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Mathematics Statistics Session Objectives Session Objectives 1. Intervals 2. Basic properties of inequalities 3. Definition and solution of linear inequation 4. Solution of modulus inequations 5. Solution of two variable inequations 6. Inequalities related to AM, GM and HM Interval (i) Open interval: ]a, b[ or (a, b) =x R:a x b a b (ii) Closed interval: [a, b] =x R : a x b a b Interval (iii) Open closed interval: (a, b] =x R : a x b a b (iv) Closed open interval: [a, b) =x R : a x b a b Note: (iii) and (iv) are also called semi-closed or semi-open intervals. 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 and am < bm for m < 0. (iv) If a > b > 0, then 1 1 . a b Basic Properties of Inequalities (v) If a1 b1, a2 b2 , ..., an bn, then a1 a2 a3 ... an b1 b2 ... bn for all positive numbers ai and bi for i = 1, 2, ... n. (vi) If a1 b1, a2 b2 , a3 b3 , ...an bn , then a1 a2 ...an b1 b2 ... bn for all positive numbers ai and bi for i = 1, 2, ... n. (vii) If a > b > 0 and n > 0, then an bn and 1 an 1 bn . Definition of Inequation A statement involving variable(s) and the sign of inequality, i.e. >, <, or is called an inequation. For example: f x 0 or f x 0 or f x 0 or f x 0 Illustrative Example Solve for x where x is non-negative integer. (i) 2x + 8 = 20 (ii) 2x + 8 < 20 (iii) 2x 8 20 Solution: (i) 2x + 8 = 20 (ii) 2x + 8 < 20 2x 20 8 12 2x 12 x6 x 6 Possible values of x are 0, 1, 2, 3, 4, 5. Illustrative Example (iii) 2x 8 20 2x 12 x6 Possible values of x are 0, 1, 2, 3, 4, 5, 6. An inequation may be linear or quadratic or cubic, etc., containing one or more variables. Solutions of Linear Inequations in One Variable It is the process of obtaining all possible solutions 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 set of the inequation x2 + 2 < 0 is the null set. Rules for Solving Inequations 1. Adding or subtracting the same number or expression from each side of an inequation does not change the inequality. 2. Multiplying or dividing each side of an inequation by the same positive number does not change the inequality. 3. Multiplying each side of an inequation by the same negative number reverse the inequality. Rules for Solving Inequations Example : Solve the inequations 12 5x 3 3x 5x 3x 3 12 2x 15 x 7.5 Answer is ] , 7.5] – – 7 . 5 0 Rules for Solving Absolute Value Function Inequations 1. Factorize the expression. 2. Get the roots of the expression or say critical point (critical points are the points where the expression becomes zero or ). Expression only changes its sign at critical value. 3. Make various interval on number line. 4. Assign the sign of each bracket in these intervals and check the sign of expression. 5. List out the intervals where expression is positive or negative separately. 6. Left as it is, when expression is positive and multiply with –1 when function is negative to make it positive. 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: – – 1 0 So intervals are , 1 and 1, . Rules for Solving Absolute Value Function Inequations Step 4: Sign of x + 1 in their interval , 1 is negative. To get the sign of x + 1 in the interval we generally a value of x which is less than –1. Say it as –1.008. Then obviously x + 1 will be negative and in the interval 1, expression x + 1 is positive. This can be realized by taking the value of x = –0.98 which lies in the interval 1, . Rules for Solving Absolute Value Function Inequations Step 5: So in , 1 , (x + 1) is negative. and in 1, , (x + 1) is positive. 1 x x 1 Step 6: x 1 x 1 x 1 Minus sign is added here as x + 1 is negative in this interval and we are interested in positive value of the expression. Rules for Solving Absolute Value Function Inequations Now solve the equation. For 1 x |x + 1| > 4 But the value of |x + 1| in this interval is x + 1. So x + 1 > 4 x>3 So the solution is 3 x For x 1 the value of |x + 1| is –(x + 1). x 1 4 – x – 1 > 4 Rules for Solving Absolute Value Function Inequations x 5 x 5 x , 5 Possible solutions are , 5 – 3, – 5 3 Some Important Results on Modulus Inequations or Absolute Value function Result 1: If ‘a’ is a positive real number, then (i) x a a x a (ii) x a a x a i.e. x a, a i.e. x a, a (iii) x a x a or x a (iv) x a x a or x a Some Important Results on Modulus Inequations or Absolute Value function Result 2: Let r be a positive real number and ‘a’ be a fixed real number. Then (i) x a r a r x a r i.e. x a r, a r (ii) x a r ar x ar i.e. x a r, a r (iii) x a r x a r or x a r (iv) x a r x a r or x a r 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. Illustrative Example Draw the graph of the inequation 2x – y < 4. Solution: Step 1: 2x – y = 4 Step 2: y x 0 (2, 0) (0, – 4) y x 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. y x 0 (2, 0) (0, – 4) y x Graphical Solution of Two Variable Algorithm (i) Draw the graph of two inequations. (ii) Get the required common region bounded by the two lines. Illustrative Example Get the required region. x – 2y < 3 2x + y >1 Solution: (0, 1) x Dotted part is for x – 2y < 3. y= –3 2 (3, 0) + 1 So the required region is that region where both cross and dots are present, i.e. the circled region. 1 –, 0 2 2x Crossed part is for 2x + y > 1. 0 – = 2y 3 Inequality related to AM, GM and HM Let ‘a’ and ‘b’ be two real positive and unequal numbers and A, G, H are arithmetic, geometric and harmonic means respectively between them. ab 2ab A , G ab and H 2 ab a b 2ab 2 Now AH ab G ab 2 A G G H ...(i) ab Again A G ab 2 Inequality related to AM, GM and HM 2 a b 2 ab a b 0 2 2 A G 0 AG ...(ii) A Again from (ii), A G 1 G G 0 A G From i , 1 G H GH H 0 ...(iii) From (ii) and (iii), A > G > H, i.e. AM > GM > HM Inequality related to AM, GM and HM Cor: If the two numbers are equal, i.e. a = b, then 2 a b A G 0 2 A G Again from (i), 1 G H AG GH Hence, A = G = H Therefore, we can write AM GM HM equally holds when a = b. Inequality related to AM, GM and HM Note: This inequality holds for n numbers also. a1 a2 a3 ... an AM of 'n' numbers n GM of 'n' numbers 1 a1 a2 a3 ...an n n and HM of 'n' numbers 1 1 1 1 ... a1 a2 a3 an 1 a1 a2 ... an n AM GM HM a1 a2 ...an n . 1 1 1 n ... a1 a2 an Class Test Class Exercise - 1 x 3x 2 5x 3 Solve . 5 4 5 Solution 5 3x 2 4 5x 3 x We have, 5 20 x 15x 10 20x 12 5 20 20x 25x 10 x x 5x 2 5 20 45x 10 10 2 x 45 9 Solution set of the given inequality is , 2 . 9 Class Exercise - 2 x 1 Solve . x5 2 Solution We have x 1 x5 2 x 1 0 x 5 2 2x x 5 0 2 x 5 2x x 5 0 2 x 5 x5 0 x 5 ...(i) Solution contd.. Here coefficient of x is positive. Now equating numerator and denominator to zero, we get x = –5 and 5 respectively. – 5 + + – 5 x5 From i, 0 x 5 Required solution region , 5 5, Class Exercise - 3 Solve the following system of inequations 5x 3x 39 2x 1 x 1 3x 1 , . 4 8 8 12 3 4 Solution The given system of inequation is 5x 3x 39 4 8 8 2x 1 x 1 3x 1 12 3 4 From (i), ...(i) ...(ii) 5x 3x 39 4 8 8 10x 3x 39 8 8 13x 39 x3 ...(iii) Solution contd... From (ii), 2x 1 x 1 3x 1 12 3 4 2x 1 4 x 1 12 3x 1 4 2x 3 3 3x 1 2x 3 3x 1 12 4 2x 9x 3 3 11x 0 x>0 ...(iv) Combining the solution (ii) and (iii) on the number line – 0 3 Solution set 3, Class Exercise - 4 Solve the following system of inequations x 1 5, x 2. Solution The given system of inequation is x 1 5 ...(i) x 2 ..(ii) From (i), x 1 5 5 x 1 5 x a a x a 5 1 x 1 1 5 1 4 x6 Solution of inequation (i) is [–4, 6]. ...(iii) Solution contd... From (ii), x 2 x2 or x , 2 x 2 2, ...(iv) Combining the solution (iii) and (iv) on the number line, we get – – 4– 2 2 4 6 The combined solution is 4, 2 2, 6 Class Exercise - 5 Solve x 1 x2 1. Solution We have x 1 x2 x 1 x 2 x2 1 0. 0 ...(ii) Now the following cases arise: Case I: When x 1 0, i. e. x 1 . In this case, we have |x – 1| = x – 1. From (i), x 1 x 2 x2 x 1 x 2 0 x2 0 Solution contd.. 3 0 x2 x20 a 0 and a 0 b 0 b x 2 But in this case x 1. Solution is 1, . ...(i) Case II: When x – 1 < 0, i.e. x < 1. In this case |x – 1| = – (x – 1). Solution contd.. Inequation (i) can be written as x 1 x 2 x2 2x 1 0 x2 0 2x 1 0 x2 2x 1 0 x2 Equating numerator and denominator to zero, we get x 1 and –2 respectively. 2 Solution contd.. Therefore, + – – 2 + 1 – – 2 Solution is , 2 1 , . 2 But x < 1, therefore 1 x , 2 , 1 2 ...(ii) Combining the solution (i) and (ii) on the number line, we get – – 2 01 1 – – 2 The solution is , 2 1 2 , . Class Exercise - 6 In the first four papers each of 100 marks, Mukesh got 83, 73, 72, 95 marks. If he wants an average of greater than or equal to 75 marks and less than 80 marks, find the range of marks he should score in the fifth paper. Solution Let Mukesh scores x marks in the fifth paper. According to the given condition, 75 83 73 72 95 x 80 5 75 323 x 80 5 375 323 x 400 52 x 77 Hence, Mukesh must score between 52 and 77 marks. Class Exercise - 7 Draw the diagram of the solution set of linear inequations, x y 1, x 2y 8, 2x y 2, x 0, y 0. Solution Here we have to draw the graph of the corresponding equations, i.e. x – y = 1, x + 2y = 8, 2x + y = 2. We get Putting (0, 0) into each inequation, we get the shaded region as the required solution. Y 4 – – – – – – 3 2 – – – – – – 1 X´ –5 –4 –3 –2 –1 O –1 –4 Y´ 3 =2 – 1 –3 2 +y x y= 1 2x –2 X 4 5 6 7 8 x +9 2y = 8 Class Exercise - 8 If x, y and z are three positive numbers, prove that (x + y + z) 1 1 1 9. x y z Solution x, y and z are positive quantities. AM of x, y and z > GM of x, y, z. 1 xyz xyz 3 3 x y z 3 xyz Again AM of 1 3 ...(i) 1 1 1 1 1 1 , , GM of , , x y z x y z 1 1 1 1 x y z 1 1 1 3 . . 3 x y z Solution contd.. 1 3 1 1 1 1 3 x y z xyz ...(ii) Multiplying of corresponding sides of (i) and (ii), we get 1 1 1 1 x y z 9 xyz 3 x y z 1 1 1 x y z 9 x y z 1 xyz 1 3 Proved. Class Exercise - 9 If a, b, c and d be four distinct positive quantities in HP, then show that (i) a + d > b + c (ii) ad > bc Solution (i) a, b, c and d are in HP. For first three terms AM > HM a, b and c are in HP ac b 2 b is HM of a and c a c 2b ...(i) and for last three terms AM > HM Again b, c and d are in HP bd c 2 c is HM of b and d b d 2c ...(ii) Solution contd.. From (i) and (ii), a + c + b + d > 2b + 2c adbc Proved (ii) For first three terms, GM > HM ac b ac b2 ...(iii) and for the last three terms bd c 2 ...(iv) From (iii) and (iv) ac bd b2c 2 bd c ad bc Pr oved. Class Exercise - 10 If a > 0, b > 0, c > 0, prove that 1 1 1 9 , where s a s b s c 2s s = a + b + c. Solution As we know that for three positive quantities, x, y and z, we have xyz 3 and A > H. A , H 1 1 1 3 x y z Here (s – a), (s – b), (s – c) are positive quantities. s a s b s c 3 3 1 1 1 sa sb sc Solution contd.. 3s a b c 2s 3 3 3 1 1 1 sa sb sc 3 1 1 1 sa sb sc 1 1 1 9 s a s b s c 2s s a b c Pr oved. Thank you