The solution set of the inequation | x − 1 | ≥ | x − 3 |, is

Here, x =1 and x = 3 are two critical points which divide the real line into three parts, namely, $x < 1, 1 ≤ x < 3$ and $x ≥ 3$. So, we discuss the following cases:

CASE I When $x <1$:

In this case, we have

$|x-1|=-(x-1)$ and $| x - 3| = -(x-3)$

$∴|x-1|≥|x-3|$

$⇒-(x-1)≥-(x-3)⇒1≥3, which is absurd.

So, the inequation has no solution for $x <1$.

CASE II When $1 ≤ x < 3$

$|x-1|=-(x-1)$ and $| x - 3| = -(x-3)$

$∴|x-1|≥|x-3|$

$x-1≥-(x-3)⇒2x-4≥0⇒x≥2$

But, $1≤x<3$. Therefore,

$1≤x<3$ and $x≥2⇒x∈[2, 3)$

CASE III When $x ≥ 3$

In this case, we have

$|x-1|=-(x-1)$ and $| x - 3| = -(x-3)$

$∴|x-1|≥|x-3|$

$⇒x-1≥x-3-1⇒1-≥-3$, which is correct.

So, the given inequation has all solutions satisfying $x ≥ 3$.

Hence, the solution set of the given inequation is $[2, ∞)$.