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Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 0 - Sample Test Problems - Number 1 – 29

Pembahasan Soal Buku Calculus 9th Purcell Chapter 0 Section 0.8

CHAPTER 0 PRELIMINARIES

Sample Test Problems


Calculus 9th Purcell Chapter 0 - 0.9

Sample Test Problems, Number 1 – 29
  1. Calculate each value for n = 1, 2, and –2.
    2. (a)
    (n + (1/n))n
    (b)
    (n2 – n + 1)2
    (c)
    43/n
    (d)
    (|1/n|)(1/n)
    💬 SOLUTION:
    (a)
    (n + (1/n))n
     
    (1 + (1/1))1 = 2;
     
    (2 + (1/2))2 = 25/4;
     
    (–2 + (1/–2))–2 = 4/25
    (b)
    (n2 – n + 1)2
     
    [(1)2 – 1 + 1]2 = 1;
     
    [(2)2 – (2) + 1]2 = 9;
     
    [(–2)2 – (–2) + 1]2 = 49
    (c)
    43/n
     
    43/1 = 64;
     
    43/2 = 8;
     
    4–3/2 = 1/8
    (d)
    (|1/n|)(1/n)
     
    (|1/1|)(1/1) = 1;
     
    (|1/2|)(1/2) = (2)(1/2)/2;
     
    (|1/–2|)(–1/2) = (2)(1/2)

  2. Simplify.
    3. (a)
    (1 +
    1
    +
    1
    ) (1 –
    1
    +
    1
    ) –1
    m
    n
    m
    n
    (b)
    2
    –
    x
     
    x + 1
    x2 – x – 2
     
    3
    –
    2
     
    x + 1
    x – 2
     
    (c)
    (t3 – 1)
     
     
    (t – 1)
     
    💬 SOLUTION:
    (a)
    (1 +
    1
    +
    1
    ) (1 –
    1
    +
    1
    ) –1
     
    m
    n
    m
    n
     
     
    =
    1 +
    1
    +
    1
     
    m
    n
     
     
    1 –
    1
    +
    1
     
    m
    n
     
     
    =
    mn + n + m
     
     
    mn + n + m
     
    (b)
    2
    –
    x
     
    x + 1
    x2 – x – 2
     
    3
    –
    2
     
    x + 1
    x – 2
     
     
    =
    2
    –
    x
     
    x + 1
    (x – 2)(x + 1)
     
    3
    –
    2
     
    x + 1
    x – 2
     
     
    =
    2(x – 2) – x
     
     
    3(x – 2) – 2(x + 1)
     
     
    =
    x – 4
     
     
    x – 8
     
    (c)
    (t3 – 1)
    =
    (t – 1)(t2 + t + 1)
    =
    t2 + t + 1
     
    (t – 1)
    (t – 1)

  3. Show that the average of two rational numbers is a rational number.
    4. 💬 SOLUTION:
    Let a, b, c, and d be integers.
     
    (a/b) + (c/d)
    =
    a
    +
    c
    =
    ad + bc
    Which is rational.
    2
    2b
    2d
    2bd

  4. Write the repeating decimal 4.1282828… as a ratio of two integers.
    5. 💬 SOLUTION:
     
    x
    =
    4.1282828…
     
    1000x
    =
    4128.2828…
     
    10x
    =
    41.282828…
     
    990x
    =
    4087
     
    x
    =
    4087
     
     
    990
     

  5. Find an irrational number between 1/2 and 13/25.
    6. 💬 SOLUTION:
    Answers will vary. Possible answer:
    √
    13
    ≈
    0.50990…
    50

  6. Calculate
    7.  
    (∛(8.15 × 104) – 1.32)2
    3.24
    💬 SOLUTION:
     
    (∛(8.15 × 104) – 1.32)2
    ≈
    545.39
    3.24

  7. Calculate
    8. [π – (2.0)1/2]2.5 – ∛2.0
    💬 SOLUTION:
    [π – (2.0)1/2]2.5 – ∛2.0 ≈ 2.66

    Semoga Bermanfaat 😁

  8. Calculate
    9. sin2 (2.45) + cos2 (2.40) – 1.00
    💬 SOLUTION:
    sin2 (2.45) + cos2 (2.40) – 1.00 ≈ –0.0495

    In Problems 9–18, find the solution set, graph this set on the real line, and express this set in interval notation.
  9. 1 – 3x > 0
    10. 💬 SOLUTION:
    1 – 3x
    > 
    0
    3x
    < 
    1
    x
    < 
    1
    3
    (–∞, 1/3)

    Calculus 9th Purcell Chapter 0 - 0.8


  10. 6x + 3 > 2x – 5
    11. 💬 SOLUTION:
    6x + 3
    > 
    2x – 5
    4x
    < 
    –8
    x
    < 
    –2
    (–2, ∞)

    Calculus 9th Purcell Chapter 0 - 0.8


  11. 3 – 2x ≤ 4x + 1 ≤ 2x + 7
    12. 💬 SOLUTION:
    3 – 2x
    ≤
    4x + 1
    ≤
    2x + 7
     
     
    3 – 2x
    ≤
    4x + 1
    and
    4x + 1
    ≤
    2x + 7
    6x
    ≥
    2
    and
    2x
    ≥
    6
    x
    ≥
    1
    and
    x
    ≤
    3;
    3
    [1/3, 3]

    Calculus 9th Purcell Chapter 0 - 0.8


  12. 2x2 + 5x – 3 < 0
    13. 💬 SOLUTION:
    2x2 + 5x – 3
    < 
    0
    (2x – 1)(x + 3)
    < 
    0
    –3 < x < ½
    ;
    (–3, 1/2)

    Calculus 9th Purcell Chapter 0 - 0.8


  13. 21t2 – 44t + 12 ≤ –3
    14. 💬 SOLUTION:
    21t2 – 44t + 12
    ≤
    –3
    21t2 – 44t + 15
    ≤
    0
    t
    =
    44 ± √[442 – 4(21)(15)]
    2(21)
    t
    =
    44 ± 26
     
    42
     
    t
    =
    3
    ,
    5
     
    7
    3
     
    (t – 3/7)(t – 5/3)
    ≤
    0
    [3/7 , 5/3]

    Calculus 9th Purcell Chapter 0 - 0.9


  14.  
    2x – 1
    > 
    0
     
    x – 2
    15. 💬 SOLUTION:
     
    2x – 1
    > 
    0;
     
    x – 2
     
    (–∞, 1/2) ∪ (2, ∞)

    Calculus 9th Purcell Chapter 0 - 0.9


    Semoga Bermanfaat 😁

  15. (x + 4)(2x – 1)2(x – 3) ≤ 0
    16. 💬 SOLUTION:
    (x + 4)(2x – 1)2(x – 3) ≤ 0; [–4, 3]

    Calculus 9th Purcell Chapter 0 - 0.9


  16. |3x – 4| < 6
    17. 💬 SOLUTION:
    |3x – 4|
    < 
    6
    –6
    < 
    3x – 4
    < 
    6
    –2
    < 
    3x
    < 
    10;
    –
    2
    < 
    x
    < 
    10
    ;
    3
    3
    (–
    2
    ,
    10
    )
    3
    3

    Calculus 9th Purcell Chapter 0 - 0.9


  17.  
    3
    ≤
    2
     
    1 – x
    18. 💬 SOLUTION:
     
    3
    ≤
    2
     
    1 – x
     
    3
    – 2
    ≤
    0
     
    1 – x
     
    3 – 2(1 – x)
    ≤
    0
     
    1 – x
     
    2x + 1
    ≤
    0;
     
    1 – x
    (–∞, –1/2] ∪ (1, ∞)

    Calculus 9th Purcell Chapter 0 - 0.9


  18. |12 – 3x| ≥ |x|
    19. 💬 SOLUTION:
     
    |12 – 3x|
    ≥
    |x|
     
    (12 – 3x)2
    ≥
    x2
     
    144 – 72x + 9x2
    ≥
    x2
     
    8x2 – 72x + 144
    ≥
    0
     
    8(x – 3)(x – 6)
    ≥
    0
    (–∞, 3] ∪ [6, ∞)

    Calculus 9th Purcell Chapter 0 - 0.9


  19. Find a value of x for which |–x| ≠ x.
    20. 💬 SOLUTION:
    For example, if x = –2, |–(–2)| = 2 ≠ –2
    |–x| ≠ x for any x < 0.

  20. For what values of x does the equation |–x| = x hold?
    21. 💬 SOLUTION:
    If |–x| = x, then |x| = x.
    x ≥ 0.

  21. For what values of t does the equation |t – 5| = 5 – t hold?
    22. 💬 SOLUTION:
    |t – 5| = |–(5 – t)| = |5 – t|.
    If |5 – t| = 5 – t, then 5 – t ≥ 0.
    t ≤ 5.

    Semoga Bermanfaat 😁

  22. For what values of a and t does the equation |t – a| = a – t hold?
    23. 💬 SOLUTION:
    |t – a| = |–(a – t)| = |a – t|.
    If |a – t| = a – t, then a – t ≥ 0.
    t ≤ a.

  23. Suppose |x| ≤ 2. Use properties of absolute values to show that
    24.  
     
     
    2x2 + 3x + 2
     
    ≤ 8
     
    x2 + 2
    💬 SOLUTION:
    If |x| ≤ 2, then
    0 ≤ |2x2 + 3x + 2| ≤ |2x2| + |3x| + 2 ≤ 8 + 6 + 2 = 16
    also |x2 + 2| ≥ 2 so
     
    1
    ≤
    1
    . Thus
     
    |x2 + 2|
    2
     
     
     
    2x2 + 3x + 2
     
    =
    |2x2 + 3x + 2|
     
     
    1
     
    ≤
    16 (1/2) = 8
     
    x2 + 2
    x2 + 2

  24. White a sentence involving the word distance to express the following algebraic sentences:
    25. (a)
    |x – 5| = 3
    (b)
    |x + 1| ≤ 2
    (c)
    |x – a| > b
    💬 SOLUTION:
    (a)
    The istance between x and 5 is 3.
    (b)
    The distance between x and –1 is less than or equal to 2.
    (c)
    The distance between x and a is greater than b.

  25. Sketch the triangle with vertices A(–2, 6), B(1, 2), and C(5, 5), and show that it is a right triangle.
    26. 💬 SOLUTION:

    Calculus 9th Purcell Chapter 0 - 0.9

    d(A, B)
    =
    √[(1 + 2)2 + (2 – 6)2]
     
     
    =
    √[9 + 16] = 5
     
    d(B, C)
    =
    √[(5 – 1)2 + (5 – 2)2]
     
     
     
    √[16 + 9] = 5
     
    d(A, C)
    =
    √[(5 + 2)2 + (5 – 6)2]
     
     
     
    √[49 + 1] = √50 = 5√2
     
    (AB)2 + (BC)2 = (AC)2, so ΔABC is a right triangle.

  26. Find the distance from (3, –6) to the midpoint of the line segment for (1, 2) to (7, 8).
    27. 💬 SOLUTION:
    Midpoint: (
    1 + 7
    ,
    2 + 8
    ) = (4, 5)
    2
    2
    d = √[(4 – 3)2 + (5 + 6)2] = √[1 + 121] = √122

  27. Find the equation of circle with diameter AB if A = (2, 0) and B = (10, 4).
    28. 💬 SOLUTION:
    center
    =
    (
    2 + 10
    ,
    0 + 4
    ) = (6, 2)
    2
    2
    center
    =
    1
    √[(10 – 2)2 + (4 – 0)2]
    2
     
    =
    1
    √[64 + 16]
    2
     
    =
    2√5
    circle
    :
    (x – 6)2 + (y – 2)2 = 20

  28. Find the center and radius of the circle with equation x2 + y2 – 8x + 6y = 0.
    29. 💬 SOLUTION:
    x2 + y2 – 8x + 6y = 0
    x2 – 8x + 16 + y2 + 6y + 9 = 16 + 9
    (x – 4)2 + (y + 3)2 = 25;
    center = (4, –3), radius = 5

  29. Find the distance between the centers of the circles with equations
    30. x2 – 2x + y2 + 2y = 2
    and
    x2 + 6x + y2 – 4y = –7
    💬 SOLUTION:
    x2 – 2x + y2 + 2y
    =
    2
    x2 – 2x + 1 + y2 + 2y + 1
    =
    2 + 1 + 1
    (x – 1)2 + (y + 1)2
    =
    4;
    center = (1, –1)
     
     
    x2 + 6x + y2 – 4y
    =
    –7
    x2 + 6x + 9 + y2 – 4y + 4
    =
    –7 + 9 + 4
    (x + 3)2 + (y – 2)2
    =
    6
    center = (–3, 2)
     
     
    d = √[(–3 – 1)2 + (2 + 1)2]
    =
    √(16 + 4) = 5

Semoga Bermanfaat 😁

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