💡First, I need to determine the number of lemons required to make one gallon of lemonade. Given that 40 lemons are needed for 50 gallons, I can divide the total number of lemons by the total number of gallons to find the lemons per gallon ratio.Once I have the lemons per gallon ratio, I can multiply it by the desired number of gallons, which is 15 gallons in this case, to find out how many lemons are needed for 15 gallons of lemonade.By following these steps, I can ensure that the ratio of lemons to gallons remains consistent, allowing me to accurately determine the number of lemons required for any given quantity of lemonade.

💡Okay, so I have this function f(x) defined as log base 1/e of (x² + 1/e) minus the absolute value of x over e. I need to find the range of x where f(x+1) is less than f(2x-1). The options are given as A: (0,2), B: (-∞,0), C: (-∞,0) ∪ (2, ∞), and D: (2, ∞). Hmm, let's try to break this down step by step.First, I should probably understand the function f(x) better. It's a logarithmic function with a base of 1/e, which is the same as e^(-1). So, log base 1/e is the same as negative natural logarithm because log_{1/a}(b) = -log_a(b). That might be useful later.So, f(x) = log_{1/e}(x² + 1/e) - |x/e|. Let me rewrite that using natural logarithms to make it easier. Since log_{1/e}(x² + 1/e) = -ln(x² + 1/e)/ln(e) = -ln(x² + 1/e). Because ln(e) is 1, so it's just -ln(x² + 1/e). So, f(x) simplifies to -ln(x² + 1/e) - |x/e|.Alright, so f(x) is a combination of a logarithmic term and an absolute value term. The logarithmic term is negative, so as x² increases, the logarithm increases, but since it's negative, the whole term decreases. The absolute value term is also negative, so as |x| increases, that term becomes more negative. So, f(x) is a combination of two decreasing functions in terms of |x|.Wait, actually, let me think again. The logarithmic term is -ln(x² + 1/e). So, as x² increases, ln(x² + 1/e) increases, so -ln(x² + 1/e) decreases. So, as |x| increases, the logarithmic term decreases. The absolute value term is -|x/e|, which also decreases as |x| increases. So, overall, f(x) is a function that decreases as |x| increases. So, it's symmetric in some way?Wait, the function f(x) is even because f(-x) = f(x). Let me check that. If I plug in -x, I get log_{1/e}((-x)^2 + 1/e) - |-x/e| = log_{1/e}(x² + 1/e) - |x/e|, which is the same as f(x). So, yes, f(x) is even. That's an important property because it tells me that the function is symmetric about the y-axis. So, whatever happens on the positive side will mirror on the negative side.Now, the problem is to find where f(x+1) < f(2x - 1). Since f is even, maybe I can relate this to the absolute values of x+1 and 2x -1? Let me think.Given that f is even, f(a) = f(|a|). So, f(x+1) = f(|x+1|) and f(2x -1) = f(|2x -1|). Therefore, the inequality f(x+1) < f(2x -1) is equivalent to f(|x+1|) < f(|2x -1|). Since f is a function that decreases as its argument increases (because both the logarithmic term and the absolute value term decrease as |x| increases), this would mean that |x+1| > |2x -1|.Wait, let me make sure. If f is decreasing, then f(a) < f(b) implies that a > b because as a increases, f(a) decreases. So, yes, if f(|x+1|) < f(|2x -1|), then |x+1| > |2x -1|.So, the problem reduces to solving the inequality |x + 1| > |2x - 1|. That seems more manageable. Let's solve this inequality.To solve |x + 1| > |2x - 1|, I can square both sides to eliminate the absolute values, since both sides are non-negative. So, (x + 1)^2 > (2x - 1)^2.Expanding both sides:Left side: (x + 1)^2 = x² + 2x + 1Right side: (2x - 1)^2 = 4x² - 4x + 1So, the inequality becomes:x² + 2x + 1 > 4x² - 4x + 1Subtracting the right side from both sides:x² + 2x + 1 - 4x² + 4x - 1 > 0Simplify:-3x² + 6x > 0Factor:-3x(x - 2) > 0Multiply both sides by -1 (remembering to reverse the inequality sign):3x(x - 2) < 0So, 3x(x - 2) < 0Since 3 is positive, we can ignore it for the inequality:x(x - 2) < 0This inequality holds when x is between 0 and 2 because the product of x and (x - 2) is negative when one is positive and the other is negative. Specifically, when x is between 0 and 2, x is positive and (x - 2) is negative, so their product is negative.Therefore, the solution is 0 < x < 2, which is interval A.But wait, let me double-check if squaring both sides introduced any extraneous solutions or missed any cases. The original inequality is |x + 1| > |2x - 1|. Squaring both sides is valid because both sides are non-negative, so the inequality direction remains the same.Another way to solve |x + 1| > |2x - 1| is to consider different cases based on the critical points where the expressions inside the absolute values change sign, which are at x = -1 and x = 1/2.Case 1: x < -1In this case, x + 1 is negative, so |x + 1| = -(x + 1) = -x -1Similarly, 2x -1 is negative because x < -1 < 1/2, so |2x -1| = -(2x -1) = -2x +1So, the inequality becomes:-x -1 > -2x +1Adding 2x to both sides:x -1 > 1Adding 1 to both sides:x > 2But in this case, x < -1, so x > 2 is not possible. Therefore, no solution in this interval.Case 2: -1 ≤ x < 1/2Here, x +1 is non-negative, so |x +1| = x +1But 2x -1 is still negative because x < 1/2, so |2x -1| = -2x +1Inequality:x +1 > -2x +1Adding 2x to both sides:3x +1 >1Subtracting 1:3x >0So, x >0But in this case, x is between -1 and 1/2. So, the solution here is 0 < x <1/2Case 3: x ≥1/2Here, both x +1 and 2x -1 are non-negative, so |x +1| =x +1 and |2x -1| =2x -1Inequality:x +1 >2x -1Subtracting x from both sides:1 >x -1Adding 1:2 >xSo, x <2But in this case, x ≥1/2, so the solution here is 1/2 ≤x <2Combining the solutions from Case 2 and Case 3:0 <x <1/2 and 1/2 ≤x <2, which together make 0 <x <2So, same result as before. Therefore, the solution is 0 <x <2, which is interval A.But wait, the options are A: (0,2), B: (-∞,0), C: (-∞,0) ∪ (2, ∞), D: (2, ∞). So, according to my calculations, the correct answer is A.But hold on, the initial analysis said that f(x) is decreasing for x >0 and increasing for x <0. So, f(x) is decreasing on (0, ∞) and increasing on (-∞,0). Also, f is even, so it's symmetric.So, if I have f(x+1) < f(2x -1), and f is decreasing for positive arguments and increasing for negative arguments, how does that affect the inequality?Wait, maybe I should consider the behavior of f(x) in different regions.Since f is even, f(a) = f(|a|). So, f(x+1) = f(|x+1|) and f(2x -1) = f(|2x -1|). Since f is decreasing for positive arguments, f(a) < f(b) implies a > b when a and b are positive. So, |x+1| > |2x -1|.So, that's consistent with my previous approach.But let me think about the function's behavior. For x >0, f(x) is decreasing, so as x increases, f(x) decreases. For x <0, f(x) is increasing, so as x increases (towards zero), f(x) decreases. Wait, that seems contradictory.Wait, if x <0, f(x) is increasing as x increases. So, for x <0, as x becomes more negative, f(x) decreases, and as x approaches zero from the left, f(x) increases.Wait, let me clarify. If f(x) is increasing for x <0, that means as x increases (from -∞ towards 0), f(x) increases. So, for x <0, f(x) is increasing, meaning f(-2) < f(-1) < f(0). Similarly, for x >0, f(x) is decreasing, so f(0) > f(1) > f(2), etc.So, f(x) has a maximum at x=0, and it decreases as |x| increases in both directions.Therefore, f(x) is symmetric about the y-axis, peaking at x=0, and decreasing as we move away from zero in either direction.Given that, when we have f(x+1) < f(2x -1), since f is decreasing for positive arguments and increasing for negative arguments, we need to consider the absolute values.But since f is even, f(x+1) = f(|x+1|) and f(2x -1) = f(|2x -1|). So, the inequality f(|x+1|) < f(|2x -1|) implies that |x+1| > |2x -1| because f is decreasing for positive arguments.Therefore, solving |x+1| > |2x -1| gives us the solution 0 <x <2, as before.But wait, let's test some values to make sure.Let me pick x=1, which is in (0,2). Then, f(x+1)=f(2), and f(2x -1)=f(1). Since f is decreasing for x>0, f(2) < f(1), so f(x+1) < f(2x -1). That works.What about x=3, which is not in the solution. Then, f(x+1)=f(4), and f(2x -1)=f(5). Since f is decreasing, f(4) > f(5), so f(x+1) > f(2x -1). So, the inequality does not hold, which is consistent.What about x=-1, which is in option B. Then, f(x+1)=f(0), and f(2x -1)=f(-3). Since f is even, f(-3)=f(3). So, f(0) is the maximum, and f(3) is less than f(0). So, f(x+1)=f(0) > f(3)=f(2x -1). So, the inequality f(x+1) < f(2x -1) does not hold for x=-1. So, x=-1 is not a solution, which is consistent with our previous result.Another test: x=0.5, which is in (0,2). Then, f(x+1)=f(1.5), f(2x -1)=f(0). Since f(1.5) < f(0), the inequality holds.x= -0.5, which is not in the solution. f(x+1)=f(0.5), f(2x -1)=f(-2). Since f is even, f(-2)=f(2). So, f(0.5) > f(2), so f(x+1) > f(2x -1). So, inequality does not hold.Wait, but according to the options, C is (-∞,0) ∪ (2, ∞). But according to my analysis, only (0,2) satisfies the inequality. So, why is option C there?Wait, maybe I missed something. Let me think again.Wait, when I considered f(x) being even, I concluded that f(x+1) < f(2x -1) is equivalent to |x+1| > |2x -1|. But is that the only case?Wait, no. Because f is decreasing for positive arguments and increasing for negative arguments. So, if |x+1| and |2x -1| are both positive, then f(|x+1|) < f(|2x -1|) implies |x+1| > |2x -1|.But what if one of them is negative? Wait, no, because f is even, so f(a) = f(|a|). So, regardless of the sign of a, f(a) depends only on |a|. So, the inequality f(x+1) < f(2x -1) is equivalent to f(|x+1|) < f(|2x -1|), which, since f is decreasing for positive arguments, implies |x+1| > |2x -1|.Therefore, the only solution is 0 <x <2.But wait, let me think about the original function f(x). It's defined as log_{1/e}(x² +1/e) - |x/e|. So, the domain of f(x) is all real numbers because x² +1/e is always positive, and |x/e| is defined for all x. So, f(x) is defined for all real x.But when we have f(x+1) and f(2x -1), we need to ensure that the arguments inside the logarithm are positive. Wait, x² +1/e is always positive, so no problem there. So, f(x) is defined for all real x.But perhaps I should consider the behavior of f(x) when x is negative. Since f(x) is increasing for x <0, maybe there's another case where f(x+1) < f(2x -1) when x is negative.Wait, let me consider x negative.Suppose x is negative, say x = -a where a >0.Then, f(x+1) = f(-a +1). Depending on whether -a +1 is positive or negative, f(-a +1) could be in the increasing or decreasing region.Similarly, f(2x -1) = f(-2a -1), which is negative, so f(-2a -1) = f(2a +1) because f is even.Wait, let me write this more clearly.Let x = -a, a >0.Then, f(x+1) = f(-a +1). If -a +1 >0, i.e., a <1, then f(-a +1) is in the decreasing region because x+1 >0. If -a +1 <0, i.e., a >1, then f(-a +1) is in the increasing region because x+1 <0.Similarly, f(2x -1) = f(-2a -1) = f(2a +1) because f is even, and 2a +1 is positive, so f(2a +1) is in the decreasing region.So, depending on whether a <1 or a >1, f(x+1) could be in the increasing or decreasing region.So, let's consider two cases for x negative:Case 1: x < -1 (a >1)Then, x+1 <0, so f(x+1) is in the increasing region. Since f is increasing for x <0, f(x+1) = f(|x+1|) but wait, no, f(x+1) is f(-a +1) where a >1, so x+1 = -a +1 <0.But f is increasing for x <0, so as x increases (towards -1), f(x) increases.Wait, this is getting a bit complicated. Maybe I should approach it differently.Since f is even, f(x+1) = f(|x+1|) and f(2x -1) = f(|2x -1|). So, regardless of the sign of x+1 and 2x -1, f(x+1) and f(2x -1) depend only on their absolute values.Therefore, the inequality f(x+1) < f(2x -1) is equivalent to f(|x+1|) < f(|2x -1|). Since f is decreasing for positive arguments, this implies |x+1| > |2x -1|.So, regardless of whether x is positive or negative, the inequality reduces to |x+1| > |2x -1|.Therefore, the solution is 0 <x <2, as previously found.But wait, let me test x = -2, which is in option C. Then, f(x+1)=f(-1)=f(1), and f(2x -1)=f(-5)=f(5). Since f is decreasing, f(1) > f(5), so f(x+1) > f(2x -1). Therefore, the inequality does not hold for x=-2.Similarly, x=3, which is in option C. f(x+1)=f(4), f(2x -1)=f(5). Since f is decreasing, f(4) > f(5), so f(x+1) > f(2x -1). So, inequality does not hold.Wait, so why is option C there? Maybe I'm missing something.Wait, perhaps when x is negative, even though f(|x+1|) < f(|2x -1|) implies |x+1| > |2x -1|, but for negative x, |x+1| and |2x -1| can behave differently.Wait, let me solve |x+1| > |2x -1| again, but considering x can be negative.We have |x+1| > |2x -1|Squaring both sides: (x+1)^2 > (2x -1)^2Which simplifies to -3x² +6x >0, as before, leading to 0 <x <2.But wait, when x is negative, squaring both sides still holds because both sides are positive. So, the solution is still 0 <x <2.Therefore, the only solution is x in (0,2), which is option A.But the initial analysis by the assistant concluded that the answer is A, but the user is now thinking about why option C is present.Wait, perhaps the assistant made a mistake in the initial analysis. Let me check.The assistant said:"When x > 0, f(x) is decreasing. When x <0, f(x) is increasing. Also, f is even. If f(x+1) < f(2x -1), then |x+1| > |2x -1|. Solving this yields 0 <x <2. Therefore, the correct choice is A."But according to my analysis, that's correct. So, why is option C there? Maybe the question is designed to test whether we consider the evenness correctly.Wait, perhaps the assistant is correct, and the answer is A. But the options include C, which is (-∞,0) ∪ (2, ∞). Maybe I need to consider another approach.Wait, let's think about the function f(x) again. It's even, so f(x) = f(-x). So, f(x+1) = f(-x -1). Similarly, f(2x -1) = f(-2x +1).So, the inequality f(x+1) < f(2x -1) can also be written as f(-x -1) < f(-2x +1).But since f is increasing for x <0, f(a) < f(b) implies a < b when a and b are negative.So, if both -x -1 and -2x +1 are negative, then f(-x -1) < f(-2x +1) implies -x -1 < -2x +1.Solving:-x -1 < -2x +1Adding 2x to both sides:x -1 <1Adding 1:x <2But in this case, both -x -1 and -2x +1 are negative. So, when is -x -1 <0 and -2x +1 <0?-x -1 <0 => x > -1-2x +1 <0 => x >1/2So, for both to be negative, x must be >1/2.So, in this case, x >1/2 and x <2, so 1/2 <x <2.But we also need to consider when one of them is negative and the other is positive.Case 1: -x -1 <0 and -2x +1 >0So, x > -1 and x <1/2In this case, f(-x -1) is in the increasing region, and f(-2x +1) is in the decreasing region because -2x +1 >0.So, f(-x -1) < f(-2x +1) would require that since f is increasing for negative arguments and decreasing for positive arguments, it's not straightforward. Maybe we need to compare f(-x -1) and f(-2x +1) directly.But since f is even, f(-x -1) = f(x +1) and f(-2x +1) = f(2x -1). So, it's the same as the original inequality.Wait, this is getting too convoluted. Maybe the initial approach was correct, that the inequality reduces to |x+1| > |2x -1|, leading to 0 <x <2.Therefore, the correct answer is A: (0,2).But why is option C present? Maybe it's a distractor to test if we consider the function's evenness incorrectly.Alternatively, perhaps the assistant made a mistake in the initial analysis by not considering the behavior for negative x.Wait, let me think again. If x is negative, say x = -a, a >0.Then, f(x+1) = f(-a +1). If -a +1 >0, i.e., a <1, then f(-a +1) = f(|-a +1|) = f(a -1). Since a <1, a -1 <0, so f(a -1) is in the increasing region.Wait, no, f(a -1) is f(|a -1|) because f is even. Wait, no, f(-a +1) = f(|-a +1|) only if f is even. But f is even, so f(-a +1) = f(|-a +1|) = f(|a -1|).Wait, this is confusing. Let me clarify.Since f is even, f(-a +1) = f(|-a +1|). So, regardless of whether -a +1 is positive or negative, f(-a +1) = f(|-a +1|).Similarly, f(2x -1) = f(|2x -1|).Therefore, the inequality f(x+1) < f(2x -1) is equivalent to f(|x+1|) < f(|2x -1|), which, since f is decreasing for positive arguments, implies |x+1| > |2x -1|.Therefore, regardless of x being positive or negative, the inequality reduces to |x+1| > |2x -1|, leading to 0 <x <2.Therefore, the correct answer is A: (0,2).So, why is option C there? Maybe it's a common mistake to think that since f is increasing for x <0, the inequality could hold for x <0 as well. But in reality, because f is even, the inequality f(x+1) < f(2x -1) depends only on the absolute values, leading to the same solution for all x.Therefore, the correct answer is A: (0,2).

💡Okay, so I have this geometry problem here. Let me try to understand it step by step. The problem says:"The line ( l ) is tangent to a circle with diameter ( AB ) at point ( C ). Let ( M ) and ( N ) be the projections of points ( A ) and ( B ) onto the line ( l ), respectively, and let ( D ) be the projection of point ( C ) onto ( AB ). Prove that ( CD^2 = AM cdot BN )."Alright, so let me visualize this. There's a circle with diameter ( AB ), so points ( A ) and ( B ) are at the ends of the diameter. The line ( l ) is tangent to this circle at point ( C ). That means ( l ) touches the circle only at ( C ) and doesn't cross it. Now, ( M ) is the projection of ( A ) onto line ( l ), and ( N ) is the projection of ( B ) onto line ( l ). So, if I imagine dropping perpendiculars from ( A ) and ( B ) onto the tangent line ( l ), their feet are ( M ) and ( N ) respectively. Also, ( D ) is the projection of ( C ) onto ( AB ). So, if I drop a perpendicular from ( C ) onto the diameter ( AB ), the foot of that perpendicular is ( D ). The goal is to prove that ( CD^2 = AM cdot BN ). Hmm, interesting. So, the square of the length of ( CD ) is equal to the product of the lengths of ( AM ) and ( BN ). Let me start by drawing a diagram to help me see this. I have circle with diameter ( AB ), tangent at ( C ), projections ( M ) and ( N ) on ( l ), and projection ( D ) on ( AB ). First, since ( AB ) is the diameter, the center of the circle ( O ) is the midpoint of ( AB ). That might come in handy. Also, since ( l ) is tangent at ( C ), the radius ( OC ) is perpendicular to ( l ). So, ( OC perp l ). Now, ( M ) is the projection of ( A ) onto ( l ), so ( AM ) is perpendicular to ( l ). Similarly, ( BN ) is perpendicular to ( l ). So, both ( AM ) and ( BN ) are perpendicular to ( l ), meaning they are parallel to each other because they are both perpendicular to the same line. Wait, so ( AM ) and ( BN ) are both perpendicular to ( l ), so they are parallel. That might help in setting up similar triangles or something. Also, ( D ) is the projection of ( C ) onto ( AB ), so ( CD ) is perpendicular to ( AB ). Let me think about the properties of projections and similar triangles here. Since ( AM ) and ( BN ) are both perpendicular to ( l ), and ( OC ) is also perpendicular to ( l ), maybe I can relate these segments somehow. Perhaps I can consider triangles involving these projections. Let me see. Since ( AM ) is perpendicular to ( l ) and ( OC ) is also perpendicular to ( l ), then ( AM ) is parallel to ( OC ). Similarly, ( BN ) is parallel to ( OC ). So, all three segments ( AM ), ( OC ), and ( BN ) are parallel. Hmm, maybe I can use similar triangles here. Let me think about triangles involving ( A ), ( M ), ( C ), ( D ), ( B ), ( N ), and ( O ). Let me consider triangle ( AMC ). Since ( AM ) is perpendicular to ( l ) and ( OC ) is also perpendicular to ( l ), then ( AM ) is parallel to ( OC ). Therefore, triangle ( AMC ) is similar to triangle ( AOC ) because they have the same angles. Wait, is that right? Let me check. If ( AM ) is parallel to ( OC ), then the corresponding angles should be equal. So, angle ( MAC ) is equal to angle ( ACO ). But ( angle ACO ) is 90 degrees because ( OC ) is perpendicular to ( l ). So, angle ( MAC ) is also 90 degrees? Wait, no, that can't be because ( AM ) is perpendicular to ( l ), but ( MAC ) is an angle at ( A ), not necessarily 90 degrees. Wait, maybe I made a mistake there. Let me think again. Since ( AM ) is perpendicular to ( l ), and ( OC ) is perpendicular to ( l ), then ( AM ) is parallel to ( OC ). So, in triangle ( AMC ), side ( AM ) is parallel to side ( OC ) in triangle ( AOC ). Therefore, by the basic proportionality theorem (Thales' theorem), the line ( AM ) divides the sides ( AC ) and ( AB ) proportionally. Wait, but ( OC ) is a radius, so ( OC = frac{AB}{2} ). Hmm, maybe I can use that. Alternatively, maybe I can use coordinate geometry. Let me assign coordinates to the points to make it easier. Let me place the circle with diameter ( AB ) on the x-axis, with center at the origin. So, let me set ( A ) at ( (-r, 0) ) and ( B ) at ( (r, 0) ), where ( r ) is the radius of the circle. Then, the center ( O ) is at ( (0, 0) ). Since ( l ) is tangent to the circle at ( C ), let me find the coordinates of ( C ). The tangent at ( C ) will have the equation ( xx_1 + yy_1 = r^2 ), where ( (x_1, y_1) ) is the point of tangency. But since ( AB ) is the diameter, ( C ) must lie somewhere on the circle above or below the x-axis. Let me assume it's above the x-axis for simplicity. Let me denote ( C ) as ( (a, b) ), lying on the circle, so ( a^2 + b^2 = r^2 ). The tangent line at ( C ) is ( ax + by = r^2 ). Now, ( M ) is the projection of ( A ) onto ( l ). The projection of a point ( (x_0, y_0) ) onto the line ( ax + by + c = 0 ) is given by a formula, but in this case, the line ( l ) is ( ax + by = r^2 ), so ( c = -r^2 ). The formula for the projection ( M ) of ( A(-r, 0) ) onto ( l ) is:[M = left( frac{a(a(-r) + b(0) - r^2)}{a^2 + b^2}, frac{b(a(-r) + b(0) - r^2)}{a^2 + b^2} right) + text{something?}]Wait, maybe I should recall the formula correctly. The projection of a point ( P(x_0, y_0) ) onto the line ( lx + my + n = 0 ) is given by:[left( x_0 - l cdot frac{l x_0 + m y_0 + n}{l^2 + m^2}, y_0 - m cdot frac{l x_0 + m y_0 + n}{l^2 + m^2} right)]In our case, the line ( l ) is ( ax + by - r^2 = 0 ), so ( l = a ), ( m = b ), ( n = -r^2 ).So, projecting ( A(-r, 0) ) onto ( l ):Compute ( l x_0 + m y_0 + n = a(-r) + b(0) - r^2 = -ar - r^2 ).Then,[x = -r - a cdot frac{-ar - r^2}{a^2 + b^2}][y = 0 - b cdot frac{-ar - r^2}{a^2 + b^2}]Simplify:[x = -r + a cdot frac{ar + r^2}{a^2 + b^2}][y = 0 + b cdot frac{ar + r^2}{a^2 + b^2}]Similarly, for point ( B(r, 0) ), projecting onto ( l ):Compute ( l x_0 + m y_0 + n = a(r) + b(0) - r^2 = ar - r^2 ).Then,[x = r - a cdot frac{ar - r^2}{a^2 + b^2}][y = 0 - b cdot frac{ar - r^2}{a^2 + b^2}]Simplify:[x = r - a cdot frac{ar - r^2}{a^2 + b^2}][y = 0 + b cdot frac{r^2 - ar}{a^2 + b^2}]Hmm, this is getting a bit messy. Maybe there's a better way. Let me think about the properties of projections and similar triangles again.Since ( AM ) and ( BN ) are both perpendicular to ( l ), and ( OC ) is also perpendicular to ( l ), then ( AM ), ( OC ), and ( BN ) are all parallel. So, triangles involving these segments might be similar.Let me consider triangles ( AMC ) and ( AOC ). Since ( AM ) is parallel to ( OC ), the triangles ( AMC ) and ( AOC ) are similar by the AA similarity criterion (they have two equal angles). Similarly, triangles ( BNC ) and ( BOC ) are also similar.Wait, is that correct? Let me check. If ( AM parallel OC ), then angle ( MAC = angle ACO ). But ( angle ACO ) is 90 degrees because ( OC perp l ). So, angle ( MAC ) is also 90 degrees? But ( AM ) is already perpendicular to ( l ), so that makes sense. Hmm, maybe not. Wait, no. ( angle MAC ) is the angle at ( A ) between ( MA ) and ( AC ). Since ( MA ) is perpendicular to ( l ), and ( AC ) is a chord of the circle, the angle ( MAC ) is not necessarily 90 degrees. Maybe I need to approach this differently. Let me consider the power of a point. Since ( l ) is tangent to the circle at ( C ), the power of point ( A ) with respect to the circle is ( AC^2 ). Similarly, the power of point ( B ) is ( BC^2 ). But the power of a point ( A ) with respect to the circle is also equal to ( AM cdot AN ), where ( AN ) is the length from ( A ) to the intersection of line ( AB ) with the circle. Wait, no, that's not quite right. The power of point ( A ) is equal to ( AM cdot AN ) if ( AN ) is the secant line, but in this case, ( l ) is tangent, so the power is ( AC^2 ). Wait, actually, the power of point ( A ) with respect to the circle is ( AC^2 ) because ( l ) is tangent at ( C ). Similarly, the power of point ( B ) is ( BC^2 ). But how does that relate to ( AM ) and ( BN )? Hmm. Alternatively, maybe I can use coordinate geometry as I started earlier. Let me try to compute ( AM ) and ( BN ) in terms of coordinates.Given that ( A(-r, 0) ) and ( M ) is the projection onto ( l ), which is ( ax + by = r^2 ). The distance from ( A ) to ( l ) is ( AM ). The formula for the distance from a point ( (x_0, y_0) ) to the line ( ax + by + c = 0 ) is ( frac{|ax_0 + by_0 + c|}{sqrt{a^2 + b^2}} ). In our case, the line ( l ) is ( ax + by - r^2 = 0 ), so the distance from ( A(-r, 0) ) to ( l ) is:[AM = frac{|a(-r) + b(0) - r^2|}{sqrt{a^2 + b^2}} = frac{|-ar - r^2|}{sqrt{a^2 + b^2}} = frac{r(a + r)}{sqrt{a^2 + b^2}}]Similarly, the distance from ( B(r, 0) ) to ( l ) is:[BN = frac{|a(r) + b(0) - r^2|}{sqrt{a^2 + b^2}} = frac{|ar - r^2|}{sqrt{a^2 + b^2}} = frac{r(r - a)}{sqrt{a^2 + b^2}}]So, ( AM = frac{r(a + r)}{sqrt{a^2 + b^2}} ) and ( BN = frac{r(r - a)}{sqrt{a^2 + b^2}} ). Therefore, the product ( AM cdot BN ) is:[AM cdot BN = left( frac{r(a + r)}{sqrt{a^2 + b^2}} right) left( frac{r(r - a)}{sqrt{a^2 + b^2}} right) = frac{r^2 (a + r)(r - a)}{a^2 + b^2}]Simplify the numerator:[(a + r)(r - a) = r^2 - a^2]So,[AM cdot BN = frac{r^2 (r^2 - a^2)}{a^2 + b^2}]But since ( C(a, b) ) lies on the circle, ( a^2 + b^2 = r^2 ). Therefore, substituting:[AM cdot BN = frac{r^2 (r^2 - a^2)}{r^2} = r^2 - a^2 = b^2]Wait, that's interesting. So, ( AM cdot BN = b^2 ). Now, what is ( CD )? ( D ) is the projection of ( C ) onto ( AB ). Since ( AB ) is the x-axis in my coordinate system, the projection of ( C(a, b) ) onto ( AB ) is ( D(a, 0) ). Therefore, the length ( CD ) is the vertical distance from ( C ) to ( D ), which is ( |b - 0| = |b| ). Since ( b ) is positive (as I assumed ( C ) is above the x-axis), ( CD = b ). Therefore, ( CD^2 = b^2 ). But earlier, I found that ( AM cdot BN = b^2 ). Therefore, ( CD^2 = AM cdot BN ), which is exactly what we needed to prove. Wait, that seems too straightforward. Did I make any mistakes? Let me double-check.1. Assigned coordinates correctly: Yes, with ( A(-r, 0) ), ( B(r, 0) ), center ( O(0, 0) ), and ( C(a, b) ) on the circle, so ( a^2 + b^2 = r^2 ).2. Equation of tangent at ( C ): Correct, ( ax + by = r^2 ).3. Distance from ( A ) to ( l ): Correct, using the distance formula. Similarly for ( B ).4. Calculated ( AM ) and ( BN ): Correct, substituted into the formula.5. Product ( AM cdot BN ): Correct, simplified to ( b^2 ).6. ( CD ): Correct, since projection onto x-axis, ( CD = b ), so ( CD^2 = b^2 ).Therefore, all steps seem correct, and the conclusion follows. So, ( CD^2 = AM cdot BN ) is proven.

💡First, I need to understand the relationship between the cost price (CP), marked price (MP), and the discount percentage (D%). The problem states that the cost price is 25% of the marked price, which means CP = 0.25 * MP.Next, the gain percent is given as 100%, indicating that the selling price (SP) is twice the cost price. Therefore, SP = 2 * CP.The selling price after applying the discount can also be expressed as SP = MP - (D% of MP). Substituting CP with 0.25 * MP in the equation SP = 2 * CP gives 2 * (0.25 * MP) = 0.5 * MP.Setting this equal to the discounted selling price equation, 0.5 * MP = MP - (D/100 * MP), and solving for D% involves isolating D/100. This leads to D/100 = 0.5, and multiplying both sides by 100 gives D = 50.Therefore, the percentage of the discount allowed is 50%.

💡First, I need to evaluate the expression (frac{6 + 6 times 3 - 3}{3}).I'll start by handling the multiplication in the numerator:[6 times 3 = 18]So, the expression becomes:[frac{6 + 18 - 3}{3}]Next, I'll perform the addition and subtraction in the numerator:[6 + 18 = 24][24 - 3 = 21]Now, the expression is:[frac{21}{3}]Finally, I'll divide 21 by 3:[frac{21}{3} = 7]The value of the expression is 7, which corresponds to option (B).

💡Alright, so I have this problem: What is the smallest prime divisor of (3^{15} + 11^{21})? Hmm, okay. Let me try to figure this out step by step.First, I know that prime divisors are prime numbers that can divide another number without leaving a remainder. The smallest prime number is 2, followed by 3, 5, 7, and so on. So, if I can determine whether 2 divides (3^{15} + 11^{21}), that would immediately tell me that 2 is the smallest prime divisor. If not, I'd have to check the next smallest primes.Let me think about the parity of (3^{15}) and (11^{21}). Both 3 and 11 are odd numbers. I remember that any power of an odd number remains odd. So, (3^{15}) is odd because 3 is odd and any exponent of an odd number is still odd. Similarly, (11^{21}) is also odd for the same reason.Now, if I add two odd numbers together, what do I get? Let me recall: odd plus odd equals even. For example, 3 + 5 = 8, which is even. So, (3^{15} + 11^{21}) must be an even number because it's the sum of two odd numbers.Since the sum is even, it means that 2 divides this sum. Therefore, 2 is a prime divisor of (3^{15} + 11^{21}). Since 2 is the smallest prime number, it must be the smallest prime divisor of the given expression.Wait, let me double-check to make sure I didn't miss anything. Is there a possibility that (3^{15} + 11^{21}) could be even but not divisible by 2? No, that doesn't make sense because if a number is even, by definition, it's divisible by 2. So, 2 is definitely a divisor here.Is there any chance that 2 isn't the smallest prime divisor? Well, 2 is the smallest prime number, so unless the number itself is 2, which it isn't because (3^{15}) and (11^{21}) are both much larger than 2, 2 will always be the smallest prime divisor if it divides the number.Just to be thorough, let me consider the actual values of (3^{15}) and (11^{21}) modulo 2. Since both 3 and 11 are congruent to 1 modulo 2 (because they're odd), raising them to any power will still give 1 modulo 2. So, (3^{15} equiv 1 mod 2) and (11^{21} equiv 1 mod 2). Adding them together, (1 + 1 = 2 equiv 0 mod 2). This confirms that the sum is indeed divisible by 2.Therefore, I'm confident that the smallest prime divisor of (3^{15} + 11^{21}) is 2.