Question

Write the solution set of the inequation |x − 1| ≥ |x − 3|.

Answer 1

\[\text{ We have }: \]
\[\left| x - 1 \right| \geq \left| x - 3 \right|\]
\[ \Rightarrow \left| x - 1 \right| - \left| x - 3 \right| \geq 0\]
\[\text{ The LHS of the inequation has two seperate modulus . Equating these to zero, we obtain x = 1, 3 as critical points } . \]
\[\text{ These points divide the real line in three regions, i . e } ( - \infty , 1], [1, 3], [3, \infty ) . \]
\[\text{ CASE 1: When } - \infty < x \leq 1, \text{ then } \left| x - 1 \right| = - (x - 1) and \left| x - 3 \right| = - (x - 3)\]
\[ \therefore \left| x - 1 \right| - \left| x - 3 \right| \geq 0\]
\[ \Rightarrow - (x - 1) - \left[ - (x - 3) \right] \geq 0\]
\[ \Rightarrow - x + 1 + x - 3 \geq 0\]
\[ \Rightarrow - 2 \geq 0\]
\[\text{ But this is not possible } . \]
\[\text{ CASE 2: When } 1 \leq x \leq 3, \text{ then } \left| x - 1 \right| = x - 1and \left| x - 3 \right| = - (x - 3)\]
\[ \therefore \left| x - 1 \right| - \left| x - 3 \right| \geq 0\]
\[ \Rightarrow x - 1 + x - 3 \geq 0\]
\[ \Rightarrow \Rightarrow 2x \geq 4\]
\[ \Rightarrow x \geq 2\]
\[ \Rightarrow x \in [2, \infty )\]
\[\text{ CASE 3: When } 3 \leq x < \infty , \text{ then } \left| x - 1 \right| = x - 1 and \left| x - 3 \right| = x - 3\]
\[ \therefore \left| x - 1 \right| - \left| x - 3 \right| \geq 0\]
\[ \Rightarrow x - 1 - x + 3 \geq 0 \]
\[ \Rightarrow 2 \geq 0\]
\[\text{ This is true } . \]
\[\text{ Hence, the solution to the given inequality comes from cases 2 and 3 } . \]
\[[2, \infty )U [3, \infty ) = [2, \infty )\]
\[ \therefore x \in [2, \infty )\]