Part I Develop your Math Techniques for IIT JEE with Dr. Vladimir LuzginA strong and solid preparation for math IIT JEE assumes that an aspirant not only studies the solutions of the previous years’ exams, which is, of course, very important, but also reviews the basic material, notions, and methods covered in the school math curriculum. But that’s not all. A successful aspirant must considerably improve his / her problem solving capabilities and be equipped with various mathematical techniques which could help crack any problem during future exams.This first part of my course is intended to equip aspirants with various and powerful mathematical weapons which will help them win the battle with math problems during exams. Follow my classes, study the material and methods discussed, complete my tests and tutorials, and you will succeed! Prepare for the IIT JEE math exams with Vlad Luzgin Sir! ABSOLUTE VALUE EQUATIONS – 1Problems involving absolute values are pretty common for IIT JEE. IIT aspirants must be familiar with the notion of absolute value and be fluent in applying various techniques while solving these types of problems. Although almost all the problems involving absolute values could be solved using the interval method, in many cases it is time consuming. Take advantage of various approaches that allow you to solve these problems effectively in a short period of time.For any real number a the absolute value or modulus of a is denoted by | a | and is defined as a, if a ≥ 0, | a | = -a, if a < 0.Example. | - 2 | = - ( - 2 ) = 2.The absolute value has the following fundamental properties: | a | ≥ 0. | a | = 0⇔ a = 0. | - a | = | a |. | a b | = | a| | b | | a / b | = | a | / | b |, b ≠ 0. = | a |.Example. = | 3 | = 3.Misconceptions: = ±, = . ↖ Wrong !!! ↗Problem 1. The sum of the roots of the equation x2 + | x – 2 | = 10 rounded to the nearest tenth is– 6.4 B) – 3.4 C) – 0.6 D) 0 E) 0.6 F) 4.0Solution. 1 interval 2 interval x ≤ 2 2 x > 2 x Interval x ≤ 2 x > 2 | x – 2 | - x + 2 x - 2 Equation x2 - x + 2 = 10, x2 - x - 8 = 0, D = (-1) 2 – 4 (1) (- 8) = 33 > 0, x1 = (1 - ) / 2 ≈ - 2.4 < 2, x2 = (1 + ) / 2 > 2. x2 + x – 2 = 10, x2 + x – 12 = 0, x3 = - 4 < 2, x4 = 3 > 2.The sum of the roots is ≈ - 2.4 + 3 = 0.6.Answer: E. Problem 2. The number of the roots of the equation | 5x - 10 | + | x | = | x - 3 | + 3 is0 B) 1 C) 2 D) 3 E) 4 F) 5Solution. I int. II int. III int. IV int. 0 2 3 x I II III IV Interval x < 0 0 ≤ x ≤ 2 2 < x < 3 x ≥ 3 | 5x - 10 | - 5x + 10 - 5x + 10 5x - 10 5x - 10 | x | - x x x x | x - 3 | - x + 3 - x + 3 - x + 3 x - 3Case 1: x < 0. Equation: - 5x + 10 – x = - x + 3 + 3, - 5x = - 4, x = 0.8 > 0 ⇒ no solutions.Case 2: 0 ≤ x ≤ 2. Equation: - 5x + 10 + x = - x + 3 + 3, - 3x = - 4, x = 1 1/3 ∈ [0, 2].Case 3: 2 < x < 3. Equation: 5x – 10 + x = - x + 3 + 3, 7x = 16, x = 2 2/7 ∈ (2, 3).Case 4: x ≥ 3. Equation: 5x – 10 + x = x – 3 + 3, x = 2 < 3 ⇒ no solutions.Equation has 2 solutions: 1 1/3 and 2 2/7.Answer: C. Scheme for the equation |f (x)| = c, c > 0. | f (x) | = c, c > 0, ⇔ f (x) = c or f (x) = - c.Example. Solve: | 2x - 1 | = 3.Solution.| 2x – 1 | = 3⇔ 2x – 1 = - 3 or 2x – 1 = 3; 2x = - 2 or 2x = 4; x = - 1 or x = 2.Answer: { - 1, 2 }.Problem 3. The sum of the roots of the equation | x2 – 4x | = 3 is 1 B) 3 C) 4 D) 5 E) 6 F) 8Solution.| x2 – 4x | = 3 ⇔ x2 – 4x = - 3 or x2 – 4x = 3. x2 – 4x + 3 = 0, x2 – 4x – 3 = 0, D = 4 > 0, D = 28 > 0, x1 + x2 = 4, x3 + x4 = 4,The sum of the roots of the equation equals 4 + 4 = 8. Answer: F.Problem 4. The sum of the roots of the equation ||x + 2| – 3| = 1 is - 12 B) - 10 C) - 8 D) - 6 E) - 2 F) 0Solution.||x + 2| – 3| = 1 ⇔ |x + 2| – 3 = - 1 or|x + 2| – 3 = 1, |x + 2| = 2, |x + 2|= 4 x + 2 = ± 2, x + 2 = ± 4, x1 = - 4, x2 = 0, x3 = - 6, x4 = 2.The sum of the roots of the equation equals - 4 + 0 + (- 6) + 2 = - 8. Answer: C. Problem 5. The sum of the roots of the equation |x2 – 1| + | x + 1 | = 0 is - 2 B) - 1 C) 0 D) 1 E) 2 F) 3 Solution. |x2 – 1| = 0, x = ± 1,|x2 – 1| + | x + 1 | = 0 ⇔ ⇔ ⇔ x = - 1. | x + 1 | = 0, x = - 1,or|x2 – 1| + | x + 1 | = 0 ⇔ | x + 1 || x – 1 | + | x + 1 | = 0, ⇔ | x + 1| (| x – 1 | + 1) = 0.Since | x – 1 | + 1 > 0, we get: | x + 1| = 0, x = - 1.Answer: B.Problem 6. The sum of the roots of the equation |x2 + x – 2| + | x – 1 | = 2 | x + 2 | + 2 is - 6 B) - 4 C) - 3 D) 2 E) 3 F) 6 Solution.|x2 + x – 2| + | x – 1 | = 2 | x + 2 | + 2,| x + 2 || x – 1 | + | x – 1 | – 2 | x + 2 | – 2 = 0,| x – 1| (| x + 2 | + 1) – 2 (| x + 2 | + 1) = 0.(| x + 2 | + 1) (| x – 1 | – 2) = 0,Since | x + 2 | + 1 > 0, we get: | x – 1| = 2, x – 1 = - 2 or x – 1 = 2,x1 = - 1, x2 = 3.The sum of the roots is - 1 + 3 = 2.Answer: D. Scheme for the equation |f (x)| = |g (x)|. |f (x)| = |g (x)| ⇔ f (x) = g (x) or f (x) = - g(x).Problem 7. The product of the roots of the equation | x2 – 5x + 3 | = | x – 5 | is – 16 B) - 8 C) – 2 D) 2 E) 4 F) 8Solution. x2 – 5x + 3 = x – 5 or x2 – 5x + 3 = - x + 5. x2 – 6x + 8 = 0, x2 – 4x – 2 = 0, D = 4 > 0, D = 24 > 0, x1 x2 = 8, x3 x4 = - 2.The product of the roots of the equation equals - 16. Answer: A. Scheme 1 for the equation |f (x)| = g (x). f (x) ≥ 0, f (x) < 0, |f (x)| = g (x) ⇔ or f (x) = g (x), - f (x) = g (x). Example. Solve the equation | 4x – 3 | = 2x2 – 1.Case 1: 4x – 3 < 0 ⇔ x < 0.75. Equation: - 4x + 3 = 2x2 – 1 , 2x2 + 4x – 4 = 0, x2 + 2x – 2 = 0, x1 = - 1 – < 0.75, x2 = - 1 + ≈ 0.73 < 0.75, 2 solutions.Case 2: 4x – 3 ≥ 0 ⇔ x ≥ 0.75. Equation: 4x – 3 = 2x2 – 1, 2x2 – 4x + 2 = 0, x2 – 2x + 1 = 0, ( x – 1 )2 = 0, x3 = 1 > 0.75.Answer: x = 1, - 1 ±. Scheme 2 for the equation |f (x)| = g (x). g (x) ≥ 0,|f (x)| = g (x) ⇔ f (x) = g (x) or f (x) = - g(x). Problem 8. The sum of the roots of the equation | 2x2 – 1 | = 4x – 3 is – 1.5 B) - 1 C) – 0.5 D) 0 E) 1 F) none of theseSolution. 4x – 3 ≥ 0 ⇒ x ≥ 0.75.2x2 – 1 = 4x - 3 or 2x2 – 1 = - 4x + 3. 2x2 – 4x + 2 = 0, 2x2 + 4x – 4 = 0, x2 – 2x + 1 = 0, x2 + 2x – 2 = 0, ( x – 1 )2 = 0, x2 = - 1 – < 0.75, x1 = 1 > 0.75. x3 = - 1 + ≈ 0.73 < 0.75, ⇒ no solutions.The sum of the roots of the equation equals 1. Answer: E.Problem 9. The sum of the roots of the equation || x – 3| – x | = 2x is 1 B) 0.75 C) 0.5 D) - 0.75 E) - 1 F) - 1.5Solution. x ≥ 0,|| x – 3| – x | = 2x ⇔ | x – 3| – x = - 2x or|x – 3| – x = 2x. Case 1: | x – 3| – x = - 2x , | x – 3| = - x, x ∈ φ (no solutions), since x ≥ 0. Case 2: |x – 3| – x = 2x, | x – 3| = 3x, x – 3 = - 3x, or x – 3 = 3x, 4x = 3, 2x = - 3, x = 0.75 > 0; x = - 1.5 < 0.Answer. B. The Triangle Inequality: | a + b | ≤ | a | + | b |, a, b ∈ R. | a + b | = | a | + | b | ⇔ a b ≥ 0.Example. | ( - 3 ) + ( - 2) | = | ( - 3 ) | + | ( - 2 ) |, ( - 3 ) x ( - 2 ) > 0, | ( - 3 ) + 2| < | ( - 3 ) | + | 2 |, ( - 3 ) x ( 2 ) < 0. Geometrical Interpretation of | a – b |:| a – b | represents the distance between the points a and b on the number line. a < b b < a a b b a | a – b| = b – a | a – b | = a – bExample. Solve the equation: | x + 1 | + | x - 2 | = 5.Solution. | x - (- 1 )| + | x - 2 | = 5. x = -2 -1 2 3 = x The sum of the distances from a point x to the points -1 and 2 equals 5. Only points x = -2 and x = 3 satisfy this condition.Answer: { - 2, 3 }.Problem 10. The equation | x – 1| + | x + 1| = 2 has 0 roots B) 1 root C) 2 roots D) 3 roots E) 4 roots F) infinitely many solutionsSolution.| - x + 1| + | x + 1| = | 2 |, |a| + |b| = |a + b| ⇔ a b ≥ 0. a = - x + 1, b = x + 1, a + b = (- x + 1) + (x + 1) = 2.Therefore, | - x + 1| + | x + 1| = 2 ⇔ (- x + 1) (x + 1) ≥ 0 ⇔ - 1 ≤ x ≤ 1. y = (- x + 1) (x + 1) – +– - 1 1 x Geometric interpretation The sum of the distances from a point x to the points -1 and 1 equals 2 if and only if x lies between -1 and 1: - 1 ≤ x ≤ 1. | x + 1| | x – 1| -1 x 1 2Answer. F.