Some answers:<p><pre><code> 1. Solve for positive x:
x (8 sqrt(1 - x) + sqrt(1 + x)) <= 11 sqrt(1 + x) - 16 sqrt(1 - x)
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The upper bound is clearly x=1 because the square roots cease to exist for x > 1 and and substituting in x = 1 - epsilon^2 gives to first order (8 e + 1) <= (11 - 16 e) which works because 8 <= 11. Does this thing have a root in [0, 1]?<p>The inverse of y = sqrt(1 + x)/sqrt(1 - x) is with some work x = (y^2 - 1)/(y^2 + 1) so when we divide through by sqrt(1-x) which we know now to be positive, and replace the above, we find:<p><pre><code> (y^2 - 1)/(y^2 + 1) * (8 + y) = 11 y - 16
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Expanding out we're looking for a root of<p><pre><code> -10 y^3 + 24*y^2 - 12 y + 8 = 0
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Dividing by -2 to normalize somewhat:<p><pre><code> 5 y^3 - 12 y^2 + 6 y - 4 = 0
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At this point I almost gave up (since it's a cubic and I had no guarantee that x was rational) but got lucky, I started trying positive integers to see where this transitions from negative (y=0) to positive (5y^3) and accidentally found that y=2 solves the equation. Sending it back through I find 3/5, so assuming that it doesn't double back somewhere in the interval [3/5, 1] that's the interval that we're looking for. Phew!<p><pre><code> 2. Find all functions F(x) : R -> R having the propery that for any x1 and x2,
F(x1) - F(x2) <= (x1 - x2)^2.
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Yeah, if you're just entering a university you're probably not going to get this one. Rewriting x1 as x + dx, x2 as x, then this says F(x + dx) - F(x) <= dx^2. With some limits and the Squeeze theorem, this restricts the functions to be differentiable with derivative zero, so they are constant functions. By inspection that is not just necessary but also sufficient.<p><pre><code> 3. Given a triangle ABC construct with a straightedge and compass a point K on
AB and a point M on BC such that AK = KM = MC.
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This seems in general impossible except for some very specific triangles -- is that true? For example if |AB| = 2 and |BC| = 1, it seems that the only point on AB which could possibly be a candidate for K is the midpoint, with M being B. But the only way that the distance from K to M is the same is if BCK is an equilateral triangle, which requires furthermore that the angle BAC is 60 degrees, no?<p><pre><code> 4. Solve 2 cubert(2y - 1) = y^3 + 1 for real y.
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So y = 1 is an obvious solution. Cubing both sides we get<p><pre><code> y^9 + 3y^6 + 3y^3 - 16 y + 9 = 0
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Then doing polynomial division by y-1 I get<p><pre><code> y^8 + y^7 + y^6 + 4 y^5 + 4 y^4 + 4 y^3 + 7 y^2 + 7 y - 9.
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At this point I got stuck and turned to automated tools, which say that this is factorizable as:<p><pre><code> (y^2 + y - 1)(y^6 + 2 y^4 + 2 y^3 + 4 y^2 + 2 y + 9)
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Graphing the right hand side it seems to be consistently positive, so that just leaves the left hand side, which is -1/2 +/- sqrt(5)/2. That's pretty difficult.