CAT 2025 Lesson : Inequalities - Basics of Modulus
5. Inequalities with Modulus
You would have learnt in school that a modulus function (also called absolute value) provides the same number if it is positive or equals 0, and provides the positive value of a number if it is negative. Therefore, the modulus function always provides a non-negative number (0 or more).
The following is the definition of the modulus function of x.
∣x∣=x, if x≥0; and =−x, if x<0
Where k is a constant,
if ∣x∣<k, then −k<x<k
if ∣x∣>k, then x<−k or x>k
Key properties of modulus involving 2 variables are as follows.
1) ∣x+y∣≤∣x∣+∣y∣
2) ∣x+y∣≥∣x∣−∣y∣
3) ∣x−y∣≥∣x∣−∣y∣
Example 7
Where x is an integer, how many values of x satisfy ∣2x+5∣<10?
Solution
Outlined initially is the standard method that is used to solve modulus functions. The alternative method is useful when the modulus is less than a constant (which would then provide a range).
Case 1: If 2x+5≥0 (i.e. x≥−2.5),
then 2x+5<10
⇒ x<2.5
The conditions in bold above when merged provides −2.5<x<2.5. 5 integers from –2 to 2 satisfy x for this condition.
Case 2: If 2x+5<0 (i.e. x<−2.5),
then −(2x+5)<10
⇒ 2x+5>−10 ⇒ x>−7.5
The conditions in bold above when merged provides −7.5<x<−2.5. 5 integers from –7 to –3 satisfy x for this condition.
Therefore, there is a total of 5+5=10integral values that x can take.
Alternatively (Recommended)
If ∣x∣<k, then −k<x<k
∴ ∣2x+5∣<10
⇒ −10<2x+5<10
Now, start moving constants directly to the two extremes of this combined inequality.
⇒ −10−5<2x<10−5
⇒ −15<2x<5
⇒ 2−15<x<25
⇒ −7.5<x<2.5
Integers between –7 and 2 (both inclusive) are the values that x can take.
Number of integers between x and y both inclusive = y−x+1
∴ Number of integers that satisfy the inequality = 2−(−7)+1=10
Answer: 10
Example 8
What is the range of x in ∣∣x−1∣−1∣<3
Solution
∣∣x−1∣−1∣<3
⇒ −3<∣x−1∣−1<3
⇒ −2<∣x−1∣<4
As modulus of ∣x−1∣≥0, we can ignore the lower limit of –2.
⇒ ∣x−1∣<4
⇒ −4<x−1<4
⇒ −3<x<5
Answer: −3<x<5
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