If $min. (2x^{2} - ax + 2) > max. (b - 1 + 2x - x^{2})$ then roots of the equation $2x^{2} + ax + (2 - b) = 0$, are
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Positive and distinct
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Negative and distinct
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Opposite in sign
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Imaginary
Min(2x^2 - ax + 2) = 2 - a^2/8. Max(b - 1 + 2x - x^2) = b - 1 + 1 = b. Condition: 2 - a^2/8 > b, so a^2/8 + b < 2. Equation 2x^2 + ax + (2 - b) = 0. Discriminant D = a^2 - 4(2)(2 - b) = a^2 - 16 + 8b = a^2 + 8b - 16. Since a^2 + 8b < 16, D < 0. Roots are imaginary.
The minimum of the expression 2x^2 - ax + 2 is found by setting its derivative to zero, giving x = a divided by 4, and its minimum value is 2 minus a^2 divided by 8. The maximum of b - 1 + 2x - x^2 is found by setting its derivative to zero, giving x = 1, and its maximum value is b. Equating the two bounds gives 2 minus a^2 divided by 8 is greater than b, which rearranges to b minus 2 plus a^2 divided by 8 is less than 0. For the new quadratic equation 2x^2 + ax + (2 - b) = 0, the discriminant D equals a^2 - 8(2 - b), which simplifies to a^2 + 8b - 16. Since a^2 divided by 8 plus b minus 2 is less than 0, multiplying by 8 yields a^2 + 8b - 16 is less than 0. Because the discriminant is strictly negative, the roots of the equation are imaginary.