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question:There exists a constant k so that the minimum value of [4x^2 - 6kxy + (3k^2 + 2) y^2 - 4x - 4y + 6]over all real numbers x and y is 0. Find k.
answer:We can write the expression as begin{align*} 4x^2 - 6kxy + (3k^2 + 2) y^2 - 4x - 4y + 6 &= x^2 - 4x + 4 + 2y^2 - 4y + 2 + 3x^2 - 6kxy + 3k^2 y^2 &= (x^2 - 4x + 4) + 2(y^2 - 2y + 1) + 3(x^2 - 2kxy + k^2 y^2) &= (x - 2)^2 + 2(y - 1)^2 + 3(x - ky)^2. end{align*}The only way that this expression can take on the value of 0 is if x = 2, y = 1, and x = ky. Thus, k = boxed{2}.
question:Determine the number of pairs (a,b) of real numbers such that 10, a, b, ab is an arithmetic progression.
answer:Since 10, a, b is an arithmetic progression, we have a = frac12 (10+b). Also, we have a+ab = 2b, and so a(1+b) = 2b. Substituting the expression for a gives (10+b)(1+b) = 4b. Solving this quadratic equation gives the solutions b = -2 and b = -5. The corresponding values for a can be found by a = frac12 (10+b), giving solutions (4,-2) left(frac{5}{2},-5 right), for a total of boxed{2} solutions.
question:Functions that aren't invertible can be made invertible by restricting their domains. For example, the function x^2 is invertible if we restrict x to the interval [0,infty), or to any subset of that interval. In that case, the inverse function is sqrt x. (We could also restrict x^2 to the domain (-infty,0], in which case the inverse function would be -sqrt{x}.) Similarly, by restricting the domain of the function f(x) = 2x^2-4x-5 to an interval, we can make it invertible. What is the largest such interval that includes the point x=0?
answer:Completing the square, we have f(x)=2(x-1)^2-7. The graph of this function is a parabola with its vertex at x=1. To the left of that point, f(x) is decreasing; to the right, it's increasing. Thus, by restricting the domain to either (-infty,1] or [1,infty), we make f invertible. The choice that includes x=0 is boxed{(-infty,1]}.
question:The sum of the first 2011 terms of a geometric sequence is 200. The sum of the first 4022 terms is 380. Find the sum of the first 6033 terms.
answer:Let the first term be a, and let the common ratio be r. Then [a + ar + ar^2 + dots + ar^{2010} = 200]and [a + ar + ar^2 + dots + ar^{4021} = 380.]Subtracting these equations, we get [ar^{2011} + ar^{2012} + dots + ar^{4021} = 180.]Then [r^{2011} (a + ar + dots + ar^{2010}) = 180,]so [r^{2011} = frac{180}{200} = frac{9}{10}.]Then the sum of the first 6033 terms is begin{align*} a + ar + ar^2 + dots + ar^{6032} &= (a + ar + ar^2 + dots + ar^{4021}) + (ar^{4022} + ar^{4023} + dots + ar^{6032}) &= 380 + r^{4022} (a + ar + dots + ar^{2010}) &= 380 + left( frac{9}{10} right)^2 cdot 200 &= boxed{542}. end{align*}