IB Mathematics: Analysis and Approaches Quiz: Function Inequalities
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Function InequalitiesQuestion 1 of 20

The equation ex=x2e^x = x^2 has a single negative solution x0≈−0.703x_0 \approx -0.703.

Let f(x)=exf(x) = e^x and g(x)=x2g(x) = x^2. Using the information provided, which interval is a subset of the solution to f(x)>g(x)f(x) > g(x)?

(−∞,−1)(-\infty, -1)
[−1,−0.8][-1, -0.8]
[−0.7,0][-0.7, 0]
The solution set is empty.
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Function Inequalities

Practice Function Inequalities in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Function Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Question 1

The equation ex=x2e^x = x^2 has a single negative solution x0≈−0.703x_0 \approx -0.703.

Let f(x)=exf(x) = e^x and g(x)=x2g(x) = x^2. Using the information provided, which interval is a subset of the solution to f(x)>g(x)f(x) > g(x)?

  1. (−∞,−1)(-\infty, -1)
  2. [−1,−0.8][-1, -0.8]
  3. [−0.7,0][-0.7, 0] (correct answer)
  4. The solution set is empty.
Explanation: We want to find where ex>x2e^x > x^2. Let h(x)=ex−x2h(x) = e^x - x^2. We are looking for where h(x)>0h(x) > 0. The given information tells us that the graphs of f(x)f(x) and g(x)g(x) intersect at x0≈−0.703x_0 \approx -0.703, meaning h(x0)=0h(x_0)=0. We can test a point on either side of x0x_0 to determine the sign of h(x)h(x).
  • Test a point x<x0x < x_0, for example x=−1x = -1. Here, f(−1)=e−1≈0.37f(-1) = e^{-1} \approx 0.37 and g(−1)=(−1)2=1g(-1) = (-1)^2 = 1. Since 0.37<10.37 < 1, f(x)<g(x)f(x) < g(x) in this region. So, intervals A and B are not subsets of the solution.
  • Test a point x>x0x > x_0, for example x=0x = 0. Here, f(0)=e0=1f(0) = e^0 = 1 and g(0)=02=0g(0) = 0^2 = 0. Since 1>01 > 0, f(x)>g(x)f(x) > g(x) in this region. This indicates that the solution set is (x0,∞)(x_0, \infty). The interval in option C, [−0.7,0][-0.7, 0], is entirely contained within (x0,∞)(x_0, \infty) since x0≈−0.703x_0 \approx -0.703. Therefore, [−0.7,0][-0.7, 0] is a subset of the solution.
A: In this interval, x2>exx^2 > e^x. B: In this interval, x2>exx^2 > e^x. C: In this interval, x>x0x > x_0, so ex>x2e^x > x^2. This is a subset of the solution. D: The solution set is not empty, as shown by the test point x=0x=0.

Question 2

The equation ex=x2e^x = x^2 has a single negative solution x0≈−0.703x_0 \approx -0.703.

Let f(x)=exf(x) = e^x and g(x)=x2g(x) = x^2. Using the information provided, which interval is a subset of the solution to f(x)>g(x)f(x) > g(x)?

  1. (−∞,−1)(-\infty, -1)
  2. [−1,−0.8][-1, -0.8]
  3. [−0.7,0][-0.7, 0] (correct answer)
  4. The solution set is empty.
Explanation: We want to find where ex>x2e^x > x^2. Let h(x)=ex−x2h(x) = e^x - x^2. We are looking for where h(x)>0h(x) > 0. The given information tells us that the graphs of f(x)f(x) and g(x)g(x) intersect at x0≈−0.703x_0 \approx -0.703, meaning h(x0)=0h(x_0)=0. We can test a point on either side of x0x_0 to determine the sign of h(x)h(x).
  • Test a point x<x0x < x_0, for example x=−1x = -1. Here, f(−1)=e−1≈0.37f(-1) = e^{-1} \approx 0.37 and g(−1)=(−1)2=1g(-1) = (-1)^2 = 1. Since 0.37<10.37 < 1, f(x)<g(x)f(x) < g(x) in this region. So, intervals A and B are not subsets of the solution.
  • Test a point x>x0x > x_0, for example x=0x = 0. Here, f(0)=e0=1f(0) = e^0 = 1 and g(0)=02=0g(0) = 0^2 = 0. Since 1>01 > 0, f(x)>g(x)f(x) > g(x) in this region. This indicates that the solution set is (x0,∞)(x_0, \infty). The interval in option C, [−0.7,0][-0.7, 0], is entirely contained within (x0,∞)(x_0, \infty) since x0≈−0.703x_0 \approx -0.703. Therefore, [−0.7,0][-0.7, 0] is a subset of the solution.
A: In this interval, x2>exx^2 > e^x. B: In this interval, x2>exx^2 > e^x. C: In this interval, x>x0x > x_0, so ex>x2e^x > x^2. This is a subset of the solution. D: The solution set is not empty, as shown by the test point x=0x=0.

Question 3

Find the solution to the inequality e2x−2ex−15<0e^{2x} - 2e^x - 15 < 0.

  1. ln⁡(−3)<x<ln⁡5\ln(-3) < x < \ln 5
  2. x<ln⁡5x < \ln 5 (correct answer)
  3. x>ln⁡5x > \ln 5
  4. x>ln⁡5x > \ln 5 or x<ln⁡(−3)x < \ln(-3)
Explanation: Let u=exu = e^x. Since ex>0e^x > 0 for all real xx, we have the condition u>0u > 0. The inequality can be rewritten as a quadratic in uu: u2−2u−15<0u^2 - 2u - 15 < 0 Factoring the quadratic: (u−5)(u+3)<0(u-5)(u+3) < 0 The roots are u=5u=5 and u=−3u=-3. The inequality holds for −3<u<5-3 < u < 5. Now we combine this with the condition u>0u > 0, which gives 0<u<50 < u < 5. Substitute back u=exu = e^x: 0<ex<50 < e^x < 5 The inequality ex>0e^x > 0 is true for all real xx. The inequality ex<5e^x < 5 gives x<ln⁡5x < \ln 5. Thus, the solution set is x<ln⁡5x < \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 e2x−2ex−15>0e^{2x} - 2e^x - 15 > 0. D: This answer incorrectly solves the quadratic in uu and also attempts to take the logarithm of a negative number.

Question 4

Find the solution for the inequality ∣x2−5∣<4|x^2 - 5| < 4.

  1. x∈(−3,−1)∪(1,3)x \in (-3, -1) \cup (1, 3) (correct answer)
  2. x∈(−3,3)x \in (-3, 3)
  3. x∈(−∞,−1)∪(1,∞)x \in (-\infty, -1) \cup (1, \infty)
  4. x∈(−1,1)x \in (-1, 1)
Explanation: The inequality ∣A∣<B|A| < B is equivalent to the compound inequality −B<A<B-B < A < B. So, we have −4<x2−5<4-4 < x^2 - 5 < 4. We can split this into two separate inequalities that must both be true:
  1. x2−5<4x^2 - 5 < 4 x2−9<0x^2 - 9 < 0 (x−3)(x+3)<0(x-3)(x+3) < 0 The solution is −3<x<3-3 < x < 3.
  2. x2−5>−4x^2 - 5 > -4 x2−1>0x^2 - 1 > 0 (x−1)(x+1)>0(x-1)(x+1) > 0 The solution is x<−1x < -1 or x>1x > 1. We need the intersection of the solutions from (1) and (2). The values of xx must be in (−3,3)(-3, 3) AND (in (−∞,−1)(-\infty, -1) or in (1,∞)(1, \infty)). The intersection is (−3,−1)∪(1,3)(-3, -1) \cup (1, 3).
A: This is the correct solution. B: This is the solution to only the first inequality, x2−9<0x^2-9<0. C: This is the solution to only the second inequality, x2−1>0x^2-1>0. D: This is the solution to x2−1<0x^2-1<0, which is a mistake from the second inequality.

Question 5

Let f(x)=1xf(x) = \frac{1}{x} and g(x)=x−2g(x) = x-2. For which values of xx is f(x)≥g(x)f(x) \ge g(x)?

  1. x∈[1−2,0)∪[1+2,∞)x \in [1-\sqrt{2}, 0) \cup [1+\sqrt{2}, \infty)
  2. x∈[1−2,1+2]x \in [1-\sqrt{2}, 1+\sqrt{2}]
  3. x∈(−∞,1−2]∪(0,1+2]x \in (-\infty, 1-\sqrt{2}] \cup (0, 1+\sqrt{2}] (correct answer)
  4. x∈(−∞,0)∪[1−2,1+2]x \in (-\infty, 0) \cup [1-\sqrt{2}, 1+\sqrt{2}]
Explanation: We need to solve the inequality 1x≥x−2\frac{1}{x} \ge x-2. Rearrange to get 0 on one side: 1x−(x−2)≥0\frac{1}{x} - (x-2) \ge 0 1−x(x−2)x≥0\frac{1 - x(x-2)}{x} \ge 0 1−x2+2xx≥0\frac{1 - x^2 + 2x}{x} \ge 0 −x2+2x+1x≥0\frac{-x^2 + 2x + 1}{x} \ge 0 Multiply by -1 and flip the inequality sign: x2−2x−1x≤0\frac{x^2 - 2x - 1}{x} \le 0 The roots of the numerator x2−2x−1=0x^2 - 2x - 1 = 0 are found using the quadratic formula: x=2±4−4(−1)2=1±2x = \frac{2 \pm \sqrt{4-4(-1)}}{2} = 1 \pm \sqrt{2}. The root of the denominator is x=0x=0. The critical values are 1−21-\sqrt{2} (approx -0.414), 0, and 1+21+\sqrt{2} (approx 2.414). We use a sign diagram for (x−(1−2))(x−(1+2))x≤0\frac{(x-(1-\sqrt{2}))(x-(1+\sqrt{2}))}{x} \le 0:
  • x<1−2x < 1-\sqrt{2}: (−)(−)(−)=+−=−\frac{(-)(-)}{(-)} = \frac{+}{-} = -. True.
  • 1−2<x<01-\sqrt{2} < x < 0: (+)(−)(−)=−−=+\frac{(+)(-)}{(-)} = \frac{-}{-} = +. False.
  • 0<x<1+20 < x < 1+\sqrt{2}: (+)(−)(+)=−+=−\frac{(+)(-)}{(+)} = \frac{-}{+} = -. True.
  • x>1+2x > 1+\sqrt{2}: (+)(+)(+)=++=+\frac{(+)(+)}{(+)} = \frac{+}{+} = +. False. The expression is zero when x=1±2x = 1 \pm \sqrt{2}. So we include these points. The solution is (−∞,1−2]∪(0,1+2](-\infty, 1-\sqrt{2}] \cup (0, 1+\sqrt{2}].
A: This is the solution to x2−2x−1x≥0\frac{x^2 - 2x - 1}{x} \ge 0. B: This ignores the asymptote at x=0x=0. C: This is the correct solution. D: This misinterprets the sign analysis intervals.

Question 6

Find the solution to the inequality e2x−2ex−15<0e^{2x} - 2e^x - 15 < 0.

  1. ln⁡(−3)<x<ln⁡5\ln(-3) < x < \ln 5
  2. x<ln⁡5x < \ln 5 (correct answer)
  3. x>ln⁡5x > \ln 5
  4. x>ln⁡5x > \ln 5 or x<ln⁡(−3)x < \ln(-3)
Explanation: Let u=exu = e^x. Since ex>0e^x > 0 for all real xx, we have the condition u>0u > 0. The inequality can be rewritten as a quadratic in uu: u2−2u−15<0u^2 - 2u - 15 < 0 Factoring the quadratic: (u−5)(u+3)<0(u-5)(u+3) < 0 The roots are u=5u=5 and u=−3u=-3. The inequality holds for −3<u<5-3 < u < 5. Now we combine this with the condition u>0u > 0, which gives 0<u<50 < u < 5. Substitute back u=exu = e^x: 0<ex<50 < e^x < 5 The inequality ex>0e^x > 0 is true for all real xx. The inequality ex<5e^x < 5 gives x<ln⁡5x < \ln 5. Thus, the solution set is x<ln⁡5x < \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 e2x−2ex−15>0e^{2x} - 2e^x - 15 > 0. D: This answer incorrectly solves the quadratic in uu and also attempts to take the logarithm of a negative number.

Question 7

Given f(x)=x2+1f(x) = x^2 + 1 and g(x)=2x−1g(x) = 2x - 1, find the solution set for f(g(x))≥g(f(x))f(g(x)) \ge g(f(x)).

  1. x∈[1−22,1+22]x \in [1 - \frac{\sqrt{2}}{2}, 1 + \frac{\sqrt{2}}{2}]
  2. x≤1−2x \le 1 - \sqrt{2} or x≥1+2x \ge 1 + \sqrt{2}
  3. x≤1−22x \le 1 - \frac{\sqrt{2}}{2} or x≥1+22x \ge 1 + \frac{\sqrt{2}}{2} (correct answer)
  4. x∈[1−2,1+2]x \in [1 - \sqrt{2}, 1 + \sqrt{2}]
Explanation: First, find the expressions for the composite functions: f(g(x))=f(2x−1)=(2x−1)2+1=(4x2−4x+1)+1=4x2−4x+2f(g(x)) = f(2x-1) = (2x-1)^2 + 1 = (4x^2 - 4x + 1) + 1 = 4x^2 - 4x + 2. g(f(x))=g(x2+1)=2(x2+1)−1=2x2+2−1=2x2+1g(f(x)) = g(x^2+1) = 2(x^2+1) - 1 = 2x^2 + 2 - 1 = 2x^2 + 1. Now, set up the inequality: 4x2−4x+2≥2x2+14x^2 - 4x + 2 \ge 2x^2 + 1 Rearrange into a standard quadratic inequality: 2x2−4x+1≥02x^2 - 4x + 1 \ge 0 To find the roots of 2x2−4x+1=02x^2 - 4x + 1 = 0, use the quadratic formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a}. x=4±(−4)2−4(2)(1)2(2)=4±16−84=4±84=4±224=1±22x = \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=2x2−4x+1y = 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≤1−22x \le 1 - \frac{\sqrt{2}}{2} or x≥1+22x \ge 1 + \frac{\sqrt{2}}{2}. A: This is the solution to 2x2−4x+1≤02x^2 - 4x + 1 \le 0. B: This answer has a calculation error in the quadratic formula, missing the division of 8\sqrt{8} by 2. C: This is the correct solution. D: This corresponds to the solution of 2x2−4x+1≤02x^2 - 4x + 1 \le 0 with the same calculation error as B.

Question 8

For x∈[0,π]x \in [0, \pi], find the solution to cos⁡(2x)<cos⁡(x)\cos(2x) < \cos(x).

  1. x∈(0,2π3)x \in (0, \frac{2\pi}{3}) (correct answer)
  2. x∈(2π3,π]x \in (\frac{2\pi}{3}, \pi]
  3. x∈(0,π3)x \in (0, \frac{\pi}{3})
  4. x∈(π3,π]x \in (\frac{\pi}{3}, \pi]
Explanation: Use the double angle identity cos⁡(2x)=2cos⁡2(x)−1\cos(2x) = 2\cos^2(x) - 1 to rewrite the inequality in terms of cos⁡(x)\cos(x). 2cos⁡2(x)−1<cos⁡(x)2\cos^2(x) - 1 < \cos(x) 2cos⁡2(x)−cos⁡(x)−1<02\cos^2(x) - \cos(x) - 1 < 0 Let u=cos⁡(x)u = \cos(x). The inequality becomes 2u2−u−1<02u^2 - u - 1 < 0. Factor the quadratic: (2u+1)(u−1)<0(2u+1)(u-1) < 0. The roots are u=−1/2u = -1/2 and u=1u = 1. The inequality holds for −1/2<u<1-1/2 < u < 1. Substitute back u=cos⁡(x)u = \cos(x): −1/2<cos⁡(x)<1-1/2 < \cos(x) < 1 We need to find the values of xx in the interval [0,π][0, \pi] that satisfy this.
  • cos⁡(x)<1\cos(x) < 1: In the interval [0,π][0, \pi], cos⁡(x)=1\cos(x) = 1 only at x=0x=0. So we need x∈(0,π]x \in (0, \pi].
  • cos⁡(x)>−1/2\cos(x) > -1/2: In the interval [0,π][0, \pi], cos⁡(x)=−1/2\cos(x) = -1/2 at x=2π/3x = 2\pi/3. Since cos⁡(x)\cos(x) is a decreasing function on [0,π][0, \pi], cos⁡(x)>−1/2\cos(x) > -1/2 when x<2π/3x < 2\pi/3. So we need x∈[0,2π/3)x \in [0, 2\pi/3). We need the intersection of (0,π](0, \pi] and [0,2π/3)[0, 2\pi/3), which is (0,2π/3)(0, 2\pi/3).
A: This is the correct solution. B: This is the solution for cos⁡(x)<−1/2\cos(x) < -1/2 in the given interval. C: This would be the solution if the critical value was x=π/3x=\pi/3 instead of x=2π/3x=2\pi/3. D: This corresponds to an incorrect sign analysis or incorrect critical value.

Question 9

Find the solution set for the inequality ∣x−4∣≤x2|x-4| \le \frac{x}{2}.

  1. x∈[4,8]x \in [4, 8]
  2. x≤83x \le \frac{8}{3} or x≥8x \ge 8
  3. x∈[83,8]x \in [\frac{8}{3}, 8] (correct answer)
  4. x∈(−∞,8]x \in (-\infty, 8]
Explanation: For the inequality to hold, the right side must be non-negative, since the absolute value on the left is always non-negative. So, x2≥0  ⟹  x≥0\frac{x}{2} \ge 0 \implies x \ge 0. With this condition, we can square both sides: (x−4)2≤(x2)2(x-4)^2 \le (\frac{x}{2})^2 x2−8x+16≤x24x^2 - 8x + 16 \le \frac{x^2}{4} Multiply by 4 to clear the fraction: 4x2−32x+64≤x24x^2 - 32x + 64 \le x^2 3x2−32x+64≤03x^2 - 32x + 64 \le 0 Find the roots of 3x2−32x+64=03x^2 - 32x + 64 = 0 using the quadratic formula or by factoring. Factoring gives (3x−8)(x−8)=0(3x-8)(x-8) = 0. No, that's 3x2−32x+643x^2 -32x +64. Let's use the formula: x=32±322−4(3)(64)6=32±1024−7686=32±2566=32±166x = \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 x1=32−166=166=83x_1 = \frac{32-16}{6} = \frac{16}{6} = \frac{8}{3} and x2=32+166=486=8x_2 = \frac{32+16}{6} = \frac{48}{6} = 8. Since the parabola y=3x2−32x+64y = 3x^2 - 32x + 64 opens upwards, the inequality ≤0\le 0 is satisfied between and including the roots: 83≤x≤8\frac{8}{3} \le x \le 8. This solution satisfies the initial condition x≥0x \ge 0. So the final answer is [83,8][\frac{8}{3}, 8]. A: This answer only considers the case when x−4≥0x-4 \ge 0. B: This is the solution to 3x2−32x+64≥03x^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.

Question 10

Determine the solution set for the inequality (x+2)2(x−1)(x−3)≤0(x+2)^2(x-1)(x-3) \le 0.

  1. x∈[1,3]x \in [1, 3]
  2. x∈[−2,1]∪[3,∞)x \in [-2, 1] \cup [3, \infty)
  3. x∈(−∞,−2]∪[1,3]x \in (-\infty, -2] \cup [1, 3]
  4. x∈[1,3]∪{−2}x \in [1, 3] \cup \{-2\} (correct answer)
Explanation: The critical values are the roots of the polynomial: x=−2x=-2, x=1x=1, and x=3x=3. The factor (x+2)2(x+2)^2 is always non-negative. It is zero at x=−2x=-2 and positive everywhere else. This means it does not cause a sign change at x=−2x=-2. The sign of the expression is therefore determined by the sign of (x−1)(x−3)(x-1)(x-3), which is a parabola opening upwards.
  • (x−1)(x−3)≤0(x-1)(x-3) \le 0 when 1≤x≤31 \le x \le 3.
  • The entire expression is also equal to 0 when (x+2)2=0(x+2)^2 = 0, which occurs at x=−2x=-2. So, the inequality (x+2)2(x−1)(x−3)≤0(x+2)^2(x-1)(x-3) \le 0 is satisfied for x∈[1,3]x \in [1, 3] and also at the isolated point x=−2x=-2. The complete solution set is the union of these, which is [1,3]∪{−2}[1, 3] \cup \{-2\}.
A: This answer correctly identifies the interval but omits the isolated solution x=−2x=-2. B: This answer results from an incorrect sign analysis. C: This answer incorrectly treats the double root at x=−2x=-2 as causing a sign change. D: This is the correct solution.

Question 11

Find the solution set for the inequality x+5x−1≤3\frac{x+5}{x-1} \le 3.

  1. x∈(1,4]x \in (1, 4]
  2. x∈[4,∞)x \in [4, \infty)
  3. x∈(−∞,1)∪[4,∞)x \in (-\infty, 1) \cup [4, \infty) (correct answer)
  4. x∈(−∞,1]∪[4,∞)x \in (-\infty, 1] \cup [4, \infty)
Explanation: To solve the inequality, rearrange it to have 0 on one side: x+5x−1−3≤0\frac{x+5}{x-1} - 3 \le 0 Combine the terms over a common denominator: x+5−3(x−1)x−1≤0\frac{x+5 - 3(x-1)}{x-1} \le 0 x+5−3x+3x−1≤0\frac{x+5 - 3x + 3}{x-1} \le 0 −2x+8x−1≤0\frac{-2x + 8}{x-1} \le 0 2(4−x)x−1≤0\frac{2(4-x)}{x-1} \le 0 The critical values are the roots of the numerator and denominator, which are x=4x=4 and x=1x=1. We use a sign diagram to test the intervals defined by these values.
  • For x<1x < 1, let x=0x=0: 2(4)−1=−8≤0\frac{2(4)}{-1} = -8 \le 0. The inequality holds.
  • For 1<x<41 < x < 4, let x=2x=2: 2(2)1=4>0\frac{2(2)}{1} = 4 > 0. The inequality does not hold.
  • For x>4x > 4, let x=5x=5: 2(−1)4=−0.5≤0\frac{2(-1)}{4} = -0.5 \le 0. The inequality holds. We must also check the endpoints. At x=4x=4, the expression is 0, so 0≤00 \le 0 is true. At x=1x=1, the expression is undefined. Therefore, the solution is x<1x < 1 or x≥4x \ge 4, which is (−∞,1)∪[4,∞)(-\infty, 1) \cup [4, \infty).
A: This is the solution for 2(4−x)x−1≥0\frac{2(4-x)}{x-1} \ge 0 with an incorrect handling of the boundary at x=1x=1. B: This answer omits the interval (−∞,1)(-\infty, 1). C: This is the correct solution set. D: This incorrectly includes the asymptote x=1x=1 in the solution set.

Question 12

Find the set of values for which x−1<12xx - 1 < \frac{12}{x}.

  1. x<−3x < -3 or 0<x<40 < x < 4 (correct answer)
  2. −3<x<4-3 < x < 4
  3. −3<x<0-3 < x < 0 or x>4x > 4
  4. x<−3x < -3 or x>4x > 4
Explanation: 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 xx, which is not allowed as xx can be negative. x−1−12x<0x - 1 - \frac{12}{x} < 0 Combine the terms over a common denominator: x2−x−12x<0\frac{x^2 - x - 12}{x} < 0 Factor the numerator: (x−4)(x+3)x<0\frac{(x-4)(x+3)}{x} < 0 The critical values are the roots of the numerator and denominator: x=−3,x=0,x=4x=-3, x=0, x=4. We test the intervals defined by these values:
  • x<−3x < -3: e.g., x=−4x=-4. (−)(−)(−)=+−<0\frac{(-)(-)}{(-)} = \frac{+}{-} < 0. True.
  • −3<x<0-3 < x < 0: e.g., x=−1x=-1. (−)(+)(−)=−−>0\frac{(-)(+)}{(-)} = \frac{-}{-} > 0. False.
  • 0<x<40 < x < 4: e.g., x=1x=1. (−)(+)(+)=−+<0\frac{(-)(+)}{(+)} = \frac{-}{+} < 0. True.
  • x>4x > 4: e.g., x=5x=5. (+)(+)(+)=++>0\frac{(+)(+)}{(+)} = \frac{+}{+} > 0. False. The solution is x<−3x < -3 or 0<x<40 < x < 4.
A: This is the correct solution. B: This answer incorrectly ignores the critical value at the asymptote x=0x=0. C: This is the solution for the inequality (x−4)(x+3)x>0\frac{(x-4)(x+3)}{x} > 0. D: This answer treats x=0x=0 as a simple root and ignores the middle interval.

Question 13

Find the set of all real numbers xx such that xx2−1≤0\frac{x}{x^2-1} \le 0.

  1. x∈(−1,0]∪(1,∞)x \in (-1, 0] \cup (1, \infty)
  2. x∈(−∞,−1)∪[0,1)x \in (-\infty, -1) \cup [0, 1) (correct answer)
  3. x∈[−1,0]∪[1,∞)x \in [-1, 0] \cup [1, \infty)
  4. x∈(−∞,−1]∪[0,1]x \in (-\infty, -1] \cup [0, 1]
Explanation: First, factor the denominator: x(x−1)(x+1)≤0\frac{x}{(x-1)(x+1)} \le 0. The critical values are the roots of the numerator and the denominator: x=0x=0, x=1x=1, and x=−1x=-1. We use a sign diagram to test the intervals.
  • x<−1x < -1 (e.g., x=−2x=-2): \frac{-}{(-)(-)_} = \frac{-}{+} < 0. True.
  • −1<x<0-1 < x < 0 (e.g., x=−0.5x=-0.5): −(−)(+)=−−>0\frac{-}{(-)(+)} = \frac{-}{-} > 0. False.
  • 0<x<10 < x < 1 (e.g., x=0.5x=0.5): +(−)(+)=+−<0\frac{+}{(-)(+)} = \frac{+}{-} < 0. True.
  • x>1x > 1 (e.g., x=2x=2): +(+)(+)=++>0\frac{+}{(+)(+)} = \frac{+}{+} > 0. False. The inequality is ≤0\le 0, so we must also include where the expression equals 0. This occurs when the numerator is 0, i.e., x=0x=0. The expression is undefined at x=1x=1 and x=−1x=-1, so these points are excluded. Combining the results, the solution is (−∞,−1)∪[0,1)(-\infty, -1) \cup [0, 1).
A: This is the solution for xx2−1≥0\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 ≥0\ge 0 and also incorrectly includes the asymptotes. D: This answer incorrectly includes the asymptotes x=−1x=-1 and x=1x=1 in the solution set.

Question 14

For which set of values of kk is the inequality x2−kx+4≥0x^2 - kx + 4 \ge 0 true for all real numbers xx?

  1. k∈(−4,4)k \in (-4, 4)
  2. k<−4k < -4 or k>4k > 4
  3. k∈[−4,4]k \in [-4, 4] (correct answer)
  4. k∈[−2,2]k \in [-2, 2]
Explanation: The quadratic function f(x)=x2−kx+4f(x) = x^2 - kx + 4 represents a parabola opening upwards (since the coefficient of x2x^2 is positive). For f(x)≥0f(x) \ge 0 for all real xx, 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, Δ=b2−4ac\Delta = b^2 - 4ac, must be less than or equal to zero. For this quadratic, a=1,b=−k,c=4a=1, b=-k, c=4. Δ=(−k)2−4(1)(4)≤0\Delta = (-k)^2 - 4(1)(4) \le 0 k2−16≤0k^2 - 16 \le 0 (k−4)(k+4)≤0(k-4)(k+4) \le 0 The roots are k=4k=4 and k=−4k=-4. Since the parabola y=k2−16y=k^2-16 opens upwards, the expression is less than or equal to zero between and including the roots. Therefore, −4≤k≤4-4 \le k \le 4. A: This is the solution for Δ<0\Delta < 0, which excludes the cases where the parabola touches the x-axis. B: This is the solution for Δ>0\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 k2−8k^2-8 or k2−4k^2-4.

Question 15

Find the solution set for the inequality x+5x−1≤3\frac{x+5}{x-1} \le 3.

  1. x∈(1,4]x \in (1, 4]
  2. x∈[4,∞)x \in [4, \infty)
  3. x∈(−∞,1)∪[4,∞)x \in (-\infty, 1) \cup [4, \infty) (correct answer)
  4. x∈(−∞,1]∪[4,∞)x \in (-\infty, 1] \cup [4, \infty)
Explanation: To solve the inequality, rearrange it to have 0 on one side: x+5x−1−3≤0\frac{x+5}{x-1} - 3 \le 0 Combine the terms over a common denominator: x+5−3(x−1)x−1≤0\frac{x+5 - 3(x-1)}{x-1} \le 0 x+5−3x+3x−1≤0\frac{x+5 - 3x + 3}{x-1} \le 0 −2x+8x−1≤0\frac{-2x + 8}{x-1} \le 0 2(4−x)x−1≤0\frac{2(4-x)}{x-1} \le 0 The critical values are the roots of the numerator and denominator, which are x=4x=4 and x=1x=1. We use a sign diagram to test the intervals defined by these values.
  • For x<1x < 1, let x=0x=0: 2(4)−1=−8≤0\frac{2(4)}{-1} = -8 \le 0. The inequality holds.
  • For 1<x<41 < x < 4, let x=2x=2: 2(2)1=4>0\frac{2(2)}{1} = 4 > 0. The inequality does not hold.
  • For x>4x > 4, let x=5x=5: 2(−1)4=−0.5≤0\frac{2(-1)}{4} = -0.5 \le 0. The inequality holds. We must also check the endpoints. At x=4x=4, the expression is 0, so 0≤00 \le 0 is true. At x=1x=1, the expression is undefined. Therefore, the solution is x<1x < 1 or x≥4x \ge 4, which is (−∞,1)∪[4,∞)(-\infty, 1) \cup [4, \infty).
A: This is the solution for 2(4−x)x−1≥0\frac{2(4-x)}{x-1} \ge 0 with an incorrect handling of the boundary at x=1x=1. B: This answer omits the interval (−∞,1)(-\infty, 1). C: This is the correct solution set. D: This incorrectly includes the asymptote x=1x=1 in the solution set.

Question 16

Find the solution to 2sin⁡2(x)>sin⁡(x)2\sin^2(x) > \sin(x) for x∈(0,2π)x \in (0, 2\pi).

  1. x∈(0,π6)∪(5π6,π)x \in (0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, \pi)
  2. x∈(π6,5π6)x \in (\frac{\pi}{6}, \frac{5\pi}{6})
  3. x∈(π,2π)x \in (\pi, 2\pi)
  4. x∈(π6,5π6)∪(π,2π)x \in (\frac{\pi}{6}, \frac{5\pi}{6}) \cup (\pi, 2\pi) (correct answer)
Explanation: Rearrange the inequality: 2sin⁡2(x)−sin⁡(x)>02\sin^2(x) - \sin(x) > 0 Factor out sin⁡(x)\sin(x): sin⁡(x)(2sin⁡(x)−1)>0\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\sin(x) > 0 and 2sin⁡(x)−1>0  ⟹  sin⁡(x)>1/22\sin(x) - 1 > 0 \implies \sin(x) > 1/2. Combining these gives sin⁡(x)>1/2\sin(x) > 1/2. For x∈(0,2π)x \in (0, 2\pi), this occurs when x∈(π6,5π6)x \in (\frac{\pi}{6}, \frac{5\pi}{6}). Case 2: Both negative. sin⁡(x)<0\sin(x) < 0 and 2sin⁡(x)−1<0  ⟹  sin⁡(x)<1/22\sin(x) - 1 < 0 \implies \sin(x) < 1/2. Combining these gives sin⁡(x)<0\sin(x) < 0. For x∈(0,2π)x \in (0, 2\pi), this occurs when x∈(π,2π)x \in (\pi, 2\pi). The total solution is the union of the solutions from both cases: x∈(π6,5π6)∪(π,2π)x \in (\frac{\pi}{6}, \frac{5\pi}{6}) \cup (\pi, 2\pi). A: This is the solution for sin⁡(x)(2sin⁡(x)−1)<0\sin(x)(2\sin(x) - 1) < 0 and sin⁡(x)>0\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.

Question 17

Let f(x)=1xf(x) = \frac{1}{x} and g(x)=x−2g(x) = x-2. For which values of xx is f(x)≥g(x)f(x) \ge g(x)?

  1. x∈[1−2,0)∪[1+2,∞)x \in [1-\sqrt{2}, 0) \cup [1+\sqrt{2}, \infty)
  2. x∈[1−2,1+2]x \in [1-\sqrt{2}, 1+\sqrt{2}]
  3. x∈(−∞,1−2]∪(0,1+2]x \in (-\infty, 1-\sqrt{2}] \cup (0, 1+\sqrt{2}] (correct answer)
  4. x∈(−∞,0)∪[1−2,1+2]x \in (-\infty, 0) \cup [1-\sqrt{2}, 1+\sqrt{2}]
Explanation: We need to solve the inequality 1x≥x−2\frac{1}{x} \ge x-2. Rearrange to get 0 on one side: 1x−(x−2)≥0\frac{1}{x} - (x-2) \ge 0 1−x(x−2)x≥0\frac{1 - x(x-2)}{x} \ge 0 1−x2+2xx≥0\frac{1 - x^2 + 2x}{x} \ge 0 −x2+2x+1x≥0\frac{-x^2 + 2x + 1}{x} \ge 0 Multiply by -1 and flip the inequality sign: x2−2x−1x≤0\frac{x^2 - 2x - 1}{x} \le 0 The roots of the numerator x2−2x−1=0x^2 - 2x - 1 = 0 are found using the quadratic formula: x=2±4−4(−1)2=1±2x = \frac{2 \pm \sqrt{4-4(-1)}}{2} = 1 \pm \sqrt{2}. The root of the denominator is x=0x=0. The critical values are 1−21-\sqrt{2} (approx -0.414), 0, and 1+21+\sqrt{2} (approx 2.414). We use a sign diagram for (x−(1−2))(x−(1+2))x≤0\frac{(x-(1-\sqrt{2}))(x-(1+\sqrt{2}))}{x} \le 0:
  • x<1−2x < 1-\sqrt{2}: (−)(−)(−)=+−=−\frac{(-)(-)}{(-)} = \frac{+}{-} = -. True.
  • 1−2<x<01-\sqrt{2} < x < 0: (+)(−)(−)=−−=+\frac{(+)(-)}{(-)} = \frac{-}{-} = +. False.
  • 0<x<1+20 < x < 1+\sqrt{2}: (+)(−)(+)=−+=−\frac{(+)(-)}{(+)} = \frac{-}{+} = -. True.
  • x>1+2x > 1+\sqrt{2}: (+)(+)(+)=++=+\frac{(+)(+)}{(+)} = \frac{+}{+} = +. False. The expression is zero when x=1±2x = 1 \pm \sqrt{2}. So we include these points. The solution is (−∞,1−2]∪(0,1+2](-\infty, 1-\sqrt{2}] \cup (0, 1+\sqrt{2}].
A: This is the solution to x2−2x−1x≥0\frac{x^2 - 2x - 1}{x} \ge 0. B: This ignores the asymptote at x=0x=0. C: This is the correct solution. D: This misinterprets the sign analysis intervals.

Question 18

Find the solution set for the inequality ∣x−4∣≤x2|x-4| \le \frac{x}{2}.

  1. x∈[4,8]x \in [4, 8]
  2. x≤83x \le \frac{8}{3} or x≥8x \ge 8
  3. x∈[83,8]x \in [\frac{8}{3}, 8] (correct answer)
  4. x∈(−∞,8]x \in (-\infty, 8]
Explanation: For the inequality to hold, the right side must be non-negative, since the absolute value on the left is always non-negative. So, x2≥0  ⟹  x≥0\frac{x}{2} \ge 0 \implies x \ge 0. With this condition, we can square both sides: (x−4)2≤(x2)2(x-4)^2 \le (\frac{x}{2})^2 x2−8x+16≤x24x^2 - 8x + 16 \le \frac{x^2}{4} Multiply by 4 to clear the fraction: 4x2−32x+64≤x24x^2 - 32x + 64 \le x^2 3x2−32x+64≤03x^2 - 32x + 64 \le 0 Find the roots of 3x2−32x+64=03x^2 - 32x + 64 = 0 using the quadratic formula or by factoring. Factoring gives (3x−8)(x−8)=0(3x-8)(x-8) = 0. No, that's 3x2−32x+643x^2 -32x +64. Let's use the formula: x=32±322−4(3)(64)6=32±1024−7686=32±2566=32±166x = \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 x1=32−166=166=83x_1 = \frac{32-16}{6} = \frac{16}{6} = \frac{8}{3} and x2=32+166=486=8x_2 = \frac{32+16}{6} = \frac{48}{6} = 8. Since the parabola y=3x2−32x+64y = 3x^2 - 32x + 64 opens upwards, the inequality ≤0\le 0 is satisfied between and including the roots: 83≤x≤8\frac{8}{3} \le x \le 8. This solution satisfies the initial condition x≥0x \ge 0. So the final answer is [83,8][\frac{8}{3}, 8]. A: This answer only considers the case when x−4≥0x-4 \ge 0. B: This is the solution to 3x2−32x+64≥03x^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.

Question 19

Find the solution set for x2−4x−2>3\frac{x^2-4}{x-2} > 3.

  1. x>1x > 1
  2. x>3x > 3
  3. x∈(1,2)∪(2,∞)x \in (1, 2) \cup (2, \infty) (correct answer)
  4. x∈(−∞,1)x \in (-\infty, 1)
Explanation: The expression on the left has a domain of x≠2x \ne 2. For values of x≠2x \ne 2, we can simplify the expression: x2−4x−2=(x−2)(x+2)x−2=x+2\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>3x+2 > 3 with the restriction that x≠2x \ne 2. Solving x+2>3x+2 > 3 gives x>1x > 1. We must now apply the restriction x≠2x \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)∪(2,∞)(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=2x=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.

Question 20

Find the set of values of xx for which ∣3x−2∣>∣x+6∣|3x - 2| > |x + 6|.

  1. −1<x<4-1 < x < 4
  2. x<−4x < -4 or x>1x > 1
  3. −4<x<1-4 < x < 1
  4. x<−1x < -1 or x>4x > 4 (correct answer)
Explanation: 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(3x - 2)^2 > (x + 6)^2 9x2−12x+4>x2+12x+369x^2 - 12x + 4 > x^2 + 12x + 36 Rearrange into a standard quadratic inequality: 8x2−24x−32>08x^2 - 24x - 32 > 0 Divide by 8: x2−3x−4>0x^2 - 3x - 4 > 0 Factor the quadratic expression: (x−4)(x+1)>0(x-4)(x+1) > 0 The roots are x=4x=4 and x=−1x=-1. Since the parabola y=x2−3x−4y = x^2 - 3x - 4 opens upwards, the expression is positive (greater than 0) outside the roots. Thus, the solution is x<−1x < -1 or x>4x > 4. A: This is the solution to the inequality (x−4)(x+1)<0(x-4)(x+1) < 0. B: This would be the solution if the factored form was (x+4)(x−1)>0(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(x+4)(x-1) < 0. D: This is the correct solution set.