Find the range of values of x which satisfies the inequality 12x2 < x + 1
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Correct Answer: Option A
Explanation:
12x2 < x + 1
12 - x - 1 < 0
12x2 - 4x + 3x - 1 < 0
4x(3x - 1) + (3x - 1) < 0
Case 1 (+, -)
4x + 1 > 0, 3x - 1 < 0
x > -\(\frac{1}{4}\)x < \(\frac{1}{3}\)
Case 2 (-, +) 4x + 1 < 0, 3x - 1 > 0
-\(\frac{1}{4}\) < x < - \(\frac{1}{3}\)
12x2 < x + 1
12 - x - 1 < 0
12x2 - 4x + 3x - 1 < 0
4x(3x - 1) + (3x - 1) < 0
Case 1 (+, -)
4x + 1 > 0, 3x - 1 < 0
x > -\(\frac{1}{4}\)x < \(\frac{1}{3}\)
Case 2 (-, +) 4x + 1 < 0, 3x - 1 > 0
-\(\frac{1}{4}\) < x < - \(\frac{1}{3}\)