AP Precalculus -1.11 Equivalent Forms of Polynomial and Rational Expressions- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.11 Equivalent Forms of Polynomial and Rational Expressions- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.11 Equivalent Forms of Polynomial and Rational Expressions- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The function $ h $ is given by $ h(x) = \frac{2x^3}{x+3} – \frac{4}{x-1} $. Which of the following statements is true?

(A) $ h $ is equivalent to $ \frac{2x^4 – 2x^3 – 4x – 12}{x^2 + 2x – 3} $ and has the same end behavior as the graph of $ y = 2x^2 $.
(B) $ h $ is equivalent to $ \frac{2x^3 – 4}{x^2 + 2x – 3} $ and has the same end behavior as the graph of $ y = 2x $.
(C) $ h $ is equivalent to $ \frac{2x^3}{x} + \frac{2x^2}{3} – \frac{4}{x-1} $ and has the same end behavior as the graph of $ y = 2x^2 $.
(D) $ h $ is equivalent to $ \frac{2x^2 – 2x – 12}{3x – 3} $ and has the same end behavior as the graph of $ y = 2x $.

▶️ Answer/Explanation

Detailed solution

First, combine into a single fraction:
Common denominator: $(x+3)(x-1) = x^2 + 2x – 3$.
\[ h(x) = \frac{2x^3}{x+3} – \frac{4}{x-1} = \frac{2x^3(x-1) – 4(x+3)}{(x+3)(x-1)}. \]
Simplify numerator:
\[ 2x^3(x-1) – 4(x+3) = 2x^4 – 2x^3 – 4x – 12. \]
So $ h(x) = \frac{2x^4 – 2x^3 – 4x – 12}{x^2 + 2x – 3} $.

For end behavior: degree(numerator) = 4, degree(denominator) = 2, so as $ x \to \pm\infty $,
\[ h(x) \approx \frac{2x^4}{x^2} = 2x^2. \]
Thus $ h $ has the same end behavior as $ y = 2x^2 $.

✅ Answer: (A)

Question

The polynomial function $ p $ is given by $ p(x) = (x + 2)^4 $. Which of the following expressions is equivalent to $ (x + 2)^4 $?

(A) $ x^4 + 2x^3 + 4x^2 + 8x + 16 $
(B) $ x^4 + 4x^3 + 6x^2 + 4x + 1 $
(C) $ x^4 + 8x^3 + 24x^2 + 32x + 16 $
(D) $ 2x^4 + 8x^3 + 12x^2 + 8x + 2 $

▶️ Answer/Explanation

Detailed solution

Using binomial expansion or Pascal’s Triangle for $ (x + 2)^4 $:
$ (x + 2)^4 = \sum_{k=0}^4 \binom{4}{k} x^{4-k} \cdot 2^{k} $
$ = \binom{4}{0}x^4\cdot 2^0 + \binom{4}{1}x^3\cdot 2^1 + \binom{4}{2}x^2\cdot 2^2 + \binom{4}{3}x^1\cdot 2^3 + \binom{4}{4}x^0\cdot 2^4 $
$ = 1\cdot x^4 \cdot 1 + 4\cdot x^3\cdot 2 + 6\cdot x^2\cdot 4 + 4\cdot x \cdot 8 + 1\cdot 1\cdot 16 $
$ = x^4 + 8x^3 + 24x^2 + 32x + 16 $.
✅ Answer: (C)

Question

The polynomial function $f$ is given by $f(x) = 2x^3 – 3x^2 – 23x + 12$. Which of the following is true about $\frac{f(x)}{x+3}$?

(A) The remainder of $\frac{f(x)}{x+3}$ is 0. The quotient of $\frac{f(x)}{x+3}$ is a quadratic polynomial that factors into two linear factors involving only real numbers.
(B) The remainder of $\frac{f(x)}{x+3}$ is 0. The quotient of $\frac{f(x)}{x+3}$ is a quadratic polynomial that does not factor into linear factors involving only real numbers.
(C) The remainder of $\frac{f(x)}{x+3}$ is a nonzero constant. The quotient of $\frac{f(x)}{x+3}$ is a quadratic polynomial that factors into two linear factors involving only real numbers.
(D) The remainder of $\frac{f(x)}{x+3}$ is a nonzero constant. The quotient of $\frac{f(x)}{x+3}$ is a quadratic polynomial that does not factor into linear factors involving only real numbers.

▶️ Answer/Explanation

Detailed solution

We can check if $x+3$ is a factor by evaluating $f(-3)$:
$f(-3) = 2(-27) – 3(9) – 23(-3) + 12 = -54 – 27 + 69 + 12 = 0$.
So $x+3$ divides $f$ with remainder 0.

Perform polynomial division: $ (2x^3 – 3x^2 – 23x + 12) \div (x+3) $ yields $ 2x^2 – 9x + 4 $.
Now factor $ 2x^2 – 9x + 4 $: it factors into $(2x – 1)(x – 4)$, both linear factors with real coefficients.

Thus: remainder is 0, quotient is quadratic and factors into real linear factors.
✅ Answer: (A)

Question

The function $ f $ is given by $ f(x) = (x + 3)^4 $. When $ f $ is rewritten in the form $ f(x) = x^4 + ax^3 + bx^2 + cx + d $, which of the following values is greatest?

(A) $ a $
(B) $ b $
(C) $ c $
(D) $ d $

▶️ Answer/Explanation

Detailed solution

Expand $ (x + 3)^4 $ using binomial theorem:
$ (x + 3)^4 = x^4 + 4\cdot 3\cdot x^3 + 6\cdot 3^2\cdot x^2 + 4\cdot 3^3\cdot x + 3^4 $
$ = x^4 + 12x^3 + 54x^2 + 108x + 81 $.
So $ a = 12,\; b = 54,\; c = 108,\; d = 81 $.
Greatest is $ c = 108 $.
✅ Answer: (C)

Question

The binomial theorem can be used to expand an expression of the form $(a + b)^n$. Which of the following is equivalent to $(x + 3y)^5$?

(A) $x^5 + (3y)^5$

(B) $x^5 + x^4(3y)^2 + x^3(3y)^3 + x^2(3y)^3 + x(3y)^4 + (3y)^5$

(D) $x^5 + 5x^4(3y)^2 + 10x^3(3y)^3 + 10x^2(3y)^3 + 5x(3y)^4 + (3y)^5$

▶️ Answer/Explanation

Detailed solution

The binomial expansion of $(x + 3y)^5$ is found using the formula $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$, with $n=5$.
The binomial coefficients for the 5th power (Row 5 of Pascal’s Triangle) are $1, 5, 10, 10, 5, 1$.
The correct theoretical expansion is: $1(x)^5 + 5(x)^4(3y)^1 + 10(x)^3(3y)^2 + 10(x)^2(3y)^3 + 5(x)^1(3y)^4 + 1(3y)^5$.
We compare this pattern with the given options by checking the coefficients.
Option (A) is missing terms, Option (B) has incorrect coefficients (all are $1$), and Option (C) uses coefficients for $n=4$ ($1, 4, 6, 4, 1$).
Option (D) correctly matches the coefficients $1, 5, 10, 10, 5, 1$. Note that while the question image contains typographical errors in the exponents, (D) is the only option with the correct binomial structure.
Therefore, the correct option is (D).

Question

The binomial theorem can be used to expand an expression of the form $(a + b)^n$. Which of the following is equivalent to $(x + 3y)^5$?

(A) $x^5 + (3y)^5$

(B) $x^5 + x^4(3y) + x^3(3y)^2 + x^2(3y)^3 + x(3y)^4 + (3y)^5$

(D) $x^5 + 5x^4(3y) + 10x^3(3y)^2 + 10x^2(3y)^3 + 5x(3y)^4 + (3y)^5$

▶️ Answer/Explanation

Detailed solution

The general binomial expansion is given by the formula: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.
In this problem, we identify the values as $a = x$, $b = 3y$, and the exponent $n = 5$.
The binomial coefficients for the power of 5 (from Pascal’s triangle) are: $1, 5, 10, 10, 5, 1$.
We apply these coefficients to the decreasing powers of $x$ and increasing powers of $(3y)$.
The expansion becomes: $1(x)^5 + 5(x)^4(3y) + 10(x)^3(3y)^2 + 10(x)^2(3y)^3 + 5(x)(3y)^4 + 1(3y)^5$.
Comparing this expanded form with the given choices, it corresponds exactly to Option (D).

Question

The binomial theorem can be used to expand an expression of the form $(a+b)^n$. Which of the following is equivalent to $(x-4y)^5$?

(A) $x^5 – (4y)^5$

(B) $x^5 – x^4(4y) + x^3(4y)^2 – x^2(4y)^3 + x(4y)^4 – y^5$

(D) $x^5 – 5x^4(4y) – 10x^3(4y)^2 – 10x^2(4y)^3 – 5x(4y)^4 – (4y)^5$

▶️ Answer/Explanation

Detailed solution

The correct option is (C).

1. According to the Binomial Theorem, $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.
2. Identify the components for substitution: $a = x$, $b = -4y$, and $n = 5$.
3. The binomial coefficients for $n=5$ (from Pascal’s triangle) are: $1, 5, 10, 10, 5, 1$.
4. Substitute these into the formula: $1(x)^5 + 5(x)^4(-4y) + 10(x)^3(-4y)^2 + 10(x)^2(-4y)^3 + 5(x)(-4y)^4 + 1(-4y)^5$.
5. Simplify the signs: odd powers of $(-4y)$ result in a negative sign, while even powers result in a positive sign.
6. The final expansion is: $x^5 – 5x^4(4y) + 10x^3(4y)^2 – 10x^2(4y)^3 + 5x(4y)^4 – (4y)^5$.

Question

The binomial theorem can be used to expand the polynomial function $p$, given by $p(x) = (x – 3)^5$. What is the coefficient of the $x^3$ term in the expanded polynomial?

(A) $(-3)^3 \cdot 10$

(B) $(-3)^2 \cdot 10$

(D) $(-3)^2 \cdot 5$

▶️ Answer/Explanation

Detailed solution

The general term of the expansion $(a + b)^n$ is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$.
For $p(x) = (x – 3)^5$, we have $n = 5$, $a = x$, and $b = -3$.
To find the $x^3$ term, we set the exponent of $x$ to $3$, so $n – k = 3$.
Solving for $k$ gives $5 – k = 3$, which means $k = 2$.
The term is $\binom{5}{2} (x)^3 (-3)^2$.
The binomial coefficient $\binom{5}{2} = \frac{5 \cdot 4}{2 \cdot 1} = 10$.
The coefficient of the $x^3$ term is therefore $10 \cdot (-3)^2$.
This matches option (B).

Question

The binomial theorem can be used to expand an expression of the form $(a + b)^n$. Which of the following is equivalent to $(x + 3y)^5$?

(A) $x^5 + (3y)^5$

(B) $x^5 + x^4(3y) + x^3(3y)^2 + x^2(3y)^3 + x(3y)^4 + (3y)^5$

(D) $x^5 + 5x^4(3y) + 10x^3(3y)^2 + 10x^2(3y)^3 + 5x(3y)^4 + (3y)^5$

▶️ Answer/Explanation

Detailed solution

The correct answer is (D).
The binomial theorem states $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k$.
For $(x + 3y)^5$, the power $n = 5$, so the expansion has $n+1 = 6$ terms.
The coefficients are found using the 5th row of Pascal’s Triangle: $1, 5, 10, 10, 5, 1$.
The powers of $x$ decrease from $5$ to $0$, while powers of $(3y)$ increase from $0$ to $5$.
Applying these coefficients yields $1x^5 + 5x^4(3y)^1 + 10x^3(3y)^2 + 10x^2(3y)^3 + 5x^1(3y)^4 + 1(3y)^5$.
This matches the expression provided in option (D).

Question

The polynomial function $ p $ is given by $ p(x) = (x + 2)^4 $. Which of the following expressions is equivalent to $ (x + 2)^4 $?

(A) $ x^4 + 2x^3 + 4x^2 + 8x + 16 $

(B) $ x^4 + 4x^3 + 6x^2 + 4x + 1 $

(D) $ 2x^4 + 8x^3 + 12x^2 + 8x + 2 $

▶️ Answer/Explanation

Answer:

(C) $ x^4 + 8x^3 + 24x^2 + 32x + 16 $

Explanation:

Using the Binomial Theorem, \[ (a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \] Substituting $ a = x $ and $ b = 2 $: \[ (x+2)^4 = x^4 + 4x^3(2) + 6x^2(2^2) + 4x(2^3) + 2^4 \] \[ = x^4 + 8x^3 + 24x^2 + 32x + 16 \] Thus, the correct answer is (C).

Detailed solution

1. Apply the binomial formula: $ (a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 $.
2. Substitute $ a = x $, $ b = 2 $.
3. Compute each term: $ 4x^3(2) = 8x^3 $.
4. Compute $ 6x^2(2^2) = 6x^2(4) = 24x^2 $.
5. Compute $ 4x(2^3) = 4x(8) = 32x $.
6. Compute $ 2^4 = 16 $.
7. Combine all terms: $ x^4 + 8x^3 + 24x^2 + 32x + 16 $.
8. Therefore, the equivalent expression is option (C).

Question

The function $ f $ is given by $ f(x) = (x + 3)^4 $. When $ f $ is rewritten in the form $ f(x) = x^4 + ax^3 + bx^2 + cx + d $, which of the following values is greatest?

(A) $ a $

(B) $ b $

(D) $ d $

▶️ Answer/Explanation

Answer: (C) $ c $

Detailed solution

1. Use the binomial expansion: $ (x+3)^4 = x^4 + 4x^3(3) + 6x^2(3^2) + 4x(3^3) + 3^4 $.
2. Compute each term: $ 4(3)=12 $, $ 6(9)=54 $, $ 4(27)=108 $, $ 3^4=81 $.
3. Therefore, $ (x+3)^4 = x^4 + 12x^3 + 54x^2 + 108x + 81 $.
4. Comparing with $ x^4 + ax^3 + bx^2 + cx + d $:
5. $ a=12 $, $ b=54 $, $ c=108 $, $ d=81 $.
6. The greatest value among $ 12, 54, 108, 81 $ is $ 108 $.
7. Hence, the greatest coefficient is $ c $.

Question

The polynomial function is $ f(x) = 2x^3 – 3x^2 – 23x + 12 $. Which of the following is true about $ \dfrac{f(x)}{x+3} $?

(A) The remainder is $0$. The quotient is a quadratic polynomial that factors into two linear factors involving only real numbers.

(B) The remainder is $0$. The quotient is a quadratic polynomial that does not factor into linear factors involving only real numbers.

(C) The remainder is a nonzero constant. The quotient is a quadratic polynomial that factors into two linear factors involving only real numbers.

(D) The remainder is a nonzero constant. The quotient is a quadratic polynomial that does not factor into linear factors involving only real numbers.

▶️ Answer/Explanation

Correct Answer: (A)

Detailed solution

1. By the Remainder Theorem, compute $ f(-3) $.
$ f(-3) = 2(-27) – 3(9) – 23(-3) + 12 = -54 – 27 + 69 + 12 = 0. $

2. Since $ f(-3) = 0 $, the remainder is $0$.

3. Perform synthetic division by $-3$: the quotient is $ 2x^2 – 9x + 4 $.

4. Compute the discriminant:
$ \Delta = (-9)^2 – 4(2)(4) = 81 – 32 = 49. $

5. Since $ \Delta = 49 > 0 $, the quadratic factors over the real numbers.

6. Factor: $ 2x^2 – 9x + 4 = (2x – 1)(x – 4). $

Therefore, the remainder is $0$ and the quotient factors into real linear factors.

Question

The functions $ g(x) = 3x^2 – 2x $ and $ f(x) = 6x^4 + 5x^3 + 3x – 5 $. Which of the following statements is true about the remainder when $ f(x) $ is divided by $ g(x) $?

(A) The remainder is $0$, so $g(x)$ is a factor of $f(x)$.

(B) The remainder is $0$, so $f(x)$ is a factor of $g(x)$.

(C) The remainder is $7x – 5$, so $g(x)$ is not a factor of $f(x)$, and the graph of $y = \dfrac{f(x)}{g(x)}$ has a slant asymptote.

(D) The remainder is $7x – 5$, so $g(x)$ is not a factor of $f(x)$, and the graph of $y = \dfrac{f(x)}{g(x)}$ does not have a slant asymptote.

▶️ Answer/Explanation

Answer:

(D)

Detailed solution

1. Divide $6x^4$ by $3x^2$ to get $2x^2$; subtract to obtain $9x^3$.
2. Divide $9x^3$ by $3x^2$ to get $3x$; subtract to obtain $6x^2$.
3. Divide $6x^2$ by $3x^2$ to get $2$; subtract to obtain remainder $7x – 5$.
4. Since the remainder is $7x – 5 \neq 0$, $g(x)$ is not a factor of $f(x)$.
5. Degree of $f(x)$ is $4$; degree of $g(x)$ is $2$.
6. The degree difference is $2$, so the quotient is quadratic (not linear).
7. A slant asymptote occurs only when the degree difference is $1$.
8. Therefore, the correct statement is (D).

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.
Unless otherwise specified, combine terms using algebraic methods and rules for exponents and logarithms where applicable.
For each part of the question, show the work that leads to your answers.

(A) The functions $ g $ and $ h $ are given by

$ g(x) = 3^{(2x)} \cdot 3^{(x+4)} $
$ h(x) = 2\tan^2 x – 1 $

(i) Solve $ g(x) = 27 $ for values of $ x $ in the domain of $ g $.
(ii) Solve $ h(x) = 5 $ for values of $ x $ in the interval $ [0, 2\pi) $.

(B) The functions $ j $ and $ k $ are given by

$ j(x) = 2\log_{10}(x+3) – \log_{10} x – \log_{10} 3 $
$ k(x) = \frac{(\tan^2 x)(\cot x)}{\csc x} $

(i) Rewrite $ j(x) $ as a single logarithm base 10 without negative exponents in any part of the expression. Your result should be of the form $ \log_{10}(\text{expression}) $.
(ii) Rewrite $ k(x) $ as a fraction involving $ \sec x $ and no other trigonometric functions.

$ m(x) = \frac{2^{(5x+3)}}{\left(2^{(x-2)}\right)^3} $

Find all input values in the domain of $ m $ that yield an output value of $ \frac{1}{16} $.

▶️ Answer/Explanation

Detailed Solution

(A)(i) Solve $ g(x) = 27 $

First, simplify the expression for $ g(x) $ using the property of exponents $ a^m \cdot a^n = a^{m+n} $.
$ g(x) = 3^{2x} \cdot 3^{x+4} = 3^{(2x + x + 4)} = 3^{(3x+4)} $
Set $ g(x) $ equal to 27 and rewrite 27 as a base of 3:
$ 3^{(3x+4)} = 27 $
$ 3^{(3x+4)} = 3^3 $
Since the bases are equal, the exponents must be equal:
$ 3x + 4 = 3 $
$ 3x = 3 – 4 $
$ 3x = -1 $
$ x = -\frac{1}{3} $

(A)(ii) Solve $ h(x) = 5 $ on $ [0, 2\pi) $

Set the expression for $ h(x) $ equal to 5:
$ 2\tan^2 x – 1 = 5 $
Add 1 to both sides:
$ 2\tan^2 x = 6 $
Divide by 2:
$ \tan^2 x = 3 $
Take the square root of both sides:
$ \tan x = \pm\sqrt{3} $
The reference angle for $ \tan \theta = \sqrt{3} $ is $ \frac{\pi}{3} $.
Since we have $ \pm\sqrt{3} $, we must consider solutions in all four quadrants within the interval $ [0, 2\pi) $:
Quadrant I: $ x = \frac{\pi}{3} $
Quadrant II: $ x = \pi – \frac{\pi}{3} = \frac{2\pi}{3} $
Quadrant III: $ x = \pi + \frac{\pi}{3} = \frac{4\pi}{3} $
Quadrant IV: $ x = 2\pi – \frac{\pi}{3} = \frac{5\pi}{3} $
Solution set: $ x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} $

(B)(i) Rewrite $ j(x) $ as a single logarithm

Given: $ j(x) = 2\log_{10}(x+3) – \log_{10} x – \log_{10} 3 $
Use the power rule $ n\log a = \log(a^n) $:
$ = \log_{10}((x+3)^2) – \log_{10} x – \log_{10} 3 $
Factor out the negative sign for the last two terms to group them:
$ = \log_{10}((x+3)^2) – (\log_{10} x + \log_{10} 3) $
Use the product rule $ \log a + \log b = \log(ab) $:
$ = \log_{10}((x+3)^2) – \log_{10}(3x) $
Use the quotient rule $ \log a – \log b = \log(\frac{a}{b}) $:
$ = \log_{10}\left(\frac{(x+3)^2}{3x}\right) $

(B)(ii) Rewrite $ k(x) $ involving $ \sec x $

Given: $ k(x) = \frac{(\tan^2 x)(\cot x)}{\csc x} $
First, simplify the numerator using $ \tan x \cdot \cot x = 1 $:
$ (\tan^2 x)(\cot x) = \tan x \cdot (\tan x \cdot \cot x) = \tan x \cdot 1 = \tan x $
Now substitute back into the expression:
$ k(x) = \frac{\tan x}{\csc x} $
Convert to sine and cosine:
$ = \frac{\frac{\sin x}{\cos x}}{\frac{1}{\sin x}} $
$ = \frac{\sin x}{\cos x} \cdot \frac{\sin x}{1} = \frac{\sin^2 x}{\cos x} $
We need the expression in terms of $ \sec x $. Use the identity $ \sin^2 x = 1 – \cos^2 x $:
$ = \frac{1 – \cos^2 x}{\cos x} $
Substitute $ \cos x = \frac{1}{\sec x} $:
$ = \frac{1 – \left(\frac{1}{\sec x}\right)^2}{\frac{1}{\sec x}} $
$ = \frac{1 – \frac{1}{\sec^2 x}}{\frac{1}{\sec x}} $
Find a common denominator for the numerator:
$ = \frac{\frac{\sec^2 x – 1}{\sec^2 x}}{\frac{1}{\sec x}} $
Multiply by the reciprocal of the denominator:
$ = \frac{\sec^2 x – 1}{\sec^2 x} \cdot \frac{\sec x}{1} $
$ = \frac{\sec^2 x – 1}{\sec x} $

(C) Find input values for $ m(x) = \frac{1}{16} $

First, simplify the expression for $ m(x) $:
$ m(x) = \frac{2^{(5x+3)}}{\left(2^{(x-2)}\right)^3} $
Simplify the denominator using the power rule $ (a^m)^n = a^{m \cdot n} $:
$ \left(2^{(x-2)}\right)^3 = 2^{3(x-2)} = 2^{3x-6} $
Now apply the quotient rule $ \frac{a^m}{a^n} = a^{m-n} $:
$ m(x) = \frac{2^{5x+3}}{2^{3x-6}} = 2^{(5x+3) – (3x-6)} $
$ = 2^{5x + 3 – 3x + 6} $
$ = 2^{2x + 9} $
Set $ m(x) = \frac{1}{16} $ and rewrite $ \frac{1}{16} $ as a power of 2:
$ \frac{1}{16} = \frac{1}{2^4} = 2^{-4} $
Equate the simplified $ m(x) $ to $ 2^{-4} $:
$ 2^{2x + 9} = 2^{-4} $
Equate the exponents:
$ 2x + 9 = -4 $
$ 2x = -13 $
$ x = -\frac{13}{2} $ or $ -6.5 $

AP Precalculus -1.11 Equivalent Forms of Polynomial and Rational Expressions- MCQ Exam Style Questions - Effective Fall 2023