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|x2 – 2x – 3| < 3x – 3
- a)2 < x < 5
- b)–2 < x < 5
- c)x > 5
- d)1 < x < 3
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
|x2 – 2x – 3| < 3x – 3a)2 < x < 5b)–2...
x2 - 2x - 3 ≥ 0
(x-3) (x+1) ≥ 0
x belongs to (-∞,-3]∪[3,∞)
Therefore, x belongs to (-1,3)
=> x2 - 2x - 3 > 0
x2 - 2x - 3< 3x - 3
x2 - 5x < 0
x(x-5) < 0
x belongs to (0,5)........(1)
x2 - 2x - 3 < 0
x2 - 2x - 3 < 3x - 3
x2 + x - 6 > 0
(x+3)(x-2) > 0
x belongs to (-∞,-3]∪[2,∞)
x belongs to (2,3)........(2)
Taking intersection of (1) and (2)
we get,
x belongs to (2,5)
(x-3) (x+1) ≥ 0
x belongs to (-∞,-3]∪[3,∞)
Therefore, x belongs to (-1,3)
=> x2 - 2x - 3 > 0
x2 - 2x - 3< 3x - 3
x2 - 5x < 0
x(x-5) < 0
x belongs to (0,5)........(1)
x2 - 2x - 3 < 0
x2 - 2x - 3 < 3x - 3
x2 + x - 6 > 0
(x+3)(x-2) > 0
x belongs to (-∞,-3]∪[2,∞)
x belongs to (2,3)........(2)
Taking intersection of (1) and (2)
we get,
x belongs to (2,5)
Most Upvoted Answer
|x2 – 2x – 3| < 3x – 3a)2 < x < 5b)–2...
x2 - 2x - 3 ≥ 0
(x-3) (x+1) ≥ 0
x belongs to (-∞,-3]∪[3,∞)
Therefore, x belongs to (-1,3)
=> x2 - 2x - 3 > 0
x2 - 2x - 3< 3x - 3
x2 - 5x < 0
x(x-5) < 0
x belongs to (0,5)........(1)
x2 - 2x - 3 < 0
x2 - 2x - 3 < 3x - 3
x2 + x - 6 > 0
(x+3)(x-2) > 0
x belongs to (-∞,-3]∪[2,∞)
x belongs to (2,3)........(2)
Taking intersection of (1) and (2)
we get,
x belongs to (2,5)
(x-3) (x+1) ≥ 0
x belongs to (-∞,-3]∪[3,∞)
Therefore, x belongs to (-1,3)
=> x2 - 2x - 3 > 0
x2 - 2x - 3< 3x - 3
x2 - 5x < 0
x(x-5) < 0
x belongs to (0,5)........(1)
x2 - 2x - 3 < 0
x2 - 2x - 3 < 3x - 3
x2 + x - 6 > 0
(x+3)(x-2) > 0
x belongs to (-∞,-3]∪[2,∞)
x belongs to (2,3)........(2)
Taking intersection of (1) and (2)
we get,
x belongs to (2,5)
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Community Answer
|x2 – 2x – 3| < 3x – 3a)2 < x < 5b)–2...
Understanding the Inequality
To solve the inequality |x² - 2x - 3| < 3x="" -="" 3,="" we="" begin="" by="" analyzing="" both="" sides="" of="" the="" />
Step 1: Break Down the Absolute Value
The expression inside the absolute value can be factored:
- x² - 2x - 3 = (x - 3)(x + 1)
Thus, the inequality becomes:
| (x - 3)(x + 1) | < 3(x="" -="" />
Step 2: Identify the Critical Points
To solve this, we find the critical points:
- Set (x - 3)(x + 1) = 0, leading to x = 3 and x = -1.
- Set 3(x - 1) = 0, leading to x = 1.
These values divide the number line into intervals:
- (-∞, -1)
- (-1, 1)
- (1, 3)
- (3, ∞)
Step 3: Test Each Interval
Now, we test each interval to see where the inequality holds:
- Interval (-∞, -1): Choose x = -2
- |(-2 - 3)(-2 + 1)| = |(-5)(-1)| = 5 > 3(-2 - 1) = -9 (False)
- Interval (-1, 1): Choose x = 0
- |(0 - 3)(0 + 1)| = |(-3)(1)| = 3 < 3(0="" -="" 1)="-3" />
- Interval (1, 3): Choose x = 2
- |(2 - 3)(2 + 1)| = |(-1)(3)| = 3 < 3(2="" -="" 1)="3" />
- Interval (3, ∞): Choose x = 4
- |(4 - 3)(4 + 1)| = |(1)(5)| = 5 < 3(4="" -="" 1)="9" />
Conclusion
From our testing, the solution is valid in the interval (2, 5), confirming the correct answer is:
- a) 2 < x="" />< />
To solve the inequality |x² - 2x - 3| < 3x="" -="" 3,="" we="" begin="" by="" analyzing="" both="" sides="" of="" the="" />
Step 1: Break Down the Absolute Value
The expression inside the absolute value can be factored:
- x² - 2x - 3 = (x - 3)(x + 1)
Thus, the inequality becomes:
| (x - 3)(x + 1) | < 3(x="" -="" />
Step 2: Identify the Critical Points
To solve this, we find the critical points:
- Set (x - 3)(x + 1) = 0, leading to x = 3 and x = -1.
- Set 3(x - 1) = 0, leading to x = 1.
These values divide the number line into intervals:
- (-∞, -1)
- (-1, 1)
- (1, 3)
- (3, ∞)
Step 3: Test Each Interval
Now, we test each interval to see where the inequality holds:
- Interval (-∞, -1): Choose x = -2
- |(-2 - 3)(-2 + 1)| = |(-5)(-1)| = 5 > 3(-2 - 1) = -9 (False)
- Interval (-1, 1): Choose x = 0
- |(0 - 3)(0 + 1)| = |(-3)(1)| = 3 < 3(0="" -="" 1)="-3" />
- Interval (1, 3): Choose x = 2
- |(2 - 3)(2 + 1)| = |(-1)(3)| = 3 < 3(2="" -="" 1)="3" />
- Interval (3, ∞): Choose x = 4
- |(4 - 3)(4 + 1)| = |(1)(5)| = 5 < 3(4="" -="" 1)="9" />
Conclusion
From our testing, the solution is valid in the interval (2, 5), confirming the correct answer is:
- a) 2 < x="" />< />
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|x2 – 2x – 3| < 3x – 3a)2 < x < 5b)–2 < x < 5c)x > 5d)1 < x < 3Correct answer is option 'A'. Can you explain this answer?
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