Mathematics

1. Mathematics

2. Statistics

3. Session Objectives

4. Session Objectives • Intervals • Basic properties of inequalities • Definition and solution of linear inequation • Solution of modulus inequations • Solution of two variable inequations • Inequalities related to AM, GM and HM

5. Interval (i) Open interval: (ii) Closed interval:

6. Interval (iii) Open closed interval: Note: (iii) and (iv) are also called semi-closed or semi-open intervals. (iv) Closed open interval:

7. (iv) If a > b > 0, then Basic Properties of Inequalities (i) If a > b, b > c, then a > c. (ii) If a > b, then a + m > b + m. (iii) If a > b, then am > bm for m > 0 andam < bm for m < 0.

8. (v) If , thenforall positivenumbers ai and bi for i = 1, 2, ... n. (vi) If , thenforall positivenumbers ai and bi for i = 1, 2, ... n. (vii)If a > b > 0 and n > 0, then Basic Properties of Inequalities

9. A statement involving variable(s) andthe sign of inequality, i.e. >, <,is called aninequation. Definition of Inequation For example:

10. Solve for x where x is non-negative integer. • 2x + 8 = 20 • 2x + 8 < 20 Illustrative Example Solution: (ii)2x + 8 < 20 (i) 2x + 8 = 20

11. Illustrative Example An inequation may be linear or quadratic or cubic,etc.,containing one or more variables.

12. Solutions of Linear Inequations in One Variable It is the process of obtaining all possiblesolutions of an inequation. Solution set The set of all possible solutions of an inequation is known as its solution set. For example: The solution set of the inequation x2 + 2 > 0 is the set of all real numbers whereas the solution setof the inequation x2 + 2 < 0 is the null set.

13. Rules for Solving Inequations • Adding or subtracting the same number or expression from each side of an inequation doesnot change the inequality. • Multiplying or dividing each side of an inequation by the same positive number does not change the inequality. • Multiplying each side of an inequation by the same negative number reverse the inequality.

14. Rules for Solving Inequations Example : Solve the inequations

15. Rules for Solving Absolute Value Function Inequations • Factorize the expression. • Get the roots of the expression or say criticalpoint (critical points are the points where the expression becomes zero or ). Expression onlychanges its sign at critical value. • Make various interval on number line. • Assign the sign of each bracket in these intervals and check the sign of expression. • List out the intervals where expression is positive or negative separately. • Left as it is, when expression is positive and multiply with –1 when function is negative to makeit positive.

16. So intervals are . Rules for Solving Absolute Value Function Inequations Example |x + 1| > 4 Step 1:Already in factorized form, i.e x + 1 Step 2:x + 1 = 0 x = –1 Step 3:

17. Step 4:Sign of x + 1 in their intervalis negative. To get the sign of x + 1 inthe interval we generally a value of xwhich is less than –1. Say it as –1.008.Then obviously x + 1 will be negativeand in the interval expression x + 1is positive. This can be realized bytaking the value of x = –0.98 whichlies in the interval Rules for Solving Absolute Value Function Inequations

18. Step 5: So in , (x + 1) is negative. and in , (x + 1) is positive. Step 6: Rules for Solving Absolute Value Function Inequations Minus sign is added here as x + 1 is negativein this interval and we are interested in positivevalue of the expression.

19. So x + 1 > 4 x > 3 the value of |x + 1| is –(x + 1). – x – 1 > 4 Rules for Solving Absolute Value Function Inequations Now solve the equation. |x + 1| > 4 But the value of |x + 1| in this interval is x + 1.

20. Rules for Solving Absolute Value Function Inequations

21. (i) (ii) (iii) (iv) Some Important Results on Modulus Inequations or Absolute Value function Result 1:If ‘a’ is a positive real number,then

22. (i) (ii) (iii) (iv) Some Important Results on Modulus Inequations or Absolute Value function Result 2:Let r be a positive real numberand ‘a’ be a fixed real number. Then

23. Graphical Solution of One or Two Variable Equations Let the equation is Ax + By + C < 0. Algorithm Step 1: Convert the inequation into equation,i.e. Ax + By + C = 0 Step 2: Draw the graph of Ax + By + C = 0. Step 3: Take any point [generally (0, 0)] or any point not on the line whose position is known to you with respect to line. Step 4: Put this point in the given inequation and check the validity of the inequation. If inequation satisfied then the corresponding side of the line where lies the chosen point is the graph of the inequation.

24. Illustrative Example Draw the graph of the inequation2x – y < 4. Solution: Step 1: 2x – y = 4 Step 2:

25. Illustrative Example Step 3: Take the point as (0, 0). Step 4: 2 × 0 – 0 < 4 0 < 4, so it is valid. Hence, the shaded side is the required solution.

26. Graphical Solution of Two Variable Algorithm (i)Draw the graph of two inequations. (ii)Get the required common region bounded by the two lines.

27. Illustrative Example Get the required region.x – 2y < 32x + y >1 Solution: Dotted part is for x – 2y < 3. Crossed part is for 2x + y > 1. So the required region is that region where both cross and dots are present, i.e. the circled region.

28. Inequality related to AM, GM and HM Let ‘a’ and ‘b’ be two real positive andunequal numbers and A, G, H arearithmetic, geometric and harmonicmeans respectively between them.

29. Inequality related to AM, GM and HM From (ii) and (iii), A > G > H, i.e. AM > GM > HM

30. Therefore, we can write equally holdswhen a = b. Inequality related to AM, GM and HM Cor:If the two numbers are equal,i.e. a = b, then Hence, A = G = H

31. Inequality related to AM, GM and HM Note: This inequality holds for nnumbers also.

32. Class Test

33. Class Exercise - 1

34. We have, Solution set of the given inequality is . Solution

35. Class Exercise - 2

36. We have Solution

37. Required solution region Solution contd.. Here coefficient of x is positive. Now equatingnumerator and denominator to zero, we getx = –5 and 5 respectively.

38. Class Exercise - 3

39. Solution The given system of inequation is

40. From (ii), Solution contd... Combining the solution (ii) and (iii) on the number line

41. Class Exercise - 4

42. Solution of inequation (i) is [–4, 6]. ...(iii) Solution The given system of inequation is

43. From (ii), The combined solution is Solution contd... Combining the solution (iii) and (iv) on thenumber line, we get

44. Class Exercise - 5

45. We have . Case I: When . Solution Now the following cases arise: In this case, we have |x – 1| = x – 1.

46. Solution contd.. Case II: When x – 1 < 0, i.e. x < 1. In this case |x – 1| = – (x – 1).

47. Inequation (i) can be written as Equating numerator and denominator to zero,we get and –2 respectively. Solution contd..

48. Therefore, But x < 1, therefore ...(ii) Solution contd.. Combining the solution (i) and (ii) onthe numberline, we get

49. Class Exercise - 6 In the first four papers each of 100marks, Mukesh got 83, 73, 72, 95marks. If he wants an average ofgreater than or equal to 75 marksand less than 80 marks, find therange of marks he should score inthe fifth paper.

50. Solution Let Mukesh scores x marks in the fifth paper. Hence, Mukesh must score between 52 and 77 marks.