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1. INEQUALITIES

2. INTRODUCTION If a and b are real numbers then we can compare their positions by the relation… Less than < Greater than > Less than or equal to ≤ Greater than or equal to ≥ For example: if x > 3 , it means x can be any value more than 3

4. y = x 1 2 3 – 1 – 2 – 3 3 2 1 – 1 – 2 – 3 y = x

5. y  x 1 2 3 – 1 – 2 – 3 3 2 1 – 1 – 2 – 3 y  x

6. y  x 1 2 3 – 1 – 2 – 3 3 2 1 – 1 – 2 – 3 y  x

7. y = x + 1 1 2 3 – 1 – 2 – 3 3 2 1 – 1 – 2 – 3 y = x + 1

8. y  x + 1 1 2 3 – 1 – 2 – 3 3 2 1 – 1 – 2 – 3 y  x + 1

9. y  x + 1 1 2 3 – 1 – 2 – 3 3 2 1 – 1 – 2 – 3 y  x + 1

10. y > x + 1 1 2 3 – 1 – 2 – 3 3 2 1 – 1 – 2 – 3 y > x + 1

11. y < x + 1 1 2 3 – 1 – 2 – 3 3 2 1 – 1 – 2 – 3 y < x + 1

12. x > 2 1 2 3 – 1 – 2 – 3 3 2 1 – 1 – 2 – 3 x > 2

13. WRITE THE EQUATION 1 2 3 – 1 – 2 – 3 3 2 1 – 1 – 2 – 3 4

14. WRITE THE EQUATION 1 2 3 – 1 – 2 – 3 3 2 1 – 1 – 2 – 3 4

16. You know 8 is greater than 4 or 8 > 4 Add 2 on both sides 8 + 2 > 4 + 2 10 > 6 TRUE EXAMPLE

17. PROPERTIES OF INEQUALITIES If ‘a’ is greater than ‘b’ If we subtract ‘c’ (any real number) then which one is greater a – c or b – c Solution: (a – c) is greater than b – c

18. You know 8 is greater than 4 or 8 > 4 Subtract 2 from both sides 8 – 2 > 4 – 2 6 > 2 TRUE EXAMPLE

19. If ‘a’ is greater than ‘b’ i.e. (a > b) If we multiply by ‘c’ (any real number) then which one is greater ac or bc ? Depends upon ‘c’ because c can be a positive or negative real number PROPERTIES OF INEQUALITIES

20. You know 8 is greater than 4 or 8 > 4 Multiply by 2 both sides 8(2) > 4(2) 16 > 8 TRUE EXAMPLE

21. You know 8 is greater than 4 or 8 > 4 Multiply by – 2 both sides 8(– 2) > 4(– 2) – 16 > – 8 FALSE EXAMPLE

22. REMEMBER If ‘a’ is greater than ‘b’ WHICH IS GREATER ac or bc If c is positive then ac > bc If c is negative then ac < bc

23. If ‘a’ is greater than ‘b’ WHICH IS GREATER or Is Greater than REMEMBER

24. REMEMBER We know that 8 is greater than 4 or 8 > 4 Take reciprocals on both sides

26. INTERVAL NOTATION Real number line graph Inequality notation

27. SOLVING LINEAR INEQUATIONS You solve linear inequalities in the same way as you would solve linear equations, but with one exception. Property: If in the process of solving an inequality, you multiply or divide the inequality by a negative number, then , you must switch the direction of the inequality. If – x > a , then x < – a .

30. What happens when the number is negative? If you divide both sides by – 3, (The inequality will change if we multiply or divide with a negative number on both sides.) CASE-3

31. Solve the followings: 1. 3x – 4 > 8 2. – 6x – 18 < – 24 Answers: 1. x > 4 2. x > 1 EXAMPLE

32. SOLVING QUADRATIC INEQUATIONS When we have an inequality with &quot; x 2 &quot; as the highest-degree term, it is called a &quot;quadratic inequality&quot;. Solve x 2 – 3 x + 2 > 0 Step 1: Change the inequality to an equation . Find x- interce p t x 2 – 3 x + 2 = 0 (x – 1) (x – 2) = 0 x = 1 or 2 Step 2 : Plot the points ( x = 1, 2) on the number line The number line is divided into the intervals (- ∞ , 1), (1, 2), and (2, ∞).  3  2  1 0 1 2 3 1 2

33. x 2 – 3 x + 2 > 0 or (x – 1) (x – 2) > 0 The number line is divided into the intervals (- ∞ , 1), (1, 2), and (2, ∞). Test-point method: Pick a point (any point) in each interval x = 3 (x - 1) is p ositive (x - 1) (x - 2) is positive (x - 3) is p ositive POSITIVE x = 1.5 (x - 1) is negative (x - 1) (x - 2) is negative (x - 3) is p ositive NEGATIVE x = 0 (x - 1) is negative (x - 1) (x - 2) is positive (x - 2) is negative POSITIVE (x – 1) (x – 2) is positive when x > 2 or x < 1 SOLVING QUADRATIC INEQUATIONS 1 2 1 2

34. WHICH ONE IS GREATER? x 2 x or You can’t say Lets find the interval where x 2 is greater than x

35. SOLUTION For what value of x ? x 2 – x ≥ 0 or (x) (x – 1) ≥ 0 Step 1: Change the inequality to an equation . Find value of x x = 0, 1 x 2 – x = 0 Step 2: Plot the points Step 3: Test point method At x = 2 At x = 0.5 At x = -1 + ─ + Step 4: Solution x > 1 or x < 0 0 1 0 1 0 1

36. You can solve some absolute-value equations using logics. For instance, you have learned that the equation | x |  8 has two solutions: 8 and  8. SOLVING ABSOLUTE-VALUE EQUATIONS To solve absolute-value equations, you can use the fact that the expression inside the absolute value symbols can be either positive or negative. Because I X I = + X if X > 0 - X If X < 0 0 if X = 0

37. SOLVING AN ABSOLUTE-VALUE EQUATION Solve | x  2 |  5 x  2 IS POSITIVE | x  2 |  5 x  7 x   3 x  2 IS NEGATIVE | x  2 |  5 | 7  2 |  | 5 |  5 |  3  2 |  |  5 |  5 The expression x  2 can be equal to 5 or  5 . x  2   5 Solve | x  2 |  5 S OLUTION x  2   5 The equation has two solutions: 7 and –3. x  2 IS POSITIVE x  2   5 The expression x  2 can be equal to 5 or  5 . x  2 IS POSITIVE | x  2 |  5 x  2   5 x  7 x  2 IS POSITIVE | x  2 |  5 x  2   5 x  7 x  2 IS NEGATIVE x  2   5 x   3 x  2 IS NEGATIVE | x  2 |  5 x  2   5 C HECK

38. Recall that  x  is the distance between x and 0. If  x  8, then any number between  8 and 8 is a solution of the inequality. You can use the following properties to solve absolute-value inequalities and equations. Recall that | x | is the distance between x and 0. If | x |  8, then any number between  8 and 8 is a solution of the inequality.  8  7  6  5  4  3  2  1 0 1 2 3 4 5 6 7 8 Recall that | x | is the distance between x and 0. If | x |  8, then any number between  8 and 8 is a solution of the inequality.

39. a x  b  c and a x  b   c. a x  b  c and a x  b   c. a x  b  c or a x  b   c. a x  b  c or a x  b   c. a x  b  c or a x  b   c. | a x  b |  c | a x  b |  c | a x  b |  c | a x  b |  c | a x  b |  c When an absolute value is less than a number, the inequalities are connected by and . When an absolute value is greater than a number, the inequalities are connected by or . S OLVING A BSOLUTE- V ALUE I NEQUALITIES S OLVING A BSOLUTE- V ALUE E QUATIONS AND I NEQUALITIES means means means means means means means means means means

40. Solve | x  4 | < 3 x  4 IS POSITIVE x  4 IS NEGATIVE | x  4 |  3 x  4   3 x  7 | x  4 |  3 x  4   3 x  1 Reverse inequality symbol. This can be written as 1  x  7 . The solution is all real numbers greater than 1 and less than 7. EXAMPLE 

41. 2 x  1   9 | 2 x  1 |  3  6 | 2 x  1 |  9 2 x   10 2 x + 1 IS NEGATIVE x   5 Solve | 2 x  1 |  3  6 and graph the solution. | 2 x  1 |  3  6 | 2 x  1 |  9 2 x  1  +9 2 x  8 2 x + 1 IS POSITIVE x  4 SOLVING AN ABSOLUTE-VALUE INEQUALITY Reverse inequality symbol. The solution is all real numbers greater than or equal to 4 or less than or equal to  5 . This can be written as the compound inequality x   5 or x  4 .  5 4 .  | 2 x  1 |  3  6 | 2 x  1 |  9 2 x  1  +9 x  4 2 x  8 | 2 x  1 |  3  6 | 2 x  1 |  9 2 x  1   9 2 x   10 x   5 2 x + 1 IS POSITIVE 2 x + 1 IS NEGATIVE  6  5  4  3  2  1 0 1 2 3 4 5 6

42. SOLVING THE INEQUALITIES WITH THE HELP OF OPTIONS We can solve all the inequality questions by going with the options. Take an example: x 2 – 7x + 10 < 0 (1) X < 2 (2) x > 5 (3) x < 5 (4) 2 < x < 5 (5) Both (1) and (2)

43. SOLUTION Since the first option is x < 2, we take x = 1 and check whether the given inequality is satisfying or not. If x = 1, 1 2 – 7(1) + 10 < 0 4 < 0 (wrong) Option (1), (3) and (5) are wrong. Now take x = 6, 6 2 – 7  6 + 10 < 0 4 < 0 (wrong) So, option (2) is wrong. So, the answer is (4).