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Math 20-1

Math 20-1 · Worksheets

Mixed review · Inequalities and sequences

Ten questions of mixed difficulty, covering Inequalities and graphs, Sequences and series. Print it, or work through it on screen — the answer key starts on its own page.

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Mixed review · Inequalities and sequences

Math 20-1 · maddyhelps.com

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Circle the best answer for each question. Show your work in the space provided.

  1. An arithmetic sequence has t₁ = 5 and d = 3. Then t₁₀ is

    1. a) 30
    2. b) 32
    3. c) 53
    4. d) 35

  2. The solution to x² > 16 is

    1. a) x > 4
    2. b) −4 < x < 4
    3. c) x < −4 or x > 4
    4. d) x > ±4

  3. The graph of y < x² − 4 is drawn with

    1. a) a dashed parabola, shaded above
    2. b) a dashed parabola, shaded below
    3. c) a solid parabola, shaded below
    4. d) a solid parabola, shaded above

  4. The sum of the first 20 terms of 3, 7, 11, … is

    1. a) 410
    2. b) 800
    3. c) 1600
    4. d) 820

  5. The inequality x² + 1 < 0 has

    1. a) no solution
    2. b) x < −1 as its solution
    3. c) x = 0 as its only solution
    4. d) all real numbers as solutions

  6. A geometric sequence has t₁ = 2 and r = 3. Then t₅ is

    1. a) 162
    2. b) 486
    3. c) 243
    4. d) 30

  7. The solution to x² − x − 6 ≤ 0 is

    1. a) x ≤ 3
    2. b) x ≤ −2 or x ≥ 3
    3. c) −3 ≤ x ≤ 2
    4. d) −2 ≤ x ≤ 3

  8. Which sequence is arithmetic?

    1. a) 5, 9, 13, 17
    2. b) 1, 1, 2, 3
    3. c) 2, 6, 18, 54
    4. d) 1, 4, 9, 16

  9. Compared with y = 2x − 6, the graph of y = |2x − 6|

    1. a) is the whole line shifted up
    2. b) is identical everywhere
    3. c) is identical for x ≥ 3 and reflected in the x-axis for x < 3
    4. d) is the whole line reflected in the x-axis

  10. The graph of y = 1/(x² − 4) has

    1. a) vertical asymptotes at x = ±2 and a local maximum at (0, −¼)
    2. b) vertical asymptotes at x = ±2 and a local minimum at (0, −¼)
    3. c) no vertical asymptotes
    4. d) a vertical asymptote at x = 0

Answer key · Mixed review · Inequalities and sequences

Math 20-1 · maddyhelps.com

  1. b) 32 — Use tₙ = t₁ + (n − 1)d: 5 + 9(3) = 32. The n − 1 matters — you take nine steps to get from the first term to the tenth, not ten.
  2. c) x < −4 or x > 4 — Rewrite as x² − 16 > 0, so (x − 4)(x + 4) > 0, with roots ±4. An upward parabola is above zero outside its roots, so the solution is two separate pieces. Taking a square root and writing x > 4 quietly loses every negative solution.
  3. b) a dashed parabola, shaded below — A strict inequality excludes the boundary, so the curve is dashed. Less than means the region below the parabola. Using a solid curve claims the points on it are solutions, and they are not.
  4. d) 820 — Use Sₙ = n/2[2t₁ + (n − 1)d] = 10[6 + 19(4)] = 10(82) = 820. You can sanity-check it: the average term is about 41, and 20 × 41 = 820.
  5. a) no solution — A square is never negative, so x² + 1 is at least 1 for every real x and can never be below zero. The discriminant agrees: 0 − 4 = −4, so the parabola never touches the axis and sits entirely above it.
  6. a) 162 — Use tₙ = t₁r^(n−1): 2 × 3⁴ = 2 × 81 = 162. Using 3⁵ gives 486 and is the same off-by-one that catches people in the arithmetic formula.
  7. d) −2 ≤ x ≤ 3 — Factor to (x − 3)(x + 2) ≤ 0, so the roots are −2 and 3. The parabola opens upward, so it is at or below zero between the roots. Picture the graph rather than trying to divide by a bracket, which is not allowed when its sign is unknown.
  8. a) 5, 9, 13, 17 — Arithmetic means a constant difference, and 5, 9, 13, 17 adds 4 each time. The squares 1, 4, 9, 16 have growing differences, 2, 6, 18, 54 is geometric, and 1, 1, 2, 3 is Fibonacci.
  9. c) is identical for x ≥ 3 and reflected in the x-axis for x < 3 — Absolute value leaves non-negative outputs alone and flips negative ones. The line is negative to the left of its x-intercept x = 3, so only that part is reflected, producing a corner at (3, 0).
  10. a) vertical asymptotes at x = ±2 and a local maximum at (0, −¼) — The denominator is zero at x = 2 and x = −2, so those are the asymptotes. At x = 0 the original is −4, so the reciprocal is −¼ — and because the original has a minimum there, the reciprocal has a maximum. Reciprocals turn minima into maxima whenever the value is negative.