Function Inequalities - IB Mathematics Analysis and Approaches

1. Find the solution set for $\frac{x^2-4}{x-2} > 3$.
   Answer: $x \in (1, 2) \cup (2, \infty)$ The expression on the left has a domain of $x \ne 2$. For values of $x \ne 2$, we can simplify the expression: $$\frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2$$ The original inequality is therefore equivalent to solving $x+2 > 3$ with the restriction that $x \ne 2$. Solving $x+2 > 3$ gives $x > 1$. We must now apply the restriction $x \ne 2$ to this solution set. The set of numbers greater than 1 but not equal to 2 is represented by the union of two intervals: $(1, 2) \cup (2, \infty)$. A: This answer is correct for the simplified inequality but ignores the removable discontinuity (hole) in the original function's graph at $x=2$. B: This answer results from incorrect algebraic manipulation, possibly subtracting 2 instead of adding it to both sides. C: This is the correct solution, accounting for both the simplified inequality and the domain restriction. D: This answer results from flipping the inequality sign.
2. Given $f(x) = x^2 + 1$ and $g(x) = 2x - 1$, find the solution set for $f(g(x)) \ge g(f(x))$.
   Answer: $x \le 1 - \frac{\sqrt{2}}{2}$ or $x \ge 1 + \frac{\sqrt{2}}{2}$ First, find the expressions for the composite functions: $f(g(x)) = f(2x-1) = (2x-1)^2 + 1 = (4x^2 - 4x + 1) + 1 = 4x^2 - 4x + 2$. $g(f(x)) = g(x^2+1) = 2(x^2+1) - 1 = 2x^2 + 2 - 1 = 2x^2 + 1$. Now, set up the inequality: $$4x^2 - 4x + 2 \ge 2x^2 + 1$$ Rearrange into a standard quadratic inequality: $$2x^2 - 4x + 1 \ge 0$$ To find the roots of $2x^2 - 4x + 1 = 0$, use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. $$x = \frac{4 \pm \sqrt{(-4)^2 - 4(2)(1)}}{2(2)} = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}$$ The parabola $y = 2x^2 - 4x + 1$ opens upwards, so the expression is greater than or equal to zero outside of and including the roots. Therefore, the solution is $x \le 1 - \frac{\sqrt{2}}{2}$ or $x \ge 1 + \frac{\sqrt{2}}{2}$. A: This is the solution to $2x^2 - 4x + 1 \le 0$. B: This answer has a calculation error in the quadratic formula, missing the division of $\sqrt{8}$ by 2. C: This is the correct solution. D: This corresponds to the solution of $2x^2 - 4x + 1 \le 0$ with the same calculation error as B.
3. Solve the inequality $\log_3(x) > \log_9(x+2)$.
   Answer: $x \in (2, \infty)$ First, establish the domain. We need $x > 0$ and $x+2 > 0 \implies x > -2$. The intersection is $x > 0$. Next, use the change of base formula to make the bases of the logarithms the same. Let's use base 3. $\log_9(x+2) = \frac{\log_3(x+2)}{\log_3(9)} = \frac{\log_3(x+2)}{2} = \frac{1}{2}\log_3(x+2) = \log_3(\sqrt{x+2})$. The inequality becomes: $$\log_3(x) > \log_3(\sqrt{x+2})$$ Since $y=\log_3(t)$ is an increasing function, we can compare the arguments: $$x > \sqrt{x+2}$$ For this to hold, $x$ must be positive, which is already required by our domain. We can square both sides: $$x^2 > x+2$$ $$x^2 - x - 2 > 0$$ $$(x-2)(x+1) > 0$$ This holds for $x < -1$ or $x > 2$. Finally, we find the intersection of this solution with our domain $x > 0$. The result is $x > 2$. A: This is the solution to $x^2 - x - 2 < 0$. B: This is the result of flipping the inequality sign at some stage. C: This is just the domain of the functions, not the solution to the inequality. D: This is the correct solution.
4. Find the solution to the inequality $e^{2x} - 2e^x - 15 < 0$.
   Answer: $x < \ln 5$ Let $u = e^x$. Since $e^x > 0$ for all real $x$, we have the condition $u > 0$. The inequality can be rewritten as a quadratic in $u$: $$u^2 - 2u - 15 < 0$$ Factoring the quadratic: $$(u-5)(u+3) < 0$$ The roots are $u=5$ and $u=-3$. The inequality holds for $-3 < u < 5$. Now we combine this with the condition $u > 0$, which gives $0 < u < 5$. Substitute back $u = e^x$: $$0 < e^x < 5$$ The inequality $e^x > 0$ is true for all real $x$. The inequality $e^x < 5$ gives $x < \ln 5$. Thus, the solution set is $x < \ln 5$. A: This answer incorrectly takes the logarithm of a negative number, which is undefined in real numbers. B: This is the correct solution. C: This is the solution for the inequality $e^{2x} - 2e^x - 15 > 0$. D: This answer incorrectly solves the quadratic in $u$ and also attempts to take the logarithm of a negative number.
5. Find the solution to $2\sin^2(x) > \sin(x)$ for $x \in (0, 2\pi)$.
   Answer: $x \in (\frac{\pi}{6}, \frac{5\pi}{6}) \cup (\pi, 2\pi)$ Rearrange the inequality: $$2\sin^2(x) - \sin(x) > 0$$ Factor out $\sin(x)$: $$\sin(x)(2\sin(x) - 1) > 0$$ This inequality holds if both factors are positive or both are negative. Case 1: Both positive. $\sin(x) > 0$ and $2\sin(x) - 1 > 0 \implies \sin(x) > 1/2$. Combining these gives $\sin(x) > 1/2$. For $x \in (0, 2\pi)$, this occurs when $x \in (\frac{\pi}{6}, \frac{5\pi}{6})$. Case 2: Both negative. $\sin(x) < 0$ and $2\sin(x) - 1 < 0 \implies \sin(x) < 1/2$. Combining these gives $\sin(x) < 0$. For $x \in (0, 2\pi)$, this occurs when $x \in (\pi, 2\pi)$. The total solution is the union of the solutions from both cases: $x \in (\frac{\pi}{6}, \frac{5\pi}{6}) \cup (\pi, 2\pi)$. A: This is the solution for $\sin(x)(2\sin(x) - 1) < 0$ and $\sin(x)>0$. B: This is only the solution from Case 1. C: This is only the solution from Case 2. D: This is the correct union of solutions from both cases.
6. Determine the solution set for the inequality $(x+2)^2(x-1)(x-3) \le 0$.
   Answer: $x \in [1, 3] \cup \{-2\}$ The critical values are the roots of the polynomial: $x=-2$, $x=1$, and $x=3$. The factor $(x+2)^2$ is always non-negative. It is zero at $x=-2$ and positive everywhere else. This means it does not cause a sign change at $x=-2$. The sign of the expression is therefore determined by the sign of $(x-1)(x-3)$, which is a parabola opening upwards. - $(x-1)(x-3) \le 0$ when $1 \le x \le 3$. - The entire expression is also equal to 0 when $(x+2)^2 = 0$, which occurs at $x=-2$. So, the inequality $(x+2)^2(x-1)(x-3) \le 0$ is satisfied for $x \in [1, 3]$ and also at the isolated point $x=-2$. The complete solution set is the union of these, which is $[1, 3] \cup \{-2\}$. A: This answer correctly identifies the interval but omits the isolated solution $x=-2$. B: This answer results from an incorrect sign analysis. C: This answer incorrectly treats the double root at $x=-2$ as causing a sign change. D: This is the correct solution.
7. Find the set of values for which $x - 1 < \frac{12}{x}$.
   Answer: $x < -3$ or $0 < x < 4$ To solve this rational inequality, we must bring all terms to one side and use a sign diagram. A common mistake is to multiply by $x$, which is not allowed as $x$ can be negative. $$x - 1 - \frac{12}{x} < 0$$ Combine the terms over a common denominator: $$\frac{x^2 - x - 12}{x} < 0$$ Factor the numerator: $$\frac{(x-4)(x+3)}{x} < 0$$ The critical values are the roots of the numerator and denominator: $x=-3, x=0, x=4$. We test the intervals defined by these values: - $x < -3$: e.g., $x=-4$. $\frac{(-)(-)}{(-)} = \frac{+}{-} < 0$. True. - $-3 < x < 0$: e.g., $x=-1$. $\frac{(-)(+)}{(-)} = \frac{-}{-} > 0$. False. - $0 < x < 4$: e.g., $x=1$. $\frac{(-)(+)}{(+)} = \frac{-}{+} < 0$. True. - $x > 4$: e.g., $x=5$. $\frac{(+)(+)}{(+)} = \frac{+}{+} > 0$. False. The solution is $x < -3$ or $0 < x < 4$. A: This is the correct solution. B: This answer incorrectly ignores the critical value at the asymptote $x=0$. C: This is the solution for the inequality $\frac{(x-4)(x+3)}{x} > 0$. D: This answer treats $x=0$ as a simple root and ignores the middle interval.
8. Let $f(x) = \frac{1}{x}$ and $g(x) = x-2$. For which values of $x$ is $f(x) \ge g(x)$?
   Answer: $x \in (-\infty, 1-\sqrt{2}] \cup (0, 1+\sqrt{2}]$ We need to solve the inequality $\frac{1}{x} \ge x-2$. Rearrange to get 0 on one side: $$\frac{1}{x} - (x-2) \ge 0$$ $$\frac{1 - x(x-2)}{x} \ge 0$$ $$\frac{1 - x^2 + 2x}{x} \ge 0$$ $$\frac{-x^2 + 2x + 1}{x} \ge 0$$ Multiply by -1 and flip the inequality sign: $$\frac{x^2 - 2x - 1}{x} \le 0$$ The roots of the numerator $x^2 - 2x - 1 = 0$ are found using the quadratic formula: $x = \frac{2 \pm \sqrt{4-4(-1)}}{2} = 1 \pm \sqrt{2}$. The root of the denominator is $x=0$. The critical values are $1-\sqrt{2}$ (approx -0.414), 0, and $1+\sqrt{2}$ (approx 2.414). We use a sign diagram for $\frac{(x-(1-\sqrt{2}))(x-(1+\sqrt{2}))}{x} \le 0$: - $x < 1-\sqrt{2}$: $\frac{(-)(-)}{(-)} = \frac{+}{-} = -$. True. - $1-\sqrt{2} < x < 0$: $\frac{(+)(-)}{(-)} = \frac{-}{-} = +$. False. - $0 < x < 1+\sqrt{2}$: $\frac{(+)(-)}{(+)} = \frac{-}{+} = -$. True. - $x > 1+\sqrt{2}$: $\frac{(+)(+)}{(+)} = \frac{+}{+} = +$. False. The expression is zero when $x = 1 \pm \sqrt{2}$. So we include these points. The solution is $(-\infty, 1-\sqrt{2}] \cup (0, 1+\sqrt{2}]$. A: This is the solution to $\frac{x^2 - 2x - 1}{x} \ge 0$. B: This ignores the asymptote at $x=0$. C: This is the correct solution. D: This misinterprets the sign analysis intervals.
9. Find the set of all real numbers $x$ such that $\frac{x}{x^2-1} \le 0$.
   Answer: $x \in (-\infty, -1) \cup [0, 1)$ First, factor the denominator: $\frac{x}{(x-1)(x+1)} \le 0$. The critical values are the roots of the numerator and the denominator: $x=0$, $x=1$, and $x=-1$. We use a sign diagram to test the intervals. - $x < -1$ (e.g., $x=-2$): \frac{-}{(-)(-)_} = \frac{-}{+} < 0. True. - $-1 < x < 0$ (e.g., $x=-0.5$): $\frac{-}{(-)(+)} = \frac{-}{-} > 0$. False. - $0 < x < 1$ (e.g., $x=0.5$): $\frac{+}{(-)(+)} = \frac{+}{-} < 0$. True. - $x > 1$ (e.g., $x=2$): $\frac{+}{(+)(+)} = \frac{+}{+} > 0$. False. The inequality is $\le 0$, so we must also include where the expression equals 0. This occurs when the numerator is 0, i.e., $x=0$. The expression is undefined at $x=1$ and $x=-1$, so these points are excluded. Combining the results, the solution is $(-\infty, -1) \cup [0, 1)$. A: This is the solution for $\frac{x}{x^2-1} \ge 0$ with incorrect handling of the boundaries. B: This is the correct solution. C: This would be the solution for $\ge 0$ and also incorrectly includes the asymptotes. D: This answer incorrectly includes the asymptotes $x=-1$ and $x=1$ in the solution set.
10. Solve the inequality $\log_2(x^2 - 6) > \log_2(x)$.
    Answer: $x \in (3, \infty)$ First, we must consider the domain of the logarithmic functions. 1. For $\log_2(x^2 - 6)$, we need $x^2 - 6 > 0 \Rightarrow x^2 > 6 \Rightarrow x < -\sqrt{6}$ or $x > \sqrt{6}$. 2. For $\log_2(x)$, we need $x > 0$. The overall domain is the intersection of these conditions, which is $x > \sqrt{6}$. Now, we solve the inequality. Since $\log_2(x)$ is an increasing function, we can compare the arguments: $$x^2 - 6 > x$$ $$x^2 - x - 6 > 0$$ $$(x-3)(x+2) > 0$$ The roots are $x=3$ and $x=-2$. The quadratic is positive outside the roots, so $x < -2$ or $x > 3$. Finally, we must find the intersection of this solution set with the domain $x > \sqrt{6}$. Since $\sqrt{6} \approx 2.45$, the intersection is $x > 3$. A: This is the solution to the quadratic inequality without considering the domain restrictions. B: This results from incorrectly combining the domain and the quadratic solution. C: This is the correct solution. D: This is the solution to the quadratic inequality $x^2 - x - 6 < 0$.
11. Find the solution set for the inequality $\frac{x+5}{x-1} \le 3$.
    Answer: $x \in (-\infty, 1) \cup [4, \infty)$ To solve the inequality, rearrange it to have 0 on one side: $$\frac{x+5}{x-1} - 3 \le 0$$ Combine the terms over a common denominator: $$\frac{x+5 - 3(x-1)}{x-1} \le 0$$ $$\frac{x+5 - 3x + 3}{x-1} \le 0$$ $$\frac{-2x + 8}{x-1} \le 0$$ $$\frac{2(4-x)}{x-1} \le 0$$ The critical values are the roots of the numerator and denominator, which are $x=4$ and $x=1$. We use a sign diagram to test the intervals defined by these values. - For $x < 1$, let $x=0$: $\frac{2(4)}{-1} = -8 \le 0$. The inequality holds. - For $1 < x < 4$, let $x=2$: $\frac{2(2)}{1} = 4 > 0$. The inequality does not hold. - For $x > 4$, let $x=5$: $\frac{2(-1)}{4} = -0.5 \le 0$. The inequality holds. We must also check the endpoints. At $x=4$, the expression is 0, so $0 \le 0$ is true. At $x=1$, the expression is undefined. Therefore, the solution is $x < 1$ or $x \ge 4$, which is $(-\infty, 1) \cup [4, \infty)$. A: This is the solution for $\frac{2(4-x)}{x-1} \ge 0$ with an incorrect handling of the boundary at $x=1$. B: This answer omits the interval $(-\infty, 1)$. C: This is the correct solution set. D: This incorrectly includes the asymptote $x=1$ in the solution set.
12. Solve the inequality $\sqrt{x+2} < x$.
    Answer: $x \in (2, \infty)$ There are two main conditions for $\sqrt{A} < B$ to be true. 1. The term inside the square root must be non-negative: $A \ge 0$. So, $x+2 \ge 0 \implies x \ge -2$. 2. The right side must be positive, since a square root (principal) cannot be negative: $B > 0$. So, $x > 0$. Combining these preliminary conditions, we require $x > 0$. Now, we can square both sides of the inequality: $$x+2 < x^2$$ $$0 < x^2 - x - 2$$ $$0 < (x-2)(x+1)$$ The roots are $x=2$ and $x=-1$. The quadratic is positive outside the roots, so $x < -1$ or $x > 2$. Finally, we find the intersection of this result with our condition $x > 0$. The intersection is $x > 2$. A: This is the solution to $x^2 - x - 2 < 0$. B: This is the correct solution. C: This is a common error from squaring both sides without first establishing that $x$ must be positive. D: This is another error from misinterpreting the result of the quadratic inequality.
13. Let $f(x) = e^x$ and $g(x) = x^2$. Using the information provided, which interval is a subset of the solution to $f(x) > g(x)$?
    Answer: $[-0.7, 0]$ We want to find where $e^x > x^2$. Let $h(x) = e^x - x^2$. We are looking for where $h(x) > 0$. The given information tells us that the graphs of $f(x)$ and $g(x)$ intersect at $x_0 \approx -0.703$, meaning $h(x_0)=0$. We can test a point on either side of $x_0$ to determine the sign of $h(x)$. - Test a point $x < x_0$, for example $x = -1$. Here, $f(-1) = e^{-1} \approx 0.37$ and $g(-1) = (-1)^2 = 1$. Since $0.37 < 1$, $f(x) < g(x)$ in this region. So, intervals A and B are not subsets of the solution. - Test a point $x > x_0$, for example $x = 0$. Here, $f(0) = e^0 = 1$ and $g(0) = 0^2 = 0$. Since $1 > 0$, $f(x) > g(x)$ in this region. This indicates that the solution set is $(x_0, \infty)$. The interval in option C, $[-0.7, 0]$, is entirely contained within $(x_0, \infty)$ since $x_0 \approx -0.703$. Therefore, $[-0.7, 0]$ is a subset of the solution. A: In this interval, $x^2 > e^x$. B: In this interval, $x^2 > e^x$. C: In this interval, $x > x_0$, so $e^x > x^2$. This is a subset of the solution. D: The solution set is not empty, as shown by the test point $x=0$.
14. For $x \in [0, \pi]$, find the solution to $\cos(2x) < \cos(x)$.
    Answer: $x \in (0, \frac{2\pi}{3})$ Use the double angle identity $\cos(2x) = 2\cos^2(x) - 1$ to rewrite the inequality in terms of $\cos(x)$. $$2\cos^2(x) - 1 < \cos(x)$$ $$2\cos^2(x) - \cos(x) - 1 < 0$$ Let $u = \cos(x)$. The inequality becomes $2u^2 - u - 1 < 0$. Factor the quadratic: $(2u+1)(u-1) < 0$. The roots are $u = -1/2$ and $u = 1$. The inequality holds for $-1/2 < u < 1$. Substitute back $u = \cos(x)$: $$-1/2 < \cos(x) < 1$$ We need to find the values of $x$ in the interval $[0, \pi]$ that satisfy this. - $\cos(x) < 1$: In the interval $[0, \pi]$, $\cos(x) = 1$ only at $x=0$. So we need $x \in (0, \pi]$. - $\cos(x) > -1/2$: In the interval $[0, \pi]$, $\cos(x) = -1/2$ at $x = 2\pi/3$. Since $\cos(x)$ is a decreasing function on $[0, \pi]$, $\cos(x) > -1/2$ when $x < 2\pi/3$. So we need $x \in [0, 2\pi/3)$. We need the intersection of $(0, \pi]$ and $[0, 2\pi/3)$, which is $(0, 2\pi/3)$. A: This is the correct solution. B: This is the solution for $\cos(x) < -1/2$ in the given interval. C: This would be the solution if the critical value was $x=\pi/3$ instead of $x=2\pi/3$. D: This corresponds to an incorrect sign analysis or incorrect critical value.
15. Determine the solution to the inequality $|\ln(x)| \ge 2$.
    Answer: $x \in (0, e^{-2}] \cup [e^2, \infty)$ First, the domain of $\ln(x)$ requires $x > 0$. The inequality $|A| \ge B$ is equivalent to $A \ge B$ or $A \le -B$. So, we have two cases: 1. $\ln(x) \ge 2$ Exponentiating both sides with base $e$ gives $x \ge e^2$. 2. $\ln(x) \le -2$ Exponentiating both sides gives $x \le e^{-2}$. Both of these solution sets are within the domain $x > 0$ (since $e^{-2} > 0$). The final solution is the union of the solutions from the two cases, restricted to the domain. This gives $0 < x \le e^{-2}$ or $x \ge e^2$. In interval notation, this is $(0, e^{-2}] \cup [e^2, \infty)$. A: This is the solution to $|\ln(x)| \le 2$. B: This answer only considers the first case, $\ln(x) \ge 2$. C: This is the correct solution. D: This answer incorrectly includes negative values, which are not in the domain of $\ln(x)$.
16. Find the solution for the inequality $|x^2 - 5| < 4$.
    Answer: $x \in (-3, -1) \cup (1, 3)$ The inequality $|A| < B$ is equivalent to the compound inequality $-B < A < B$. So, we have $-4 < x^2 - 5 < 4$. We can split this into two separate inequalities that must both be true: 1. $x^2 - 5 < 4$ $x^2 - 9 < 0$ $(x-3)(x+3) < 0$ The solution is $-3 < x < 3$. 2. $x^2 - 5 > -4$ $x^2 - 1 > 0$ $(x-1)(x+1) > 0$ The solution is $x < -1$ or $x > 1$. We need the intersection of the solutions from (1) and (2). The values of $x$ must be in $(-3, 3)$ AND (in $(-\infty, -1)$ or in $(1, \infty)$). The intersection is $(-3, -1) \cup (1, 3)$. A: This is the correct solution. B: This is the solution to only the first inequality, $x^2-9<0$. C: This is the solution to only the second inequality, $x^2-1>0$. D: This is the solution to $x^2-1<0$, which is a mistake from the second inequality.
17. Find the solution set for the inequality $|x-4| \le \frac{x}{2}$.
    Answer: $x \in [\frac{8}{3}, 8]$ For the inequality to hold, the right side must be non-negative, since the absolute value on the left is always non-negative. So, $\frac{x}{2} \ge 0 \implies x \ge 0$. With this condition, we can square both sides: $$(x-4)^2 \le (\frac{x}{2})^2$$ $$x^2 - 8x + 16 \le \frac{x^2}{4}$$ Multiply by 4 to clear the fraction: $$4x^2 - 32x + 64 \le x^2$$ $$3x^2 - 32x + 64 \le 0$$ Find the roots of $3x^2 - 32x + 64 = 0$ using the quadratic formula or by factoring. Factoring gives $(3x-8)(x-8) = 0$. No, that's $3x^2 -32x +64$. Let's use the formula: $x = \frac{32 \pm \sqrt{32^2 - 4(3)(64)}}{6} = \frac{32 \pm \sqrt{1024 - 768}}{6} = \frac{32 \pm \sqrt{256}}{6} = \frac{32 \pm 16}{6}$. The roots are $x_1 = \frac{32-16}{6} = \frac{16}{6} = \frac{8}{3}$ and $x_2 = \frac{32+16}{6} = \frac{48}{6} = 8$. Since the parabola $y = 3x^2 - 32x + 64$ opens upwards, the inequality $\le 0$ is satisfied between and including the roots: $\frac{8}{3} \le x \le 8$. This solution satisfies the initial condition $x \ge 0$. So the final answer is $[\frac{8}{3}, 8]$. A: This answer only considers the case when $x-4 \ge 0$. B: This is the solution to $3x^2 - 32x + 64 \ge 0$. C: This is the correct solution. D: This is an answer that could result from an algebraic error when dealing with the inequality.
18. Find the values of $m$ for which the line $y = mx - 1$ does not intersect the parabola $y = 2x^2 + 3x$.
    Answer: $3 - 2\sqrt{2} < m < 3 + 2\sqrt{2}$ For the line and parabola to not intersect, the equation $mx - 1 = 2x^2 + 3x$ must have no real solutions. We can rearrange this equation into a standard quadratic form $ax^2+bx+c=0$: $$2x^2 + 3x - mx + 1 = 0$$ $$2x^2 + (3-m)x + 1 = 0$$ For this quadratic equation to have no real solutions, its discriminant ($\Delta = b^2 - 4ac$) must be negative. Here, $a=2$, $b = 3-m$, and $c=1$. $$\Delta = (3-m)^2 - 4(2)(1) < 0$$ $$(3-m)^2 - 8 < 0$$ $$(3-m)^2 < 8$$ Taking the square root of both sides: $$|3-m| < \sqrt{8}$$ $$|3-m| < 2\sqrt{2}$$ This absolute value inequality is equivalent to: $$-2\sqrt{2} < 3-m < 2\sqrt{2}$$ Subtract 3 from all parts: $$-3 - 2\sqrt{2} < -m < -3 + 2\sqrt{2}$$ Multiply by -1 and reverse the inequality signs: $$3 + 2\sqrt{2} > m > 3 - 2\sqrt{2}$$ Rewriting in standard order gives $3 - 2\sqrt{2} < m < 3 + 2\sqrt{2}$. A: This is the solution for $\Delta = 0$, where the line is tangent to the parabola. B: This is the solution for $\Delta > 0$, where the line intersects the parabola at two distinct points. C: This would result from a calculation error in the discriminant, such as $4ac = 16$. D: This is the correct solution.
19. For which set of values of $k$ is the inequality $x^2 - kx + 4 \ge 0$ true for all real numbers $x$?
    Answer: $k \in [-4, 4]$ The quadratic function $f(x) = x^2 - kx + 4$ represents a parabola opening upwards (since the coefficient of $x^2$ is positive). For $f(x) \ge 0$ for all real $x$, the parabola must always be on or above the x-axis. This means it can have at most one real root (i.e., one root or no real roots). The condition for this is that the discriminant, $\Delta = b^2 - 4ac$, must be less than or equal to zero. For this quadratic, $a=1, b=-k, c=4$. $$\Delta = (-k)^2 - 4(1)(4) \le 0$$ $$k^2 - 16 \le 0$$ $$(k-4)(k+4) \le 0$$ The roots are $k=4$ and $k=-4$. Since the parabola $y=k^2-16$ opens upwards, the expression is less than or equal to zero between and including the roots. Therefore, $-4 \le k \le 4$. A: This is the solution for $\Delta < 0$, which excludes the cases where the parabola touches the x-axis. B: This is the solution for $\Delta > 0$, which corresponds to two distinct real roots. C: This is the correct solution. D: This would result from an arithmetic error, such as calculating the discriminant as $k^2-8$ or $k^2-4$.
20. Find the set of values of $x$ for which $|3x - 2| > |x + 6|$.
    Answer: $x < -1$ or $x > 4$ Since both sides of the inequality are non-negative, we can square both sides without changing the inequality direction: $$(3x - 2)^2 > (x + 6)^2$$ $$9x^2 - 12x + 4 > x^2 + 12x + 36$$ Rearrange into a standard quadratic inequality: $$8x^2 - 24x - 32 > 0$$ Divide by 8: $$x^2 - 3x - 4 > 0$$ Factor the quadratic expression: $$(x-4)(x+1) > 0$$ The roots are $x=4$ and $x=-1$. Since the parabola $y = x^2 - 3x - 4$ opens upwards, the expression is positive (greater than 0) outside the roots. Thus, the solution is $x < -1$ or $x > 4$. A: This is the solution to the inequality $(x-4)(x+1) < 0$. B: This would be the solution if the factored form was $(x+4)(x-1) > 0$, indicating a factoring error. C: This would be the solution if the factored form was $(x+4)(x-1) < 0$. D: This is the correct solution set.

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