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Find the range of values of x which satisfies the inequality 12x2 < x + 1....


Question

Find the range of values of x which satisfies the inequality 12x2 < x + 1.

Options

A) -\(\frac{1}{4}\) < x < \(\frac{1}{3}\)

B) \(\frac{1}{4}\) < x < -\(\frac{1}{3}\)

C) -\(\frac{1}{3}\) < x < \(\frac{1}{4}\)

D) -\(\frac{1}{4}\) < x < - \(\frac{1}{3}\)

The correct answer is 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 < - \(\frac{1}{3}\)


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Dicussion (1)

  • 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 < - \(\frac{1}{3}\)

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