∣x∣={x,x≥0−x,x<0,x∈R
If a∈(0,∞), then
I. ∣x∣>a⇔x<−a, or x>a
i.e. x∈(−∞,−a)∪(a,∞)
II.
∣x∣⩾a⇔x⩽−a or x⩾a
i.e. x∈(−∞,−a]∪[a,∞)
Let r be positive real number and a be a fixed real number then
(i) $|x-a|
i. e. x∈(a−r,a+r)
(ii) ∣x−a∣⩽r⇔a−r⩽x⩽a+r i.e. x∈[a−r,a+r]
(iii) $|x-a|>r \Leftrightarrow x
(iv) ∣x−a∣⩾r⇔x⩽a−r or x⩾a+r
Let a,b∈(0,∞), then
(i) $a<|x|
(ii) a⩽∣x∣⩽b⇔x∈[−b,−a]∪[a,b]
(iii) $a<|x-c|
(vi) a⩽∣x−c∣⩽b⇔x∈[−b+c−a+c]∪[a+c,b+c]
Solve ∣3x−2∣⩽21,x∈R
Solution
∣x∣≤a
⇒−a≤x≤a
∴∣3x−2∣≤21
⇒−21≤3x−2⩽21
⇒−21+2⩽3x⩽21+2
⇒23⩽3x⩽25
⇒23×31⩽33x⩽25×31
⇒21⩽x⩽65
⇒x∈[21,65]
Solve ∣x−2∣⩾5,x∈R.
Solution
∣x−2∣⩾5
we know that
∣x∣⩾a
⇒x⩽−a or x⩾a
∴∣x−2∣⩾5
⇒(x−2)⩽−5 or (x−2)⩾5
⇒(x−2)⩽−5 or (x−2)⩾5
⇒x−2+2⩽−5+2 or x−2+2⩾5+2
⇒x⩽−3 or x⩾7
⇒x∈(−∞,−3] or x∈[7,∞)
⇒x∈(−∞,−3]∪[7,∞)
Solve the system of inequations
∣x−1∣⩽5,∣x∣⩾2
Solution
Given: ∣x−1∣⩽5 ⟶(i) &
∣x∣⩾2 ⟶(ii)
from (i), ∣x−1∣⩽5
⇒−5⩽(x−1)⩽5
∣x∣⩽a⇔−a⩽x⩽0
⇒−5+1⩽x⩽5+1 ⇒−4⩽x⩽6⟶(i)
from (ii)x∈[−4,6]
∣x∣⩾2
∣x∣⩾a⇔x⩽a or x⩾a
⇒x⩽−2 or x⩾2
⇒x∈(−∞,−2]∪[2,∞)
from (i) ∣x−1∣⩽5⇒x∈[−4,6]
from (ii) ∣x∣⩾2⇒x∈(−∞,−2]∪[2,∞)
Combining solution (i) & (ii), we get
Solution set =[−4,−2]∪[2,6]
Solve 1⩽∣x−2∣⩽3
Solution
We know that
a⩽∣x−c∣⩽b
⇔x∈[−b+c,−a+c]∪[a+c,b+c]
Given inequation
1≤∣x−2∣⩽3
⇒x∈[−3+2,−1+2]∪[1+2,3+2]
⇒x∈[−1,1]∪[3,5]
∴ solution set for the given inequation
1⩽∣x−2∣⩽3 is
x∈[−1,1]∪[3,5]
Solve ∣x∣−2∣x∣−1⩾0,x∈R, x=±2
Solution Given : ∣x∣−2∣x∣−1⩾0 Let ∣x∣=z
⇒z−2z−1⩾0
⇒z⩽1 or z>2
(i) $ \frac{x-a}{x-b} \geqslant 0, a
⇒x⩽a a x⩾b
(ii) $ \frac{x-a}{x-b} \leqslant 0, a
⇒a⩽x⩽b
⇒∣x∣⩽1 or ∣x∣>2
⇒(−1⩽x⩽1) or (x⩽−2 or x>2)
⇒x∈[−1,1] or x∈(−∞,−2)∪(2,∞)
On the basis of solution represented on number line
Solution set =[−1,1]∪(−∞,−2)∪(2,∞)
Solve: ∣x∣−2−1⩾1,x∈R, x=±2
Solution Given: ∣x∣−2−1⩾1
Let ∣x∣=z
⇒z−2−1⩾1
⇒−1⩾z−2
⇒−1(−1)⩽−1(z−2)
⇒1⩽2−z
⇒2−z≥1
⇒z−2⩽−1
⇒z−2+2⩽−1+2
⇒z⩽1
⇒∣x∣⩽1
⇒−1⩽x⩽1
∴ Solution for the given inequalities
x∈[−1,1]
Solve the inequation:
x−42>1,x=4
Solution
Given inequation
∣x−42∣>1
⇒∣x−4∣2>1
⇒2>∣x−4∣
⇒∣x−4∣<2
$ \Rightarrow 4-2
$\quad|x-a|
$ \Rightarrow 2
⇒x∈(2,6), but x=4
⇒x∈(2,4)∪(4,6)
Solution Set : (2,4)∪(4,6)
Solve: ∣x−1∣+∣x−2∣⩾4
Solution
Given inequation, ∣x−1∣+∣x−2∣⩾4
put x−1=0 & x−2=0
⇒x=1,2
We will discuss in three intervals
i.e. (−∞,1),[1,2),[2,∞)
Case I: When x∈(−∞,1)
∣x−1∣=−(x−1)
∣x−2∣=−(x−2)
∴∣x−1∣+∣x−2∣⩾4
⇒−(x−1)−(x−2)⩾4
∣x∣=x,x⩾0
=−x,x<0
⇒−2x+3⩾4
⇒−2x⩾4−3
⇒−2x⩾1
⇒2x⩽−1
⇒x⩽−21⇒x∈(−∞,−21]
⇒ but x∈(−∞,1)
⇒x∈(−∞,−21] -(i)
Case II: when x∈[1,2)
∣x−1∣=(x−1)
∣x−2∣=−(x−2)
∴∣x−1∣+∣x−2∣⩾4
⇒(x−1)−(x−2)⩾4
⇒1⩾4,
Which is absurd result
∴ No solution exist for the given inequation in [1,2).
Case-III: When x∈[2,∞)
∴∣x−1∣=(x−1)
∣x−2∣=(x−2)
∴∣x−1∣+∣x−2∣⩾4
⇒(x−1)+(x−2)⩾4
⇒2x−3⩾4
⇒2x⩾7
⇒x⩾27
⇒x∈[27,∞)
Case I: x∈(−∞,−21]
Case-II: No Solution-
Case-III: x∈[27,∞)
Combining cases I, II & III we get
x∈(−∞,−21]∪[27,∞)
∴ Solution set =(−∞,−21]∪[27,∞)