

Detailed solution

To find the inverse:
Set $ y = 3\cos(\pi x) $, swap $ x $ and $ y $:
$ x = 3\cos(\pi y) $.
Solve for $ y $:
$ \cos(\pi y) = \frac{x}{3} $
$ \pi y = \cos^{-1}\left( \frac{x}{3} \right) $
$ y = \frac{1}{\pi} \cos^{-1}\left( \frac{x}{3} \right) $.

Domain of $ f^{-1} $ is the range of $ f $:
$ f(x) = 3\cos(\pi x) $ has range $ [-3, 3] $ because $\cos(\pi x)$ ranges from $ -1 $ to $ 1 $.
So $ -3 \leq x \leq 3 $.

✅ Answer: (D)

Question

Consider the graph of the polar function $ r = f(\theta) $, where $ f(\theta) = \frac{3\theta^5 + 50}{2\theta^4 + 5} $, in the polar coordinate system. Which of the following is true?

(A) Because $ \lim_{\theta \to \infty} f(\theta) = -\infty $, points on the graph of $ r = f(\theta) $ will be arbitrarily close to the origin for sufficiently large values of $ \theta $.
(B) Because $ \lim_{\theta \to \infty} f(\theta) = 0 $, points on the graph of $ r = f(\theta) $ will be arbitrarily close to the origin for sufficiently large values of $ \theta $.
(C) Because $ \lim_{\theta \to \infty} f(\theta) = \frac{3}{2} $, points on the graph of $ r = f(\theta) $ will be arbitrarily close to the polar curve $ r = \frac{3}{2} $ for sufficiently large values of $ \theta $.
(D) Because $ \lim_{\theta \to \infty} f(\theta) = \infty $, points on the graph of $ r = f(\theta) $ will be increasingly distant from the origin for sufficiently large values of $ \theta $.

▶️ Answer/Explanation

Detailed solution

Step 1: Analyze the function

The polar function is $ r = f(\theta) $, where
$ f(\theta) = \dfrac{3\theta^5 + 50}{2\theta^4 + 5} $.

Step 2: Find the limit as $ \theta \to \infty $

Compare the highest powers of $ \theta $ in the numerator and denominator.
The numerator has degree $5$ and the denominator has degree $4$.

Divide the numerator and denominator by $ \theta^4 $:

$ f(\theta) = \dfrac{3\theta + \dfrac{50}{\theta^4}}{2 + \dfrac{5}{\theta^4}} $.

As $ \theta \to \infty $, both $ \dfrac{50}{\theta^4} \to 0 $ and
$ \dfrac{5}{\theta^4} \to 0 $.

Thus, $ f(\theta) \sim \dfrac{3\theta}{2} $, which increases without bound.

Therefore, $ \lim_{\theta \to \infty} f(\theta) = \infty $.

Step 3: Interpret the result

Since $ r = f(\theta) $ and $ r \to \infty $ as $ \theta \to \infty $,
points on the graph move increasingly far from the origin.

$\boxed{\text{Correct answer: (D)}}$

Question

Which of the following is the graph of the polar function $ r = f(\theta) $, where $ f(\theta) = 3 \cos \theta + 2 $, in the polar coordinate system for $ 0 \leq \theta \leq 2\pi $?

▶️ Answer/Explanation

Detailed solution

$ f(\theta) = 3\cos\theta + 2 $:
– At $ \theta = 0 $: $ r = 3(1) + 2 = 5 $ (max).
– At $ \theta = \pi/2 $: $ r = 3(0) + 2 = 2 $.
– At $ \theta = \pi $: $ r = 3(-1) + 2 = -1 $, which plots as $ r = 1 $ in opposite direction (point at (1, π) in polar).
– At $ \theta = 3\pi/2 $: $ r = 2 $.
– At $ \theta = 2\pi $: back to 5.

Shape: Symmetric about polar axis. Inner loop when $ r < 0 $ for $ \pi/2 < \theta < 3\pi/2 $, but here $ r $ becomes negative only between the two zeros of $ 3\cos\theta + 2 = 0 $, i.e., $ \cos\theta = -2/3 $ → two angles where $ r = 0 $, giving a limaçon with an inner loop because $ 2 < 3 $ (since $ a = 3, b = 2 $, $ a > b $).

✅ Answer: (D)

Question

The graph of the polar function $ r = f(\theta) $ is given in the polar coordinate system. Which of the following defines $ f(\theta) $ for $ 0 \leq \theta \leq 2\pi $?

(A) $ 3 + \cos(3\theta) $
(B) $ 3 \cos(3\theta) $
(C) $ 3 + \sin(3\theta) $
(D) $ 3 \sin(3\theta) $

▶️ Answer/Explanation

Detailed solution

The graph has 3 petals (rose curve) and max radius 3.
General form for a 3-petal rose: $ r = a \cos(3\theta) $ or $ r = a \sin(3\theta) $.
From the graph, at $ \theta = 0 $, $ r = 3 $, which matches $ 3\cos(3\cdot 0) = 3 $.
If it were $ 3\sin(3\theta) $, at $ \theta = 0 $, $ r = 0 $, so not match.
Also, $ 3 + \cos(3\theta) $ ranges from 2 to 4, not 0 to 3 symmetrically.
Thus correct form is $ r = 3\cos(3\theta) $.
✅ Answer: (B)

Question

The polar function $ r = f(\theta) $, where $ f(\theta) = \frac{2}{\theta} $, is defined for $ \theta \geq 1 $. Which of the following describes the graph of $ r = f(\theta) $ in the polar coordinate system?

(A) The graph of $ r = f(\theta) $ is a line along the polar axis (x-axis) for which increasing angle measures correspond to decreasing radii.
(B) The graph of $ r = f(\theta) $ is a line along the polar axis (x-axis) for which increasing angle measures correspond to increasing radii.
(C) The graph of $ r = f(\theta) $ is a spiral for which increasing angle measures correspond to decreasing radii.
(D) The graph of $ r = f(\theta) $ is a spiral for which increasing angle measures correspond to increasing radii.

▶️ Answer/Explanation

Detailed solution

$ r = \frac{2}{\theta} $:
As $ \theta $ increases, $ r $ decreases → spiral getting closer to origin.
Not a line because $ r $ depends on $ \theta $ non-linearly.
Thus: spiral with decreasing radii.
✅ Answer: (C)

Question

Let $ r = f(\theta) $ be a polar function, where $ f(\theta) = 5 $. The graph of $ r = f(\theta) $ for $ 0 \leq \theta \leq \pi $ appears as a semicircle in the polar coordinate system. The graph of which of the following polar functions in the polar coordinate system is the same semicircle?

(A) $ r = g(\theta) $, where $ g(\theta) = -5 $ for $ 0 \leq \theta \leq \pi $
(B) $ r = h(\theta) $, where $ h(\theta) = \frac{1}{5} $ for $ 0 \leq \theta \leq \pi $
(C) $ r = k(\theta) $, where $ k(\theta) = -5 $ for $ \pi \leq \theta \leq 2\pi $
(D) $ r = m(\theta) $, where $ m(\theta) = \frac{1}{5} $ for $ \pi \leq \theta \leq 2\pi $

▶️ Answer/Explanation

Detailed solution

Original: $ r = 5 $ for $ 0 \leq \theta \leq \pi $ → upper semicircle (radius 5, center origin, above polar axis).
If $ r = -5 $ for $ 0 \leq \theta \leq \pi $, points plot opposite direction → lower semicircle (not same).
If $ r = -5 $ for $ \pi \leq \theta \leq 2\pi $, the angle range corresponds to lower half of plane, but negative radius flips direction to upper half → gives upper semicircle, same as original.
Thus same semicircle obtained with $ r = -5 $ for $ \pi \leq \theta \leq 2\pi $.
✅ Answer: (C)

Question

The graph of the polar function $ r = g(\theta) $, where $ g(\theta) = 4 \cos (\theta + \frac{\pi}{3}) $, is given in the polar coordinate system. Which of the following descriptions is true?

(A) Values of $ \theta $ for $ \frac{\pi}{2} < \theta < \pi $ correspond to the portion of the graph of $ r = g(\theta) $ in Quadrant I.
(B) Values of $ \theta $ for $ \frac{\pi}{2} < \theta < \pi $ correspond to the portion of the graph of $ r = g(\theta) $ in Quadrant II.
(C) Values of $ \theta $ for $ \frac{\pi}{2} < \theta < \pi $ correspond to the portion of the graph of $ r = g(\theta) $ in Quadrant III.
(D) Values of $ \theta $ for $ \frac{\pi}{2} < \theta < \pi $ correspond to the portion of the graph of $ r = g(\theta) $ in Quadrant IV.

▶️ Answer/Explanation

Detailed solution

$ r = 4\cos\left( \theta + \frac{\pi}{3} \right) $.
Test interval $ \frac{\pi}{2} < \theta < \pi $:
Let $ \theta = \frac{2\pi}{3} $ (midpoint). Then $ \theta + \frac{\pi}{3} = \pi $, $ r = 4\cos\pi = -4 $, negative.
Plotting: angle = $ 2\pi/3 $ (Quadrant II), but negative radius → point in opposite direction: Quadrant IV.
Check another: $ \theta = \frac{3\pi}{4} $, then $ \theta + \frac{\pi}{3} = \frac{13\pi}{12} $ > $ \pi $, cosine negative, radius negative → direction same as terminal ray angle $ 3\pi/4 $ (Quadrant II)? Wait: negative radius flips to opposite quadrant: Quadrant IV again.
Thus for this entire $ \theta $ range, points lie in Quadrant IV.
✅ Answer: (D)

Question

The graph of the function $ y = f(x) $ is shown in the $ xy $-plane. Which of the following is the graph of the polar function $ r = f(\theta) $ in the polar coordinate system?

▶️ Answer/Explanation

Detailed solution

The Cartesian graph given is of $ y = 2\cos x + 1 $.
In polar form, $ r = 2\cos\theta + 1 $.
Key points: at $ \theta = 0 $, $ r = 3 $; at $ \theta = \frac{2\pi}{3} $, $ r = 2\cos\frac{2\pi}{3} + 1 = 2(-\frac12) + 1 = 0 $; at $ \theta = \pi $, $ r = 2(-1)+1 = -1 $ (plots as $ r=1$ in opposite direction, i.e., at $ \theta=\pi $); at $ \theta = \frac{4\pi}{3} $, $ r = 2\cos\frac{4\pi}{3} + 1 = 2(-\frac12) + 1 = 0 $; at $ \theta = 2\pi $, back to 3.
The polar graph is symmetric about polar axis, has a loop (since $ a=2, b=1, a > b $), and passes through origin at two angles.
Matching this shape to options, the correct polar graph is option B.
✅ Answer: (B)

Question

A portion of the graph of the polar function $r=f(\theta)$, where $f(\theta)=2-4~cos~\theta,$ is shown in the polar coordinate system for $a\le\theta\le b.$ If $0\le a<b<2\pi,$ which of the following are the values for $a$ and $b$?

(A) $a=0$ and $b=\frac{\pi}{3}$
(B) $a=0$ and $b=\frac{\pi}{6}$
(C) $a=\pi$ and $b=\frac{4\pi}{3}$
(D) $a=\pi$ and $b=\frac{7\pi}{6}$

▶️ Answer/Explanation

Detailed solution

1. Analyze the Start Point ($\theta = 0$):
$f(0) = 2 – 4\cos(0) = 2 – 4 = -2$.
The graph starts at radius -2 (on the negative x-axis).

2. Find where the graph passes through the Pole ($r=0$):
$2 – 4\cos\theta = 0 \implies \cos\theta = \frac{1}{2}$.
The angles where $\cos\theta = \frac{1}{2}$ are $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$.

3. Determine the Interval:
The loop typically shown for this limaçon goes from the minimum radius to the pole. The interval starting at 0 and ending at the first pole crossing is $0 \le \theta \le \frac{\pi}{3}$.

✅ Answer: (A)

Question

The figure shows the graph of the polar function $ r = f(\theta) $, where $ f(\theta) = 4\cos(2\theta) $, in the polar coordinate system for $ 0 \leq \theta \leq 2\pi $. There are five points labeled $ A, B, C, D, $ and $ E $. If the domain of $ f $ is restricted to $ 0 \leq \theta \leq \frac{\pi}{2} $, the portion of the given graph that remains consists of two pieces. One of those pieces is the portion of the graph in Quadrant I from $ C $ to $ E $. Which of the following describes the other remaining piece?

(A) The portion of the graph in Quadrant I from $ E $ to $ B $
(B) The portion of the graph in Quadrant II from $ E $ to $ A $
(C) The portion of the graph in Quadrant III from $ E $ to $ A $
(D) The portion of the graph in Quadrant III from $ E $ to $ D $

▶️ Answer/Explanation

Detailed solution

The polar function $ r = 4\cos(2\theta) $ for $ 0 \leq \theta \leq \frac{\pi}{2} $ corresponds to two “petals” of the four-leaved rose.
One petal lies in Quadrant I (from $ C $ to $ E $). The other petal for this domain lies in Quadrant III, because when $ \theta $ increases from 0 to $ \pi/2 $, $ r $ becomes negative for certain intervals, plotting points in the opposite direction (Quadrant III).
From the answer key, the other piece is the portion in Quadrant III from $ E $ to $ D $.
✅ Answer: (D)

Question

Given the equation $r^2 = 25 \cos 2\theta$, which graph best corresponds to the equation for when $0 \le \theta < 2\pi$.

▶️ Answer/Explanation

Detailed solution

The equation $r^2 = 25 \cos 2\theta$ represents a lemniscate curve.
When $\theta = 0$, $r^2 = 25 \cos(0) = 25$, so $r = \pm 5$, indicating intercepts on the polar axis.
The maximum value of $r$ is $5$, which occurs at $\theta = 0$ and $\theta = \pi$.
The graph is symmetric with respect to the pole, the polar axis, and the line $\theta = \frac{\pi}{2}$.
The loops exist only where $\cos 2\theta \ge 0$, which occurs in the intervals $[-\frac{\pi}{4}, \frac{\pi}{4}]$ and $[\frac{3\pi}{4}, \frac{5\pi}{4}]$.
Comparing the given options, graph d correctly shows the horizontal lemniscate with a maximum radius of $5$.
Correct Option: d

Question

Which of the following is the graph of the polar function (r = f(\theta)), where (f(\theta) = 4 \cos(2\theta)) in the polar coordinate system for (\pi \le \theta \le \frac{3\pi}{2})?

▶️ Answer/Explanation

Detailed solution

The correct graph corresponds to option (B).

To determine the correct graph, we evaluate the function $r = 4 \cos(2\theta)$ at the starting value of the given interval, $\theta = \pi$.

Substitute $\theta = \pi$ into the equation:
$r = 4 \cos(2(\pi)) = 4 \cos(2\pi)$

Since $\cos(2\pi) = 1$, we have:
$r = 4(1) = 4$

This corresponds to the polar point $(4, \pi)$. In the polar coordinate system, this point is located 4 units away from the pole (origin) along the ray $\theta = \pi$, which lies on the negative x-axis.

Question

The polar function $r = f(\theta)$, where $f(\theta) = 1 – 2\cos(-\theta)$, is graphed in the polar coordinate system. As $\theta$ varies from $\theta = \frac{\pi}{2}$ to $\theta = \pi$, how is the distance between the origin and the point with polar coordinates $(f(\theta), \theta)$ changing?

(A) The distance remains constant.

(B) The distance is decreasing.

(D) The distance increases, then the distance decreases.

▶️ Answer/Explanation

Detailed solution

The correct answer is (C) The distance is increasing.

Step 1: Simplify the function.
Recall that the cosine function is an even function, which means $\cos(-\theta) = \cos(\theta)$.
Substitute this into the given equation: $r = f(\theta) = 1 – 2\cos(\theta)$.

Step 2: Define distance from the origin.
In polar coordinates, the distance between the origin and a point $(r, \theta)$ is given by the absolute value of the radius, $|r|$.
We need to analyze the behavior of $r$ on the interval $\theta \in [\frac{\pi}{2}, \pi]$.

Step 3: Analyze the rate of change using the derivative.
Differentiate $r$ with respect to $\theta$:
$\frac{dr}{d\theta} = \frac{d}{d\theta}(1 – 2\cos(\theta)) = -2(-\sin(\theta)) = 2\sin(\theta)$.

Step 4: Evaluate the sign of the derivative in the given interval.
For $\theta$ in the interval $(\frac{\pi}{2}, \pi)$ (the second quadrant), $\sin(\theta)$ is positive ($\sin(\theta) > 0$).
Therefore, $\frac{dr}{d\theta} = 2\sin(\theta)$ is positive, indicating that $r$ is strictly increasing.

Step 5: Confirm the sign of $r$.
At $\theta = \frac{\pi}{2}$, $r = 1 – 2(0) = 1$.
At $\theta = \pi$, $r = 1 – 2(-1) = 3$.
Since $r$ starts at 1 and increases to 3, $r$ is always positive. Thus, the distance $|r| = r$.
Because $r$ is increasing, the distance from the origin is increasing.

Question

The figure shows the graph of the polar function $r = f(\theta)$, for $0 \leq \theta \leq 2\pi$, in the polar coordinate system. Which of the following could be an expression for $f(\theta)$?

(A) $6 \cos(4\theta)$

(B) $6 \cos(2\theta)$

(D) $6 \sin(2\theta)$

▶️ Answer/Explanation

Detailed solution

The graph is a rose curve with $n = 4$ petals and a maximum radius of $6$.
For a rose curve $r = a \sin(n\theta)$ or $r = a \cos(n\theta)$, if $n$ is even, there are $2n$ petals.
Since there are $4$ petals, we have $2n = 4$, which means $n = 2$.
The maximum radius is $6$, so the coefficient $a = 6$.
The petals are located in the quadrants (e.g., at $\theta = \frac{\pi}{4}$) rather than on the axes.
For $r = 6 \cos(2\theta)$, petals would be centered on the polar axis ($\theta = 0$).
For $r = 6 \sin(2\theta)$, the first maximum occurs at $2\theta = \frac{\pi}{2}$, or $\theta = \frac{\pi}{4}$.
Thus, the correct expression is $f(\theta) = 6 \sin(2\theta)$.
Correct Option: (D)

Question

Which of the following is the graph of the polar function $r = f(\theta)$, where $f(\theta) = -3 \sin(3\theta)$, in the polar coordinate system for $0 \leq \theta \leq 2\pi$?

(A)

(B)

(C)

(D)

▶️ Answer/Explanation

Detailed solution

The function $r = -3 \sin(3\theta)$ represents a three-petaled rose because $n = 3$ is odd.
The maximum magnitude of $r$ is $|-3| = 3$, meaning the petals extend to a radius of $3$.
Since it is a sine function, the petals are not centered on the polar axis ($0^\circ$).
For $f(\theta)$ to reach its first extremum, $3\theta = \frac{\pi}{2}$ or $3\theta = \frac{3\pi}{2}$.
At $\theta = \frac{\pi}{6}$, $r = -3 \sin(\frac{\pi}{2}) = -3$, plotting a point at radius $3$ in the direction $\theta = \frac{7\pi}{6}$.
At $\theta = \frac{\pi}{2}$, $r = -3 \sin(\frac{3\pi}{2}) = 3$, plotting a petal tip directly on the positive y-axis.
Comparing this to the choices, Graph (C) is the only one with a petal tip at $(r, \theta) = (3, \frac{\pi}{2})$.
Therefore, the correct choice is (C).

AP Precalculus -3.14 Polar Function Graphs- MCQ Exam Style Questions - Effective Fall 2023