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CAT 2025 Lesson : Inequalities - Basics of Modulus

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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
xx.

x=x| x | = \bm{x}, if x0x \ge 0; and
     
=x = \bm{-x}, if x<0x \lt 0

Where k is a constant,
if
x<k| x | \lt k, then k<x<k\bm{-k \lt x \lt k}
if
x>k| x | \gt k, then x<k\bm{x \lt -k} or x>k\bm{x \gt k}

Key properties of modulus involving 2 variables are as follows.
1)
x+yx+y| x + y | \le | x | + | y |
2)
x+yxy| x + y | \ge | x | - | y |
3)
xyxy| x - y | \ge | x | - | y |

Example 7

Where xx is an integer, how many values of xx satisfy 2x+5<10| 2x + 5 | \lt 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+502x + 5 \ge 0 (i.e. x2.5\bm{x \ge -2.5}),

then
2x+5<102x + 5 \lt 10

x<2.5\bm{x \lt 2.5}

The conditions in bold above when merged provides
2.5<x<2.5-2.5 \lt x \lt 2.5.
5 integers from –2 to 2 satisfy
xx for this condition.

Case 2: If
2x+5<02x + 5 \lt 0 (i.e. x<2.5\bm{x < -2.5}),

then
(2x+5)<10-(2x + 5) \lt 10

2x+5>102x + 5 \gt -10x>7.5\bm{x \gt -7.5}

The conditions in bold above when merged provides
7.5<x<2.5-7.5 \lt x \lt -2.5.
5 integers from –7 to –3 satisfy
xx for this condition.

Therefore, there is a total of
5+5=105 + 5 = \bm{10} integral values that xx can take.

Alternatively (Recommended)

If
x<k| x | \lt k, then k<x<k\bm{-k \lt x \lt k}

2x+5<10| 2x + 5 | \lt 10

10<2x+5<10-10 \lt 2x + 5 \lt 10

Now, start moving constants directly to the two extremes of this combined inequality.
105<2x<105-10-5 \lt 2x \lt 10-5

15<2x<5-15 \lt 2x \lt 5

152<x<52\dfrac{-15}{2} \lt x \lt \dfrac{5}{2}

7.5<x<2.5\bm{-7.5 \lt x \lt 2.5}

Integers between –
77 and 22 (both inclusive) are the values that xx can take.
Number of integers between
xx and yy both inclusive = yx+1y - x + 1
∴ Number of integers that satisfy the inequality =
2(7)+1=102 - (-7) + 1 = \bm{10}

Answer:
1010


Example 8

What is the range of xx in x11<3||x-1|-1| \lt 3

Solution

x11<3||x-1|-1| \lt 3

3<x11<3-3 \lt |x-1|-1 \lt 3

2<x1<4-2 \lt |x-1| \lt 4

As modulus of
x10|x-1| \ge 0, we can ignore the lower limit of 2–2.

x1<4|x-1| \lt 4

4<x1<4-4 \lt x-1 \lt 4

3<x<5 -3 \lt x \lt 5

Answer:
3<x<5-3 \lt x \lt 5


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