Math 1 Quiz: Justifying Solution Paths

What this quiz covers

This quiz focuses on Justifying Solution Paths, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Find several examples of numbers divisible by 6 and check if they're divisible by 12.
2. Prove the statement using the fact that 12 is a multiple of 6.
3. Find one counterexample: a number divisible by 6 but not by 12. (correct answer)
4. Show that any number divisible by 6 must have all prime factors of 12.

Question 2

1. The argument fails to consider that $\sqrt{x^2}$ might not be defined for negative values of $x$, making the proof incomplete for the entire real number domain.
2. The argument incorrectly assumes that $|x|^2 = x^2$, which is only true when $x \geq 0$, so the reasoning breaks down for negative $x$ values.
3. The argument lacks explicit verification that $|x|$ is indeed the unique non-negative solution, since other expressions like $\sqrt{x^2}$ could also satisfy the given conditions. (correct answer)
4. The argument correctly establishes necessary conditions but doesn't demonstrate sufficiency by showing that $|x|$ actually equals $\sqrt{x^2}$ through substitution or cases.

Question 3

1. The analogy fails because logarithm properties are derived from exponent rules, not from arithmetic operations, so the inverse relationship doesn't transfer to logarithmic expressions.
2. The reasoning is flawed because $\log_b(x - y)$ is undefined when $x - y \leq 0$, while $\log_b(x) - \log_b(y)$ can exist for positive $x$ and $y$.
3. The analogy incorrectly assumes that logarithm properties follow arithmetic patterns, when actually $\log_b(x/y) = \log_b(x) - \log_b(y)$, not a subtraction inside the logarithm. (correct answer)
4. The student's reasoning fails because the logarithm product rule requires $b > 1$, but the proposed subtraction rule would require $0 < b < 1$ for the analogy to work.

Question 4

1. The first student is correct because the definition of absolute value directly gives two cases for any equation $|A| = B$ where $B > 0$, without needing to analyze signs first. (correct answer)
2. The second student is correct because proper absolute value solving requires determining where $2x - 3 \geq 0$ and $2x - 3 < 0$ to handle the cases systematically.
3. Both approaches are mathematically valid, but the second student's method provides better insight into the geometric meaning of absolute value as distance on a number line.
4. The first student's method is more efficient and equally rigorous, while the second student's approach is unnecessarily complicated for this type of absolute value equation.

Question 5

1. Using the formal definition of even integers to translate the problem into algebraic expressions that can be manipulated using arithmetic properties of integers. (correct answer)
2. Recognizing that the closure property of integers under addition ensures that $k + j$ remains an integer, preserving the even form of the sum.
3. Applying the distributive property correctly to factor out the common factor of 2, which directly establishes the even nature of the resulting expression.
4. Ensuring that the chosen variables $k$ and $j$ are explicitly stated to be integers, which validates the use of integer properties throughout the proof.

Question 6

1. The student failed to check the domain restriction $x^2 - 4 \geq 0$, which requires $x \leq -2$ or $x \geq 2$, making $x = 2$ a boundary case needing special consideration.
2. When squaring both sides of an equation involving square roots, potential extraneous solutions can be introduced, requiring verification that solutions satisfy the original equation's implicit restrictions.
3. The algebraic manipulation is incorrect because $(x-2)^2$ should equal $x^2 - 4x + 4$, but the student wrote the expansion incorrectly in the simplification step.
4. The student should have considered that $\sqrt{x^2 - 4} \geq 0$ always, so solutions are only valid when $x - 2 \geq 0$, meaning $x \geq 2$, which wasn't verified. (correct answer)

Question 7

1. The student incorrectly treats the indeterminate form $\frac{0}{0}$ as if it equals 1, when indeterminate forms require special techniques like L'Hôpital's rule to evaluate properly.
2. The reasoning is circular because claiming $\sin x \approx x$ for small $x$ essentially assumes the conclusion that $\frac{\sin x}{x} \approx 1$, which is what we're trying to prove. (correct answer)
3. The student fails to provide a rigorous definition of what $\sin x \approx x$ means mathematically, making the argument imprecise and not suitable for a formal limit proof.
4. The argument doesn't address why the approximation $\sin x \approx x$ becomes more accurate as $x$ approaches 0, which is essential for justifying the limit evaluation.

Question 8

1. The student failed to establish that $p$ and $q$ are both even, which is necessary to show that the fraction wasn't actually in lowest terms as assumed.
2. The reasoning doesn't explain why $2q^2$ necessarily has an odd number of factors of 2, since $q^2$ could have any number of factors of 2 depending on $q$.
3. The student should have used the fact that $p^2$ is even to conclude $p$ is even, then substituted $p = 2k$ to show both $p$ and $q$ are even, contradicting lowest terms. (correct answer)
4. The argument incorrectly assumes that $2q^2$ and $p^2$ must have different numbers of factors of 2, when they could be equal if $q$ contains the appropriate factors.

Question 9

1. The proof is logically valid because it correctly connects the existence of a maximum value to the downward orientation, then to the sign of the leading coefficient. (correct answer)
2. The proof is invalid because it assumes that quadratic functions with maximum values must be parabolas, when other types of functions can also have maximum values.
3. The proof is valid in its logical structure but lacks rigor because it doesn't formally define what 'maximum value' means in the context of quadratic functions.
4. The proof is circular because it assumes that having a maximum value means opening downward, which presupposes knowledge about the relationship between $a$ and the parabola's orientation.

Question 10

1. The algebraic method provides independent confirmation and shows the systematic process of finding inverses, while composition verifies the final result works correctly. (correct answer)
2. Composition checking can miss errors in domain restrictions, while the algebraic method naturally reveals where the inverse function might be undefined or problematic.
3. The algebraic method is more rigorous mathematically, while composition checking is more computational, so using both ensures complete mathematical validity.
4. Different methods can reveal different types of errors: algebraic mistakes versus computational mistakes, making the verification more thorough and reliable overall.

Question 11

1. The proof is circular because it uses the co-function relationship to prove a co-function identity, without demonstrating why tangent and cotangent are co-functions. (correct answer)
2. The student failed to verify that both sides of the identity are defined for the same values of $x$, which is necessary for any trigonometric identity proof.
3. The reasoning lacks mathematical rigor because it doesn't show the algebraic steps using sine and cosine definitions to derive the identity from first principles.
4. The proof is incomplete because it doesn't address the domain restrictions where $\tan(\frac{\pi}{2} - x)$ and $\cot(x)$ might be undefined or discontinuous.

Question 12

1. The reasoning is valid because negative discriminants always indicate complex roots, which correspond to no real $x$-intercepts on the coordinate plane. (correct answer)
2. The reasoning is invalid because the discriminant formula only applies to quadratic equations, not quadratic functions used for graphing purposes.
3. The reasoning is invalid because when $a > 0$ and the discriminant is negative, the parabola opens downward and crosses the $x$-axis twice.
4. The reasoning is valid only when $c > 0$, since negative $y$-intercepts can still produce real zeros even with negative discriminants in certain cases.

Question 13

1. The student failed to check whether $x = -2$ makes any denominator zero, which would make the original equation undefined and invalidate the solution.
2. The algebraic step from $\frac{x-1}{x+2} = \frac{3}{x+2}$ to $x - 1 = 3$ is only valid when $x + 2 \neq 0$, but this condition wasn't verified beforehand. (correct answer)
3. The verification process is flawed because the student should have substituted $x = 4$ into the original equation before simplifying the fractions to check for accuracy.
4. The student incorrectly assumed that equal denominators allow direct equation of numerators without considering that the equation might have no solution or infinitely many solutions.

Question 14

1. Converting the limit condition into an absolute value inequality that directly relates to the definition of limit, establishing the framework for the epsilon-delta argument.
2. Recognizing that $|3x - 6| = 3|x - 2|$ creates a direct connection between the function's deviation from the limit and the input's deviation from the approach point. (correct answer)
3. Choosing $\delta = \frac{\epsilon}{3}$ ensures that the epsilon condition is satisfied while maintaining the logical structure required by the definition of limit.
4. Working backwards from the desired inequality $3|x - 2| < \epsilon$ to find the appropriate constraint on $|x - 2|$, demonstrating strategic problem-solving in formal proofs.

Question 15

1. Use $x = \frac{\pi}{6}$: $\sin(2 \cdot \frac{\pi}{6}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$, while $2\sin(\frac{\pi}{6}) = 2 \cdot \frac{1}{2} = 1$, showing the values differ because sine is nonlinear. (correct answer)
2. Use $x = 0$: $\sin(2 \cdot 0) = \sin(0) = 0$ and $2\sin(0) = 2 \cdot 0 = 0$, demonstrating that while this case works, the general principle fails for most angles.
3. Use $x = \frac{\pi}{4}$: $\sin(2 \cdot \frac{\pi}{4}) = \sin(\frac{\pi}{2}) = 1$, while $2\sin(\frac{\pi}{4}) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}$, proving the student's algebraic reasoning is fundamentally incorrect.
4. Use $x = \pi$: $\sin(2\pi) = 0$ and $2\sin(\pi) = 2 \cdot 0 = 0$, but explain that this equality is coincidental since trigonometric functions don't follow linear distribution rules.

Question 16

1. The proof should verify that the function is defined for all real numbers, since the definition of odd functions requires the domain to be symmetric about the origin.
2. The algebraic manipulation $(-x)^3 = -x^3$ should be justified by citing the properties of exponents and the fact that 3 is an odd integer.
3. The proof should include verification by testing specific values like $f(1)$ and $f(-1)$ to confirm that $f(-1) = -f(1)$ holds numerically.
4. The student should explicitly state that this verification must hold for all $x$ in the domain, not just demonstrate the algebraic relationship symbolically. (correct answer)

Question 17

1. The multiplication property of inequality for positive numbers. (correct answer)
2. The reflexive property of equality to show that $c = c$.
3. The transitive property of inequality to connect the steps.
4. The distributive property to justify the multiplication step.

Question 18

1. It confirms that the system has a unique solution and validates both solution techniques. (correct answer)
2. It proves that the system is consistent and that their arithmetic calculations were accurate.
3. It demonstrates that different algebraic methods always produce identical results for linear systems.
4. It shows that the coefficient matrix is invertible and the system has exactly one solution.

Question 19

1. She must prove that $\angle B = 60°$ to show the triangle is equilateral, not just isosceles.
2. She must demonstrate that the sides opposite the equal angles are congruent using the Isosceles Triangle Theorem.
3. She must verify that her angle measurements are accurate by showing her calculation work.
4. She must state the converse of the Isosceles Triangle Theorem: if two angles are congruent, then the sides opposite those angles are congruent. (correct answer)

Question 20

1. Direct proof using the definition of rational numbers.
2. Constructive proof by explicitly finding values for $p$ and $q$.
3. Proof by induction using a base case and inductive step.
4. Proof by contradiction using the assumption of the opposite conclusion. (correct answer)