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Modulus – Practise Problems
Find the range of x for which
𝑥2 − 5𝑥 + 5 ≤ 𝑥
Answer :[1,5]](https://image.slidesharecdn.com/inequalitiesandmodulussession2-160811061308/85/Inequalities-and-modulus-session-2-29-320.jpg)
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Inequalities and modulus session 2
This document discusses inequalities and modulus. It begins by explaining how to find the maximum value of expressions like xp yq when ax + by = k. It then discusses how this relates to finding minimum values. It also covers the definition and properties of modulus/absolute value, including how to solve inequalities and equations involving modulus. It gives examples of finding maximum and minimum values of functions involving multiple modulus expressions.
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Inequalities and modulus session 2
- 1. Inequalities and ModulusSession 2 1 www.georgeprep.com
- 2. Maxima and Minima– (a+b) to ab model If ax + by = k, maximum value of xp yq , when a, b, x, y, p and q are positive real numbers. Example: If 3x + 4y = 20, find the maximum possible value of x2 y3 , when x, y >0 2 www.georgeprep.com Generalization If ax + by = k, maximum value of xpyq is when 𝑎𝑥 𝑝 = 𝑏𝑦 𝑞
- 3. If 5p +6q = 42, what is the maximum possible value of p5 q2 when p, q > 0? 1.2562 2.6523 3.3523 4.None of these 3 www.georgeprep.com (a+b) to ab model - Problems Answer : Option 4
- 4. If xp yq= k, minimum value of ax + by , when a, b, x, y, p and q are positive real numbers. Example: If x5 y3 =55 28 , what is the minimum possible value of 4x + 6y when x, y > 0? 4 www.georgeprep.com Generalization If xpyq = k, minimum value of ax + by is when 𝑎𝑥 𝑝 = 𝑏𝑦 𝑞 Maxima and Minima – ab to (a + b) model
- 5. If p q2r3 =33 28 , what is the minimum possible value of p + q + r when p, q > 0? 1.15 2.20 3.12 4.18 5 www.georgeprep.com ab to (a + b) model - Problems Answer : option 3
- 6. Absolute Value (Modulus) www.georgeprep.com 6
- 7. If x isthe coordinate of a point on a real number line, then the distance of x from the origin is represented by 𝑥 . x is called Absolute value of x or modulus of x. 7 www.georgeprep.com Modulus and Distance
- 8. Definition of Absolutevalue 8 www.georgeprep.com Modulus and Distance
- 9. Properties of Modulus 1.𝑥 ≥ 0 2. − 𝑥 ≤ 0 3. 𝑥 + 𝑦 ≤ 𝑥 + 𝑦 4. 𝑥 − 𝑦 ≤ 𝑥 − 𝑦 5. 𝑥 ∗ 𝑦 = 𝑥 ∗ 𝑦 6. 𝑥 𝑦 = 𝑥 𝑦 ( as long as y≠0) 7. 𝑥 2 = 𝑥2 9 www.georgeprep.com Modulus
- 10. 𝑥 = 3 𝑥< 3 𝑥 > 3 10 www.georgeprep.com Modulus and Distance
- 11. 11 www.georgeprep.com Modulus and Distance Whatif you have to measure a distance from a point other than 0? Distance of a coordinate x from a is given by 𝑥 − 𝑎 Then what does 𝑥 + 𝑎 represent? Distance of x from -a
- 12. 𝑥 − 1= 3 𝑥 − 1 < 3 𝑥 − 1 > 3 12 www.georgeprep.com Modulus and Distance
- 13. Find the valueof x is 3𝑥 + 2 = 5 Answer : 1 or − 7 3 13 www.georgeprep.com Modulus
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- 16. 16 www.georgeprep.com Modulus - Problems Solvefor x if 2𝑥 − 1 < 3 Answer : (-1, 2) Solve for x if 7 − 3𝑥 > 2 Answer : −∞, 5 3 ∪ (3, ∞)
- 17. 17 www.georgeprep.com Modulus How about solvingfor x for the expression below? 𝑥 − 4 = 3𝑥 − 8
- 18. 18 www.georgeprep.com Modulus - Problems Solvefor x for the expression below? 3𝑥 − 4 = 𝑥 + 5
- 19. 19 www.georgeprep.com Modulus – Maximumand Minimum Values What is the maximum value of 𝑓 𝑥 = 15 − 9 + 𝑥 ? Answer : 15, when x = -9
- 20. 20 www.georgeprep.com Modulus – Maximumand Minimum Values Maximum and Minimum values when more than one modulus functions are given. What is the minimum value of the function 𝑥 − 2 + 𝑥 − 9 + 𝑥 + 4 Answer: 13 ( 9+4)
- 21. 21 www.georgeprep.com Modulus – Maximumand Minimum Values Maximum and Minimum values when more than one modulus functions are given. What is the minimum value of the function 𝑥 − 2 + 2 𝑥 − 9 + 3 𝑥 + 4 Answer: 32
- 22. 22 www.georgeprep.com Modulus – Maximumand Minimum Values Maximum and Minimum values when more than one modulus functions are given. What is the minimum value of the function 𝑥 − 2 + 3 𝑥 − 9 + 2 𝑥 + 4 Answer: 33
- 23. 23 www.georgeprep.com Modulus – Maximumand Minimum Values Maximum and Minimum values when more than one modulus functions are given. What is the minimum value of the function 𝑝(𝑥) = 𝑥 − 3 + 3|𝑥 − 4| + 2|𝑥 − 5| + 3|𝑥 − 7| Answer: 11
- 24. 24 www.georgeprep.com Modulus – Maximumand Minimum Values How many integer solutions are possible for the inequality 𝑥 − 6 + 𝑥 − 8 + 𝑥 + 4 < 11 1.1 2.2 3.0 4.Infinitely many Answer : 0, as minimum value of LHS is 8+4 = 12
- 25. 25 www.georgeprep.com Modulus – PractiseProblems Find the range of x for which |4𝑥 + 1| < |3𝑥| Answer :−1 < 𝑥 < −1/7
- 26. 26 www.georgeprep.com Modulus – PractiseProblems Find the range of x for which |𝑥 + 3| ≥ |𝑥 − 1| Answer :x ≥ −1
- 27. 27 www.georgeprep.com Modulus – PractiseProblems Find the range of x for which 𝑥2 − 7𝑥 + 12 ≤ 𝑥2 − 4𝑥 Answer : [ 3 2 , ∞)
- 28. 28 www.georgeprep.com Modulus – PractiseProblems Find the range of x for which |𝑥2 − 3𝑥 + 1| < 1 Answer : (0,1) ∪ (2,3)
- 29. 29 www.georgeprep.com Modulus – PractiseProblems Find the range of x for which 𝑥2 − 5𝑥 + 5 ≤ 𝑥 Answer :[1,5]
Editor's Notes
- #3 Solution : x = 8/3 and y = 3 Answer: 192
- #4 Answer: 6^5 2^2
- #5 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #6 Answer: 12
- #8 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #9 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #10 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #11 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #12 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #13 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #14 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #15 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #16 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #17 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #18 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #19 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #20 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #21 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #22 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #23 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #24 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #25 Solution: 2x/5 = 3y/3 X = 5 Answer : 16
- #26 ℎ𝑡𝑡𝑝𝑠://𝑚𝑎𝑡ℎ.𝑠𝑡𝑎𝑐𝑘𝑒𝑥𝑐ℎ𝑎𝑛𝑔𝑒.𝑐𝑜𝑚/𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛𝑠/896917/ℎ𝑜𝑤−𝑡𝑜−𝑠𝑜𝑙𝑣𝑒−𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑖𝑒𝑠−𝑤𝑖𝑡ℎ−𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒−𝑣𝑎𝑙𝑢𝑒𝑠−𝑜𝑛−𝑏𝑜𝑡ℎ−𝑠𝑖𝑑𝑒𝑠?𝑟𝑞=1
- #27 ℎ𝑡𝑡𝑝𝑠://𝑚𝑎𝑡ℎ.𝑠𝑡𝑎𝑐𝑘𝑒𝑥𝑐ℎ𝑎𝑛𝑔𝑒.𝑐𝑜𝑚/𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛𝑠/1905592/ℎ𝑜𝑤−𝑡𝑜−𝑠𝑜𝑙𝑣𝑒−𝑎𝑛−𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦−𝑤𝑖𝑡ℎ−𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒−𝑣𝑎𝑙𝑢𝑒𝑠−𝑜𝑛−𝑏𝑜𝑡ℎ−𝑠𝑖𝑑𝑒𝑠?𝑛𝑜𝑟𝑒𝑑𝑖𝑟𝑒𝑐𝑡=1&𝑙𝑞=1
- #28 𝑥 2 −7𝑥+12 ≤ 𝑥 2 −4𝑥 𝑥−3 𝑥−4 ≤ 𝑥 𝑥−4 𝑥−3 ≤ 𝑥 𝑥−3 2 ≤ 𝑥 2
- #29 | 𝑥 2 −3𝑥+1|<1 𝑥 2 −3𝑥+1<1 or 𝑥 2 −3𝑥+1>−1 The common ranges is the answer. You can use the inequality: 𝐴 < 𝐵 ⟺ 𝐴 2 < 𝐵 2 ⟺ 𝐴−𝐵 𝐴+𝐵 <0 Let 𝐴= 𝑥 2 −3𝑥+1,𝐵=1 ⇒ 𝑥 2 −3𝑥+1−1 𝑥 2 −3𝑥+1+1 <0 ⇒ 𝑥 2 −3𝑥 𝑥2−3𝑥+2 <0 ⇒𝑥 𝑥−3 𝑥−1 𝑥−2 <0 𝐴= 𝑥 2 −3𝑥+1,𝐵=1⇒ 𝑥 2 −3𝑥+1−1 𝑥 2 −3𝑥+1+1 <0 ⇒ 𝑥 2 −3𝑥 𝑥 2 −3𝑥+2 <0 ⇒𝑥(𝑥−3)(𝑥−1)(𝑥−2)<0. Can you finish it? Take a number 𝑥=4𝑥=4, and plug it into the left side we see that the left side >0>0, thus the solution is: (0,1)∪(2,3)
- #30 ℎ𝑡𝑡𝑝𝑠://𝑚𝑎𝑡ℎ.𝑠𝑡𝑎𝑐𝑘𝑒𝑥𝑐ℎ𝑎𝑛𝑔𝑒.𝑐𝑜𝑚/𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛𝑠/1511384/ℎ𝑜𝑤−𝑑𝑜−𝑖−𝑠𝑜𝑙𝑣𝑒−𝑎−𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐−𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦−𝑤𝑖𝑡ℎ−𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒−𝑣𝑎𝑙𝑢𝑒−𝑢𝑠𝑖𝑛𝑔−𝑐𝑎𝑠𝑒𝑠?𝑟𝑞=1