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The set of values of \(x\) which satisfy the inequality \(0.7^{(2x^2 - 3x + 4)} < 0.343\) is

Question and options

The set of values of xx which satisfy the inequality 0.7(2x2−3x+4)<0.3430.7^{(2x^2 - 3x + 4)} < 0.343 is
  1. (12,1)\left( \frac{1}{2}, 1 \right)
  2. (12,∞)\left( \frac{1}{2}, \infty \right)
  3. (−∞,12)\left( -\infty, \frac{1}{2} \right)
  4. (−∞,12)∪(1,∞)\left( -\infty, \frac{1}{2} \right) \cup \left( 1, \infty \right)

Answer and explanation

Correct answer: D - \(\left( -\infty, \frac{1}{2} \right) \cup \left( 1, \infty \right)\)
0.343 = 0.7^3 and base < 1 reverses the inequality: 2x^2-3x+4 > 3, i.e. (2x-1)(x-1) > 0, so x < 1/2 or x > 1.

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