0E: Exercises
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- Dec 21, 2020
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The real number system
Exercise 0E.1
Answer True or False for each of the statements below. If the statement is false, try to provide an example to demonstrate this.
1. The sum of any two integers is an integer.
2. The sum of any natural numbers is a natural number.
3. The difference between any two natural numbers is a natural number.
4. The sum of any two rational number is a rational number.
5. The sum of any two irrational number is an irrational number.
6. If the sum of the digits of an integer is divisible by 3, then the integer is divisible by 3.
7. If an integer has units digit 0 or 5 ( ends in either 0 or 5 ), then the integer is divisible by 5.
8. If n is an integer then
a. 2n−1 and 2n+1 are both odd integers.
b. 2n is an even integer.
c. 3n is an odd integer.
9. For any real number x, |x|≤0.
10. For any real number x, |x|≥0.
11. For any real number x, and y, |x|+|y|=|x+y|.
12. For any real number x, and y, |x|+|y|≥|x+y|.
13. For any real number x, it follows that √x is again real number.
a. √x2=x.
b. √x2=−x if x<0.
- Answer
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1. True.
2. True.
3. False. 5−7=−2 is not a natural number.
4. True. (Note: The rational numbers include the integers.)
5. False. √2+−√2=0.
6. True.
7. True.
8. a. True.
b. True.
c. False. 3×2=6 is even.
9. False. |3|=3>0.
10. True.
11. False. |2|+|−2|=2+2=4 but |2−2|=0.
12. False. Use same example as above.
Note: The correct statement is |x+y|≤|x|+|y|.
13. a. False. √(−2)2=√4=2≠−2.
b. True.
Exercise 0E.2
For each of the following terms, identify (a) its coefficient and (b) its degree in the variables which occur.
1. −2x5
2. 7x3y
3. 5
4. −4x2y3z4
- Answer
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1. Coefficient: −2 ; degree 5.
2. Coefficient: 7; degree 4.
3. Coefficient: 5; degree 0.
4. Coefficient: −4; degree 9.
Exercise 0E.3
Classify each of the expressions below as polynomial expression, rational expression or neither. If the expression is a polynomial, determine the coefficients of each of its terms.
1. 2x4−3x3+7x2−14x−19
2. 3x7−13x3−x3/2−21
3. 1x2−5x+31
4. −28x−1
5. 5x2+3x−17
6. 2x−1x−1
7. 3x+13√x
8. 17
9. 0
10. −42x2−x+1
11. x2−5x+11−2x−3x2
12. (x−3)(x+5)(2x−1)
13. 1x−32−1x
14. √(3x+1)x2−x−2
15. 1−x3+5x
16. |x|
17. {2x−1x,x>0x−1,x≤0
18. 3x+21x
19. 7x23√xx2/3,x≠0.
20. x−2√x−1
- Answer
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1. Polynomial; degree 4; coefficients 2, −3, 7, −14, 19.
2. Neither.
3. Neither.
4. Neither.
5. Polynomial; degree 2; coefficient 57, 37, −17.
6. Neither.
7. Neither.
8. Polynomial; degree 0; coefficient 17.
9. Polynomial; degree 0; coefficient 0.
10. Rational Expression.
11. Rational Expression.
12. Neither.
13. Neither.
14. Neither.
15. Rational Expression.
16. Neither.
17. Neither.
18. Neither.
19. Neither.
20. Neither.
Exercise 0E.4
Identify each of the following numbers as natural numbers, integers, whole numbers, rational numbers, or irrational numbers.
−34,−15,89,−4−23,−√9,√13,310,−√2
- Answer
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−34: rational number.
−15: integer, rational number.
89: natural number, whole number, integer, rational number.
−4−23: rational number.
−√9=−3: integer, rational number.
√13: irrational number.
310: rational number.
−√2: irrational number.
Exercise 0E.5
A collection of real numbers is represented geometrically in each diagram below. In each case, give the interval representation.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
- Answer
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1. [4,∞)
2. (−∞,−7)
3. (−∞,1]
4. (−2,7]
5. (−∞,8)∪[15,∞)
6. (−∞,7)∪(7,∞)
7. (−1,2)
8. [−1,2]
9. [−10,−3)∪[1,5]
10. [1,2)∪(2,3]
11. (−∞,−1)∪(−1,3)∪[5,∞)
12. 1,3,5
13. (−∞,−1)∪(−1,2)∪(2,∞)
14. (−2,1]∪(10,∞)
15. [1,5)∪(5,∞)
Exercise 0E.6
Use long division (not calculators) to obtain a decimal representation of the following fractions.
1. 38
2. 27
3. 23
4. 17
5. 29
6. 37
7. 34
8. 45
9. 47
10. 57
11. 67
12. 59
13. 89
14. 79
- Answer
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1. 38=0.375
2. 27=0.¯285714
3. 23=0.¯6
4. 17=0.¯142857
5. 29=0.¯2
6. 37=0.¯428571
7. 34=0.75
8. 45=0.2
9. 47=0.¯571428
10. 57=0.¯714285
11. 67=0.¯857142
12. 59=0.¯5
13. 89=0.¯8
14. 79=0.¯7
Exercise 0E.7
Express each number given below as a product of its prime factors (in exponential form):
1. 81
2. 24
3. 1080
- Answer
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1. 34
2. 22⋅3
3. 23⋅33⋅5
Exercise 0E.8
Evaluate each of the following using exponential laws:
1. 26
2. (−3)5
3. 17−3
4. 9−3
5. 252−2
6. 3−23−1
7. (−2−3)4
8. (2x5y)3
9. (23)2
10. (5−2)−1
11. 51057
12. (34)−2
13. (5167)0
- Answer
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1. 26=64
2. (−3)5=−243
3. 17−3=73=343
4. 9−3=193=1729
5. 252−2=25−(−2)=27=128
6. 3−23−1=3−2−(−1)=3−1=13
7. (−23)4=(−2)434=1681
8. (2x5y)3=(2x)3(5y)3=8x3125y3
9. (23)2=23⋅2=26=64
10. (5−2)−1=5(−2)⋅(−1)=52=25
11. 51057=510−7=53=125
12. (34)−2=4232=169
13. (5167)0=1
Exercise 0E.9
Express each of the following in radical form:
1. x1/3
2. 1x1/2
3. x−2/5
4. 1x−3/4
5. x−1/9
- Answer
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1. x1/3=3√x
2. 1x1/2=1√x=√xx
3. x−2/5=1x2/5=15√x2=5√x3x
4. 1x−3/4=x3/4=4√x3
5. x−1/9=1x1/9=19√x=9√x8x
Exercise 0E.10
Write the following radical expressions in exponent form:
1. 4√x
2. 3√x7
3. 14√x3
4. √x3√x
5. 3√x5√x√x,x>0
6. 13√x−2
- Answer
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1. 4√x=x1/4
2. 3√x7=x7/3
3. 14√x3=x−3/4
4. √x3√x=x1/2−1/3=x1/6
5. 3√x5√x√x=x(1/3+1/5)−1/2=x1/30
6. 13√x−2=x2/3
Exercise 0E.11
Determine a real number value of each of the following if one exists. If one does not exist, explain why.
1. 3√−64
2. 3√215
3. 1√26
4. √−400
5. 3√(−2)3
6. 4√−16
- Answer
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1. 3√−64=−4
2. 3√215=215/3=25=32
3. 1√26=126/2=123=18
4. √−400 is undefined because the square root of a negative number is undefined.
5. 3√(−2)3=−2
6. 4√−16 is undefined because the fourth root of a negative number is undefined.
Exercise 0E.12
Simplify each of the following:
1. −3√98+5√12−3√24+√27−2√54+4√50
2. (2√5−3√2)(2√5+3√2)
3. (2−√3)(7+2√3)
4. (3√5−2√7)2−(2√5+√7)(√5−√7)
5. 5√3−103√3
6. √128√8
7. √72√8
- Answer
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1. −3√98+5√12−3√24+√27−2√54+4√50=−21√2+10√3−6√6+3√3−6√6+20√2=−√2+13√3−12√6
2. (2√5−3√2)(2√5+3√2)=(2√5)2−(3√2)2=4⋅5−9⋅2=2
3. (2−√3)(7+2√3)=14+4√3−7√3−2(3)=8−3√3
4. (3√5−2√7)2−(2√5+√7)(√5−√7)=9⋅5−2(3√5)(2√7)+4⋅7−2⋅5+2√35−√35+7=70−11√35
5. 5√3−103√3=5⋅3√3√3(−10)=−32
6. √128√8=8√22√2=4
7. √72⋅√8=3√8⋅√8=3⋅8=24
Exercise 0E.13
Use order of operations to evaluate each of the following:
1. −31+35/7−(2)(24)+70
2. 32−(−4)21−52+724
- Answer
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1. −4
2. 712
Basic algebra
Exercise 0E.1
Simplify each of the following expressions:
1. 5x3y2−21xy+17x3y2−19xy+6
2. −32xz+53xz2+72xz−193xz2+4xz−19
3. (3x3y2−7x2y+5y3−2)+(−5x3y2+4x2y+11)
4. 2x2y2((y−2+24x−2y−11x)−(2xy−2−4x−2y−11x+1))
5. 3xy(5x−7y)
6. 3x2y4(2x−1+9y−2+4xy)
7. (3x−2y)(4x2−y2)
8. (x+y−2z)(x−y+2z)
9. (3x−1y)(5x−2y)
10. 2x−3y−12x+18y
- Answer
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1. 22x3y2−40xy+6
2. 6xz−143xz2−19
3. −2x3y2−3x2y+5y3+9
4. 2x2+56y3−4x3−2x2y2
5. 15x2y−21xy2
6. 6xy4+27x2y2+12x3y5
7. 12x3−3xy2−8x2y+2y3
8. x2−y2+4yz−4z2
9. (3x−1y)(5x−2y)=(3y−xxy)(5y+2xxy)=(5y+2x)(3−x)(xy)2
10. 2x−3y−12x+18y=2y−3xxy−12y+18xxy=(2y−3x)(xy)(−12y+18x)(xy)=2y−3x−6(2y−3x)=−16,2y≠3x,xy≠0
Factoring polynomials with some applications
Exercise 0E.1
Factor each of the expressions given. It may be necessary to apply the Factorization theorem.
1. 15x3y2−45xy4
2. 12xz−4x2z2+36x
3. 2x2−50
4. 7y3−56
5. (3x−2)2−(4y+3)2
6. x2−x−56
7. x4−8x2−9
8. x4−10x2+9
9. x6+8y6
10. z2+64
11. 6x2+19x+15
12. x3+x2−4x−4
13. yz−2ty+z2−2tz
14. 2ac−4ad+3bc−6bd
15. 4x3+500y3
16. (x−2y)2−2(x−2y)−8
17. 48−6t3
18. x8−256
19. y2−4xy−45x2
20. 2x2−11x−21
21. 24x2−98x−45
22. 24x2−13x−45
23. x6+19x3−216
24. x5−9x3+8x2−72
25. 3(x−2)2(x+4)5+5(x+4)4(x−2)3
26. x4+x2+1
27. x4−12x2+16
- Answer
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1. 15x3y2−45xy4=15xy2(x2−y2)=15xy2(x−y)(x+y)
2. 12xz−4x2z2+36x=4x(3z−xz2+9)
3. 2x2−50=2(x2−25)=2(x−5)(x+5)
4. 7y3−56=7(y3−8)=7(y−2)(y2+2y+4)
5. (3x−2)2−(4y+3)2=((3x−2)−(4y+3))((3x−2)+(4y+3))=(3x−4y−5)(3x+4y+1)
6. x2−x−56=(x−8)(x+7)
7. x4−8x2−9=(x2−9)(x2+1)=(x−3)(x+3)(x2+1)
8. x4−10x2+9=(x2−9)(x2−1)=(x−3)(x+3)(x−1)(x+1)
9. x6+8y6=(x2)3+(2y)3=(x2+2y2)(x4−2x2y2+y4)
10. z2+64 does not factor.
11. 6x2+19x+15=(3x+5)(2x+3)
12. x3+x2−4x−4=x2(x+1)−4(x+1)=(x+1)(x2−4)=(x+1)(x−2)(x+2)
13. yz−2ty+z2−2tz=(y+z)(z−2t)
14. 2ac−4ad+3bc−6bd=(c−2d)(2a+3b)
15. 4x3+500y3=4(x+5y)(x2−5xy+25y2)
16. (x−2y)2−2(x−2y)−8=((x−2y)−4)((x−2y)+2)
17. 48−6t3=6(8−t3)=6(2−t)(4+2t+t2)
18. x8−256=(x4−16)(x4+16)=(x2−4)(x2+4)(x4+16)=(x−2)(x+2)(x2+4)(x4+16)
19. y2−4xy−45x2=(y−9x)(y+5x)
20. 2x2−11x−21=(2x+3)(x−7)
21. 24x2−98x−45=(2x−9)(12x+5)
22. 24x2−13x−45=(3x−5)(8x+9)
23. x6+19x3−216=(x3−8)(x3+27)=(x−2)(x2+2x+4)(x+3)(x2−3x+9)
24. x5−9x3+8x2−72=x3(x2−9)+8(x2−9)=(x3+8)(x2−9)=(x−2)(x2+2x+4)(x−3)(x+3)
25. 3(x−2)2(x+4)5+5(x+4)4(x−2)3=(x−2)2(x+4)4(3(x+4)+5(x−2))=2(x−2)2(x+4)4(4x+1)
26. x4+x2+1=x4+2x2+1−x2=(x2+1)2−x2=(x2+x+1)(x2−x+1)
27. x4−12x2+16=x4−8x2+16−4x2=(x2−4)2−4x2=(x2−2x−4)(x2+2x−4)
Exercise 0E.2
Evaluate each of the following:
1. 25+414−710
2. 6⋅(−23)
3. 1736−124+518
4. 219÷538
5. 29−5+12
6. (1752)(2634)
7. −247−849
- Answer
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1. −170
2. −4
3. 1724
4. 45
5. −7718
6. 14
7. 21
Exercise 0E.3
Simplify each of the expressions below:
1. x2−4x2−2x−8−1x−4
2. (x3−27x2−6x+9)(x2−7x+6(x2+3x+9)(x−6))
3. 2x2−82x2+3x+1−x2−2x−352x−6÷2x2−13x−72x2−4x−6
4. 3(x−3)2−xx2−9
5. −1x−3x−12x−5x2+3x−10
6. 1x−1+3(x+2)2−xx+34x−5+2x+2
7. 2(x+h)−13(x+h)+1−2x−13x+1h
- Answer
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1. x2−4x2−2x−8−1x−4=(x−2)(x+2)(x−4)(x+2)−1x−4=x−2+1x−4=x−1x−4, x≠−2
2. x3−27x2−6x+9⋅x2−7x+6(x2+3x+9)(x−6)=(x−3)(x2+3x+9)(x−6)(x−1)(x−3)2(x2+3x+9)(x−6)=x−1x−3,x≠6
3. 2x2−82x2+3x+1−x2−x−352x−6÷2x2−13x−72x2−4x−6=2x2−8(2x+1)(x+1)−(x−7)(x+5)2(x−3)×2(x−3)(x+1)(2x+1)(x−7)=2x2−8(2x+1)(x+1)−(x+5)(x+1)2x+1=(2x2−8)−(x+5)(x+1)2(2x+1)(x+1)=2x2−8−(x3+7x2+11x−5)(2x+1)(x+1)=−x3−5x2−11x−3(2x+1)(x+1)
This cannot be reduced any further as 2x+1 and x+1 are not factors of the numerator. We can use the remainder theorem to check:
Let p(x)=−x3−5x2−11x−3. p(−1/2)=19/8 and p(−1)=4. Since neither of them are zero, then x=−1/2,−1 are not zeros of p(x) and therefore 2x+1 and x+1 are not factors of p(x).
4. 3(x−3)2−xx2−9=3(x+3)−x(x−3)(x−3)2(x+3)=9+6x−x2(x−3)2(x+3)
5. 1x−3x−12x−5x2+3x−10=(x−1)−(3x)x(x−1)2x−5(x+5)(x−2)=−(2x+1)(x+5)(x−2)x(x−1)(2x−5)
6. 1x−1+3(x+2)2−xx+34x−5+2x+2=(x+2)2(x+3)+3(x−1)(x+3)−x(x−1)(x+2)2(x−1)(x+2)2(x+3)4(x+2)+2(x−5)(x−5)(x+2)=−x4−2x3+10x2+26x+3(x−1)(x+2)2(x+3)⋅(x+2)(x−5)6x−2=(−x4−2x3+10x2+26x+3)(x−5)(x−1)(x+2)(x+3)(6x−2),x≠−2
7. 2(x+h)−13(x+h)+1−2x−13x+1h=1h((2x+2h−1)(3x+1)−(2x−1)(3x+3h+1)(3x+3h+1)(3x+1))=(6x2+6xh−3x+2x+2h−1)−(6x2+6xh+2x−3x−3h−1)h(3x+3h+1)(3x+1)=5hh(3x+3h+1)(3x+1)=5(3x+3h+1)(3x+1),h≠0
Exercise 0E.4
Simplify each of the expressions below:
1. x2−4x2−2x−8−1x−4
2. (x3−27x2−6x+9)(x2−7x+6(x2+3x+9)(x−6))
3. 2x2−82x2+3x+1−x2−2x−352x−6÷2x2−13x−72x2−4x−6
4. 3(x−3)2−xx2−9
5. −1x−3x−12x−5x2+3x−10
6.1x−1+3(x+2)2−xx+34x−5+2x+2
7. 2(x+h)−13(x+h)+1−2x−13x+1h
- Answer
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Under construction.
Exercise 0E.5
Express your answer in its simplest form.
1. √(2x+3)−1x2−x−2
2. 1√(x+1)−1x2+2x
3. √(2x2+6x−7)−13x2+2x−5
4. √(3x+13)−√(17−x)1−x2
5. 1√x−1√2x2+2x−8
6. 1x+1−1x√(4x2+3)−2
7. √(2x+2h+3)−√(2x+3)h
8. 1√(4−h)−12h
- Answer
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1. √2x+3−1x2−x−2=√2x+3−1(x−2)(x+1)⋅√2x+3+1√2x+3+1=(√2x+3)2−(1)2(x−2)(x+1)(√2x+3+1)=2x+2(x−2)(x+1)(√2x+3+1)=2(x−2)(√2x+3+1)
2. 1√x+1−1x2+2x=√x+1−(x+1)√x+1x(x+2)⋅√x+1+1)√x+1+1)=(√x+1)2−(1)2x(x+2)√x+1(√x+1+1))=xx(x+2)√x+1(√x+1+1)=−1(x+2)√x+1(√x+1+1),x≠0
3. √2x2+6x−7−13x2+2x−5=√2x2+6x−7−13x2+2x−5⋅√2x2+6x−7+1√2x2+6x−7+1=2x2+6x−7−1(3x+5)(x−1)(√2x2+6x−7+1)=2(x+4)(x−1)(3x+5)(x−1)(√2x2+6x−7+1)=2(x+4(3x+5)(√2x2+6x−7+1),x≠1
4. √3x+13−√17−x1−x2=√3x+13−√17−x1−x2⋅√3x+13+√17−x√3x+13+√17−x=(3x+13)−(17−x)(1−x)(1+x)(√3x+13+√17−x)=4(x−1)(1−x)(1+x)(√3x+13+√17−x)=−4(1+x)(√3x+13+√17−x),x≠1
5. 1√x−1√2x2+2x−8=√xx−√22(x+4)(x−2)=2√x−√2x2x(x+4)(x−2)=2√x−√2x2x(x+4)(x−2)⋅2√x+√2x2√x+√2x=4x−2x22x(x+4)(x−2)(2√x+√2x)=−1(x+4)(2√x+√2x),x≠0,2
6. 1x+1+1x√4x2+3−2=2x+1x(x+1)(√4x2+3−2)=2x+1x(x+1)(√4x2+3−2)⋅√4x2+3+2√4x2+3+2=(2x+1)(√4x2+3+2)x(x+1)(4x2+3−4)=(2x+1)(√4x2+3+2)x(x+1)(2x−1)(2x+1)=√4x2+3+2x(x+1)(2x−1),x≠−1/2
7. √2x+2h+3−√2x+3h=√2x+2h+3−√2x+3h⋅√2x+2h+3+√2x+3√2x+2h+3+√2x+3=(2x+2h+3)−(2x+3)h(√2x+2h+3+√2x+3)=2hh(√2x+2h+3+√2x+3)=2√2x+2h+3+√2x+3,h≠0
8. 1√4−h−12h=2−√4−h2h√4−h=2−√4−h2h√4−h⋅2+√4−h2+√4−h=4−(4−h)2h√4−h(2+√4−h)=12√4−h(2+√4−h),h≠0
Solving equations
Exercise 0E.1
Solve the following equations:
1. 3x−17=0
2. −14z+21=0
3. −12x=76
4. 3x−187=0
5. −5x=0
6. −23x=75
- Answer
-
1. x=173
2. z=32
3. x=−73
4. x=67
5. x=0
6. x=−2110
Exercise 0E.2
Solve each of the folwing quadratic equations by (a)factoring, (b)using the method of completing the square, (c) using the quadratic formula.
1. x2−x−20=0
2. x2−12x+36=0
3. 4x2+3x=0
4. 10x2+x−3=0
5. 4x2−4x+1=0
6. 2x2−5x−3=0
- Answer
-
1. x=5,−4
2. x=6
3. x=0,−34
4. x=12,−35
5. x=12
6. x=3,−12
Exercise 0E.3
Use any method to solve the equations given below.
1. 3t2−14t=0
2. x2+1=0
3. x2+x+1=0
4. 2x2+x+5=0
5. 36x2−25=0
6. 6x2−31x+18=0
- Answer
-
1. t=0,14/3
2. There are no solutions: x2=−1 only has complex solutions.
3. There are no solutions: Using the quadratic formula, we get x=−1±√−32, which is a complex number.
4. There are no solutions: Using the quadratic formula, we get x=−1±√−394, which is a complex number.
5. x=±56
6. x=92,23
Exercise 0E.4
Solve each of the following equations given for x.
1. √(3x+1)=2
2. √(2x−3)−√(4x+1)=−2
3. √5x−27x+4=23
4. √(3x−5)−4x+9=0
5. √(2x−7)=x−5
6. x3−8=0
7. 8x3+27=0
8. 6x4−5x2−6=0
9. x2/3−5x1/3−6=0
10. x−√x−12=0
11. 4x2/5−4x1/5+1=0
12. x3/2−x3/4−6=0
13. t4−9t2+8=0
14. y4−2y2−8=0
15. x4−10x2+9=0
16. x6−26x23−27=0
17. x−√x−2=0
18. x3−17x2+30x=0
19. t2/3−t1/3−6=0
20. (x−1)(x+5)(2x+3)=0
21. x+10x+13−2x3x−1=0
22. 2x2+7x−15x2+1=0
23. x2+1x2+31=14
24. 2x−2−9x+2=0
25. √(3x+15)−x=5
26. √(2x+14)−5x=−1
27. √(x+8)+3=0
28. √(x+4)=√(2x+1)+1
29. |4x−3|=5
30. |x−1|=|7−x|
31. |2x+3|−3x+5=0
- Answer
-
1.
2. √2x−3−√4x+1=−2√2x−3=√4x+1−22x−3=(4x+1)−4√4x+1+42x−3=4x+5−4√4x+14√4x+1=2x+816(4x+1)=4x2+32x+64x2−8x+12=0x=6,2
Check answers:
x=6:√9−√25=−2√
x=2:√1−√9=−2√3. √5x−27x+4=235x−27x+4=499(5x−2)=4(7x+4)17x=34x=2
Check answer:
√818=√49=23√4. √3x−5−4x+9=0√3x−5=4x−93x−5=16x2−72x+8116x2−75x+86=0(16x−43)(x−2)=0x=2,43/16
Check answer:
x=2:√1−8+9=2→ Not a solution
x=43/16:√49/16−43/4+9=−36/4+9=0√5. √2x−7=x−52x−7=x2−10x+25x2−12x+32=(x−8)(x−4)=0x=8,4
Check answers:
x=8:LHS=√9=3,RHS=8−5=3√
x=4:LHS=√1=1,RHS=4−5=−1→ Not a solution.6. x3−8=(x−2)(x2+2x+4)=0
x=2.7. 8x3+27=(2x+3)(2x2−6x+9)=0
x=−3/28. 6x4−5x2−6=(3x2+2)(2x2−3)=(3x2+2)(√2x−√3)(√2x+√3)=0
x=±√329. Let u=x1/3
u2−5u−6=(u−6)(u+1)=0.
Therefore, x1/3=u=−1,6→x=−1,216.10. Let u=√x.
u2−u−12=(u−4)(u+3)=0.
Therefore, √x=u=4,−3→x=2.11. Let u=x1/5.
4u2−4u+1=(2u−1)2=0
Therefore, x1/5=u=1/2→x=(1/2)5=1/32.12. Let u=x3/4. Then u2=(x3/4)2=x(3/4)∗2=x3/2.
u2−u−6=(u−3)(u+2)=0.
Therefore, x3/4=u=3,−2→x=34/3.
Note: x3/4=−2 is not a solution since x3/4≥0.13. t4−9t2+8=(t2−8)(t2−1)=(t−2√2)(t+2√2)(t−1)(t+1)=0
t=±2√2,±114. y4−2y2−8=(y2−4)(y2+2)=(y−2)(y+2)(y2+2)=0
y=±215. x4−10x2+9=(x2−9)(x2−1)=(x−3)(x+3)(x−1)(x+1)=0
x=±1,±316. x6−26x3−27(x3−27)(x3+1)=(x−3)(x2+3x+9)(x+1)(x2+x+1)=0
x=3,−117. Let u=√x.
u2−u−2=(u−2)(u+1)=0→√x=u=2,−1→x=418. x3−17x2+30x=x(x2−17x+30)=x(x−15)(x−2)=0→x=0,15,2
19. Let u=t1/3.
u2−u−6=(u−3)(u+2)=0→t1/3=u=3,−2→t=27,−820. x=1,−5,−3/2
21. Note: x≠−13,1/3
x+10x+13−2x3x−1=0x+10x+13=2x3x−1(x+10)(3x−1)=2x(x+13)3x2+29x−10=2x2+26xx2+3x−10=0(x+5)(x−2)=0x=−5,222. 2x2+7x−15x2+1=02x2+7x−15=0(2x−3)(x+5)=0x=3/2,−5
23. x2+1x2+31=144(x2+1)=x2+313x2−27=03(x−3)(x+3)=0x=±3
24. Note: x≠2,−2
2x−2−9x+2=02x−2=9x+22(x+2)=9(x−2)x=22/725. √3x+15−x=5√3x+15=x+53x+15=x2+10x+25x2+7x+10=0(x+5)(x+2)=0x=−5,−2
Check answers:
x=−5:√0+5=5√
x=−2:√9+2=3+2=5√26. √2x+14−5x=−1√2x+14=5x−12x+14=25x2−10x+125x2−12x−13=0(25x+13)(x−1)=0x=−13/25,1
Check answers:
x=−13/25:√324/25+13/5=18/25+13/5=31/5≠−1.
Therefore, x=−13/25 is not a solution.
x=1:√16−5=4−5=−1√.27. √x+8+3=0√x+8=−3
No solutions because √x+8>0.
28. √x+4=√2x+1+1x+4=(2x+1)+2√2x+1+12√2x+1=2−x4(2x+1)=4−4x+x2x2−12x=0x(x−12)=0x=0,12
Check answers:
x=0:LHS=√4=2;RHS=√1+1=2√
x=12:LHS=√16=4;RHS=√25+1=6→LHS≠RHS,x=12 is not a solution.29. Note: Use |a|2=a2=|a2|.
|4x−3|=5(4x−3)2=2516x2−24x−16=02x2−3x−2=0(2x+1)(x−2)=0x=−1/2,2
30. |x−1|=|7−x|x2−2x+1=49−14x+x212x−48=0x=4
31. |2x+3|−3x+5=0|2x+3|=3x−5 (Note: 3x−5≥0→x≥5/3)4x2+12x+9=9x2−30x+255x2−42x+16=0(x−8)(5x−2)=0x=8,2/5
x=2/5<5/3. Therefore, it is not a solution. The solution set is x=8.
Exercise 0E.5
Solve each of the following systems of equations given below.
1. 3x−4y=−52x+5y=12
2. 2x+y=1x+2y=−4
3. 12x−23y=1−2x+y=2
4. x+34y=124x−y=1
5. xy=1x2+y2=4
Leave your answer in radical form.
6. x2−y2=9x2+y2=21
Leave your answer in radical form
7. 2x+y=3(x−3y)2−(x−3y)−8=0
- Answer
-
1. Solve for y (or x, it doesn't matter) in one equation and substitute it
into the other equation. This creates an equation with one unknown. Solve for
the unknown.If we solve for y in the first equation, we get =y=(14)(3x+5). Substitute y
into the second equation 2x+5(14)(3x+5))=12. Simplify to (23/4)x=23/4. Therefore,
x=1 and y=1/4(3⋅1+5)=8/4=2.2. Solve for y in equation 1 and substitute it into equation 2. Solve for x.
y=1−2xx+2(1−2x)=−4−3x+2=−4x=2y=−3
3. Solve for y in equation 2 and substitute it into equation 1. Solve for x.
y=2+2x12x−23(2+2x)=1x=−145y=−185
4. Solve for x in equation 1 and substitute it into equation 2. Solve for y.
x=12−34y4(12−34y)−y=1y=14x=516
5. Solve for y in equation 1 and substitute it into equation 2. Solve for x.
y=1xx2+(1x)2=4x4+1x2=4x4+1=4x2x4−4x2+1=0x2=4±√122x2=2±√3
Case 1: x2=2+√3
x=±√2+√3,y=±1√2+√3
Case 2: x2=2−√3
x=±√2−√3,y=±1±√2−√36. Equation 2 - Equation 1 yields 2y2=12.
Solve for y: y=±√6.
Since x2=21−y2=21−6=15, x=±√15.7. Solve for y in the first equation: y=3−2x.
Substitute y into the second equation: (x−3(3−2x))2−2(x−3(3−2x))−8=(7x−9)2−2(7x−9)−8=0.
Let u=7x−9: u2−2u−8=(u−4)(u+2)=0.
Therefore, 7x−9=u=4,−2. \\
In conclusion, x=1,y=1 or x=13/7,y=−5/7.
Inequalities
Exercise 0E.1
Answer true or false in each of the following statements. If the conclusion of the statement is not correct, make the correct conclusion.
1. a<b and c<b implies that a<c.
2. a<c and c<0 implies that a<0.
3. a<b implies that 1a<1b.
4. a<b implies that −a<−b
5. a<b implies that a2<b2
6. If a<b and both a and b are positive real numbers, then a2<b2
- Answer
-
1. False. a<b and b<c implies that a<c.
2. True.
3. False. a<b implies that 1b<1a.
4. False. a<b implies that −a>−b.
5. False. No correct conclusion.
6. True.
Exercise 0E.2
For each of the following, write without absolute value:
1. |−29|
2. |45|
3. |2x+5|2
4. (5x−1)2−|1−5x|2
5. |x|−xx if x<0
6. |x−2|−x+2x−2 if x>2
- Answer
-
1. 29
2. 45
3. |2y+5|2=(2y+5)2=4y2+20y+25
4. (5x−1)2−|1−5x|2=(5x−1)2−(1−5x)2=0
5. |x|−xx=−x−xx=−2,x≠0
6. |x−2|−x+2x−2=x−2−x+2x−2=0
Exercise 0E.3
Solve each of the following inequalities:
1. x2−3x≤4
2. 2x2−5x+2>0
3. 4−3x≥0
4. 4x2−28x+49≤0
5. (x−3)(x2+1)<0
6. x2+24<0
7. x2−x−20x3(x+1)≤0
8. 2x2−x−1x2−16<0
9. x2+1(x−1)2(2−x)<0
10. (2x−1)2(x2+1)(3x−2)(x−2)(x2−x+4)≤0
11. |2x+5|>2
12. |1−x|≤|x+1|
13. |2x−1|≤|5x−1|
14. |2+x||2−x|>1
15. |4x−1|<x
16. |2x−3|≥x
- Answer
-
1. Isolate all of the terms on one side of the inequality: x2−3x−4≤0.
Factor: (x−4)(x+1)≤0.
Make a sign chart:
Solution: x∈[−1,4].
2. Factor: (2x−1)(x−2).
Make a sign chart:
Solution: x∈(−∞,1/2)∪(2,∞).
3. Since 4−3x is linear, solve for x: x≤4/3
4. Factor: (4x−7)2.
Anything squared is always positive. Therefore, 4x2−28x+49 is never negative but it is zero.
Solution: x=7/4
5. Make a sign chart:
Solution: x∈(−∞,3).
6. x2+24 is always positive. Therefore, there is no solution.
7. Note: x≠0,−1
Factor: (x−5)(x+4)x3(x+1).
Make a sign chart:
Solution: x∈[−4,−1)∪(0,5].
8. Note: x≠±4
Factor: (2x+1)(x−1)(x−4)(x+4).
Make a sign chart:
Solution: x∈(−4,−1/2)∪(1,4).
9. Note: x≠1,2
Make a sign chart:
Solution: x∈(2,∞).
10. Note: x≠2 and x2−x+4 is not factorable.
Make a sign chart:
Solution: x∈[2/3,2).
11. Case 1: If 2x+5≥0, then |2x+5|=2x+5
2x+5>2→x>−3/2
Case 2: If 2x+5<0, then |2x+5|=−(2x+5)
−2x−5>2→x<−7/2
Solution: x∈(−∞,−7/2)∪(−3/2,∞)
12. |1−x|≤|x+1|1−2x+x2≤x2+2x+10≤4x0≤x
13. |2x−1|≤|5x−1|4x2−4x+1≤25x2−10x+10≤16x2+6x=2x(8x+3)
Make a sign chart to get the solution x∈(−∞,−3/8]∪[0,∞).
14. Note: x≠2 and (2−x)2>0, therefore we can multiply both sides of the inequality by it.
|2+x2−x|>1 Square both sides4+4x+x24−4x+x2>1 Since (2−x)2 is positive we can cross multiply4+4x+x2>4−4x+x28x>0
Solution: x>0
15. Note: x≥0
0≤|4x−1|<x16x2−8x+1<x215x2−8x+1<0(3x−1)(5x−1)<0
Make a sign chart to get the solution x∈(1/5,1/3).
16. |2x−3|≥x4x2−12x+9≥x23x2−12x+9≥03(x−1)(x−3)≥0
Make a sign chart to get the solution x∈(−∞,1)∪(3,∞).
Analytic geometry
Exercise 0E.1
In each cases, Given points P and Q, determine
a)the slope of the line segment PQ,
b) the midpoint, M, of the line segment PQ and
c) the lengh of the line segment PQ.
1. P=(−2,9),Q=(2,6)
2. P=(0,0),Q=(−6,8)
3. P=(5,−3),Q=(7,−3)
4. P=(7,−2),Q=(7,−9)
5. P=(11,0),Q=(0,−60)
6. P=(6,−5),Q=(−6,0)
7. P=(13,−2),Q=(5,13)
- Answer
-
1. a) −34, b) (0,152), c) 5
2. a) −43, b) (−3,4), c) 10
3. a) 0, b) (6,−3), c) 2
4. a) undefined, b) (7,−112), c) 7
5. a) 6011, b) (112,−30), c) 61
6. a) −512, b) (0,−52), c) 13
7. a) −158, b) (9,112), c) 17
Exercise 0E.2
1. Find the equation of the straight line passing through (1,−2) and which has a slope of −2.
2. Find the equation of the straight line passing through (2,−3) that is
a) parallel to the line 2x−5y+3=0.
b) perpendicular to the line 2x−5y+3=0.
- Answer
-
1. y=−2x
2. a) y=25x−195
b) y=−52x+2
Exercise 0E.3
1. Find the equation of the straight line passing through the midpoint of the line segment PQ and perpendicular to PQ given that P=(−1,3), and Q=(5,−7).
2. Find the equation of the straight line which contains both of the points A=(4,−2), and B=(7,−2).
3. Find the equation of the straight line which contains the point (−1,5) and is parallel to the x−axis.
4. Find the equation of the straight line which contains the point (10,−1) and is parallel to the y−axis.
- Answer
-
1. y=35x−165
2. y=−2
3. y=5
4. x=10
Exercise 0E.4
Which of the following pairs of lines is parallel?, perpendicular?, neither parallel nor perpendicular?
1. 4x+3y−2=02x=7y−1
2. 12x−34y+1=09y−6x+3=0
3. x+11=02y−7=0
4. 3x+7y+1=035x−15y−2=0
- Answer
-
1. Neither parallel nor perpendicular.
2. Parallel.
3. Perpendicular.
4. Perpendicular.
Exercise 0E.5
Sketch the graph of each of the following straight lines:
1. 3x−6y−2=0
2. 7x+5y−35=0
3. 2x−y=0
4. 3x+4y=0
5. 2y−11=0
6. 3x+2=0
- Answer
-
1.
2.
3.
4.
5.
6.
Exercise 0E.6
Determine the point of intersection of each of the following pairs of lines, if they exists.
1. 2x+5y=−11 and 7x−4y=26
2. 2x+−3y=1 and 12x−34y=−14
- Answer
-
1. x=2,y=−3
2. The lines are parallel. Therefore, there is no point of intersection.
Exercise 0E.7
1. Given that A=(3,4),B=(−2,3), and C=(2,−1), show that the triangle ABC is isoceles. (An isoceles triangle is a triangle in which two of the sides have the same length.)
2. Given that A=(5,3),B=(−2,2), and C=(2,−1), show that the triangle ABC has a right angle at C. (Hint: Recall Theorem of Pythagoras'.)
3. Show that the points P=(1,−1),Q=(0,2) and R=(−1,5) are collinear (lie on the same line).
4. Use the distance formula to show that the triangle ABC is an equilateral triangle, given that A=(1,0),B=(5,0), and C=(3,2/3). (An equilateral triangle is triangle in which all sides are of equal length).
5. If ABCD is parallelogram with A=(7,−3),B=(8,3), and C=(3,5), determine the coordinates of the vertex D.
- Answer
-
1. ¯AB=(−5,−1) and |¯AB|=√(−5)2+(−1)2=√26
¯AC=(−1,−5) and |¯AC|=√(−1)2+(−5)2=√26Since two of the lines of the triangle have equal length, the three points form an isoceles triangle.
2. ¯AB=(−7,−1) and |¯AB|=√(−7)2+(−1)2=5√2
¯AC=(−3,−4) and |¯AC|=√(−3)2+(−4)2=5
¯BC=(4,−3) and |¯BC|=√(4)2+(−3)2=5Pythagoras' theorem is only true for right angled triangles:
|¯AC|2+|¯BC|2=(5)2+(5)2=50. If the triangle is right angled, we would expect that √50=5√2 would equal |¯AB|. It does!
3. Find the equation of the line containing the points P and Q.
Slope = =−3.
Line: y=−3x+2.
Check if R lies on the line: If x=−1, y=−3(−1)+2=5.
Therefore, the point R=(−1,5) is on the line and the three points are collinear.4. ¯AB=(4,0) and |¯AB|=√(4)2=4
¯AC=(2,2√3 and |¯AB|=√(2)2+(2√3)2=4
¯BC=(2,2√3) and |¯AB|=√(−2)2+(2√3)2=4Since |¯AB|=|¯AC|=|¯BC|, it is an equilateral triangle.
5. Let D=(h,k) so that ABCD is a parallelogram.
m¯AD=m¯BC⇒k+3h−7=5−33−8−5(k+3)=2(h−7)2h+5k=−1(1)
m¯AB=m¯CD⇒k−5h−3=68−7k−5=6(h−3)6h−k=13(2)
To determine h and k solve (1) and (2) simultaneously to obtain h=2 and k=−1.
Thus, D=(2,−1).
Exercise 0E.8
Find an equation of the circle with center and radius as indicated:
1. Center (4,3), radius =9.
2. Center (−1,2), radius =√5.
3. Center (2,0), radius =1.
4. Center (0,0), radius =1.
5. Center (−7,−3), radius =√172.
6. Center (0,−4), radius =3.
- Answer
-
1. (x−4)2+(y−3)2=81
2. (x+1)2+(y−2)2=5
3. (x−2)2+y2=1
4. x2+y2=1
5. (x+7)2+(y+3)2=17/4
6. x2+(y+4)2=9
Exercise 0E.9
Find the center and radius of the following circles:
1. (x−1)2+(y+3)2=16
2. (x+2)2+(y+1)2=24
3. x2+y2=3
4. (x−7)2+y2=64
5. x2+(y+2)2=1
6. x2+y2−2x+2y+1=0
7. x2+y2+8x−6y+17=0
8. x2+y2−4y+1=0
9. 4x2+4y2−4x+16y+13=0
10. 9x2+9y2+18x+8=0
- Answer
-
1. Centre at (1,−3), radius = 4.
2. Centre at (−2,−1), radius = 2√6.
3. Centre at (0,0), radius = √3.
4. Centre at (7,0), radius = 8.
5. Centre at (0,−2), radius = 1.
6. Complete the square to find the centre and radius.
(x−1)2+(y+1)2=1
Centre at (1,−1), radius = 1.7. (x+4)2+(y−3)2=8
Centre at (−4,3), radius = 2√2.8. x2+(y−2)2=3
Centre at (0,2), radius = √3.9. (x−1/2)2+(y+2)2=1
Centre at (1/2,−2), radius = 1.10. (x+1)2+y2=1/9
Centre at (−1,0), radius = 13.
Exercise 0E.10
1. Find an equation of the circle centered at (1,5) passing through the point (7,−1).
2. Determine the equation of the circle with diameter PQ, where P=(−1,1) and Q=(2,0).
3. Determine the equation of the circle with center C=(−3,5) and tangent to
a) the x−axis
b) the y−axis
4. Determine the equation of the circle with center C=(3,−5) and tangent to
a) the x−axis
b) the y−axis
- Answer
-
1. We know that the equation will be (x+1)2+(y−5)2=r2, where r is the radius.
Substitute (7,−1) into the equation and solve for r.
82+(−6)2=64+36=100=r2→r=10
Therefore, the equation of the circle is (x+1)2+(y−5)2=100.2. The radius is half the distance between the two points: √10/2.
The centre will be the midpoint between the two points: (12,−12).
(x−1/2)2+(y+1/2)2=5/23. a) (x+3)2+(y−5)2=25
b) (x+3)2+(y−5)2=9
4. a) (x−3)2+(y+5)2=25
b) (x−3)2+(y+5)2=9
Exercise 0E.11
1. Find an equation of the circle with center at (2,−3) and radius of length 4 units.
2. Find an equation of the circle which has center at the point (4,−7) and which contains the point (2,1).
3. Find an equation of the circle which has diameter AB, where A=(2,3) and B=(6,−1).
4. Find an equation of the circle which is tangent to the x−axis at the point (−2,0), and is tangent to the y−axis at the point (0,2).
- Answer
-
1. (x−2)2+(y+3)2=4
2. (x−4)2+(y+7)2=68
3. (x−4)2+(y−1)2=32
4. (x+2)2+(y−2)2=4
Exercise 0E.12
Classify each equation given as being an equation of a parabola, an ellipse, a circle, a hyperbola or none of these.
1. 4x2−y2+8x+4y+3=0
2. 4x2+5y2−8x+15y=1
3. x2+y2+12x−8y=0
4. 4x2+y−3=0
5. 4x+y2−2y+4=0
6. 4x2+y2+3y−4=0
7. 2x2−3y2−6x+12y=1
8. 3x+2y−5=0
9. x2+5y3−3x+2y−1=0
10. x3+y3=1
11. (x−1)2+(y+3)2=22
12. (x−1)2−(y+3)2=22
- Answer
-
The general form of a conic is Ax2+By2+Cxy+Dx+Ey+F=0.
If the conic is not degenerate, then we can use the general form to classify the conic.
If A=B, then the conic is a circle.
If A≠B, then the conic is an ellipse if A and B are both positive, the conic is a hyperbola if A and B are opposite signs, and the conic is a parabola if A or B is zero.1. Hyperbola.
2. Ellipse.
3. Circle.
4. Parabola.
5. Parabola.
6. Parabola.
7. Hyperbola.
8. Degenerate conic.
9. Not a conic.
10. Not a conic.
11. Circle.
12. Hyperbola.
Trignometry
Exercise 0E.1
1. In each of the triangle, determine the values of sin(α),cos(α),tan(α),sec(α),csc(α), and cot(α).
a) 
b) 
2. Use the information provided in the diagram to determine cot(t),sin(t),sec(t).
- Answer
-
1. a) The unknown side is 6.
sinα=3/5,cosα=4/5,tanα=3/4,secα=5/4,cscα=5/3,cotα=4/3.b) The unknown side is 15.
sinα=15/17,cosα=8/17,tanα=15/8,secα=17/8,cscα=17/15,cotα=8/15.2. The unknown side is 60.
cott=60/11,sint=11/61,sect=61/60.
Exercise 0E.2
1. If cos(α)=35, and if α+β=90∘, find cot(β).
2. If sec(α)=2, and if α+β=90∘, find sin(α),csc(β) and tan(β).
- Answer
-
1. Since α+β=90∘, β is the third angle of the triangle.
cotβ=4/3.2. sinα=√3,cscβ=2,tanβ=√33
Exercise 0E.3
Convert each of the angles given in degree measure to radian measure:
1. 225∘
2. 450∘
3. −210∘
4. 175∘
5. 1∘
- Answer
-
1. 5π4
2. 5π2
3. −7π6
4. 35π36
5. π180
Exercise 0E.4
Convert each of the angles given in radian measure to degree measure:
1. 2
2. π5
3. −π12
4. 7π24
5. 11π6
6. 15π8
- Answer
-
1. 360∘π
2. 36∘
3. −15∘
4. 52.5∘
5. 330∘
6. 337.5∘
Exercise 0E.5
Find the reference angle for each of the following angles.
1. 60∘
2. 225∘
3. 120∘
4. −120∘
5. π4
6. 4π3
7. 7π6
8. 5π6
9. 11π6
10. 3000∘
- Answer
-
1. 60∘
2. 45∘
3. 60∘
4. 60∘
5. π4
6. π3
7. π6
8. π6
9. π6
10. 60∘
Exercise 0E.6
Complete the table below:
- Answer
-
Exercise 0E.7
Determine the exact values of each of the following:
1. sin(3π2)
2. cos(11π6)
3. sec(420∘)
4. tan(225∘)
5. tan(29π3)
6. sin(7π6)
7. csc(−120∘)
8. sin(11π3)
- Answer
-
1. −1
2. √32
3. 2
4. 1
5. −√3
6. −2√33
7. −2√33
8. −√32
Exercise 0E.8
1. If sin(α)=−35, and α is an angle in the third quadrant, then find
a) cos(α)
b) tan(α)
2. If sec(α)=175, and \(270^{\circ}<\alpha < 360^{\circ} \(, then find
a) cos(α)
b) sin(α)
c) tan(α)
3.If cos(t)=−1213, and tan(t)>0, find
a) sin(t)
b) tan(t)
4.If cot(β)=43, and sin(β)>0, find
a) cos(β)
b) csc(β)
5. If tan(t)=1√3, and cos(t)<0, find
a) sin(t)
b) cos(t)
6. If csc(t)=2, and cos(t)>0, find
a) sin(t)
b) tan(t)
c) cos(t)
- Answer
-
1. a) cosα=−4/5, b) tanα=3/4.
2. a) cosα=15/17, b) sinα=−8/17, c) tanα=−8/15.
3. a) sint=−5/13, b) tant=5/12.
4. a) cosβ=4/5, b) cscβ=5/3.
5. a) sint=−1/2, b) cost=−√3/2.
6. a) sint=1/2, c) tant=−√3/3, c) cost=−√3/2.
Exercise 0E.9
Use the sum of difference formulas to determine the exact values of each of the following:
1. sec(15∘)
2. sin(75∘)
3. tan(75∘)
4. csc(5π12)
5. cos(105∘)
6. sin(165∘)
7. sin(π12)
- Answer
-
1. cos(15∘)=cos(45∘−30∘)=cos(45∘)cos(30∘)+sin(45∘)sin(30∘)=(√22)(√32)+(√22)(12)=√2(√3+1)4
Therefore, sec(15∘)=4√2(√3+1).
2. sin(75∘)=sin(45∘+30∘)=√2(√3+1)4
3. tan(75∘)=√6+22√3−√2
4. sin(5π12)=sin(3π12+2π12)=sin(π4+π6)=√2(√3+1)4
Therefore, csc(5π12)=4√2(√3+1).
5. cos(105∘)=√2(√3−1)4
6. sin(165∘)=sin(180∘−15∘)=sin(180∘)cos(15∘)−cos(180circ)sin(15∘)=sin(15∘)=√2(√3−1)4
7. sin(π12)=√2(√3−1)4
Exercise 0E.10
1. If sin(α)=−35, and α is in the third quadrant, then determine cos(2α),sin(2α), and tan(2α).
2. If sec(α)=175, and 270∘<α<360∘, then find tan(2α).
3. If tan(β)=1√3, and cos(β)<0, find sin(2β) and cos(2β).
4. If csc(t)=2, and if π2<t<π, find sin(2t),csc(2t),sec(2t) and tan(2t).
5. If sin(α)=−45, and α is in the third quadrant, then find cos(α/2),sin(α/2), and tan(α/2).
6. If tan(t)=1√3, and \cos(t)<0, find \sin(t/2),\cos(t/2) and \tan(t/2).
7. If \csc (t)=-2, and if \displaystyle \frac{3\pi}{2} < t < 2\pi, find \cos(t/2), and \tan(t/2).
8. If \cot (\alpha)=-1, and if \sin(\alpha) >0 , then find \cos( \alpha/2), \sin( \alpha/2), and \tan( \alpha/2).
9. If \sin (\alpha)=\frac{11}{61}, and \alpha is in the first quadrant, then find \csc( \alpha/2), and \sec( \alpha/2).
- Answer
-
1. \cos (2\alpha) = \cos^2 \alpha - \sin^2 \alpha = 7/25
\sin (2\alpha) = 2\sin \alpha \cos \alpha = 24/25
\displaystyle{\tan (2\alpha) = \displaystyle \frac{2\tan \alpha}{1-\tan^2 \alpha}} = 24/72. \tan (2\alpha) = -240/161
3. \sin (2\beta) = \sqrt{3}/2
\cos (2\beta) = 1/24. \sin (2t) = -\sqrt{3}/2
\cos (2t) = 1/2
\sec (2t) = 2
\tan (2t) = - \sqrt{3}5. \displaystyle{\cos (\alpha /2) = \pm \sqrt{\displaystyle \frac{1+\cos \alpha}{2}} = -\displaystyle \frac{2\sqrt{5}}{5}}, `-' because the angle is in the third quadrant.
\displaystyle{\sin (\alpha /2) = \pm \sqrt{\displaystyle \frac{1-\cos \alpha}{2}} = - \displaystyle \frac{4\sqrt{5}}{5} }
\displaystyle{\tan (\alpha /2) = \displaystyle \frac{1-\cos \alpha}{\sin \alpha} = -2}6. \displaystyle{\sin (t/2) = -\sqrt{\displaystyle \frac{2 + \sqrt{3}}{4}}}
\displaystyle{\cos (t/2) = -\sqrt{\displaystyle \frac{2 - \sqrt{3}}{4}}}
\displaystyle{\tan (t/2) = -2- \sqrt{3}}7. \displaystyle{\cos (t/2) = \sqrt{\displaystyle \frac{2 + \sqrt{3}}{4}}}
\displaystyle{\tan (t/2) = -2 + \sqrt{3}}8. \displaystyle{\sin (\alpha/2) = \sqrt{\displaystyle \frac{2 + \sqrt{2}}{4}}}
\displaystyle{\cos (\alpha/2) = -\sqrt{\displaystyle \frac{2 - \sqrt{2}}{4}}}
\displaystyle{\tan (\alpha/2) = 1 + \sqrt{2}}9. \displaystyle{\csc (\alpha/2) = \sqrt{122}}
\displaystyle{\sec (\alpha/2) = \sqrt{\displaystyle \frac{122}{121}}}
Exercise \PageIndex{11}
For each of the following, determine the amplitude, phase shift and period. Use this information to sketch the graph.
1. y=4\sin(2x+\pi/3)
2. y=\displaystyle \frac{1}{2}\cos(5x-\pi/4)
3. y=-2\cos(x/3)
4. y=2\tan(3x+\pi/2)
5. y=-3\sin(4x)
6. y=\displaystyle \frac{4}{3}\cos(2x)
- Answer
-
1. Amplitude: 4; Phase shift: \displaystyle\frac{\pi}{3}; Period: \pi.
2. Amplitude: \displaystyle\frac{1}{2}; Phase shift: \displaystyle\frac{\pi}{4}; Period: \displaystyle\frac{2\pi}{5}.
3. Amplitude: 2; Phase shift: none; Period: 6\pi.
4. Amplitude: none; Phase shift: \displaystyle\frac{\pi}{2}; Period: \displaystyle\frac{\pi}{3}.
5. Amplitude: 3; Phase shift: none; Period: \displaystyle\frac{\pi}{4}.
6. Amplitude: \displaystyle\frac{4}{3}; Phase shift: none; Period: \displaystyle\frac{\pi}{2}.
Exercise \PageIndex{12}
Solve each of the following trigonometric equations in the interval indicated.
1. 2 \cos (x) +1=0, \,\, [0^{\circ},360^{\circ})
2. 2 \tan(x) \sin(x)+ 2 \sin(x)-\tan(x)-1=0,\,\, [0,\pi]
3. \csc^2 (t) -4=0, \,\, [0^{\circ},180^{\circ}]
4. \sin^2 (t) -3=0, \,\, [0^{\circ},360^{\circ})
5. 2 \cos^2(x)-3 \cos(x)+1=0,\,\, [0, 2 \pi)
6. 2\sin^2 (x) +3 \sin(x)+1=0, \,\, [0^{\circ},360^{\circ})
7. (2\sin (x) -1)(\tan(x)+1)=0, \,\, [0^{\circ},360^{\circ})
8. \sin(x) (2\cos (x) +\sqrt{3})(\tan(x)-\sqrt{3})=0, \,\, [0^{\circ},360^{\circ})
9. \sin(2x)=0,\, \cos(2x)=0,\, [0,2 \pi)
- Answer
-
1. x = 120^{\circ}, 240^{\circ}
2. Factor by grouping to get \tan x + 1 = 0 or 2\sin x -1 = 0.
\tan x + 1 = 0 yields the solution x = 3\pi/4 and 2\sin x -1 = 0 yields the solutions x = \pi/6, 5\pi /6.3. Need to solve \csc t = \pm 2.
The solutions are t = 60^{\circ}, 150^{\circ}.4. Need to solve \sin t = \pm\sqrt{3}/2.
The solutions are t = 60^{\circ}, 120^{\circ}, 240^{\circ}, 300^{\circ}.5. Need to solve \cos x = 1, 1/2.
The solutions are x = 0, \pi/3, 5\pi/3.6. Need to solve \sin x = -1, -1/2.
The solutions are x = 270^{\circ}, 210^{\circ}, 330^{\circ}.7. Need to solve \sin x = 1/2 and \tan x = -1.
The solutions are x = 30^{\circ}, 150^{\circ}, 135^{\circ}, 315^{\circ}.8. Need to solve \sin x = 0, \cos x = -\sqrt{3}/2, \tan x = \sqrt{3}.
The solutions are x = 0^{\circ}, 180^{\circ}, 150^{\circ}, 210^{\circ}, 60^{\circ}, 240^{\circ}.9. The solutions are x = 0, \pi/2, \pi, 3\pi/2, \pi/4, 3\pi/4, 5\pi/4, 7\pi/4.
Exercise \PageIndex{13}
Given that \tan(\alpha/2)=t, express each of the following in terms of t:
1. \sin( \alpha/2)
2. \cos( \alpha/2)
3. \tan( \alpha/2)
4. \sin( \alpha)
5. \cos( \alpha)
6. \tan( \alpha)
- Answer
-
Construct a right-angled triangle with \alpha/2 as one of the angles.
Then the opposite side it t and the adjacent side is 1. Using Pythagoras' theorem, we obtain a hypotenuse of \sqrt{t^2+1}.Therefore, \displaystyle{\sin (\alpha/2) = \displaystyle \frac{t}{\sqrt{t^2+1}}} and \displaystyle{\cos (\alpha/2) = \displaystyle \frac{1}{\sqrt{t^2+1}}}.
Using \sin (\alpha) = 2\sin(\alpha/2) \cos (\alpha/2), we obtain \displaystyle{\sin (\alpha) = \displaystyle \frac{2t}{t^2+1}}.
Using \cos (\alpha) = 2\cos^2 x -1, we obtain \displaystyle{\cos (\alpha) = \displaystyle \frac{-t^2+2t-1}{t^2+1}}.
Therefore, \displaystyle{\tan(\alpha) = \displaystyle \frac{2t}{-t^2+2t-1}}.
Note: This solution assumes 0 \leq t <\pi/2. If t was in another quadrant, how does this effect the solution?
Exercise \PageIndex{14}
In each of the following, find other information about the triangle ABC, using the information provided:
1. A=130^{\circ},\, B=20^{\circ}, \, b=10
2. B=120^{\circ},\, C=30^{\circ}, \, a=16
3. C=30^{\circ},\, a=24, \, b=34
4. C=108^{\circ},\, a=10, \, b=15
5. a=12,\,b=17,\,c=19
6. A=40^{\circ},\, a=7, \, c=10
7. A=40^{\circ},\, a=12, \, c=10
- Answer
-
1. C = 108^{\circ} - 130^{\circ} - 20^{\circ} = 30^{\circ}
a = \displaystyle \frac {b\sin(130^{\circ})}{\sin(20^{\circ})} = 22.398
c = \displaystyle \frac {b\sin(30^{\circ})}{\sin(20^{\circ})} = 14.619
2. A = 180 ^{\circ} - 120 ^{\circ} - 30 ^{\circ} = 30 ^{\circ}
b = \displaystyle \frac {a\sin(120^{\circ})}{\sin30^{\circ})} = 27.713
c = \displaystyle \frac {a\sin(30^{\circ})}{sin(30^{\circ})} = 16
3. c = \sqrt{a^2 + b^2 - 2ab\cos(30^{\circ})} = 17.851
A = \arccos\left(\displaystyle \frac {c^2 + b^2 - a^2}{2cb}\right) = 107.76^{\circ}
B = \arccos\left(\displaystyle \frac {c^2 + a^2 - b^2}{2ca}\right) = 42.24^{\circ}
4. c = \sqrt{a^2 + b^2 - 2ab\cos(108^{\circ})} = 20.438^{\circ}
A = \arccos\left(\displaystyle \frac {c^2 + b^2 - a^2}{2cb}\right) = 27.73^{\circ}
B = \arccos\left(\displaystyle \frac {c^2 + a^2 - b^2}{2ca}\right) = 44.27^{\circ}
5. A = \arccos\left(\displaystyle \frac {c^2 + b^2 - a^2}{2cb}\right) = 38.44^{\circ}
B = \arccos\left(\displaystyle \frac {c^2 + a^2 - b^2}{2ca}\right) = 61.73^{\circ}
C = \arccos\left(\displaystyle \frac {a^2 + b^2 - c^2}{2ab}\right) = 79.84^{\circ}
6. C = \arcsin\left(\displaystyle \frac {c\sin(40^{\circ})}{a}\right) = 66.67^{\circ}
B = 180^{\circ} - 40^{\circ} - 66.67^{\circ} = 73.33^{\circ}
b = \displaystyle \frac {a\sin(73.33^{\circ})}{\sin(40^{\circ})} = 10.432
7. C = \arcsin\left(\displaystyle \frac {c\sin(40^{\circ})}{a}\right) = 32.39^{\circ}
B = 180^{\circ} - 40^{\circ} - 32.39^{\circ} = 107.61^{\circ}
b = \displaystyle \frac {a\sin(107.61^{\circ})}{\sin(40^{\circ})} = 17.794
Exercise \PageIndex{15}
1. A ladder 10 meters long is leaning against the side of a building. It makes an angle of 75^{\circ} with the ground. How far up the side of the building does the ladder reach?
2. A man 6 ft tall on the ground 12 ft away from a lamp-post. The light come from the lamp casts a shadow which is 8 ft long. Determine the height of the lamp-post.
3. From the top of a light house, 40 metres above sea-level, the angle of depression to a boat is observed to be 25^{\circ}. Find the horizontal distance between the boat and the light house.
- Answer
-
1. The ladder reaches up 10 \sin (75^{\circ}) m up the wall.
2. Using similar triangles, the lamppost is 15 ft high.
3. The horizontal distance is 40 \tan (25^{\circ}) m.
Logarithms and Exponents
Exercise \PageIndex{1}
Which of the following functions are exponential functions? If the function is exponential, determine its base.
1. f(x)=6^x
2. f(x)=x^6
3. f(x)=2^{-3x}
4. f(x)=\sqrt{(5^{3x})}
- Answer
-
1. The function is exponential and its base is 6.
2. The function is not exponential.
3. The function is exponential and its base is 2.
4. The function is exponential and its base is 5.
Exercise \PageIndex{2}
Sketch the graphs of each of the following:
1. y=3^x
2. y=3^{-2x}
3. y=5^x-7
4. y=2^{x-4}+1
- Answer
-
1.
2.
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3.
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4.
%252B1.JPG?revision=1)
Exercise \PageIndex{3}
Solve each of the following exponential equations:
1. 3^{(2x-1)}=27^{7/3}
2. 8^{-2x}=\displaystyle \frac{1}{64}
3. 81^{x}=21^{2-x}
4. 7^{x-2}=\displaystyle \frac{1}{343}
5. 2^{7x}=\displaystyle \frac{1}{\sqrt{2}}
6. 10^{2x-1}=5
7. 3^{x}=17
8. 4^{x-2}=2^{1-x}
9. 2^{x}=21
10. (2^{3x}-8)(2^{5x}-4)=0
11. 3^{x}+3^{-x}=2
12. 3^{x}+3^{-x}=\displaystyle \frac{10}{3}
- Answer
-
1. Write each side of the equation as a same base (in this case 3).
3^{2x-1} = 3^{3(7/3)} = 3 ^ 7
x = 42. Since we have 8^{-2x} = 8^{-2}, then x = 1
3. Since we have 3^{4x} = 3^{6-3x}, then x = 6/7.
4. Since we have 7^{x-2} = 7^{-3}, then x = -1.
5. Since we have 2^{7x} = 2^{-1/2}, then x = -1/14.
6. Since 10 and 5 cannot be neatly written in terms of the same base, we will use logarithms to solve.
\begin{eqnarray*} 10^{2x-1} &=& 5 \\ \log (10^{2x-1}) &=& \log (5)\\ 2x - 1 &=& \log (5) \\ x &=& (\log (5) +1)/2 \end{eqnarray*}7. x= \log_3 (17)
8. Since 2^{2x+4} = 2^{1-x}, then x = 1.
9. x = \log_{2} (21)
10. Since 2^{3x} = 8 or 2^{5x} = 4, then x = 1 or \displaystyle \frac{2}{5}.
11. Multiply both sides by 3^{x} to obtain 3^{2x} + 1 = 2\cdot 3^{x}. Bring all the terms to one side and recognize that this is a hidden quadratic.
Therefore, let 3^x = 1 and x = 0.
12. x = -1, 1
Exercise \PageIndex{4}
Express each of the following in logarithmic form:
1. 2^{x}=y
2. 5^{-3}=\displaystyle \frac{1}{125}
3. 4^{x+y}=15
4. 29^{0}=1
5. x^{3}=2y
6. 4^{1}=4
- Answer
-
1. x = \log_2 (y)
2. \displaystyle{-3 = \log_5 (\displaystyle \frac{1}{125})}
3. x+y = \log_4(15)
4. 0 = \log_{29} (1)
5. 3 = \log_{x} (2y)
6. 1 = \log_4 (4)
Exercise \PageIndex{5}
Express each of the following in exponential form:
1. x= \log_2(y)
2. y= \log_x(4)
3. \log_5(\displaystyle \frac{1}{125})=-3
4. \log_7(7)=1
5. \log_3(1)=0
6. \log(y)=2
7. \log(.001)=-3
8. \log_{1/3}(81)=-4
9. \log_3(y+z)=x+y
10. \log(x-5)=2
- Answer
-
1. 2^x = y
2. 4 = x^y
3. \displaystyle{\displaystyle \frac{1}{125} = 5^{-3}}
4. 7 = 7^1
5. 1 = 3^{0}
6. y = 10^2
7. 0.001 = 10^{-3}
8. 81 = (1/3)^{-4}
9. y+z = 3^{x+y}
10. x-5 = 10^2
Exercise \PageIndex{6}
Sketch each of the following graphs and compare them to the graphs of the exponential function given.
1. y= \log_4(x) compare with y=4^x.
2. y= \log(x) compare with y=10^x.
- Answer
-
1. y= \log_4(x)
\textcolor{red}{y=4^x}
2. \textcolor{blue}{y= \log(x)}
\textcolor{green}{y=10^x}
Exercise \PageIndex{7}
Express as a single logarithm:
1. 2 \log_2(x)-5 \log_2(x-2)+\displaystyle \frac{1}{2} \log_2(x^2+1)
2. \displaystyle \frac{1}{3} \log(2x-1)-52\log(x)-7 \log (1+x)+\displaystyle \frac{2}{3} \log(5x+12)
- Answer
-
1. \displaystyle{\log_2 \left ( \displaystyle \frac{x^2\sqrt{x^2+1}}{(x-2)^5} \right )}
2. \displaystyle{\log \left ( \displaystyle \frac{\sqrt[3]{(2x-1)(5x+12)^2}}{x^2(1+x)^7}\right )}
Exercise \PageIndex{8}
Express as a sum or difference of logarithms:
1. \log_2 \displaystyle \frac {x^2 \sqrt{(x^2+1)}}{(x-2)^5}
2. \log \displaystyle \frac {\sqrt[3]{(2x-1)}\, \sqrt[3]{(5x+12)^2}}{x^2(1+x)^7}
3. \log_5\displaystyle \frac {x(1-x)^2}{(2-x)^4}
4. \log_{1/2}\displaystyle \frac {x^2(1+x)^{2/7}}{4(x-2)^3}
- Answer
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1. \displaystyle{2\log_2 (x) + \displaystyle \frac{1}{2}\log_2 (x^2+1) - 5\log_2(x-2)}
2. \displaystyle{\displaystyle \frac{1}{3}\log (2x-1) + \displaystyle \frac{2}{3} \log (5x+12) - 2\log (x) - 7\log (1+x)}
3. \displaystyle{\log_5 x + 2\log_5 (1-x) - 4\log_5 (2-x)}
4. \displaystyle{2\log_{1/2}x + \displaystyle \frac{2}{7}\log_{1/2}(1+x) -2 - 3\log_{1/2}(x-2)}
Exercise \PageIndex{9}
Simplify each of the following:
1. \log_232
2. \log (0.001)
3. \log_x x^3
4. \log_2 4-\log_4 16
- Answer
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1. \log_2(32) = \log _2(2^5) = 5
2. \log(0.001) = \log (10^{-3}) = -3
3. \log_x (x^3) = 3
4. \log_2 (4) - \log_4(16) = \log_2(2^2) - \log_4(4^2) = 0
Exercise \PageIndex{10}
Solve each of the following equations:
1. \log_3 x=-2
2. \log 1000=x
3. \log_x 16=4
4. \log (0,1)=2x
5. \log_2 (3 x+1)- \log_2(x-8)=-1
6. \log_x 4=-1
7. \log x- \log(2x-1)=0
8. \log_5 x- \log_5(x+7)=\log_5 2- \log_5(x-3)
9. 2\log_2 x- \log_2 \left(x-\displaystyle\frac{1}{3}\right)=\log_2 3 -1
10. \log_{3 \sqrt{2}} x+ \log_{3 \sqrt{2}} (x-3)=2
11. \log_{3 \sqrt{2}} x+ \log_{3 \sqrt{2}} (x+3)=2
- Answer
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1. x= 3^{-2} = \displaystyle \frac {1}{9}
2. x=-3
3. x = 2, since 2^4 = 16
4. x=\displaystyle \frac{-1}{2}
5. \begin{eqnarray*} \log_2(3x+1) - \log_2(x-8) &=& -1 \\ \log_2 \left ( \displaystyle \frac{3x+1}{x-8} \right )&=& -1 \\ \displaystyle \frac{3x+1}{x-8} &=& 2^{-1} \\ 6x+2 &=& x-8 \\ x &=& -2 \end{eqnarray*}
But x = -2 is extraneous because if we x=-1 in the original equation, we get the logarithm of a negative number which is undefined. Therefore, the equation has no solution.
6. x=\displaystyle \frac {1}{4}
7. \begin{eqnarray*} \log x - \log (2x-1) &=& 0 \\ \log x &=& \log (2x-1) \\ x &=& 2x-1 \\ x &=& 1 \end{eqnarray*}
8. \begin{eqnarray*} \log_5(x) - \log_5(x+7) &=& \log_5 (2) - \log_5 (x-3) \\ \log_5 \left ( \displaystyle \frac{x}{x+7} \right ) &=& \left ( \displaystyle \frac{2}{x-3} \right ) \\ \displaystyle \frac{x}{x+7} &=& \displaystyle \frac{2}{x-3} \\ x(x-3) &=& 2(x+7) \\ x &=& 7, -2 \end{eqnarray*}
x= -2 is an extraneous solution.
Solution: x = 7.9. \begin{eqnarray*} 2\log_2 (x) - \log_2(x-1/3) &=& \log_2 (3) -1 \\ 2\log_2 (x) - \log_2(x-\displaystyle \frac{1}{3}) -\log_2 (3) &=& -1 \\ \log_2 \left ( \displaystyle \frac{x^2}{3(x-\displaystyle \frac{1}{3})} \right ) &=& -1 \\ \displaystyle \frac{x^2}{3x-1} &=& \displaystyle \frac{1}{2} \\ x &=& 1, 1/2 \end{eqnarray*}
10.
\begin{eqnarray*} \log_{3\sqrt{2}} (x) + \log_{3\sqrt{2}} (x-3) &=& 2 \\ \log_{3\sqrt{2}} (x(x-3)) &=& 2 \\ x(x-3) &=& 18 \\ x &=& -3, 6 \end{eqnarray*}
x = -3 is an extraneous solution.
Solution: x = 6.11. \begin{eqnarray*} \log_{3\sqrt{2}} (x) + \log_{3\sqrt{2}} (x+3) &=& 2 \\ \log_{3\sqrt{2}} (x(x+3)) &=& 2 \\ x(x+3) &=& 18 \\ x &=& 3, -6 \end{eqnarray*}
x = -6 is an extraneous solution.
Solution: x = 3.
