# Topic 1.1: Modulus Functions

Information about Topic 1.1: Modulus Functions

Published on January 5, 2009

Author: geakuan

Source: authorstream.com

Modulus functions : Modulus functions |4| = |-4| = |a| = b => |x - a| = b => |a| = |b| => |x - a| = |b| => 4 4 a = b (x – a) = b Modulus inequalities : Modulus inequalities |x| < 1 => |x| > 1 => |a| = b => + a = b -1 < x < 1 x < -1 or x > 1 Quadratic inequalities : Quadratic inequalities Quadratic inequality (x - a)(x – b) < 0 => where a is smaller than b (x – a)(x – b) > 0 = > a < x < b, x > b or x < a. < > Solve |x – 3| < 7 : Solve |x – 3| < 7 +(x – 3) < 7 (or) –(x – 3) < 7 x < 10 -4 10 –x + 3 < 7 -x < 4 x > -4 Answer is -4 < x < 10. Solve |2x – 1| > 7 : Solve |2x – 1| > 7 (or) +(2x – 1) > 7 2x > 7 + 1 2x > 8 x > 4 -(2x – 1) > 7 -2x + 1 > 7 -2x > 7 - 1 -2x > 6 x < -3 -3 4 Answer is x < -3 (or) x > 4 Solve 3x + 1 > |x – 5| : Solve 3x + 1 > |x – 5| (or) 3x + 1 > - (x – 5) 3x + 1 > - x + 5 4x > 4 x > 1 3x + 1 > +(x – 5) 3x + 1 > +x – 5 2x > - 6 x > - 3 0 2 CHECK FOR x = 0 3x + 1 > |x – 5| 3(0) + 1 > |0 – 5| 1 > 5 (FALSE) Since it’s wrong, therefore Reject x > -3 Therefore answer is x > 1 Solve |x – 3| > 2x + 1 : Solve |x – 3| > 2x + 1 +(x – 3) > 2x + 1 + x – 3 > 2x + 1 -x > 4 x < -4 (or) -(x – 3) > 2x + 1 - x + 3 > 2x + 1 -3x > - 2 x < -4 2 3 0 -5 CHECK FOR x = 0 |x – 3| > 2x + 1 |0 – 3| > 2(0) + 1 3 > 1 (TRUE) Since it’s correct, therefore x < 2 3 Solve |2x - 3| < |x + 1| : Solve |2x - 3| < |x + 1| |a| = |b| => < Solve |2x - 3| > |x + 1| : Solve |2x - 3| > |x + 1| |a| = |b| => >