Question:

Let $x$ be a real number such that $\dfrac{x-3}{x-2} \geq 1$. Then the solution set of the inequality is:

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Always simplify rational inequalities before applying sign analysis.
Updated On: Apr 24, 2026
  • $(-\infty,3)$
  • $(-\infty,2)$
  • $[0,\infty)$
  • $(-9,\infty)$
  • $(0,8)$
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The Correct Option is B

Solution and Explanation

Concept:
• Solve rational inequalities by bringing all terms to one side

Step 1:
Simplify inequality
\[ \frac{x-3}{x-2} - 1 \geq 0 \] \[ \frac{x-3 - (x-2)}{x-2} \geq 0 = \frac{-1}{x-2} \geq 0 \]

Step 2:
Solve inequality
\[ \frac{-1}{x-2} \geq 0 \Rightarrow x-2<0 \Rightarrow x<2 \]

Step 3:
Check restriction
\[ x \neq 2 \] Final Conclusion:
\[ (-\infty,2) \]
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