Commerce Exam > Commerce Questions > x - 1)(x² - 5x + 7) < (x - 1), then x...
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x - 1)(x² - 5x + 7) < (x - 1), then x belongs to
- a)(1, 2) ∪ (3, ∞)
- b)(2, 3)
- c)(-∞, 1) ∪ (2, 3)
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
x - 1)(x² - 5x + 7) < (x - 1), then x belongs toa)(1, 2) ∪...
Understanding the Inequality
To solve the inequality (x - 1)(x² - 5x + 7) < (x="" -="" 1),="" we="" start="" by="" simplifying="" both="" sides.="" />Step 1: Rearranging the Inequality
Subtract (x - 1) from both sides:
(x - 1)(x² - 5x + 7) - (x - 1) < 0="" this="" can="" be="" factored:="" (x="" -="" 1)(x²="" -="" 5x="" +="" 6)="" />< 0="" />Step 2: Factoring the Quadratic
Now, we factor the quadratic expression:
x² - 5x + 6 = (x - 2)(x - 3)
So, the inequality can be rewritten as:
(x - 1)(x - 2)(x - 3) < 0="" />Step 3: Finding Critical Points
The critical points from the factors are x = 1, x = 2, and x = 3. These points divide the number line into intervals:
1. (-∞, 1)
2. (1, 2)
3. (2, 3)
4. (3, ∞)
Step 4: Testing Intervals
We test each interval to determine where the product is negative:
- In (-∞, 1): Choose x = 0 → (0-1)(0-2)(0-3) = negative
- In (1, 2): Choose x = 1.5 → (1.5-1)(1.5-2)(1.5-3) = negative
- In (2, 3): Choose x = 2.5 → (2.5-1)(2.5-2)(2.5-3) = positive
- In (3, ∞): Choose x = 4 → (4-1)(4-2)(4-3) = positive
Step 5: Conclusion
The inequality (x - 1)(x - 2)(x - 3) < 0="" holds="" for="" the="" intervals="" (-∞,="" 1)="" and="" (2,="" 3).="" hence,="" the="" solution="" is:="" />Final Answer: x belongs to (-∞, 1) ∪ (2, 3), which corresponds to option 'C'.
To solve the inequality (x - 1)(x² - 5x + 7) < (x="" -="" 1),="" we="" start="" by="" simplifying="" both="" sides.="" />Step 1: Rearranging the Inequality
Subtract (x - 1) from both sides:
(x - 1)(x² - 5x + 7) - (x - 1) < 0="" this="" can="" be="" factored:="" (x="" -="" 1)(x²="" -="" 5x="" +="" 6)="" />< 0="" />Step 2: Factoring the Quadratic
Now, we factor the quadratic expression:
x² - 5x + 6 = (x - 2)(x - 3)
So, the inequality can be rewritten as:
(x - 1)(x - 2)(x - 3) < 0="" />Step 3: Finding Critical Points
The critical points from the factors are x = 1, x = 2, and x = 3. These points divide the number line into intervals:
1. (-∞, 1)
2. (1, 2)
3. (2, 3)
4. (3, ∞)
Step 4: Testing Intervals
We test each interval to determine where the product is negative:
- In (-∞, 1): Choose x = 0 → (0-1)(0-2)(0-3) = negative
- In (1, 2): Choose x = 1.5 → (1.5-1)(1.5-2)(1.5-3) = negative
- In (2, 3): Choose x = 2.5 → (2.5-1)(2.5-2)(2.5-3) = positive
- In (3, ∞): Choose x = 4 → (4-1)(4-2)(4-3) = positive
Step 5: Conclusion
The inequality (x - 1)(x - 2)(x - 3) < 0="" holds="" for="" the="" intervals="" (-∞,="" 1)="" and="" (2,="" 3).="" hence,="" the="" solution="" is:="" />Final Answer: x belongs to (-∞, 1) ∪ (2, 3), which corresponds to option 'C'.
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x - 1)(x² - 5x + 7) < (x - 1), then x belongs toa)(1, 2) ∪...
(x - 1)(x² - 5x + 7) < (x - 1)
⇒ (x - 1)(x² - 5x + 6) < 0
⇒ (x - 1)(x - 2)(x - 3) < 0
∴ x ∈ (-∞, 1) ∪ (2, 3)
⇒ (x - 1)(x² - 5x + 6) < 0
⇒ (x - 1)(x - 2)(x - 3) < 0
∴ x ∈ (-∞, 1) ∪ (2, 3)
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