💡Okay, so I have this problem here: Given the function f(x) = sin(x) + 2x, where x is a real number, and it's given that f(2a) < f(a - 1). I need to find the range of a. Hmm, okay, let me try to figure this out step by step.First, let me write down the function again to make sure I have it right: f(x) = sin(x) + 2x. So, it's a combination of a sine function and a linear function. I remember that sine functions are periodic and oscillate between -1 and 1, while linear functions like 2x just keep increasing or decreasing depending on the coefficient. In this case, since the coefficient is positive, 2x is increasing.Now, the problem states that f(2a) is less than f(a - 1). So, I need to compare the values of the function at two different points: 2a and a - 1. To do this, maybe I should first analyze the function f(x) itself to see if it's increasing or decreasing. If I can determine whether f(x) is increasing or decreasing, that might help me compare f(2a) and f(a - 1).To check if f(x) is increasing or decreasing, I can find its derivative. The derivative of f(x) with respect to x is f'(x) = cos(x) + 2. Hmm, cos(x) oscillates between -1 and 1, so cos(x) + 2 will oscillate between 1 and 3. That means the derivative is always positive because the smallest it can be is 1, which is still greater than 0. So, f'(x) > 0 for all x. This tells me that f(x) is an increasing function on the entire real line. That's good to know because if the function is increasing, then if I have two inputs, say x1 and x2, and x1 < x2, then f(x1) < f(x2).So, since f(x) is increasing, the inequality f(2a) < f(a - 1) implies that 2a < a - 1. Because if the function is increasing, the input with the smaller value will have a smaller output. So, if f(2a) is less than f(a - 1), then 2a must be less than a - 1.Let me write that down: 2a < a - 1. Now, I can solve this inequality for a. Subtract a from both sides: 2a - a < -1, which simplifies to a < -1. So, it seems that the range of a is all real numbers less than -1.Wait, let me double-check that. If a is less than -1, then 2a is less than a - 1. Let me pick a value for a that's less than -1, say a = -2. Then 2a = -4, and a - 1 = -3. So, -4 < -3, which is true. Then, f(-4) should be less than f(-3). Let me compute f(-4) and f(-3) to verify.f(-4) = sin(-4) + 2*(-4) = -sin(4) - 8. Similarly, f(-3) = sin(-3) + 2*(-3) = -sin(3) - 6. Now, sin(4) is approximately sin(4 radians) which is about -0.7568, so f(-4) ≈ -(-0.7568) - 8 ≈ 0.7568 - 8 ≈ -7.2432. Similarly, sin(3) is approximately 0.1411, so f(-3) ≈ -0.1411 - 6 ≈ -6.1411. So, f(-4) ≈ -7.2432 and f(-3) ≈ -6.1411. Therefore, f(-4) < f(-3), which is consistent with our earlier conclusion.What if a is equal to -1? Then 2a = -2, and a - 1 = -2. So, f(-2) and f(-2) are equal, so f(2a) = f(a - 1). But the problem states f(2a) < f(a - 1), so a cannot be equal to -1. Similarly, if a is greater than -1, say a = 0, then 2a = 0 and a - 1 = -1. So, f(0) = sin(0) + 0 = 0, and f(-1) = sin(-1) + 2*(-1) ≈ -0.8415 - 2 ≈ -2.8415. So, f(0) = 0 and f(-1) ≈ -2.8415. Therefore, f(0) > f(-1), which contradicts the given inequality f(2a) < f(a - 1). So, a cannot be greater than or equal to -1.Therefore, the range of a is all real numbers less than -1, which can be written as (-∞, -1).Wait, just to make sure I haven't missed anything, let me think again. The function f(x) is strictly increasing because its derivative is always positive. So, if f(2a) < f(a - 1), then it must be that 2a < a - 1, which simplifies to a < -1. There doesn't seem to be any other condition or constraint here because the function is defined for all real numbers, and the inequality is straightforward once we know the function is increasing.I think that's solid. I don't see any mistakes in my reasoning. So, the range of a is all real numbers less than -1.**Final Answer**The range of ( a ) is boxed{(-infty, -1)}.

💡Okay, so I have this problem here: Let Sₙ be the sum of the first n terms of an arithmetic sequence {aₙ}. If S₇ = 35, then what is a₄? The options are A: 8, B: 7, C: 6, D: 5.Hmm, arithmetic sequences. I remember that in an arithmetic sequence, each term increases by a constant difference. So, if I denote the first term as a₁ and the common difference as d, then the nth term aₙ can be written as a₁ + (n-1)d.Now, the sum of the first n terms, Sₙ, is given by the formula Sₙ = n/2 * (a₁ + aₙ). Alternatively, since aₙ = a₁ + (n-1)d, the sum can also be written as Sₙ = n/2 * [2a₁ + (n-1)d]. In this problem, we're given that S₇ = 35. So, plugging n = 7 into the sum formula, we get S₇ = 7/2 * (a₁ + a₇) = 35. Let me write that down:7/2 * (a₁ + a₇) = 35.I can simplify this equation. First, multiply both sides by 2 to get rid of the denominator:7 * (a₁ + a₇) = 70.Then, divide both sides by 7:a₁ + a₇ = 10.Okay, so the sum of the first and seventh terms is 10. Now, I need to find a₄, which is the fourth term of the sequence.Since it's an arithmetic sequence, the terms are equally spaced. So, the fourth term is actually the average of the first and seventh terms. That makes sense because in an arithmetic sequence, the middle term is the average of the terms equidistant from it. Since 7 is odd, the fourth term is right in the middle.So, a₄ = (a₁ + a₇)/2. We already found that a₁ + a₇ = 10, so:a₄ = 10/2 = 5.Therefore, the fourth term is 5, which corresponds to option D.Wait, let me double-check my reasoning. I used the property that in an arithmetic sequence, the middle term is the average of the terms around it. For seven terms, the fourth term is indeed the middle one. So, if the sum of the first and seventh terms is 10, their average is 5, which is the fourth term. That seems correct.Alternatively, I could have used the sum formula in terms of a₁ and d. Let's try that approach to confirm.We know that S₇ = 35, so using the sum formula:S₇ = 7/2 * [2a₁ + (7-1)d] = 35.Simplify:7/2 * (2a₁ + 6d) = 35.Multiply both sides by 2:7*(2a₁ + 6d) = 70.Divide both sides by 7:2a₁ + 6d = 10.Simplify:a₁ + 3d = 5.But wait, a₁ + 3d is exactly the fourth term, a₄. Because a₄ = a₁ + (4-1)d = a₁ + 3d.So, a₁ + 3d = 5, which means a₄ = 5. Yep, that confirms it.So, both methods lead me to the same answer, which is 5. That must be the correct choice, D.I think I'm confident with this answer now. It's always good to check using different methods to make sure I didn't make a mistake. Both approaches—the average method and expanding the sum formula—give me the same result, so I feel pretty sure that D is the right answer.**Final Answer**boxed{D}

💡Okay, so I have this problem here about transforming a sine function and finding a symmetry axis. Let me try to figure this out step by step. The original function is ( y = sin(4x - frac{pi}{6}) ). The first transformation is to stretch the horizontal coordinates to twice their original length. I remember that stretching horizontally by a factor of 2 means replacing ( x ) with ( frac{x}{2} ) in the function. So, let me apply that.Starting with ( y = sin(4x - frac{pi}{6}) ), if I stretch horizontally by 2, the new function becomes ( y = sin(4(frac{x}{2}) - frac{pi}{6}) ). Simplifying inside the sine function: ( 4 times frac{x}{2} = 2x ), so the function becomes ( y = sin(2x - frac{pi}{6}) ). Wait, is that right? Stretching horizontally by 2 should make the period longer, which would mean the coefficient of ( x ) should be smaller. Since the original coefficient is 4, stretching by 2 should make it 2. Yeah, that seems correct.Next, the problem says to shift the graph to the left by ( frac{pi}{4} ) units. Shifting left by ( frac{pi}{4} ) means replacing ( x ) with ( x + frac{pi}{4} ) in the function. So, applying that to ( y = sin(2x - frac{pi}{6}) ), we get ( y = sin(2(x + frac{pi}{4}) - frac{pi}{6}) ).Let me simplify that. First, distribute the 2: ( 2x + 2 times frac{pi}{4} = 2x + frac{pi}{2} ). So now the function is ( y = sin(2x + frac{pi}{2} - frac{pi}{6}) ). Combining the constants: ( frac{pi}{2} - frac{pi}{6} ). To subtract these, I need a common denominator. ( frac{pi}{2} = frac{3pi}{6} ), so ( frac{3pi}{6} - frac{pi}{6} = frac{2pi}{6} = frac{pi}{3} ). So the function simplifies to ( y = sin(2x + frac{pi}{3}) ).Okay, so after both transformations, the function is ( y = sin(2x + frac{pi}{3}) ). Now, I need to find the equation of one symmetry axis of this resulting graph.I remember that the sine function has symmetry about its midline, which is usually the x-axis, but when there's a vertical shift, it changes. However, in this case, there's no vertical shift, so the midline is still the x-axis. But symmetry axis might refer to the vertical line around which the sine wave is symmetric. For a standard sine function ( y = sin(x) ), it's symmetric about the line ( x = frac{pi}{2} + kpi ) for any integer ( k ). These are the peaks and troughs.But in our transformed function, the sine wave has been stretched and shifted. So, the symmetry axis would be shifted accordingly. Let me recall that for a function ( y = sin(Bx + C) ), the phase shift is ( -frac{C}{B} ). So, in our case, ( B = 2 ) and ( C = frac{pi}{3} ), so the phase shift is ( -frac{pi/3}{2} = -frac{pi}{6} ). That means the graph is shifted to the left by ( frac{pi}{6} ).But wait, we already shifted it to the left by ( frac{pi}{4} ) earlier. Hmm, maybe I need to think differently. The symmetry axis of a sine function is usually at the midpoints between its maximum and minimum points. So, perhaps I should find the maximum and minimum points of the transformed function and then find the midpoint between them.Let me find the critical points. The function is ( y = sin(2x + frac{pi}{3}) ). The maximum value of sine is 1, and the minimum is -1. The maximum occurs when ( 2x + frac{pi}{3} = frac{pi}{2} + 2pi k ), and the minimum occurs when ( 2x + frac{pi}{3} = frac{3pi}{2} + 2pi k ) for any integer ( k ).Let me solve for ( x ) in both cases.For maximum:( 2x + frac{pi}{3} = frac{pi}{2} + 2pi k )Subtract ( frac{pi}{3} ):( 2x = frac{pi}{2} - frac{pi}{3} + 2pi k )Find a common denominator for ( frac{pi}{2} - frac{pi}{3} ):( frac{3pi}{6} - frac{2pi}{6} = frac{pi}{6} )So, ( 2x = frac{pi}{6} + 2pi k )Divide by 2:( x = frac{pi}{12} + pi k )For minimum:( 2x + frac{pi}{3} = frac{3pi}{2} + 2pi k )Subtract ( frac{pi}{3} ):( 2x = frac{3pi}{2} - frac{pi}{3} + 2pi k )Common denominator:( frac{9pi}{6} - frac{2pi}{6} = frac{7pi}{6} )So, ( 2x = frac{7pi}{6} + 2pi k )Divide by 2:( x = frac{7pi}{12} + pi k )So, the maximum points are at ( x = frac{pi}{12} + pi k ) and the minimum points are at ( x = frac{7pi}{12} + pi k ).Now, the symmetry axis should be exactly halfway between a maximum and the next minimum. Let's take ( k = 0 ) for simplicity. So, the maximum is at ( x = frac{pi}{12} ) and the minimum is at ( x = frac{7pi}{12} ).The midpoint between ( frac{pi}{12} ) and ( frac{7pi}{12} ) is:( frac{frac{pi}{12} + frac{7pi}{12}}{2} = frac{frac{8pi}{12}}{2} = frac{frac{2pi}{3}}{2} = frac{pi}{3} )Wait, that's ( frac{pi}{3} ). But looking at the options, ( frac{pi}{3} ) is option C. But I thought the phase shift was ( -frac{pi}{6} ). Maybe I need to consider that.Alternatively, perhaps the symmetry axis is at the phase shift plus a quarter period. Let me recall that for ( y = sin(Bx + C) ), the phase shift is ( -C/B ), and the period is ( frac{2pi}{B} ). So, the first maximum after the phase shift would be at ( x = text{phase shift} + frac{pi}{2B} ).Wait, let me think. The standard sine function ( y = sin(x) ) has its first maximum at ( frac{pi}{2} ). So, for ( y = sin(Bx + C) ), the first maximum occurs when ( Bx + C = frac{pi}{2} ), so ( x = frac{pi/2 - C}{B} ). Similarly, the first minimum occurs when ( Bx + C = frac{3pi}{2} ), so ( x = frac{3pi/2 - C}{B} ).So, the midpoint between the first maximum and first minimum would be:( frac{frac{pi/2 - C}{B} + frac{3pi/2 - C}{B}}{2} = frac{frac{2pi - 2C}{B}}{2} = frac{pi - C}{B} )In our case, ( C = frac{pi}{3} ) and ( B = 2 ), so:( frac{pi - frac{pi}{3}}{2} = frac{frac{2pi}{3}}{2} = frac{pi}{3} )So, that confirms the symmetry axis is at ( x = frac{pi}{3} ), which is option C.But wait, earlier I thought the phase shift was ( -frac{pi}{6} ). Let me see. The phase shift is indeed ( -frac{C}{B} = -frac{pi/3}{2} = -frac{pi}{6} ). So, the graph is shifted to the left by ( frac{pi}{6} ). But the symmetry axis is at ( frac{pi}{3} ), which is to the right of the phase shift. That makes sense because the symmetry axis is halfway between a maximum and a minimum, which are spread out over the period.Alternatively, maybe I can think about the graph. The original function ( y = sin(4x - frac{pi}{6}) ) has a certain period and phase shift. After stretching horizontally by 2, the period becomes longer, and then shifting left by ( frac{pi}{4} ). So, the combined effect is a phase shift that's the sum of the original phase shift and the shift from the transformation.Wait, let me calculate the total phase shift. The original function is ( y = sin(4x - frac{pi}{6}) ), which can be written as ( y = sin(4(x - frac{pi}{24})) ). So, the original phase shift is ( frac{pi}{24} ) to the right. Then, we stretch horizontally by 2, which would affect the phase shift. Stretching by 2 would multiply the phase shift by 2, so ( frac{pi}{24} times 2 = frac{pi}{12} ). Then, we shift left by ( frac{pi}{4} ), which is equivalent to subtracting ( frac{pi}{4} ) from the phase shift. So, total phase shift is ( frac{pi}{12} - frac{pi}{4} = frac{pi}{12} - frac{3pi}{12} = -frac{2pi}{12} = -frac{pi}{6} ). So, that matches the earlier calculation.But the symmetry axis is not the same as the phase shift. The phase shift is where the sine wave starts, but the symmetry axis is the vertical line around which the wave is symmetric. For a sine wave, this is typically at the midpoint between a peak and the next trough.Given that, and having calculated the midpoint between the first maximum at ( frac{pi}{12} ) and the first minimum at ( frac{7pi}{12} ) as ( frac{pi}{3} ), that should be the symmetry axis.But let me double-check. If I plot the function ( y = sin(2x + frac{pi}{3}) ), it should have its first maximum at ( x = frac{pi}{12} ) and first minimum at ( x = frac{7pi}{12} ). The midpoint is indeed ( frac{pi}{3} ), so that's the line of symmetry.Alternatively, I can think about the general form of the sine function and its symmetry. For ( y = sin(Bx + C) ), the function is symmetric about the line ( x = frac{pi}{2B} - frac{C}{B} ). Wait, is that correct? Let me see.The standard sine function ( y = sin(x) ) is symmetric about ( x = frac{pi}{2} ). For ( y = sin(Bx + C) ), the first maximum is at ( x = frac{pi/2 - C}{B} ), and the first minimum is at ( x = frac{3pi/2 - C}{B} ). The midpoint between these two is ( frac{pi - C}{B} ), which is the line of symmetry.So, in our case, ( B = 2 ) and ( C = frac{pi}{3} ), so the symmetry axis is at ( x = frac{pi - frac{pi}{3}}{2} = frac{frac{2pi}{3}}{2} = frac{pi}{3} ). That confirms it again.Therefore, the equation of one symmetry axis is ( x = frac{pi}{3} ), which is option C.But wait, the options given are:A: ( x = frac{pi}{12} )B: ( x = frac{pi}{6} )C: ( x = frac{pi}{3} )D: ( x = -frac{pi}{12} )So, ( frac{pi}{3} ) is option C. That seems to be the correct answer.Just to make sure, let me consider another approach. The symmetry axis of a sine function is also where the function reaches its midline and is either increasing or decreasing. For ( y = sin(2x + frac{pi}{3}) ), the midline is y=0. The function crosses the midline at points where ( 2x + frac{pi}{3} = kpi ), so ( x = frac{kpi - frac{pi}{3}}{2} ).For ( k = 1 ), ( x = frac{pi - frac{pi}{3}}{2} = frac{frac{2pi}{3}}{2} = frac{pi}{3} ). So, at ( x = frac{pi}{3} ), the function crosses the midline. Since the sine function is symmetric about this point, it's a line of symmetry.Therefore, I'm confident that the correct answer is ( x = frac{pi}{3} ), which is option C.

💡First, let's denote the side length of the second square as ( s ). Since the side length of the first square is double that of the second square, the side length of the first square will be ( 2s ).Next, we calculate the areas of both squares. The area of a square is given by the formula ( text{side length}^2 ).- The area of the first square (( A_1 )) is: [ A_1 = (2s)^2 = 4s^2 ]- The area of the second square (( A_2 )) is: [ A_2 = s^2 ]To find the ratio of the area of the first square to the area of the second square, we divide ( A_1 ) by ( A_2 ):[text{Ratio} = frac{A_1}{A_2} = frac{4s^2}{s^2} = 4]Therefore, the ratio of the area of the first square to the area of the second square is ( 4 ).

💡Alright, so I have this problem where there are n friends, and they take r distinct photos. Each photo has at least one person, and no two photos are exactly the same. The key condition here is that for every pair of photos, there must be at least one person who appears in both. I need to find the maximum value of r.Okay, let me try to understand this better. Each photo is essentially a subset of the n friends, right? And since each photo has at least one person, we're talking about non-empty subsets. The total number of possible non-empty subsets is 2^n - 1 because each friend can either be in a photo or not, giving us 2^n subsets, and we subtract 1 to exclude the empty set.But we have this additional condition that every pair of photos must share at least one common person. So, if I think about it, this means that no two photos can be completely disjoint; they must overlap in at least one person. That makes me think about something called an intersecting family in combinatorics. I remember that an intersecting family is a set system where any two sets intersect, meaning they share at least one common element.In this case, our sets are the photos, and the elements are the friends. So, we're looking for the largest intersecting family of subsets of an n-element set, where each subset is non-empty. I think there's a theorem related to this called Erdos-Ko-Rado theorem, but I'm not entirely sure about the specifics. Maybe I should look that up or try to recall.Wait, the Erdos-Ko-Rado theorem is about intersecting families of k-element subsets, right? It states that for n ≥ 2k, the maximum size of an intersecting family of k-element subsets is C(n-1, k-1). But in our case, the subsets can be of any size, not just a fixed size k. So, maybe the theorem doesn't apply directly here.Hmm, maybe I need a different approach. Let's think about the problem differently. If every pair of photos must share at least one person, then we can't have two photos that are complements of each other. Because if one photo is, say, {A, B}, then its complement would be {C, D, E, ..., N}, and these two would be disjoint. So, to satisfy the condition, we can't have both a subset and its complement in our family of photos.Therefore, the maximum number of photos would be half of the total number of non-empty subsets, because for every subset, we can include either the subset or its complement, but not both. The total number of non-empty subsets is 2^n - 1, so half of that would be (2^n - 1)/2. But wait, (2^n - 1)/2 isn't necessarily an integer. For example, if n=2, 2^2 -1 = 3, and 3/2 is 1.5, which isn't an integer. So, maybe we need to take the floor of that value.But actually, let's think about it more carefully. If n is at least 1, then 2^n is even when n ≥ 1, right? Because 2^n is always even for n ≥ 1. So, 2^n -1 is odd, and (2^n -1)/2 would be a half-integer. But since we can't have half a photo, we need to take the floor of that value, which would be 2^{n-1} - 1/2, but that still doesn't make sense because we need an integer.Wait, maybe I'm approaching this incorrectly. Let's consider that for each non-empty subset, we can pair it with its complement, which is also non-empty because the original subset is non-empty. So, there are (2^n -1)/2 such pairs. Since we can't have both subsets in a pair, we can choose at most one from each pair. Therefore, the maximum number of photos would be (2^n -1)/2. But again, this is a fractional value when n is such that 2^n -1 is odd.Hmm, perhaps I need to think about it differently. Maybe instead of trying to pair subsets with their complements, I can fix a particular person and ensure that this person is in every photo. That way, every pair of photos will share this common person, satisfying the condition. How many subsets include this fixed person? Well, if we fix one person, say person A, then the number of subsets that include A is 2^{n-1}, because for each of the remaining n-1 people, they can either be included or not.But wait, 2^{n-1} includes the subset that only contains person A, right? And all other subsets that include A and some combination of the other n-1 people. So, that gives us 2^{n-1} subsets. But in our problem, we need non-empty subsets, and since we're fixing person A, all these subsets are non-empty. So, does that mean the maximum number of photos is 2^{n-1}?But earlier, I thought it might be (2^n -1)/2, which is approximately 2^{n-1} - 0.5. But since we can't have half a photo, maybe 2^{n-1} is the correct answer. Let me test this with small values of n.Let's take n=1. There's only one person, so the only possible photo is {A}. There's only one photo, so r=1. According to the formula 2^{n-1}, that would be 2^{0}=1, which matches.Now, n=2. The possible non-empty subsets are {A}, {B}, {A,B}. If we fix person A, the subsets including A are {A}, {A,B}. So, r=2. According to 2^{n-1}, that's 2^{1}=2, which matches. But according to (2^n -1)/2, that's (4-1)/2=1.5, which isn't an integer, but we can take the floor, which would be 1, but that's less than 2. So, in this case, fixing a person gives us a better result.Wait, but in n=2, can we actually have r=2? Let's see. If we have photos {A} and {A,B}, then any pair shares A. But if we try to add {B}, then {A} and {B} don't share any common person, which violates the condition. So, indeed, r=2 is the maximum for n=2.Similarly, for n=3. The non-empty subsets are {A}, {B}, {C}, {A,B}, {A,C}, {B,C}, {A,B,C}. If we fix person A, the subsets including A are {A}, {A,B}, {A,C}, {A,B,C}, so r=4. According to 2^{n-1}, that's 2^{2}=4, which matches. If we try to include any subset not containing A, like {B}, then {A} and {B} don't share any common person, which violates the condition. So, r=4 is indeed the maximum.Wait, but what if we don't fix a particular person? Is there a way to have more than 2^{n-1} photos without having any two photos being disjoint? Let me think about n=3. If I try to include all subsets except those that don't contain A, that would be 4 subsets. But if I try to include another subset that doesn't contain A, say {B}, then {A} and {B} are disjoint, which is not allowed. So, I can't include more than 4 subsets.Similarly, for n=4, fixing a person would give us 8 subsets, which seems to be the maximum. If we try to include any subset not containing that fixed person, we'd have to exclude some subsets to maintain the intersection property, but I don't think we can get more than 8 subsets.Therefore, it seems that fixing a particular person and including all subsets that contain that person gives us the maximum number of photos, which is 2^{n-1}. This satisfies the condition that every pair of photos shares at least one common person, and it's the largest possible family of subsets with this property.I think I've convinced myself that the maximum r is 2^{n-1}. Let me just recap:- Each photo is a non-empty subset of n friends.- To ensure every pair of photos shares at least one person, we can fix one person to be in all photos.- The number of subsets containing this fixed person is 2^{n-1}.- Trying to include more subsets would require including subsets that don't contain the fixed person, which could lead to disjoint subsets, violating the condition.Therefore, the maximum r is 2^{n-1}.

💡First, I observe that there are two equations with two variables, ( x ) and ( y ). This suggests that I can solve for both variables by manipulating the equations.I notice that if I add the two equations together, the ( y ) terms will cancel out, allowing me to solve for ( x ) first. Adding the equations gives me ( 2x = 26 ), so ( x = 13 ).With the value of ( x ) known, I can substitute it back into one of the original equations to find ( y ). Using ( x + y = 18 ), I substitute ( x = 13 ) and solve for ( y ), resulting in ( y = 5 ).This method efficiently solves the system of equations by eliminating one variable at a time, leading to the correct value of ( y ).