Can anybody briefly explain this picture? Brother or Sister
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Answers
Note :
• The modules function (or absolute value function) always gives the non-negative output .
• This function is denoted by |x| .
• It can be defined in following ways :
1) f(x) = |x| = x , if x ≥ 0
= -x , if x < 0
2) f(x) = |x| = max.( x , -x )
• Dom(f) = R
• Range(f) = [0,∞)
Given properties :
1) If |x| < a , then -a < x < a .
2) If |x| ≤ a , then -a ≤ x ≤ a .
3) If |x| > a , then x < -a or x > a (a > 0)
4) If |x| ≥ a , then x ≤ -a or x ≥ a (a > 0)
Explanation :
1) Given : |x| < a
Case1 : Let x > 0 , then |x| = x
Thus , |x| < a
=> x < a
AND
Case2 : Let x < 0 , then |x| = -x
Thus , |x| < a
=> -x < a
=> x > -a
Thus , from both the cases we get ,
x < a and x > -a
=> -a < x < a
=> x € (-a,a)
2) Given : |x| ≤ a
Case1 : Let x ≥ 0 , then |x| = x
Thus , |x| ≤ a
=> x ≤ a
AND
Case2 : Let x < 0 , then |x| = -x
Thus , |x| ≤ a
=> -x ≤ a
=> x ≤ -a
Thus , from both the cases we get ,
x ≤ a and x ≤ -a
=> -a ≤ x ≤ a
=> x € [-a,a]
3) Given : |x| > a
Clearly , the solution of (3) will be the complement of the solution found in (2).
Thus , here x € R - [-a,a]
=> x € (-∞,a)U(a,∞)
4) Given : |x| ≥ a
Clearly , the solution of (4) will be the complement of the solution found in (1).
Thus , here x € R - (-a,a)
=> x € (-∞,a]U[a,∞)