answer：First, I note that Jordan blocked 4 shots in the first period.In the second period, he blocked twice as many shots as the first period, which is 4 multiplied by 2, resulting in 8 shots.For the third period, he blocked three fewer shots than in the second period. Therefore, he blocked 8 minus 3, which equals 5 shots.Adding up the shots blocked in the first three periods: 4 (first period) + 8 (second period) + 5 (third period) = 17 shots.Since the total number of shots blocked in the game is 21, the number of shots blocked in the fourth period is 21 minus 17, which equals 4 shots.

answer：Alright, so I have this inequality: (x^{2} + 25 + left|x^{3} - 5 x^{2}right| geq a x), and I need to find the range of real numbers (a) such that this inequality holds for all (x) in the interval ([1, 12]). Three students have given their approaches, and I need to figure out which one is correct or maybe combine them somehow. Let me think through this step by step.First, let me understand the inequality. On the left side, there's a combination of (x^2), a constant 25, and an absolute value term (|x^3 - 5x^2|). On the right side, it's a linear term (a x). The goal is to find all (a) such that no matter what (x) we pick in ([1, 12]), the left side is always greater than or equal to the right side.So, Student A says: "We just need the smallest value on the left side to be no less than the largest value on the right side." Hmm, that sounds a bit vague. Let me parse that. So, if I can find the minimum of the left side over the interval and the maximum of the right side over the same interval, then setting the minimum left >= maximum right would ensure the inequality holds everywhere. That seems plausible, but I need to check if that's the right approach.Student B suggests: "Transform the inequality into a function of (x) on the left side, with only a constant on the right side, and find the extremum of the function." Okay, so Student B is saying to rearrange the inequality so that we have something like (f(x) geq a), where (f(x)) is the left side minus (a x). Then, to find the minimum of (f(x)) over the interval and set that minimum to be greater than or equal to zero. That makes sense because if the smallest value of (f(x)) is non-negative, then the inequality holds for all (x). So, Student B is suggesting to define (f(x) = x^2 + 25 + |x^3 - 5x^2| - a x) and find the minimum of (f(x)) over ([1, 12]), then set that minimum >= 0 to solve for (a). That seems like a solid approach.Student C says: "Consider both sides of the inequality as functions of (x) and plot their graphs." So, plotting (y = x^2 + 25 + |x^3 - 5x^2|) and (y = a x), and then seeing for which (a) the first graph is always above the second. That also sounds reasonable, but plotting might not be precise enough without calculus. Maybe it's more of a visual approach, but perhaps not the most rigorous.So, considering these approaches, I think Student B's method is the most precise and mathematical. It involves calculus, finding the extremum (minimum) of the function (f(x)) over the interval, which will give the tightest constraint on (a). Student A's idea is somewhat similar but not as precise because the maximum of the right side isn't necessarily the only constraint; it's more about the pointwise comparison across the interval. Student C's approach is more of a graphical method, which can be helpful but might not give an exact answer without further analysis.Alright, so I'll proceed with Student B's approach. Let me define (f(x) = x^2 + 25 + |x^3 - 5x^2| - a x). To ensure (f(x) geq 0) for all (x in [1, 12]), I need to find the minimum value of (f(x)) over this interval and set it to be greater than or equal to zero.First, let's analyze the absolute value term (|x^3 - 5x^2|). Let me factor that expression inside the absolute value:(x^3 - 5x^2 = x^2(x - 5)).So, the expression inside the absolute value is (x^2(x - 5)). Since (x^2) is always non-negative, the sign of the expression depends on (x - 5). Therefore, for (x < 5), (x - 5) is negative, so the expression inside the absolute value is negative, and for (x geq 5), it's non-negative.Therefore, we can write:(|x^3 - 5x^2| = begin{cases}5x^2 - x^3, & text{if } 1 leq x < 5, x^3 - 5x^2, & text{if } 5 leq x leq 12.end{cases})So, the function (f(x)) can be written piecewise:For (1 leq x < 5):(f(x) = x^2 + 25 + (5x^2 - x^3) - a x = x^2 + 25 + 5x^2 - x^3 - a x = 6x^2 + 25 - x^3 - a x).For (5 leq x leq 12):(f(x) = x^2 + 25 + (x^3 - 5x^2) - a x = x^2 + 25 + x^3 - 5x^2 - a x = x^3 - 4x^2 + 25 - a x).So, now we have two functions to consider:1. For (1 leq x < 5): (f_1(x) = -x^3 + 6x^2 - a x + 25)2. For (5 leq x leq 12): (f_2(x) = x^3 - 4x^2 - a x + 25)Our goal is to find the minimum of (f_1(x)) on ([1, 5)) and the minimum of (f_2(x)) on ([5, 12]), then set both minima to be greater than or equal to zero. The smallest (a) that satisfies both will give the range for (a).Alternatively, since (f(x)) is continuous at (x = 5) (we can check that), we can consider the entire interval ([1, 12]) but with different expressions on each subinterval.Let me first check continuity at (x = 5):Compute (f_1(5)):(f_1(5) = -125 + 6*25 - 5a + 25 = -125 + 150 - 5a + 25 = 50 - 5a).Compute (f_2(5)):(f_2(5) = 125 - 4*25 - 5a + 25 = 125 - 100 - 5a + 25 = 50 - 5a).So, both expressions give the same value at (x = 5), which is good. Therefore, (f(x)) is continuous at (x = 5).Now, to find the minima of (f_1(x)) on ([1, 5)) and (f_2(x)) on ([5, 12]), we can take derivatives and find critical points.Starting with (f_1(x)) on ([1, 5)):(f_1(x) = -x^3 + 6x^2 - a x + 25)Compute the derivative:(f_1'(x) = -3x^2 + 12x - a)Set derivative equal to zero to find critical points:(-3x^2 + 12x - a = 0)Multiply both sides by -1:(3x^2 - 12x + a = 0)This is a quadratic equation in (x):(3x^2 - 12x + a = 0)Let me compute the discriminant:Discriminant (D = (-12)^2 - 4*3*a = 144 - 12a)For real critical points, we need (D geq 0), so (144 - 12a geq 0) => (a leq 12).So, if (a leq 12), there are real critical points in the interval ([1, 5)). Let's find them:(x = [12 pm sqrt{144 - 12a}]/(2*3) = [12 pm sqrt{144 - 12a}]/6 = [6 pm sqrt{36 - 3a}]/3 = 2 pm sqrt{12 - a}/3)Wait, let's compute that again:Wait, discriminant is (144 - 12a), so sqrt(D) = sqrt(144 - 12a). Then,x = [12 ± sqrt(144 - 12a)] / (6) = [12 ± sqrt(12*(12 - a))]/6 = [12 ± 2*sqrt(3*(12 - a))]/6 = [6 ± sqrt(3*(12 - a))]/3 = 2 ± sqrt(3*(12 - a))/3.Wait, that seems a bit messy. Maybe I should just keep it as:x = [12 ± sqrt(144 - 12a)] / 6 = [12 ± sqrt(12*(12 - a))]/6 = [12 ± 2*sqrt(3*(12 - a))]/6 = [6 ± sqrt(3*(12 - a))]/3 = 2 ± sqrt(3*(12 - a))/3.Hmm, maybe it's better to just write it as:x = [12 ± sqrt(144 - 12a)] / 6 = 2 ± sqrt(144 - 12a)/6.Wait, sqrt(144 - 12a)/6 = sqrt(144 - 12a)/sqrt(36) = sqrt((144 - 12a)/36) = sqrt(4 - (a)/3).Wait, that seems incorrect because sqrt(144 - 12a)/6 is equal to sqrt((144 - 12a)/36) = sqrt(4 - (a)/3). Wait, no:Wait, sqrt(144 - 12a)/6 = sqrt(144 - 12a)/sqrt(36) = sqrt((144 - 12a)/36) = sqrt(4 - (a)/3). Yes, that's correct.So, x = 2 ± sqrt(4 - a/3).Wait, that seems a bit confusing, but okay.So, the critical points are at x = 2 + sqrt(4 - a/3) and x = 2 - sqrt(4 - a/3). But since x must be in [1, 5), we need to check whether these critical points lie within that interval.But before that, let's note that for the critical points to be real, we need sqrt(4 - a/3) to be real, so 4 - a/3 >= 0 => a <= 12, which is consistent with the discriminant condition.So, for a <= 12, we have critical points at x = 2 ± sqrt(4 - a/3). Let's see if these are within [1, 5).Compute x1 = 2 + sqrt(4 - a/3). Since sqrt(4 - a/3) is non-negative, x1 >= 2. But 2 + sqrt(4 - a/3) <= 5?sqrt(4 - a/3) <= 3 => 4 - a/3 <= 9 => -a/3 <= 5 => a >= -15. But since a is a real number, and we don't have a lower bound yet, but in our problem, a is likely positive because the right side is a x, and the left side is positive as well.Wait, but let's not get ahead of ourselves. Let's just note that x1 can be up to 2 + sqrt(4 - 0) = 2 + 2 = 4, since when a = 0, sqrt(4 - 0) = 2. So, x1 is in [2, 4], which is within [1,5). Similarly, x2 = 2 - sqrt(4 - a/3). Since sqrt(4 - a/3) <= 2, x2 >= 0, but we need x2 >=1.So, 2 - sqrt(4 - a/3) >=1 => sqrt(4 - a/3) <=1 => 4 - a/3 <=1 => -a/3 <= -3 => a >=9.So, when a >=9, x2 >=1, otherwise, x2 <1, which is outside our interval [1,5). So, for a >=9, we have two critical points in [1,5): x1 and x2. For a <9, only x1 is in [1,5), and x2 is less than 1, so we can ignore it.Therefore, for f1(x), the critical points in [1,5) are:- If a >=9: x = 2 + sqrt(4 - a/3) and x = 2 - sqrt(4 - a/3). But wait, when a increases, sqrt(4 - a/3) decreases, so x1 decreases and x2 increases.Wait, maybe I need to think about this again.Wait, when a increases, 4 - a/3 decreases, so sqrt(4 - a/3) decreases, so x1 = 2 + sqrt(4 - a/3) decreases, and x2 = 2 - sqrt(4 - a/3) increases.So, when a = 9:sqrt(4 - 9/3) = sqrt(4 - 3) = sqrt(1) =1So, x1 = 2 +1=3, x2=2 -1=1.So, when a=9, x2=1, which is the left endpoint.For a >9, sqrt(4 - a/3) becomes imaginary? Wait, no, wait, when a >12, discriminant becomes negative, so no real critical points.Wait, earlier we had discriminant D=144 -12a, which is non-negative when a <=12.So, for a <=12, we have real critical points.But when a >9, x2 =2 - sqrt(4 - a/3). Let's compute when a=12:sqrt(4 -12/3)=sqrt(4 -4)=0, so x1=2 +0=2, x2=2 -0=2. So, both critical points converge to x=2.So, for a in (9,12), x2=2 - sqrt(4 - a/3) is greater than 1, since sqrt(4 - a/3) <1 when a>9.Wait, let's check when a=10:sqrt(4 -10/3)=sqrt(4 -3.333)=sqrt(0.666)=approx 0.816.So, x2=2 -0.816≈1.184, which is >1.Similarly, when a=12: x2=2 -0=2.So, for a in [9,12], x2 is in [1,2], and x1 is in [2,3].For a in [0,9), x2=2 - sqrt(4 - a/3) <1, so only x1 is in [1,5).So, to summarize:- For a <=9: only one critical point at x1=2 + sqrt(4 - a/3) in [1,5).- For a >9: two critical points, x1 and x2, both in [1,5).But wait, when a=9, x2=1, which is the endpoint.So, to find the minima of f1(x) on [1,5), we need to evaluate f1 at critical points and endpoints.Similarly, for f2(x) on [5,12], let's compute its derivative.f2(x)=x^3 -4x^2 -a x +25Derivative:f2’(x)=3x^2 -8x -aSet derivative equal to zero:3x^2 -8x -a =0Quadratic equation: 3x^2 -8x -a=0Discriminant D=64 +12aSince D is always positive for any real a, there are always two real roots.Compute roots:x=(8 ± sqrt(64 +12a))/6=(8 ± sqrt(64 +12a))/6=(4 ± sqrt(16 +3a))/3So, the critical points are at x=(4 + sqrt(16 +3a))/3 and x=(4 - sqrt(16 +3a))/3.We need to check if these critical points lie within [5,12].Compute x1=(4 + sqrt(16 +3a))/3 and x2=(4 - sqrt(16 +3a))/3.Since sqrt(16 +3a) >=4, x1=(4 + something >=4)/3 >= (4 +4)/3=8/3≈2.666, but since we are considering x >=5, we need to see if x1 >=5.Compute when x1 >=5:(4 + sqrt(16 +3a))/3 >=5 => 4 + sqrt(16 +3a) >=15 => sqrt(16 +3a) >=11 => 16 +3a >=121 =>3a >=105 =>a >=35.Similarly, x2=(4 - sqrt(16 +3a))/3. Since sqrt(16 +3a) >=4, 4 - sqrt(16 +3a) <=0, so x2 <=0, which is outside our interval [5,12]. Therefore, the only critical point in [5,12] is x1=(4 + sqrt(16 +3a))/3, but only when a >=35, otherwise, x1 <5.So, for a <35, x1=(4 + sqrt(16 +3a))/3 <5.Let me check when a=0:x1=(4 + sqrt(16))/3=(4 +4)/3=8/3≈2.666 <5.When a=35:x1=(4 + sqrt(16 +105))/3=(4 + sqrt(121))/3=(4 +11)/3=15/3=5.So, for a >=35, x1 >=5, otherwise, x1 <5.Therefore, for f2(x) on [5,12], the critical points are:- If a >=35: x1=(4 + sqrt(16 +3a))/3 in [5,12].- If a <35: no critical points in [5,12], so the extrema occur at endpoints.But wait, actually, for a <35, x1=(4 + sqrt(16 +3a))/3 <5, so in [5,12], the function f2(x) has no critical points, so its extrema are at the endpoints x=5 and x=12.Wait, but let me think again. If a <35, then x1 <5, so on [5,12], f2(x) is increasing or decreasing?Compute the derivative at x=5:f2’(5)=3*(25) -8*5 -a=75 -40 -a=35 -a.If a <35, then f2’(5)=35 -a >0, so the function is increasing at x=5. Since the derivative is a quadratic opening upwards (coefficient 3>0), and the critical point is at x1 <5, so on [5,12], the derivative is increasing because the function is convex and the critical point is to the left of 5. Therefore, f2’(x) is increasing on [5,12], starting from f2’(5)=35 -a >0, so f2(x) is increasing on [5,12] when a <35.Therefore, for a <35, f2(x) is increasing on [5,12], so its minimum is at x=5, and maximum at x=12.For a >=35, f2(x) has a critical point at x1=(4 + sqrt(16 +3a))/3 in [5,12], so we need to check whether that critical point is a minimum or maximum.Since f2’(x)=3x^2 -8x -a, which is a quadratic opening upwards, so the critical point at x1 is a minimum.Therefore, for a >=35, f2(x) has a minimum at x1 in [5,12], and the function is decreasing before x1 and increasing after x1.Wait, but since f2’(x) is a quadratic opening upwards, the critical point x1 is a minimum.So, for a >=35, f2(x) has a minimum at x1, and for a <35, f2(x) is increasing on [5,12], so minimum at x=5.Okay, so now, putting this together, to find the minimum of f(x) over [1,12], we need to consider:1. For f1(x) on [1,5): - If a <=9: minimum occurs either at critical point x1=2 + sqrt(4 - a/3) or at endpoints x=1 or x=5. - If a >9: minimum occurs at one of the critical points x1 or x2, or at endpoints.2. For f2(x) on [5,12]: - If a <35: minimum at x=5. - If a >=35: minimum at x1=(4 + sqrt(16 +3a))/3.But since we're looking for the overall minimum over [1,12], we need to compare the minima from both intervals.But this seems complicated. Maybe instead of trying to handle all cases, I can instead consider that the minimum of f(x) over [1,12] occurs either at a critical point or at an endpoint.So, let's compute f(x) at the critical points and endpoints, and find the minimum.But given the complexity, perhaps it's better to consider the function f(x) over [1,12] and find its minimum by checking critical points and endpoints.Alternatively, perhaps we can find the minimum of f(x) by considering where the derivative changes sign.Wait, maybe another approach: since the inequality must hold for all x in [1,12], the minimum of the left side minus a x must be >=0. So, the minimal value of (x^2 +25 + |x^3 -5x^2|) -a x must be >=0.Therefore, the minimal value occurs where the derivative is zero or at endpoints.So, perhaps I can find the minimum of (x^2 +25 + |x^3 -5x^2|) -a x over [1,12], set it to be >=0, and solve for a.But since the absolute value complicates things, maybe I can split it into two cases as I did before: x in [1,5) and x in [5,12].So, for x in [1,5):f1(x) = -x^3 +6x^2 -a x +25We found its critical points at x=2 ± sqrt(4 -a/3), but only x=2 + sqrt(4 -a/3) is in [1,5) when a <=9, and both when a >=9.Similarly, for x in [5,12]:f2(x)=x^3 -4x^2 -a x +25Critical point at x=(4 + sqrt(16 +3a))/3, which is in [5,12] when a >=35.So, to find the minimum of f(x) over [1,12], we need to evaluate f(x) at:- x=1- x=5- x=12- Critical points in [1,5): x1=2 + sqrt(4 -a/3) and x2=2 - sqrt(4 -a/3) (if a >=9)- Critical point in [5,12]: x1=(4 + sqrt(16 +3a))/3 (if a >=35)But this seems quite involved. Maybe instead, I can consider that the minimal value occurs either at x=5 or somewhere else.Wait, let me compute f(x) at x=5:f(5)=5^2 +25 + |5^3 -5*5^2| -a*5=25 +25 + |125 -125| -5a=50 +0 -5a=50 -5a.So, f(5)=50 -5a.Similarly, at x=1:f(1)=1 +25 + |1 -5| -a*1=26 +4 -a=30 -a.At x=12:f(12)=144 +25 + |1728 - 720| -12a=169 +1008 -12a=1177 -12a.So, f(12)=1177 -12a.Now, let's think about the behavior of f(x) in each interval.For x in [1,5):f1(x) = -x^3 +6x^2 -a x +25We can analyze its behavior.At x=1: f1(1)= -1 +6 -a +25=30 -a.At x=5: f1(5)= -125 +150 -5a +25=50 -5a.So, as x increases from 1 to 5, f1(x) goes from 30 -a to 50 -5a.Depending on the value of a, this function could be increasing or decreasing.Similarly, for x in [5,12]:f2(x)=x^3 -4x^2 -a x +25At x=5: f2(5)=125 -100 -5a +25=50 -5a.At x=12: f2(12)=1728 - 576 -12a +25=1177 -12a.So, as x increases from 5 to12, f2(x) goes from 50 -5a to 1177 -12a.Again, depending on a, this could be increasing or decreasing.But to find the minimum of f(x) over [1,12], we need to find the minimal value among all these points.But perhaps the minimum occurs at x=5, x=1, or somewhere else.Wait, let's compute f(x) at critical points.For f1(x) in [1,5):Critical point at x=2 + sqrt(4 -a/3). Let's denote this as x_c1.Compute f1(x_c1):Since x_c1 is a critical point, f1’(x_c1)=0 => -3x_c1^2 +12x_c1 -a=0 => a= -3x_c1^2 +12x_c1.So, f1(x_c1)= -x_c1^3 +6x_c1^2 -a x_c1 +25.But since a= -3x_c1^2 +12x_c1, substitute:f1(x_c1)= -x_c1^3 +6x_c1^2 - (-3x_c1^2 +12x_c1) x_c1 +25= -x_c1^3 +6x_c1^2 +3x_c1^3 -12x_c1^2 +25= ( -x_c1^3 +3x_c1^3 ) + (6x_c1^2 -12x_c1^2 ) +25= 2x_c1^3 -6x_c1^2 +25.So, f1(x_c1)=2x_c1^3 -6x_c1^2 +25.Similarly, for f2(x) in [5,12):Critical point at x=(4 + sqrt(16 +3a))/3, denote as x_c2.Compute f2(x_c2):Since f2’(x_c2)=0 =>3x_c2^2 -8x_c2 -a=0 =>a=3x_c2^2 -8x_c2.So, f2(x_c2)=x_c2^3 -4x_c2^2 -a x_c2 +25= x_c2^3 -4x_c2^2 - (3x_c2^2 -8x_c2)x_c2 +25= x_c2^3 -4x_c2^2 -3x_c2^3 +8x_c2^2 +25= (-2x_c2^3) +4x_c2^2 +25.So, f2(x_c2)= -2x_c2^3 +4x_c2^2 +25.So, now, we have expressions for f at critical points in terms of x_c1 and x_c2.But this seems a bit abstract. Maybe instead, I can consider that for f1(x), the minimal value is either at x=1, x=5, or at x_c1, and for f2(x), the minimal value is either at x=5, x=12, or at x_c2.But perhaps it's better to consider that the minimal value of f(x) occurs at x=5, x=1, or at some critical point.Wait, let's compute f(x) at x=5: 50 -5a.At x=1:30 -a.At x=12:1177 -12a.So, depending on a, these could be the minima.But to find the overall minimum, we need to compare these values and the values at critical points.But this is getting complicated. Maybe I can instead consider that the minimal value of f(x) occurs at x=5, since both f1 and f2 meet there, and perhaps it's the point where the function changes its behavior.Alternatively, perhaps the minimal value is at x=5, so setting f(5)=50 -5a >=0 => a <=10.But wait, let's check.If a=10, then f(5)=50 -50=0.But we need to ensure that f(x) >=0 for all x in [1,12]. So, if a=10, f(5)=0, but what about other points?At x=1: f(1)=30 -10=20 >=0.At x=12: f(12)=1177 -120=1057 >=0.But what about the critical points?For a=10, let's compute f1(x) critical point:x_c1=2 + sqrt(4 -10/3)=2 + sqrt(4 -3.333)=2 + sqrt(0.666)=2 + ~0.816=2.816.Compute f1(x_c1)=2*(2.816)^3 -6*(2.816)^2 +25.Compute (2.816)^2≈7.927, (2.816)^3≈22.34.So, f1(x_c1)=2*22.34 -6*7.927 +25≈44.68 -47.56 +25≈22.12 >=0.Similarly, for f2(x) at a=10, since a=10 <35, f2(x) is increasing on [5,12], so minimum at x=5, which is 0.Therefore, at a=10, f(x) >=0 for all x in [1,12].But what if a>10? Let's say a=11.Then, f(5)=50 -55=-5 <0, which violates the inequality. So, a cannot be greater than 10.Wait, but what about a=10. Is that the maximum a can be?Wait, let's check another point. Suppose a=10, then f(5)=0, which is okay, but what about x=2.816, which was the critical point for f1(x). We saw that f1(x_c1)=~22.12 >=0.Similarly, for x in [5,12], since a=10 <35, f2(x) is increasing, so minimum at x=5, which is 0.Therefore, a=10 is the maximum value for a such that f(x)>=0 for all x in [1,12].But wait, let's check a=10 at x=5: f(5)=0, which is okay, but what about x=2.816: f(x)=22.12>0, and x=12:1177 -120=1057>0.So, a=10 is acceptable.But what about a=10.1? Then f(5)=50 -5*10.1=50 -50.5=-0.5 <0, which violates the inequality.Therefore, the maximum value of a is 10.But wait, let's check another critical point. Suppose a=10, then for f1(x), the critical point is at x≈2.816, and f1(x_c1)=~22.12>0.But what if a is slightly less than 10? For example, a=9.Then, f(5)=50 -45=5>0.f1(x_c1)=2*(2 + sqrt(4 -9/3))^3 -6*(2 + sqrt(4 -9/3))^2 +25.Compute sqrt(4 -3)=sqrt(1)=1.So, x_c1=2 +1=3.f1(3)=2*(27) -6*(9) +25=54 -54 +25=25>0.Similarly, for f2(x) at a=9, since a=9 <35, f2(x) is increasing on [5,12], so minimum at x=5, which is 5>0.Therefore, a=9 is acceptable.Wait, but if a=10 is acceptable, and a=10.1 is not, then the maximum a is 10.But let's check another point. Suppose a=10, and x=5: f(x)=0.But what about x approaching 5 from the left and right?From the left (x approaching 5-):f1(x)= -x^3 +6x^2 -10x +25.At x=5, f1(5)= -125 +150 -50 +25=0.From the right (x approaching 5+):f2(x)=x^3 -4x^2 -10x +25.At x=5, f2(5)=125 -100 -50 +25=0.So, it's continuous at x=5.But what about the behavior around x=5?For x just less than 5, f1(x) is decreasing towards 0.For x just more than 5, f2(x) is increasing from 0.Therefore, the minimal value is indeed at x=5, which is 0 when a=10.Therefore, the maximum a is 10.But wait, let's check another point. Suppose a=10, and x=2.816, which was the critical point for f1(x). We saw that f1(x_c1)=~22.12>0.So, even though a=10 makes f(5)=0, the function is still positive elsewhere.Therefore, the maximum a is 10.But wait, let's check a=10 at x=1:f(1)=30 -10=20>0.At x=12:f(12)=1177 -120=1057>0.So, all points are positive except at x=5, which is zero.Therefore, a=10 is the maximum value.But what about the minimal value of a? Since the inequality is x^2 +25 +|x^3 -5x^2| >=a x, and the left side is always positive, a can be as small as possible, but we need to ensure that the inequality holds for all x in [1,12].But actually, since the left side is positive, and the right side is a x, which can be negative if a is negative. But since x is in [1,12], if a is negative, a x is negative, and the left side is positive, so the inequality will always hold. Therefore, the lower bound of a is negative infinity.But wait, let me think again. If a is negative, then a x is negative, and the left side is positive, so the inequality x^2 +25 +|x^3 -5x^2| >=a x will always hold because positive >= negative.Therefore, the range of a is (-infty, 10].But let me confirm this.Suppose a is very negative, say a=-1000.Then, the inequality becomes x^2 +25 +|x^3 -5x^2| >=-1000 x.Since the left side is positive and the right side is negative, the inequality holds.Therefore, the lower bound is indeed negative infinity, and the upper bound is 10.Therefore, the range of a is all real numbers less than or equal to 10.So, the correct range is a <=10.But let me double-check.Suppose a=10, as before, the inequality holds with equality at x=5.If a>10, say a=11, then at x=5, f(5)=50 -55=-5<0, which violates the inequality.If a=10, f(x)>=0 for all x in [1,12].If a<10, f(x)>=0 as well, because f(x) would be larger.Therefore, the range of a is (-infty,10].So, the answer is a <=10.**Final Answer**The range of the real number (a) is (boxed{(-infty, 10]}).

answer：Okay, so I have this problem here: When ( -2 leq x leq 2 ), the maximum value of the function ( y = x^2 - 2x - 5 ) is ___. Hmm, I need to find the maximum value of this quadratic function within the given interval. Let me think about how to approach this.First, I remember that quadratic functions have either a maximum or a minimum value, depending on the coefficient of the ( x^2 ) term. In this case, the coefficient is 1, which is positive. That means the parabola opens upwards, so it has a minimum point, not a maximum. Wait, so does that mean the maximum value occurs at one of the endpoints of the interval? Because if the parabola opens upwards, the vertex is the minimum, and the function increases as you move away from the vertex in both directions. So, on a closed interval, the maximum should be at one of the endpoints.Let me double-check that. The function is ( y = x^2 - 2x - 5 ). The general form of a quadratic is ( y = ax^2 + bx + c ), so here ( a = 1 ), ( b = -2 ), and ( c = -5 ). The vertex of a parabola is at ( x = -frac{b}{2a} ). Plugging in the values, that would be ( x = -frac{-2}{2*1} = frac{2}{2} = 1 ). So the vertex is at ( x = 1 ). Since the parabola opens upwards, this is the minimum point. Therefore, the maximum must occur at one of the endpoints of the interval ( x = -2 ) or ( x = 2 ).Alright, so I need to evaluate the function at both endpoints and see which one gives a higher value. Let me compute ( y ) when ( x = -2 ) and when ( x = 2 ).Starting with ( x = -2 ):( y = (-2)^2 - 2*(-2) - 5 )Calculating each term:( (-2)^2 = 4 )( -2*(-2) = 4 )So, ( y = 4 + 4 - 5 = 8 - 5 = 3 )So, at ( x = -2 ), ( y = 3 ).Now, for ( x = 2 ):( y = (2)^2 - 2*(2) - 5 )Calculating each term:( (2)^2 = 4 )( -2*(2) = -4 )So, ( y = 4 - 4 - 5 = 0 - 5 = -5 )So, at ( x = 2 ), ( y = -5 ).Comparing the two values, ( y = 3 ) at ( x = -2 ) and ( y = -5 ) at ( x = 2 ). Clearly, 3 is greater than -5, so the maximum value of the function on the interval ( [-2, 2] ) is 3.Wait, let me make sure I didn't make any calculation errors. Let me recalculate both.For ( x = -2 ):( (-2)^2 = 4 )( -2*(-2) = 4 )So, ( 4 + 4 - 5 = 8 - 5 = 3 ). That seems correct.For ( x = 2 ):( 2^2 = 4 )( -2*2 = -4 )So, ( 4 - 4 - 5 = 0 - 5 = -5 ). That also seems correct.Hmm, so 3 is definitely the higher value. So, the maximum is 3 at ( x = -2 ).But just to be thorough, let me also check the value at the vertex, which is at ( x = 1 ). Maybe it's a minimum, but just to see.Calculating ( y ) at ( x = 1 ):( y = (1)^2 - 2*(1) - 5 = 1 - 2 - 5 = -6 ). So, that's the minimum value, as expected.Therefore, the function reaches its minimum at ( x = 1 ) and its maximum at one of the endpoints, which is ( x = -2 ) with a value of 3.Wait, just to visualize, the parabola opens upwards, so it curves upwards on both sides. So, on the interval from -2 to 2, the function decreases from ( x = -2 ) to ( x = 1 ) and then increases from ( x = 1 ) to ( x = 2 ). But since it's opening upwards, the sides go up, but on the interval, the left side is higher than the right side because the right side is still below the left endpoint.Wait, actually, when ( x ) increases from 1 to 2, the function increases from -6 to -5, which is an increase, but it's still lower than the value at ( x = -2 ). So, yeah, the maximum is at ( x = -2 ).I think that's solid. So, the maximum value is 3.**Final Answer**The maximum value is boxed{3}.

answer：Okay, so I have this problem about systematic sampling. Let me try to understand it step by step. There are 100 products numbered from 00 to 99. We need to select 5 products using systematic sampling. After grouping, products are sampled from each group at the same interval. Hmm, okay.Wait, the problem says "after grouping, products are sampled from each group at the same interval." I'm a bit confused about what that means. Maybe it's referring to dividing the 100 products into groups and then selecting every nth product from each group? Or perhaps it's about determining the interval for the systematic sampling.Let me recall what systematic sampling is. It's a method where you select every kth element from a population. The value of k is determined by dividing the population size by the sample size. So, in this case, the population size is 100, and the sample size is 5. So, k would be 100 divided by 5, which is 20. So, the interval k is 20.That means we would select every 20th product. But wait, how does that work exactly? I think you start by randomly selecting a number between 1 and k as the starting point, and then you add k each time to get the next sample. But in this problem, it's mentioned that after grouping, products are sampled from each group at the same interval. Maybe grouping refers to dividing the 100 products into 5 groups of 20 each?So, if we have 5 groups, each group would have 20 products. Then, from each group, we sample at the same interval. Hmm, so if the interval is 20, but each group is 20 products, then maybe the interval within each group is smaller? Or perhaps the interval is the same across all groups, so we pick every 20th product across the entire population.Wait, maybe I'm overcomplicating it. Let's go back. The key point is that it's systematic sampling with 5 products selected from 100. So, the interval k is 20, as I calculated before. So, the first product is randomly selected between 00 and 19, and then every 20th product after that.But the problem mentions grouping. Maybe the population is divided into 5 groups, each of size 20, and then one product is selected from each group using the same interval. So, if the interval is 20, then starting from a random point between 00 and 19, the next product would be 20 units apart.Wait, but if we have groups of 20, then the interval within each group would be 1? Because each group is 20 products, so if we sample at the same interval, it might be every 1st product, but that doesn't make sense because then we would be selecting all products.I'm getting confused. Let me try to think differently. Maybe the grouping is not about dividing into 5 groups, but rather about the systematic sampling process itself. In systematic sampling, you can think of the population as being arranged in a list, and then you select every kth element. So, in this case, k is 20.So, if we have 100 products, numbered 00 to 99, and we need to select 5, then k is 20. So, we randomly pick a starting point between 00 and 19, say s, and then our selected products would be s, s+20, s+40, s+60, s+80.But the problem says that after grouping, products are sampled from each group at the same interval. Maybe grouping here refers to dividing the population into k groups, each of size 100/k, which would be 5 groups of 20 each. Then, from each group, you sample one product at the same interval.Wait, so if we have 5 groups, each of 20 products, and we sample one from each group, the interval would be the same across groups. So, if we pick the first product from group 1, then the next from group 2, and so on, each time moving by the same interval.But in that case, the interval would be 1, because we're moving from one group to the next. Hmm, I'm not sure.Wait, maybe the interval is the number of groups. If we have 5 groups, then the interval could be 5, but that doesn't seem right either.Wait, let's think about the fifth group selected being 94. So, the fifth product is 94. If we're using systematic sampling with interval k, then the fifth product would be s + 4k = 94. So, if we can find s and k, we can figure out the second product.But earlier, I thought k was 20, because 100 divided by 5 is 20. So, if k is 20, then s + 4*20 = s + 80 = 94. So, s would be 94 - 80 = 14. So, the starting point is 14. Then, the products selected would be 14, 34, 54, 74, 94. So, the second product would be 34.But wait, the problem mentions grouping. So, is the grouping affecting the interval? Maybe the interval is not 20, but something else.Alternatively, maybe the population is divided into 5 groups, each of 20 products, and then from each group, a product is selected at the same interval. So, if the interval is, say, m, then the first product is s, the second is s + m, the third is s + 2m, etc., but within each group.But that might not make sense because each group is 20 products, so m would have to be less than 20.Wait, maybe the interval is the same across all groups, so if you have 5 groups, each of 20, and you pick one from each group, the interval between the groups is 20. So, the first product is s, the next is s + 20, then s + 40, etc. So, that would align with the systematic sampling method.So, if the fifth product is 94, then s + 4*20 = 94, so s = 14. Then, the second product would be 14 + 20 = 34.But wait, the problem says "after grouping, products are sampled from each group at the same interval." So, maybe the interval is within each group. So, if each group has 20 products, and we sample from each group at the same interval, say m, then the first product is s, then s + m, s + 2m, etc., but only within each group.But since we're only selecting one product from each group, the interval would be such that m is 20? Wait, that doesn't make sense.Alternatively, maybe the interval is the same across groups, so the interval is 20, as before.I think I'm overcomplicating it. Let's go back to the basics.In systematic sampling, you have a population of N, sample size n, interval k = N/n. So, here N=100, n=5, so k=20. Then, you randomly select a starting point between 1 and k, say s, and then select s, s+k, s+2k, etc.But in this problem, it's mentioned that after grouping, products are sampled from each group at the same interval. So, maybe the grouping is just another way of saying dividing the population into k groups, each of size n. So, 5 groups of 20 each.Then, from each group, you sample one product at the same interval. So, if the interval is 1, you'd sample the first product from each group, but that would give you 5 products: 00, 20, 40, 60, 80. But the fifth product is 94, which is not 80. So, that can't be.Alternatively, if the interval is 2, you'd sample every second product, but again, that might not align with the fifth product being 94.Wait, maybe the interval is determined by the group size. If each group has 20 products, and we need to sample one from each group, the interval within each group would be 20? That doesn't make sense because each group is 20 products.Wait, perhaps the interval is the same across all groups, so the interval is 20, meaning that the first product is s, then s+20, s+40, etc. So, if the fifth product is 94, then s + 4*20 = 94, so s = 14. Then, the second product is 14 + 20 = 34.But the problem mentions grouping, so maybe the grouping is just a way to explain the systematic sampling process, where each group is a segment of the population, and you sample one from each segment at the same interval.So, in that case, the interval is 20, and the starting point is 14, so the second product is 34.Alternatively, maybe the interval is 5, since there are 5 groups. So, if the interval is 5, then the fifth product would be s + 4*5 = s + 20 = 94, so s = 74. But that would make the first product 74, which is not possible because the starting point should be less than k, which would be 20 in this case. Wait, no, if k is 5, then starting point is between 1 and 5.Wait, I'm getting confused again. Let me clarify.If we have 100 products and we want to select 5, the interval k is 20. So, starting point s is between 0 and 19. Then, the selected products are s, s+20, s+40, s+60, s+80.Given that the fifth product is 94, which is s + 80 = 94, so s = 14. Therefore, the selected products are 14, 34, 54, 74, 94. So, the second product is 34.But the problem mentions grouping. Maybe the grouping is just a way to explain that the population is divided into 5 groups of 20, and from each group, one product is selected at the same interval. So, the interval between the groups is 20, meaning that the first product is from group 1, the next from group 2, etc., each 20 apart.So, in that case, the fifth product is 94, which is in group 5 (80-99). So, starting from group 1, the first product is s, then group 2 is s+20, group 3 is s+40, group 4 is s+60, group 5 is s+80. So, s+80=94, so s=14. Therefore, the second product is s+20=34.So, I think the answer is 34.But let me double-check. If the starting point is 14, then the products are 14, 34, 54, 74, 94. Yes, that makes sense. The fifth product is 94, so the second is 34.Alternatively, if the interval was different, say 5, then the fifth product would be s + 4*5 = s + 20 = 94, so s=74. But that would mean the starting point is 74, which is beyond the first group (00-19). So, that doesn't make sense because the starting point should be within the first group.Therefore, the interval must be 20, and the starting point is 14, so the second product is 34.**Final Answer**The product number of the 2nd group selected is boxed{34}.

answer：To determine how many natural numbers between 6 and 60 are relatively prime to 15, I start by identifying the range of numbers we're considering. The numbers greater than 6 and less than 60 are from 7 to 59, which gives us a total of 53 numbers.Next, I need to find how many of these numbers are relatively prime to 15. Two numbers are relatively prime if their greatest common divisor (GCD) is 1. Since 15 factors into 3 and 5, any number that is not divisible by 3 or 5 will be relatively prime to 15.I'll use the principle of inclusion and exclusion to calculate this. First, I count the numbers divisible by 3 in the range. Dividing 59 by 3 gives approximately 19.666, so there are 19 numbers divisible by 3. Similarly, dividing 59 by 5 gives 11.8, so there are 11 numbers divisible by 5.However, some numbers are divisible by both 3 and 5, which means they are divisible by 15. Dividing 59 by 15 gives about 3.933, so there are 3 numbers divisible by both.Using inclusion and exclusion, the total number of numbers not relatively prime to 15 is 19 (divisible by 3) plus 11 (divisible by 5) minus 3 (divisible by both), which equals 27.Finally, subtracting this from the total number of numbers in the range, 53 minus 27, gives 26 numbers that are relatively prime to 15.

answer：Alright, so I've got this problem here where Xiao Ming is trying to estimate the number of yellow balls in a bag. The bag has 10 red balls and an unknown number of yellow balls. Instead of just dumping them out and counting, he's using a method where he draws 10 balls each time, calculates the ratio of red balls to the total drawn (which is 10), and then puts them back. He does this 20 times and finds that the average ratio is 0.4. Now, I need to figure out how many yellow balls are in the bag based on this information.Hmm, okay. So, let me break this down. The bag has 10 red balls and some yellow balls. Let's denote the number of yellow balls as Y. So, the total number of balls in the bag is 10 + Y.Xiao Ming is drawing 10 balls each time and calculating the ratio of red balls to 10. So, each time he draws, he's looking at how many red balls he got out of those 10, and then he's taking that number and dividing it by 10 to get the ratio. Then he does this 20 times and takes the average of those ratios, which is 0.4.So, the average ratio is 0.4. That means, on average, in each draw of 10 balls, he got 0.4 red balls. Wait, 0.4 red balls? That doesn't make much sense because you can't have a fraction of a ball. But, maybe it's the average over 20 trials, so it's an average ratio.Wait, actually, let me think. If he draws 10 balls each time, the number of red balls he can get each time is an integer between 0 and 10. So, the ratio would be that integer divided by 10. So, the ratio is a decimal between 0 and 1. So, when he says the average ratio is 0.4, that means on average, he got 4 red balls out of 10 each time.Wait, hold on. If the ratio is 0.4, then 0.4 times 10 is 4. So, on average, he got 4 red balls each time he drew 10. So, the average number of red balls per draw is 4.So, how does this relate to the total number of red and yellow balls?Well, this seems like a problem that can be approached using probability. Specifically, the expected value of the number of red balls drawn in each trial.When you draw balls from a bag without replacement, the probability of drawing a red ball on each draw is proportional to the number of red balls in the bag. So, in each draw of 10 balls, the expected number of red balls should be equal to the total number of red balls divided by the total number of balls in the bag, multiplied by the number of balls drawn.In formula terms, the expected number of red balls in a draw of 10 is:E = (Number of red balls / Total number of balls) * Number of balls drawnSo, plugging in the numbers we have:E = (10 / (10 + Y)) * 10But we also know that the average number of red balls drawn over 20 trials is 4. So, the expected value E should be equal to 4.So, setting up the equation:(10 / (10 + Y)) * 10 = 4Simplify that:100 / (10 + Y) = 4Then, solving for Y:100 = 4 * (10 + Y)100 = 40 + 4YSubtract 40 from both sides:60 = 4YDivide both sides by 4:Y = 15So, according to this, the number of yellow balls is 15.Wait, let me verify that again.We have 10 red balls and Y yellow balls. The total number of balls is 10 + Y.When drawing 10 balls, the expected number of red balls is (10 / (10 + Y)) * 10.Given that the average number of red balls drawn was 4, so:(10 / (10 + Y)) * 10 = 4Which simplifies to 100 / (10 + Y) = 4So, 10 + Y = 100 / 4 = 25Therefore, Y = 25 - 10 = 15Yes, that seems correct.Alternatively, another way to think about it is using proportions. If the ratio of red balls in the bag is 10 / (10 + Y), then when you draw 10 balls, you would expect that same proportion to be red. So, 10 / (10 + Y) = 4 / 10, since on average, 4 out of 10 drawn are red.So, setting up the proportion:10 / (10 + Y) = 4 / 10Cross-multiplying:10 * 10 = 4 * (10 + Y)100 = 40 + 4YAgain, 60 = 4Y, so Y = 15.Yep, that's consistent.So, the number of yellow balls is 15.But wait, just to make sure I didn't make any mistakes in my reasoning. Let me think again.He drew 10 balls each time, and on average, 4 were red. So, the proportion of red balls in the sample is 4/10, which is 0.4. Since he's sampling with replacement each time (because he puts the balls back after each draw), the proportion of red balls in each draw should reflect the proportion in the entire bag.Therefore, the proportion of red balls in the bag is equal to the proportion observed in the samples. So, 10 / (10 + Y) = 0.4Solving for Y:10 / (10 + Y) = 0.4Multiply both sides by (10 + Y):10 = 0.4 * (10 + Y)10 = 4 + 0.4YSubtract 4:6 = 0.4YDivide both sides by 0.4:Y = 6 / 0.4 = 15Yes, same result.So, the number of yellow balls is 15.Therefore, the approximate number of yellow balls in the bag is 15.**Final Answer**The estimated number of yellow balls in the bag is boxed{15}.