Question:
If $(x-1)(x^2 - 5x + 7) < (x-1)$, then $x$ belongs to
If $(x-1)(x^2 - 5x + 7) < (x-1)$, then $x$ belongs to
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Factorize completely before solving inequalities.
Updated On: Apr 30, 2026
- $(-\infty,-1)\cup(2,3)$
- $(-\infty,-1]\cup[2,3]$
- $(-\infty,1)\cup(2,3)$
- $(-\infty,1)\cup[2,3]$
- $(-\infty,1]\cup(2,3]$
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The Correct Option is C
Solution and Explanation
Concept:
Solve inequality.
Step 1: Bring to one side
\[ (x-1)(x^2 - 5x + 7) - (x-1) < 0 \] \[ (x-1)(x^2 - 5x + 6) < 0 \] \[ (x-1)(x-2)(x-3) < 0 \]
Step 2: Sign analysis
Negative in: \[ (-\infty,1)\cup(2,3) \] Final Conclusion:
Option (C)
Step 1: Bring to one side
\[ (x-1)(x^2 - 5x + 7) - (x-1) < 0 \] \[ (x-1)(x^2 - 5x + 6) < 0 \] \[ (x-1)(x-2)(x-3) < 0 \]
Step 2: Sign analysis
Negative in: \[ (-\infty,1)\cup(2,3) \] Final Conclusion:
Option (C)
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