Presentation Transcript
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|
|x| > 1 => |a| = b
=> + a = b -1 1
Quadratic inequalities :Quadratic inequalities Quadratic inequality
(x - a)(x – b)
where a is smaller
than b
(x – a)(x – b) > 0 = > a b or x
Solve |x – 3| < 7 :Solve |x – 3| -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 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 2x + 1
- x + 3 > 2x + 1
-3x > - 2
x 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| <
Solve |2x - 3| > |x + 1| :Solve |2x - 3| > |x + 1| |a| = |b| => >