CAT 2025 Lesson : Inequalities - Basics of Modulus

Where $x$ is an integer, how many values of $x$ satisfy $| 2x + 5 | \lt 10$?

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 $2x + 5 \ge 0$ (i.e. $\bm{x \ge -2.5}$),

then $2x + 5 \lt 10$

⇒ $\bm{x \lt 2.5}$

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

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

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

⇒ $2x + 5 \gt -10$ ⇒ $\bm{x \gt -7.5}$

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

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

Alternatively (Recommended)

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

∴ $| 2x + 5 | \lt 10$

⇒ $-10 \lt 2x + 5 \lt 10$

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

⇒ $-15 \lt 2x \lt 5$

⇒ $\dfrac{-15}{2} \lt x \lt \dfrac{5}{2}$

⇒ $\bm{-7.5 \lt x \lt 2.5}$

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 = $2 - (-7) + 1 = \bm{10}$

Answer: $10$

What is the range of $x$ in $||x-1|-1| \lt 3$

Solution

$||x-1|-1| \lt 3$

⇒ $-3 \lt |x-1|-1 \lt 3$

⇒ $-2 \lt |x-1| \lt 4$

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

⇒ $|x-1| \lt 4$

⇒ $-4 \lt x-1 \lt 4$

⇒ $-3 \lt x \lt 5$

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