Algebra

Let f(x) = 2x^3 + 3x + k. The derivative f'(x) = 6x^2 + 3 is always positive, meaning the function is strictly increasing. A strictly increasing function can have at most one real root, so it cannot have two distinct real roots in [0, 1].

Substitute x = -3/2 into x^2 + x - m = 0. (-3/2)^2 + (-3/2) - m = 0. 9/4 - 6/4 - m = 0. 3/4 - m = 0, so m = 3/4.

Let y = x^3. The equation becomes y^2 - 7y - 8 = 0. Factoring gives (y - 8)(y + 1) = 0, so y = 8 or y = -1. This means x^3 = 8 (x = 2) or x^3 = -1 (x = -1). Both are real roots, so there are 2 real roots.

Solving the system: 1) 2x + 5y - z = 27, 2) x + y + z = 6, 3) 2y + 5z = -4. From (2), z = 6 - x - y. Substitute into (1) and (3) to solve for x and y. Testing option B: 2(5) + 5(3) - (-2) = 10 + 15 + 2 = 27 (Correct). 5 + 3 + (-2) = 6 (Correct). 2(3) + 5(-2) = 6 - 10 = -4 (Correct).

If roots are 3 + 2i and 3 - 2i, the sum of roots = 6 and product = 3^2 + 2^2 = 13. The equation is x^2 - (sum)x + (product) = 0, which is x^2 - 6x + 13 = 0.

Adding the equations: (x^2 + x + 3) + (2x^2 + 6x + 24) = 3x^2 + 7x + 27 = 0. Comparing this to ax^2 + bx + c = 0, we get a = 3 and b = 7. The product ab = 3 * 7 = 21.

If coefficients are in AP, then 2 * C(n, r) = C(n, r-1) + C(n, r+1). Expanding these binomial coefficients and simplifying leads to the quadratic equation in n: n^2 - n(4r + 1) + 4r^2 - 2 = 0.

For a polynomial with real coefficients, complex roots come in conjugate pairs. However, this equation has complex coefficients (1-i). Using the sum of roots = -b/a = 1, if one root is 1+i, the other is 1 - (1+i) = -i.

The roots of x^2 - 2mx + m^2 - 1 = 0 are (2m +/- sqrt(4m^2 - 4(m^2 - 1)))/2 = m +/- 1. Roots are m-1 and m+1. We need -2 < m-1 and m+1 < 4. This gives m > -1 and m < 3. Thus, -1 < m < 3.

A quadratic equation has imaginary roots if the discriminant D = b^2 - 4ac < 0. For 5x^2 + 2x + 4 = 0, D = 4 - 4(5)(4) = 4 - 80 = -76.

Since all terms are even powers of x (x^10, x^6, x^4, x^2) with positive coefficients, 3x^10 + 7x^6 + 5x^4 + 2x^2 + 1 is always >= 1 for any real x. Therefore, there are no real roots.

For roots to be imaginary, the discriminant D < 0. D = k^2 - 4(1)(14) = k^2 - 56. k^2 - 56 < 0 => k^2 < 56. The integers k satisfying this are -7 to 7. The greatest integer is 7.