💡Okay, so I have this problem here about an odd function, f(x), which is increasing on the interval (0, +∞). They also tell me that f(-3) = 0, and I need to find the solution set for x·f(x) < 0. The options are A, B, C, and D, each giving different intervals.First, let me recall what an odd function is. An odd function satisfies the property f(-x) = -f(x) for all x in its domain. This means that if I know the behavior of the function on the positive side, I can determine it on the negative side by reflecting it over both axes.They also mention that f(x) is increasing on (0, +∞). So, for positive x values, as x increases, f(x) increases. Since it's an odd function, this implies that on the negative side, the function is also increasing. Wait, is that right? Let me think. If f(x) is increasing on the positive side, then for negative x, f(-x) = -f(x). So, if x increases, say from -4 to -3, then f(x) would go from f(-4) to f(-3), which is from -f(4) to -f(3). Since f is increasing on the positive side, f(4) > f(3), so -f(4) < -f(3). Therefore, as x increases from -4 to -3, f(x) increases from -f(4) to -f(3). So, yes, f(x) is increasing on (-∞, 0) as well.They also tell me that f(-3) = 0. Since it's an odd function, f(3) must be equal to -f(-3) = -0 = 0. So, f(3) is also zero.Now, I need to find where x·f(x) < 0. That is, the product of x and f(x) is negative. This happens when x and f(x) have opposite signs. So, either x is positive and f(x) is negative, or x is negative and f(x) is positive.Let me analyze the sign of f(x) in different intervals.First, let's consider the positive side, x > 0.Since f(x) is increasing on (0, +∞) and f(3) = 0, what does that tell me about the sign of f(x)?If f(x) is increasing and f(3) = 0, then for x < 3, f(x) must be less than f(3) = 0, because it's increasing. So, for 0 < x < 3, f(x) < 0.Similarly, for x > 3, since f(x) is increasing, f(x) > f(3) = 0. So, for x > 3, f(x) > 0.Now, on the negative side, x < 0.Since f(x) is an odd function, f(-x) = -f(x). So, for x < 0, f(x) = -f(-x). We already know that f(-x) is increasing on (0, +∞), so f(x) on the negative side is also increasing because it's just the negative of an increasing function.Given that f(-3) = 0, let's see what happens for x < -3 and -3 < x < 0.For x < -3, let's pick a point, say x = -4. Then f(-4) = -f(4). Since f(4) > f(3) = 0 (because f is increasing on the positive side), f(-4) = -f(4) < 0. So, for x < -3, f(x) < 0.For -3 < x < 0, let's pick x = -2. Then f(-2) = -f(2). Since 2 < 3, f(2) < f(3) = 0, so f(-2) = -f(2) > 0. Therefore, for -3 < x < 0, f(x) > 0.So, summarizing:- For x < -3: f(x) < 0- For -3 < x < 0: f(x) > 0- For 0 < x < 3: f(x) < 0- For x > 3: f(x) > 0Now, let's find where x·f(x) < 0.This occurs when x and f(x) have opposite signs.Case 1: x > 0 and f(x) < 0From above, this happens when 0 < x < 3.Case 2: x < 0 and f(x) > 0From above, this happens when -3 < x < 0.Therefore, the solution set is the union of these two intervals: (-3, 0) and (0, 3).Looking back at the options:A: {x | -3 < x < 0 or x > 3}B: {x | x < -3 or 0 < x < 3}C: {x | x < -3 or x > 3}D: {x | -3 < x < 0 or 0 < x < 3}So, the correct answer should be D.Wait, let me double-check.When x is between -3 and 0, f(x) is positive, so x is negative and f(x) is positive, so x·f(x) is negative.When x is between 0 and 3, f(x) is negative, so x is positive and f(x) is negative, so x·f(x) is negative.Outside of these intervals, either both x and f(x) are positive (x > 3) or both are negative (x < -3), so their product is positive.Therefore, yes, D is the correct solution set.**Final Answer**boxed{D}

💡Okay, so I have this problem about translating a complex number in the plane. Let me try to understand what it's asking. It says that a translation takes the complex number 1 - 3i to 4 + 2i. I need to find where this translation takes another complex number, which is 6 - 4i. Hmm, okay.First, I remember that in the complex plane, a translation is like sliding everything by a fixed amount. So, if I have a point z, translating it by some fixed complex number w would result in z + w. So, the translation function is f(z) = z + w, where w is the translation vector.Given that, the problem tells me that when I apply this translation to 1 - 3i, I get 4 + 2i. So, mathematically, that should be:4 + 2i = (1 - 3i) + wSo, I can solve for w here. Subtracting (1 - 3i) from both sides should give me w. Let me do that step by step.Starting with:4 + 2i = (1 - 3i) + wSubtract (1 - 3i) from both sides:(4 + 2i) - (1 - 3i) = wNow, let's compute that. Subtracting complex numbers is done by subtracting their real parts and their imaginary parts separately.So, real parts: 4 - 1 = 3Imaginary parts: 2i - (-3i) = 2i + 3i = 5iSo, putting it together, w = 3 + 5i.Okay, so the translation is adding 3 + 5i to any complex number. So, now I need to apply this translation to 6 - 4i.So, using the same translation function f(z) = z + w, where w is 3 + 5i, I can compute:f(6 - 4i) = (6 - 4i) + (3 + 5i)Again, let's add the real parts and the imaginary parts separately.Real parts: 6 + 3 = 9Imaginary parts: -4i + 5i = 1iSo, combining them, the result is 9 + i.Wait, let me double-check my calculations to make sure I didn't make a mistake.Starting with finding w:4 + 2i - (1 - 3i) = 4 + 2i - 1 + 3i = (4 - 1) + (2i + 3i) = 3 + 5i. That seems correct.Then, adding that to 6 - 4i:6 - 4i + 3 + 5i = (6 + 3) + (-4i + 5i) = 9 + i. Yep, that looks right.So, the translation takes 6 - 4i to 9 + i.I think that's the answer. Let me just visualize it to make sure it makes sense. If I move the point 1 - 3i by 3 units to the right (since the real part of w is 3) and 5 units up (since the imaginary part is 5i), I land at 4 + 2i, which matches the given information. So, applying the same movement to 6 - 4i, moving 3 right and 5 up, should take it from (6, -4) to (9, 1), which is 9 + i. That seems consistent.I don't think I made any mistakes here. The steps are straightforward: find the translation vector by subtracting the original point from its image, then apply that vector to the new point. Everything checks out.**Final Answer**The translation takes (6 - 4i) to boxed{9 + i}.

💡Alright, so I have this problem where I need to find the greatest number ( A ) such that ( A leq |r_2| ) for any function ( f ) satisfying certain properties. The function is given by ( f(x) = x^2 - r_2 x + r_3 ), and there's a sequence ( { g_n } ) defined by ( g_0 = 0 ) and ( g_{n+1} = f(g_n) ). The sequence has some specific behaviors: it alternates increasing and decreasing for the first 2012 terms, then eventually starts increasing and becomes unbounded.First, I need to understand the behavior of the sequence ( { g_n } ). The first condition says that for all ( 0 leq i leq 2011 ), ( g_{2i} < g_{2i+1} ) and ( g_{2i+1} > g_{2i+2} ). So, the sequence alternates between increasing and decreasing for the first 2012 terms. That suggests that the function ( f ) is such that applying it alternately increases and decreases the value of ( g_n ).The second condition says that after some point ( j ), the sequence becomes strictly increasing. So, after a certain number of terms, the sequence no longer decreases but keeps increasing. The third condition is that the sequence is unbounded, meaning it goes to infinity as ( n ) increases.Given that ( f(x) = x^2 - r_2 x + r_3 ), it's a quadratic function. Quadratic functions have a parabola as their graph, opening upwards since the coefficient of ( x^2 ) is positive. The vertex of this parabola is at ( x = frac{r_2}{2} ), and the minimum value of ( f(x) ) is ( fleft(frac{r_2}{2}right) = left(frac{r_2}{2}right)^2 - r_2 cdot frac{r_2}{2} + r_3 = -frac{r_2^2}{4} + r_3 ).Since the sequence ( { g_n } ) is defined recursively by ( g_{n+1} = f(g_n) ), starting from ( g_0 = 0 ), the behavior of the sequence depends heavily on the function ( f ). The alternating increase and decrease in the first 2012 terms suggests that the function ( f ) is oscillating around some fixed point or value, causing the sequence to alternate.To analyze this, I think about fixed points of the function ( f ). A fixed point is a value ( x ) such that ( f(x) = x ). Solving ( x^2 - r_2 x + r_3 = x ) gives ( x^2 - (r_2 + 1)x + r_3 = 0 ). The roots of this quadratic equation are the fixed points.The discriminant of this quadratic is ( D = (r_2 + 1)^2 - 4r_3 ). For real fixed points, the discriminant must be non-negative, so ( (r_2 + 1)^2 geq 4r_3 ). The roots are ( x = frac{r_2 + 1 pm sqrt{(r_2 + 1)^2 - 4r_3}}{2} ).Since the sequence alternates for the first 2012 terms, it suggests that the function ( f ) is such that applying it alternately increases and decreases the value. This behavior is typical near a repelling fixed point or in the vicinity of a point where the function crosses the line ( y = x ).Given that the sequence eventually becomes increasing and unbounded, it suggests that after some point, the function ( f ) causes the sequence to escape to infinity. This usually happens when the function's growth dominates, meaning that for large ( x ), ( f(x) ) is much larger than ( x ), which is true for quadratics as they grow faster than linear functions.But before that, the sequence alternates. So, perhaps the function ( f ) has two fixed points, and the sequence oscillates around them. The fact that it alternates for a large number of terms (2012) suggests that the function is designed such that each application alternately takes the sequence above and below some critical value.To get a better handle on this, I can consider the function ( f(x) - x ). This function is ( x^2 - (r_2 + 1)x + r_3 ). The roots of this function are the fixed points of ( f ). Since the sequence alternates, it must be crossing the line ( y = x ) each time, implying that ( f(x) - x ) changes sign each time. Therefore, ( f(x) - x ) must have two real roots, say ( a ) and ( b ), with ( a < b ).So, ( f(x) - x = (x - a)(x - b) ). Therefore, ( f(x) = (x - a)(x - b) + x ).Now, analyzing the behavior around these fixed points. For ( x < a ), ( f(x) - x > 0 ), so ( f(x) > x ). For ( a < x < b ), ( f(x) - x < 0 ), so ( f(x) < x ). For ( x > b ), ( f(x) - x > 0 ), so ( f(x) > x ).Given that ( g_0 = 0 ), let's see where 0 is in relation to ( a ) and ( b ). If ( 0 < a ), then ( f(0) = 0^2 - r_2 cdot 0 + r_3 = r_3 ). So, ( g_1 = r_3 ). If ( g_1 ) is greater than ( g_0 = 0 ), then ( r_3 > 0 ). Given that ( g_1 > g_0 ), and then ( g_2 = f(g_1) ). If ( g_1 ) is between ( a ) and ( b ), then ( f(g_1) < g_1 ), so ( g_2 < g_1 ). Then ( g_3 = f(g_2) ). If ( g_2 ) is less than ( a ), then ( f(g_2) > g_2 ), so ( g_3 > g_2 ). This alternation continues as long as the terms stay between ( a ) and ( b ) or below ( a ).But the sequence eventually becomes increasing and unbounded. So, at some point, ( g_n ) must exceed ( b ), after which ( f(g_n) > g_n ), causing the sequence to increase without bound.Therefore, the key is that the sequence alternates for a while, then escapes to infinity. The function ( f ) must be such that it allows this behavior.Now, to relate this to ( r_2 ). The vertex of the parabola ( f(x) ) is at ( x = frac{r_2}{2} ). The minimum value of ( f(x) ) is at this vertex. So, ( fleft(frac{r_2}{2}right) = left(frac{r_2}{2}right)^2 - r_2 cdot frac{r_2}{2} + r_3 = -frac{r_2^2}{4} + r_3 ).For the sequence to eventually escape to infinity, the function ( f ) must eventually produce values that are large enough so that ( f(g_n) > g_n ) for all subsequent ( n ). This requires that ( f(x) ) grows faster than ( x ) for large ( x ), which it does because it's quadratic.But the critical part is ensuring that the sequence doesn't get stuck oscillating indefinitely. The fact that it eventually starts increasing suggests that after some point, ( g_n ) exceeds ( b ), the larger fixed point, and then ( f(g_n) > g_n ) for all larger ( n ).So, the function ( f(x) ) must have two fixed points ( a ) and ( b ), with ( a < b ), and the sequence starts at 0, alternates around these fixed points for a while, then escapes beyond ( b ) to infinity.Now, to find the relationship between ( r_2 ) and ( r_3 ). Since ( f(x) - x = (x - a)(x - b) ), expanding this gives ( x^2 - (a + b)x + ab ). Comparing to ( f(x) - x = x^2 - (r_2 + 1)x + r_3 ), we get:1. ( a + b = r_2 + 1 )2. ( ab = r_3 )So, ( r_2 = a + b - 1 ) and ( r_3 = ab ).Now, the sequence starts at 0, then goes to ( g_1 = f(0) = r_3 ). So, ( g_1 = ab ). Then, ( g_2 = f(g_1) = f(ab) = (ab)^2 - r_2 ab + r_3 ).But ( f(ab) = (ab)^2 - (a + b - 1)ab + ab = (ab)^2 - (a + b)ab + ab + ab = (ab)^2 - (a + b)ab + 2ab ).Wait, that seems complicated. Maybe I should think differently.Given that the sequence alternates for 2012 terms, it must be that each time it alternates, it crosses the fixed points. So, starting from 0, which is less than ( a ) (assuming ( a > 0 )), then ( g_1 = f(0) = r_3 ). If ( r_3 > a ), then ( g_1 ) is between ( a ) and ( b ), so ( g_2 = f(g_1) < g_1 ). Then, ( g_2 ) could be less than ( a ), so ( g_3 = f(g_2) > g_2 ), and so on.But if ( g_1 ) is greater than ( b ), then ( g_2 = f(g_1) > g_1 ), which would contradict the alternation. So, ( g_1 ) must be between ( a ) and ( b ).Therefore, ( a < g_1 = r_3 < b ).Similarly, ( g_2 = f(g_1) < g_1 ), so ( g_2 ) must be less than ( g_1 ). If ( g_2 ) is less than ( a ), then ( g_3 = f(g_2) > g_2 ), and so on.So, the sequence alternates between being above and below ( a ) or ( b ). But since it eventually escapes to infinity, at some point, ( g_n ) must exceed ( b ), after which it keeps increasing.Therefore, the key is that the function ( f ) must have two fixed points, and the sequence alternates around them for a while, then escapes.Now, to relate this to ( r_2 ). Since ( r_2 = a + b - 1 ), and ( r_3 = ab ). We need to find the minimal possible ( |r_2| ) such that the sequence can alternate for 2012 terms and then escape.But actually, the problem asks for the greatest number ( A ) such that ( A leq |r_2| ) for any function ( f ) satisfying these properties. So, we need to find the minimal possible ( |r_2| ) such that the function ( f ) can satisfy all three conditions, and then ( A ) is that minimal value.Wait, no. It says ( A ) is the greatest number such that ( A leq |r_2| ) for any function ( f ) satisfying the properties. So, ( A ) is the infimum of ( |r_2| ) over all such functions ( f ). Therefore, ( A ) is the minimal possible ( |r_2| ) such that the function can satisfy the conditions, and ( A ) is the greatest lower bound.So, to find ( A ), we need to find the minimal ( |r_2| ) such that ( f ) can have the sequence ( { g_n } ) alternate for 2012 terms, then escape to infinity.To find this, we can consider the behavior of the function near the fixed points. The function ( f(x) ) must have two fixed points ( a ) and ( b ), with ( a < b ). The sequence alternates around these fixed points, so the function must be such that it crosses the line ( y = x ) at these points.The key is that the function must have a certain "width" between the fixed points to allow the alternation for 2012 terms. The larger the distance between ( a ) and ( b ), the more alternations can occur before escaping.But since we're looking for the minimal ( |r_2| ), we need to find the smallest possible ( |r_2| ) such that the function can still allow 2012 alternations before escaping.To analyze this, we can consider the function ( f(x) ) and its iterates. The sequence alternates because ( f(x) ) alternately lies above and below the line ( y = x ). For the sequence to alternate 2012 times, the function must be such that each application flips the side relative to ( y = x ).This suggests that the function ( f(x) ) must have a certain property where each application inverts the position relative to the fixed points. This is similar to a function with a negative derivative at the fixed points, causing oscillations.However, since ( f(x) ) is quadratic, it's convex, so the derivative at the fixed points can be positive or negative. If the derivative at the fixed points is less than 1 in absolute value, the fixed points are attracting; if greater, they are repelling.But in our case, the sequence alternates, suggesting that the fixed points are such that the function alternates around them. This might mean that the fixed points are such that the function crosses the line ( y = x ) with a certain slope.But perhaps a better approach is to consider the function ( f(x) - x ) and its roots. Since ( f(x) - x = (x - a)(x - b) ), and the sequence alternates around these roots, the distance between ( a ) and ( b ) affects how many times the sequence can alternate before escaping.To allow 2012 alternations, the distance between ( a ) and ( b ) must be sufficiently large. The larger the distance, the more alternations can occur before the sequence escapes to infinity.But since we're looking for the minimal ( |r_2| ), we need to find the smallest ( |r_2| ) such that the distance between ( a ) and ( b ) is large enough to allow 2012 alternations.The distance between ( a ) and ( b ) is ( b - a ). From the quadratic equation, ( b - a = sqrt{(r_2 + 1)^2 - 4r_3} ). So, ( b - a = sqrt{(r_2 + 1)^2 - 4r_3} ).But ( r_3 = ab ), so substituting, we get ( b - a = sqrt{(r_2 + 1)^2 - 4ab} ).But since ( r_2 = a + b - 1 ), we can write ( r_2 + 1 = a + b ). Therefore, ( b - a = sqrt{(a + b)^2 - 4ab} = sqrt{a^2 + 2ab + b^2 - 4ab} = sqrt{a^2 - 2ab + b^2} = sqrt{(a - b)^2} = |a - b| ). But since ( b > a ), ( b - a = b - a ). So, this just confirms that the distance is consistent.But we need to relate this to the number of alternations. The number of alternations depends on how the function behaves near the fixed points. For the sequence to alternate many times, the function must be such that each application doesn't move the sequence too far from the fixed points, allowing many oscillations before escaping.This is similar to the concept of the function having a certain "period" or "oscillation" around the fixed points. The closer the fixed points are, the fewer alternations can occur before escaping. Therefore, to allow 2012 alternations, the fixed points must be sufficiently far apart.But how does this relate to ( r_2 )? Since ( r_2 = a + b - 1 ), and ( b - a ) is the distance between the fixed points, we can express ( r_2 ) in terms of ( a ) and ( b ).Let me denote ( d = b - a ). Then, ( r_2 = a + b - 1 = (a + b) - 1 ). Since ( b = a + d ), we have ( r_2 = a + (a + d) - 1 = 2a + d - 1 ).But we also know that ( r_3 = ab = a(a + d) = a^2 + ad ).Now, the function ( f(x) - x = (x - a)(x - b) = x^2 - (a + b)x + ab ). Comparing to ( f(x) - x = x^2 - (r_2 + 1)x + r_3 ), we have:1. ( a + b = r_2 + 1 )2. ( ab = r_3 )So, ( r_2 = a + b - 1 ) and ( r_3 = ab ).Now, to find the minimal ( |r_2| ), we need to express ( r_2 ) in terms of ( a ) and ( d ), and then find the minimal value.But perhaps a better approach is to consider the function's behavior. Since the sequence alternates for 2012 terms, the function must be such that each application doesn't move the sequence too far from the fixed points. This suggests that the function has a certain "slope" near the fixed points that allows for oscillations.The derivative of ( f(x) ) at a fixed point ( x = c ) is ( f'(c) = 2c - r_2 ). For the fixed point to be repelling, the absolute value of the derivative must be greater than 1. If it's less than 1, the fixed point is attracting.In our case, since the sequence alternates, it suggests that the fixed points are such that the function alternates around them, which might mean that the fixed points are repelling, causing the sequence to oscillate.Therefore, for both fixed points ( a ) and ( b ), we must have ( |f'(a)| > 1 ) and ( |f'(b)| > 1 ).Calculating the derivatives:1. ( f'(a) = 2a - r_2 )2. ( f'(b) = 2b - r_2 )Given that ( r_2 = a + b - 1 ), substituting:1. ( f'(a) = 2a - (a + b - 1) = a - b + 1 )2. ( f'(b) = 2b - (a + b - 1) = b - a + 1 )So, ( f'(a) = -(b - a) + 1 ) and ( f'(b) = (b - a) + 1 ).For both derivatives to have absolute values greater than 1:1. ( |a - b + 1| > 1 )2. ( |b - a + 1| > 1 )But ( b > a ), so ( b - a > 0 ). Therefore:1. ( |-(b - a) + 1| > 1 )2. ( |(b - a) + 1| > 1 )Simplifying:1. ( |1 - (b - a)| > 1 )2. ( |1 + (b - a)| > 1 )Since ( b - a > 0 ), let's denote ( d = b - a ). Then:1. ( |1 - d| > 1 )2. ( |1 + d| > 1 )The second inequality ( |1 + d| > 1 ) is always true for ( d > 0 ), since ( 1 + d > 1 ).The first inequality ( |1 - d| > 1 ) implies that either ( 1 - d > 1 ) or ( 1 - d < -1 ).1. ( 1 - d > 1 ) implies ( -d > 0 ) or ( d < 0 ), which contradicts ( d > 0 ).2. ( 1 - d < -1 ) implies ( -d < -2 ) or ( d > 2 ).Therefore, ( d > 2 ).So, the distance between the fixed points ( a ) and ( b ) must be greater than 2.Since ( d = b - a > 2 ), and ( r_2 = a + b - 1 ), we can express ( r_2 ) in terms of ( a ) and ( d ):( r_2 = a + (a + d) - 1 = 2a + d - 1 ).But we need to find the minimal ( |r_2| ). Since ( r_2 = 2a + d - 1 ), and ( d > 2 ), we can try to minimize ( |2a + d - 1| ).However, ( a ) and ( d ) are related through the function. Let's see if we can express ( a ) in terms of ( d ).From ( r_3 = ab = a(a + d) = a^2 + ad ).But ( r_3 ) is also equal to ( g_1 = f(0) = r_3 ). Since ( g_1 > g_0 = 0 ), ( r_3 > 0 ).So, ( a^2 + ad > 0 ). Since ( d > 2 ), and ( a ) is a real number, ( a ) can be positive or negative.But let's consider the sequence starting at 0. If ( a ) is negative, then ( g_1 = r_3 = a(a + d) ). If ( a ) is negative and ( d > 2 ), ( a + d ) could be positive or negative depending on ( a ).But perhaps it's simpler to assume ( a ) is positive. Let's see.If ( a > 0 ), then ( r_3 = a(a + d) > 0 ), which is consistent.If ( a < 0 ), then ( a + d ) could be positive or negative. If ( a + d > 0 ), then ( r_3 = a(a + d) ) could be negative if ( a ) is negative and ( a + d ) is positive. But ( r_3 > 0 ) because ( g_1 > g_0 = 0 ). Therefore, if ( a < 0 ), ( a + d ) must also be negative to make ( r_3 = a(a + d) > 0 ).So, if ( a < 0 ), then ( a + d < 0 ), which implies ( d < -a ). But ( d > 2 ), so ( -a > 2 ), meaning ( a < -2 ).But let's see which case gives a smaller ( |r_2| ).Case 1: ( a > 0 ). Then, ( r_2 = 2a + d - 1 ). Since ( d > 2 ), ( r_2 > 2a + 2 - 1 = 2a + 1 ). To minimize ( r_2 ), we need to minimize ( a ). The smallest ( a ) can be is approaching 0 from the positive side. Then, ( r_2 ) approaches ( 0 + d - 1 ). Since ( d > 2 ), ( r_2 > 2 - 1 = 1 ). But we need to ensure that the sequence alternates for 2012 terms, so ( d ) must be sufficiently large to allow that many alternations.Wait, but if ( d ) is just slightly larger than 2, say ( d = 2 + epsilon ), then ( r_2 = 2a + (2 + epsilon) - 1 = 2a + 1 + epsilon ). If ( a ) is small, say approaching 0, then ( r_2 ) approaches ( 1 + epsilon ). But we need to ensure that the sequence can alternate 2012 times.However, if ( d ) is just slightly larger than 2, the distance between the fixed points is just over 2, which might not be enough to allow 2012 alternations. The number of alternations depends on how the function behaves near the fixed points. If the fixed points are too close, the sequence might escape too quickly.Therefore, perhaps ( d ) needs to be significantly larger than 2 to allow 2012 alternations. But since we're looking for the minimal ( |r_2| ), we need to find the smallest ( d ) such that the sequence can alternate 2012 times.Alternatively, maybe the minimal ( |r_2| ) is 2, regardless of the number of alternations, because of the properties of the quadratic function.Wait, let's think about the function ( f(x) = x^2 - 2x + c ). If ( r_2 = 2 ), then ( f(x) = x^2 - 2x + c ). The fixed points are solutions to ( x^2 - 3x + c = 0 ). The discriminant is ( 9 - 4c ). For real fixed points, ( 9 - 4c geq 0 ), so ( c leq 9/4 ).If ( c = 1 ), then the fixed points are ( x = [3 pm sqrt{5}]/2 ), approximately 2.618 and 0.382. The distance between them is ( sqrt{5} approx 2.236 ), which is greater than 2.In this case, the sequence starting at 0 would go to ( g_1 = f(0) = 1 ), which is between the fixed points. Then ( g_2 = f(1) = 1 - 2 + 1 = 0 ), which is less than the lower fixed point. Then ( g_3 = f(0) = 1 ), and so on. This would cause the sequence to alternate between 0 and 1 indefinitely, never escaping to infinity. But our problem requires that after some point, the sequence becomes increasing and unbounded. So, this function doesn't satisfy condition (ii).Therefore, ( r_2 = 2 ) might not be sufficient because the sequence could get stuck oscillating between two values.But wait, in the problem, the sequence alternates for 2012 terms, then becomes increasing. So, perhaps with ( r_2 = 2 ), the sequence can alternate for a certain number of terms before escaping.Wait, let's try ( r_2 = 2 ) and see. Let ( f(x) = x^2 - 2x + c ). Let's choose ( c ) such that the fixed points are such that the sequence can alternate for 2012 terms before escaping.But I'm not sure. Maybe a better approach is to consider the function's behavior near the fixed points. If the fixed points are such that the function has a certain slope, the sequence can oscillate around them for a number of terms before escaping.But perhaps the minimal ( |r_2| ) is 2 because of the properties of the quadratic function. If ( |r_2| < 2 ), the function might not allow the sequence to alternate sufficiently before escaping.Wait, let's consider ( r_2 = 2 ). Then, ( f(x) = x^2 - 2x + c ). The fixed points are at ( x = [2 + 1 pm sqrt{(2 + 1)^2 - 4c}]/2 = [3 pm sqrt{9 - 4c}]/2 ). So, the distance between the fixed points is ( sqrt{9 - 4c} ).To have the distance greater than 2, we need ( sqrt{9 - 4c} > 2 ), so ( 9 - 4c > 4 ), which implies ( 4c < 5 ), so ( c < 5/4 ).If ( c = 1 ), as before, the distance is ( sqrt{5} approx 2.236 ), which is greater than 2. So, with ( r_2 = 2 ) and ( c = 1 ), the sequence alternates between 0 and 1 indefinitely. But we need the sequence to eventually escape to infinity.Therefore, perhaps ( r_2 ) needs to be greater than 2 to allow the sequence to escape after some alternations.Wait, let's consider ( r_2 = 2 + epsilon ) for a small ( epsilon > 0 ). Then, the fixed points are solutions to ( x^2 - (2 + epsilon + 1)x + c = x^2 - (3 + epsilon)x + c = 0 ). The distance between the fixed points is ( sqrt{(3 + epsilon)^2 - 4c} ).To have the distance greater than 2, we need ( (3 + epsilon)^2 - 4c > 4 ), so ( 9 + 6epsilon + epsilon^2 - 4c > 4 ), which simplifies to ( 4c < 5 + 6epsilon + epsilon^2 ), so ( c < (5 + 6epsilon + epsilon^2)/4 ).If we choose ( c ) such that the sequence can alternate for 2012 terms before escaping, then ( r_2 ) can be slightly greater than 2.But the problem asks for the greatest number ( A ) such that ( A leq |r_2| ) for any function ( f ) satisfying the properties. So, ( A ) is the infimum of ( |r_2| ) over all such functions. Therefore, ( A ) is the minimal possible ( |r_2| ) such that the function can satisfy the conditions.From the earlier analysis, we saw that ( d = b - a > 2 ). Since ( r_2 = a + b - 1 ), and ( d = b - a ), we can express ( r_2 ) in terms of ( a ) and ( d ):( r_2 = a + (a + d) - 1 = 2a + d - 1 ).To minimize ( |r_2| ), we need to minimize ( |2a + d - 1| ). Since ( d > 2 ), and ( a ) can be positive or negative, let's consider both cases.Case 1: ( a > 0 ). Then, ( r_2 = 2a + d - 1 ). To minimize ( r_2 ), we need to minimize ( a ). The smallest ( a ) can be is approaching 0. Then, ( r_2 ) approaches ( d - 1 ). Since ( d > 2 ), ( r_2 > 1 ). But we need to ensure that the sequence can alternate for 2012 terms. If ( d ) is just slightly larger than 2, say ( d = 2 + epsilon ), then ( r_2 ) approaches ( 1 + epsilon ). However, with ( d ) only slightly larger than 2, the sequence might not alternate enough times before escaping.Case 2: ( a < 0 ). Then, ( r_2 = 2a + d - 1 ). Since ( a < 0 ), ( 2a ) is negative. To minimize ( |r_2| ), we need to make ( 2a + d - 1 ) as close to 0 as possible. Let's set ( 2a + d - 1 = 0 ), so ( a = (1 - d)/2 ). But since ( a < 0 ), ( 1 - d < 0 ), so ( d > 1 ). But we already have ( d > 2 ), so this is possible.Substituting ( a = (1 - d)/2 ) into ( r_3 = ab = a(a + d) ):( r_3 = a(a + d) = left(frac{1 - d}{2}right)left(frac{1 - d}{2} + dright) = left(frac{1 - d}{2}right)left(frac{1 - d + 2d}{2}right) = left(frac{1 - d}{2}right)left(frac{1 + d}{2}right) = frac{(1 - d)(1 + d)}{4} = frac{1 - d^2}{4} ).Since ( r_3 > 0 ), we have ( 1 - d^2 > 0 ), so ( d^2 < 1 ). But ( d > 2 ), which contradicts ( d^2 < 1 ). Therefore, this case is not possible.Thus, the minimal ( |r_2| ) occurs when ( a ) is as small as possible in the positive case, making ( r_2 ) approach ( d - 1 ). Since ( d > 2 ), the minimal ( r_2 ) approaches 1 as ( d ) approaches 2 from above. However, we need to ensure that the sequence can alternate for 2012 terms, which requires that ( d ) is sufficiently large to allow that many alternations.But how does ( d ) relate to the number of alternations? The number of alternations depends on how the function behaves near the fixed points. If ( d ) is just slightly larger than 2, the sequence might alternate a certain number of times before escaping. However, to allow 2012 alternations, ( d ) must be large enough to allow the sequence to oscillate that many times before the function's growth causes it to escape.But since we're looking for the minimal ( |r_2| ), we need to find the smallest ( d ) such that the sequence can alternate 2012 times. However, without a specific relationship between ( d ) and the number of alternations, it's difficult to determine the exact minimal ( d ).But perhaps there's a known result or a standard approach for such problems. In quadratic functions, the critical parameter is often related to the distance between the fixed points or the slope at the fixed points. For the sequence to alternate many times, the function must have a certain property, such as the fixed points being far apart or the function having a certain shape.Alternatively, considering the function ( f(x) = x^2 - r_2 x + r_3 ), the critical point is at ( x = r_2 / 2 ). The minimum value of ( f(x) ) is at this point, which is ( f(r_2 / 2) = (r_2 / 2)^2 - r_2 (r_2 / 2) + r_3 = -r_2^2 / 4 + r_3 ).For the sequence to eventually escape to infinity, the minimum value of ( f(x) ) must be such that after some point, ( f(g_n) > g_n ). This requires that the minimum value is less than some threshold, allowing the sequence to eventually exceed the fixed point ( b ).But perhaps a better approach is to consider the function's behavior near the fixed points. If the fixed points are such that the function alternates around them, the sequence can oscillate for a number of terms before escaping.Given that the problem allows for 2012 alternations, which is a large number, the function must be such that the fixed points are far apart enough to allow that many oscillations. However, since we're looking for the minimal ( |r_2| ), we need to find the smallest ( |r_2| ) such that the function can still allow 2012 alternations.But without a specific formula relating ( d ) to the number of alternations, it's challenging. However, from the earlier analysis, we saw that ( d > 2 ) is necessary for the fixed points to be repelling, allowing oscillations. Therefore, the minimal ( |r_2| ) is related to the minimal ( d ), which is just over 2.But in the case where ( d = 2 ), the fixed points are exactly 2 units apart, and the function might not allow oscillations. Therefore, ( d ) must be greater than 2, implying ( |r_2| ) must be greater than 2.Wait, earlier I thought that ( r_2 ) could approach 1, but that was under the assumption that ( d ) approaches 2. However, if ( d ) must be greater than 2, then ( r_2 = 2a + d - 1 ). If ( a ) is approaching 0, then ( r_2 ) approaches ( d - 1 ). Since ( d > 2 ), ( r_2 > 1 ). But to allow 2012 alternations, ( d ) must be sufficiently large, which would make ( r_2 ) larger.However, the problem asks for the greatest number ( A ) such that ( A leq |r_2| ) for any function ( f ) satisfying the properties. So, ( A ) is the infimum of ( |r_2| ) over all such functions. Therefore, ( A ) is the minimal possible ( |r_2| ) such that the function can satisfy the conditions.From the earlier analysis, we saw that ( d > 2 ) is necessary, and ( r_2 = 2a + d - 1 ). To minimize ( |r_2| ), we can set ( a ) to be as small as possible. If ( a ) approaches 0, then ( r_2 ) approaches ( d - 1 ). Since ( d > 2 ), ( r_2 > 1 ). However, to allow 2012 alternations, ( d ) must be sufficiently large, which would make ( r_2 ) larger than 2.Wait, perhaps the minimal ( |r_2| ) is 2 because if ( |r_2| < 2 ), the function might not allow the sequence to alternate sufficiently before escaping. Let me test this.If ( r_2 = 2 ), then ( f(x) = x^2 - 2x + c ). The fixed points are ( x = [2 + 1 pm sqrt{(2 + 1)^2 - 4c}]/2 = [3 pm sqrt{9 - 4c}]/2 ). The distance between the fixed points is ( sqrt{9 - 4c} ). To have the distance greater than 2, ( sqrt{9 - 4c} > 2 ), so ( 9 - 4c > 4 ), which gives ( c < 5/4 ).If ( c = 1 ), then the fixed points are ( x = [3 pm sqrt{5}]/2 approx 2.618 ) and ( 0.382 ). The sequence starting at 0 goes to ( g_1 = 1 ), which is between the fixed points. Then ( g_2 = f(1) = 1 - 2 + 1 = 0 ), which is less than the lower fixed point. Then ( g_3 = f(0) = 1 ), and so on. This causes the sequence to alternate between 0 and 1 indefinitely, never escaping to infinity. Therefore, ( r_2 = 2 ) with ( c = 1 ) doesn't satisfy condition (ii).But perhaps with a different ( c ), ( r_2 = 2 ) can allow the sequence to alternate for 2012 terms before escaping. Let's choose ( c ) such that the fixed points are such that the sequence can alternate many times before escaping.Suppose ( c = 0.5 ). Then, the fixed points are ( x = [3 pm sqrt{9 - 2}]/2 = [3 pm sqrt{7}]/2 approx 2.822 ) and ( 0.178 ). The distance between them is ( sqrt{7} approx 2.645 ), which is greater than 2.Now, the sequence starts at 0, ( g_1 = f(0) = 0.5 ), which is between the fixed points. Then ( g_2 = f(0.5) = 0.25 - 1 + 0.5 = -0.25 ), which is less than the lower fixed point. Then ( g_3 = f(-0.25) = 0.0625 + 0.5 + 0.5 = 1.0625 ), which is greater than the upper fixed point. Then ( g_4 = f(1.0625) = (1.0625)^2 - 2(1.0625) + 0.5 approx 1.1289 - 2.125 + 0.5 approx -0.4961 ), which is less than the lower fixed point. Then ( g_5 = f(-0.4961) approx 0.246 + 0.9922 + 0.5 approx 1.7382 ), which is greater than the upper fixed point. This alternation continues, with the sequence oscillating between values above and below the fixed points.However, each time the sequence exceeds the upper fixed point, it gets mapped to a value below the lower fixed point, and vice versa. This causes the sequence to oscillate but with increasing magnitude. Eventually, the sequence will escape to infinity because the function's quadratic term dominates, causing ( f(g_n) ) to grow without bound.Therefore, with ( r_2 = 2 ) and ( c = 0.5 ), the sequence alternates for a number of terms before escaping to infinity. The number of alternations depends on how the sequence's magnitude grows with each oscillation.Given that the problem allows for 2012 alternations, which is a large number, the function must be such that the sequence's growth is slow enough to allow that many oscillations before escaping. However, with ( r_2 = 2 ), the function's growth might be sufficient to allow the sequence to alternate many times before escaping.Therefore, it's possible that ( |r_2| = 2 ) is the minimal value that allows the sequence to alternate for any number of terms, including 2012, before escaping to infinity. Hence, the greatest number ( A ) such that ( A leq |r_2| ) for any function ( f ) satisfying the properties is ( A = 2 ).

💡First, I need to determine the number of mystery book shelves Rachel has.I know that each shelf holds exactly 9 books.Rachel has a total of 72 books.She has 2 shelves of picture books, which means she has 2 × 9 = 18 picture books.To find out how many books are mystery books, I subtract the number of picture books from the total number of books: 72 - 18 = 54 mystery books.Finally, to find the number of mystery book shelves, I divide the number of mystery books by the number of books per shelf: 54 ÷ 9 = 6 shelves.

💡First, I need to determine the slope of the line passing through points A(2, 5) and B(4, 3). The slope formula is:[m = frac{y_2 - y_1}{x_2 - x_1}]Substituting the coordinates of points A and B:[m = frac{3 - 5}{4 - 2} = frac{-2}{2} = -1]Next, I'll use the slope to find the angle of inclination. The relationship between the slope ( m ) and the angle ( theta ) is given by:[tan(theta) = m]Since the slope is (-1), we have:[tan(theta) = -1]The angle ( theta ) must satisfy ( 0 leq theta < pi ). The tangent of ( frac{3pi}{4} ) is (-1), so:[theta = frac{3pi}{4}]Therefore, the angle of inclination of line ( l ) is ( frac{3pi}{4} ).

💡First, I need to identify which numbers between 1 and 50 contain the digit 3. These numbers are 3, 13, 23, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, and 43.Next, I'll determine how replacing the digit 3 with 2 affects each of these numbers:- For numbers where 3 is in the units place (3, 13, 23, 43), replacing 3 with 2 decreases each number by 1.- For numbers where 3 is in the tens place (30, 31, 32, 33, 34, 35, 36, 37, 38, 39), replacing 3 with 2 decreases each number by 10.Calculating the total decrease:- There are 4 numbers with 3 in the units place, resulting in a total decrease of 4 * 1 = 4.- There are 10 numbers with 3 in the tens place, resulting in a total decrease of 10 * 10 = 100.Adding these decreases together gives a total decrease of 104. Therefore, Star's sum is 104 larger than Emilio's sum.However, upon reviewing the calculations, I realize there was a miscalculation in counting the numbers with 3 in the units place. There are actually 5 such numbers (3, 13, 23, 33, 43), leading to a total decrease of 5 * 1 = 5. This corrects the total decrease to 105.Thus, the correct conclusion is that Star's sum is 105 larger than Emilio's sum.