answer：First, I need to determine the range of the function ( f(x) = (x-1)^2 + 1 ) for the given domain ( x in {-1, 0, 1, 2, 3} ).I'll start by substituting each value of ( x ) into the function and calculating the corresponding ( f(x) ) values.1. For ( x = -1 ): [ f(-1) = (-1 - 1)^2 + 1 = (-2)^2 + 1 = 4 + 1 = 5 ]2. For ( x = 0 ): [ f(0) = (0 - 1)^2 + 1 = (-1)^2 + 1 = 1 + 1 = 2 ]3. For ( x = 1 ): [ f(1) = (1 - 1)^2 + 1 = 0^2 + 1 = 0 + 1 = 1 ]4. For ( x = 2 ): [ f(2) = (2 - 1)^2 + 1 = 1^2 + 1 = 1 + 1 = 2 ]5. For ( x = 3 ): [ f(3) = (3 - 1)^2 + 1 = 2^2 + 1 = 4 + 1 = 5 ]After calculating all the function values, I have the set of results: ( {5, 2, 1, 2, 5} ). To find the range, I need to identify the unique values from this set.The unique values are ( 1, 2, ) and ( 5 ). Therefore, the range of the function is ( {1, 2, 5} ).

answer：Alright, so I have this problem here with two sets, A and B, and I need to find their intersection. Let's see what each set is defined as.Set A is given by A = {x ∈ Z | 3^x > 1}. Okay, so this is the set of all integers x where 3 raised to the power of x is greater than 1. Hmm, 3^x is an exponential function, right? So when is 3^x greater than 1? Well, 3^0 is 1, and for any positive exponent, 3^x will be greater than 1. For negative exponents, 3^x becomes a fraction less than 1. So, x has to be greater than 0. Therefore, set A is all positive integers: 1, 2, 3, and so on.Now, set B is defined as B = {x ∈ R | x^2 - 5x - 6 < 0}. This is a quadratic inequality. To solve this, I need to find the values of x for which the quadratic expression x^2 - 5x - 6 is less than zero. First, let's factor the quadratic. Looking at x^2 - 5x - 6, I need two numbers that multiply to -6 and add up to -5. Hmm, -6 and +1 work because (-6) * (+1) = -6 and (-6) + (+1) = -5. So, the factored form is (x - 6)(x + 1) < 0.Now, to solve the inequality (x - 6)(x + 1) < 0, I need to find the intervals where the product is negative. The roots of the equation are x = 6 and x = -1. These roots divide the real number line into three intervals: (-∞, -1), (-1, 6), and (6, ∞). To determine where the product is negative, I'll test each interval:1. For x < -1, say x = -2: (-2 - 6)(-2 + 1) = (-8)(-1) = 8, which is positive.2. For -1 < x < 6, say x = 0: (0 - 6)(0 + 1) = (-6)(1) = -6, which is negative.3. For x > 6, say x = 7: (7 - 6)(7 + 1) = (1)(8) = 8, which is positive.So, the inequality (x - 6)(x + 1) < 0 holds true for -1 < x < 6. Therefore, set B is the interval (-1, 6).Now, to find A ∩ B, which is the intersection of set A and set B. Set A is all positive integers: 1, 2, 3, 4, 5, 6, 7, and so on. Set B is all real numbers between -1 and 6. So, the intersection would be the integers that are in both sets.Looking at set A, the positive integers start at 1 and go upwards. Set B includes all numbers greater than -1 and less than 6. So, the integers that satisfy both conditions are 1, 2, 3, 4, and 5. Wait, does 6 count? Set B is up to but not including 6, so 6 is not included. Therefore, the intersection A ∩ B is {1, 2, 3, 4, 5}.Let me double-check to make sure I didn't miss anything. Set A is definitely all positive integers because 3^x > 1 only when x > 0. Set B is the interval (-1, 6), which doesn't include -1 and 6 but includes all numbers in between. So, the integers within that interval are 1 through 5. Yep, that seems right.I think I've got it. The intersection is the set {1, 2, 3, 4, 5}.

answer：Alright, so I've got this triangle problem here, and I need to find the area of triangle ABC. Let me read it again to make sure I understand all the given information.In triangle ABC, the sides opposite angles A, B, and C are a, b, and c respectively. It's given that c = 2, sin A = 2 sin C, and cos B = 1/4. I need to find the area S of triangle ABC, and the options are A: 1, B: 2√15, C: √15, D: √15/4.Okay, let's break this down step by step. First, I know that in any triangle, the sum of the angles is 180 degrees, so A + B + C = 180°. But I don't know any of the angles yet, only some relationships between their sines and cosines.Given that sin A = 2 sin C, that seems important. Maybe I can use the Law of Sines here. The Law of Sines states that a/sin A = b/sin B = c/sin C. Since I know c = 2, maybe I can relate a and c through their sines.So, from the Law of Sines, a/sin A = c/sin C. But sin A = 2 sin C, so substituting that in, we get a/(2 sin C) = 2/sin C. Hmm, that simplifies to a/2 = 2, so a = 4. Okay, so side a is 4. That's useful.Now, I also know that cos B = 1/4. Since I need to find the area, which often involves sine of an angle, maybe I can find sin B from cos B. I remember that sin² B + cos² B = 1, so sin B = sqrt(1 - cos² B). Plugging in cos B = 1/4, we get sin B = sqrt(1 - (1/4)²) = sqrt(1 - 1/16) = sqrt(15/16) = √15 / 4. So sin B is √15 / 4.Now, to find the area of the triangle, I can use the formula (1/2)ab sin C, but I need to know two sides and the included angle. Alternatively, since I have sides a and c, and angle B is between them? Wait, no, actually, in triangle ABC, side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C. So, sides a, b, c are opposite angles A, B, C respectively.So, if I have sides a and c, and angle B, is angle B between sides a and c? Let me visualize the triangle. If angle B is at vertex B, then sides a and c are connected at vertex B. So yes, angle B is between sides a and c. Therefore, the area can be calculated as (1/2) * a * c * sin B.Given that a = 4, c = 2, and sin B = √15 / 4, plugging these into the formula: Area = (1/2) * 4 * 2 * (√15 / 4). Let's compute that step by step.First, 1/2 * 4 is 2. Then, 2 * 2 is 4. Then, 4 * (√15 / 4) is √15. So the area is √15. Looking at the options, that's option C.Wait, let me double-check to make sure I didn't make any mistakes. So, a = 4, c = 2, sin B = √15 / 4. So, (1/2)*4*2*(√15 /4). The 4 in the numerator and the 4 in the denominator cancel out, leaving (1/2)*2*√15, which is √15. Yep, that seems correct.Alternatively, I could have used the formula (1/2)*a*c*sin B, which is the same thing. So, that's another way to think about it. Either way, the area comes out to √15.I think that's solid. I don't see any errors in my reasoning. So, I'm confident that the area is √15, which is option C.**Final Answer**boxed{C}

answer：Okay, so I have this math problem here, and I need to find the value of (frac{x cdot z}{y cdot w}). The problem gives me some relationships between the variables: (x) is four times (y), (y) is three times (z), and (z) is five times (w). Hmm, that sounds a bit complicated at first, but I think I can break it down step by step.First, let me write down what each relationship means in terms of equations. If (x) is four times (y), that translates to (x = 4y). Similarly, (y) is three times (z), so (y = 3z). And (z) is five times (w), which means (z = 5w). Okay, so now I have these three equations:1. (x = 4y)2. (y = 3z)3. (z = 5w)I need to find (frac{x cdot z}{y cdot w}). To do that, I think I should express all the variables in terms of one variable so that I can substitute them into the expression. It seems like (w) is the smallest variable here, so maybe I can express everything in terms of (w). Let me try that.Starting with (z = 5w), that's straightforward. Then, since (y = 3z), I can substitute (z) with (5w). So, (y = 3 times 5w = 15w). Got that. Now, moving up to (x), which is (4y). Since (y = 15w), substituting that in gives (x = 4 times 15w = 60w). Okay, so now I have all variables expressed in terms of (w):- (x = 60w)- (y = 15w)- (z = 5w)Great, now I can substitute these into the expression (frac{x cdot z}{y cdot w}). Let's do that step by step.First, calculate the numerator: (x cdot z). Substituting the values, that's (60w times 5w). Multiplying those together, I get (60 times 5 = 300) and (w times w = w^2), so the numerator is (300w^2).Next, the denominator: (y cdot w). Substituting the values, that's (15w times w). Multiplying those, I get (15 times 1 = 15) and (w times w = w^2), so the denominator is (15w^2).Now, putting it all together, the expression becomes (frac{300w^2}{15w^2}). Hmm, I notice that both the numerator and the denominator have (w^2), so I can simplify that by canceling them out. That leaves me with (frac{300}{15}).Now, dividing 300 by 15. Let me do that calculation. 15 times 20 is 300, so (300 div 15 = 20). Therefore, the value of (frac{x cdot z}{y cdot w}) is 20.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the relationships:1. (x = 4y)2. (y = 3z)3. (z = 5w)Expressing each in terms of (w):- (z = 5w)- (y = 3 times 5w = 15w)- (x = 4 times 15w = 60w)Substituting into the expression:[frac{x cdot z}{y cdot w} = frac{60w times 5w}{15w times w} = frac{300w^2}{15w^2} = frac{300}{15} = 20]Yes, that seems correct. I don't see any errors in my calculations. So, the final answer should be 20.

answer：Okay, so I have this problem here: Can the sum of the areas of (n) squares, each with side lengths (1, 3, 5, ldots, (2n-1)) cm, equal (100) m²? Hmm, interesting. Let me try to figure this out step by step.First, I need to understand what exactly is being asked. We have squares with side lengths that are consecutive odd numbers starting from 1 cm up to ((2n-1)) cm. The question is whether the total area of all these squares can add up to (100) square meters.Wait, the side lengths are in centimeters, but the total area is in square meters. I should probably convert everything to the same unit to make things easier. Since (1) meter is (100) centimeters, (1) square meter is (100 times 100 = 10,000) square centimeters. So, (100) square meters would be (100 times 10,000 = 1,000,000) square centimeters. Okay, so I need the sum of the areas of these squares to equal (1,000,000) cm².Now, the area of each square is the side length squared. So, the areas are (1^2, 3^2, 5^2, ldots, (2n-1)^2) cm². Therefore, the total area is the sum of these squares:[1^2 + 3^2 + 5^2 + ldots + (2n-1)^2]I remember there's a formula for the sum of the squares of the first (n) odd numbers. Let me recall it. I think it's related to the sum of squares formula. The sum of the squares of the first (n) natural numbers is (frac{n(n+1)(2n+1)}{6}). But we're dealing with odd numbers here.Wait, maybe I can express the sum of the squares of the first (n) odd numbers in terms of the sum of squares of the first (2n) natural numbers minus the sum of squares of the first (n) even numbers. Let me write that down:[sum_{k=1}^{n} (2k-1)^2 = sum_{k=1}^{2n} k^2 - sum_{k=1}^{n} (2k)^2]Yes, that makes sense because the first (2n) natural numbers include both the odd and even numbers, and if I subtract the sum of the squares of the even numbers, I'm left with the sum of the squares of the odd numbers.So, let's compute each part separately. First, the sum of the squares of the first (2n) natural numbers:[sum_{k=1}^{2n} k^2 = frac{2n(2n+1)(4n+1)}{6}]And the sum of the squares of the first (n) even numbers:[sum_{k=1}^{n} (2k)^2 = 4 sum_{k=1}^{n} k^2 = 4 times frac{n(n+1)(2n+1)}{6}]So, plugging these back into our expression for the sum of the squares of the odd numbers:[sum_{k=1}^{n} (2k-1)^2 = frac{2n(2n+1)(4n+1)}{6} - 4 times frac{n(n+1)(2n+1)}{6}]Let me simplify this expression. First, factor out (frac{2n(2n+1)}{6}) from both terms:[sum_{k=1}^{n} (2k-1)^2 = frac{2n(2n+1)}{6} left[(4n+1) - 2(n+1)right]]Simplify inside the brackets:[(4n + 1) - 2(n + 1) = 4n + 1 - 2n - 2 = 2n - 1]So now, the sum becomes:[sum_{k=1}^{n} (2k-1)^2 = frac{2n(2n+1)(2n - 1)}{6}]I can simplify this further by canceling out the 2 in the numerator and denominator:[sum_{k=1}^{n} (2k-1)^2 = frac{n(2n+1)(2n - 1)}{3}]Okay, so that's the formula for the sum of the squares of the first (n) odd numbers. Now, we set this equal to (1,000,000) cm²:[frac{n(2n+1)(2n - 1)}{3} = 1,000,000]Multiply both sides by 3 to eliminate the denominator:[n(2n+1)(2n - 1) = 3,000,000]Let me denote (2n) as (m) for simplicity, so (m = 2n). Then, the equation becomes:[frac{m}{2}(m + 1)(m - 1) = 3,000,000]Wait, that might not be necessary. Alternatively, notice that ( (2n+1)(2n - 1) = (2n)^2 - 1 = 4n^2 - 1 ). So, substituting back:[n(4n^2 - 1) = 3,000,000]Which simplifies to:[4n^3 - n = 3,000,000]Or:[4n^3 - n - 3,000,000 = 0]This is a cubic equation in terms of (n). Solving cubic equations can be tricky, but maybe I can estimate the value of (n) that satisfies this equation.Let me try plugging in some large values of (n) to see where it might land.First, let's approximate. The dominant term is (4n^3), so:[4n^3 approx 3,000,000 implies n^3 approx 750,000 implies n approx sqrt[3]{750,000}]Calculating the cube root of 750,000. I know that (90^3 = 729,000) and (91^3 = 753,571). So, (n) is approximately between 90 and 91.Let me test (n = 90):[4(90)^3 - 90 = 4(729,000) - 90 = 2,916,000 - 90 = 2,915,910]Which is less than 3,000,000.Now, (n = 91):[4(91)^3 - 91 = 4(753,571) - 91 = 3,014,284 - 91 = 3,014,193]That's more than 3,000,000.So, the value of (n) lies between 90 and 91. But (n) has to be an integer because it's the number of squares. Therefore, there's no integer (n) that satisfies the equation exactly.Wait, but maybe my approximation is off? Let me check more precisely.Compute (4n^3 - n) for (n = 90):[4(90)^3 - 90 = 4(729,000) - 90 = 2,916,000 - 90 = 2,915,910]Difference from 3,000,000: (3,000,000 - 2,915,910 = 84,090).For (n = 91):[4(91)^3 - 91 = 4(753,571) - 91 = 3,014,284 - 91 = 3,014,193]Difference: (3,014,193 - 3,000,000 = 14,193).So, between (n = 90) and (n = 91), the function crosses 3,000,000. Since (n) must be an integer, and there's no integer between 90 and 91, there is no integer solution for (n).Therefore, it's impossible for the sum of the areas of (n) squares with side lengths (1, 3, 5, ldots, (2n-1)) cm to equal (100) m².But wait, let me double-check my calculations to make sure I didn't make a mistake.First, the formula for the sum of squares of the first (n) odd numbers is correct:[sum_{k=1}^{n} (2k-1)^2 = frac{n(2n+1)(2n - 1)}{3}]Yes, that seems right.Setting that equal to (1,000,000):[frac{n(2n+1)(2n - 1)}{3} = 1,000,000]Multiplying both sides by 3:[n(2n+1)(2n - 1) = 3,000,000]Which simplifies to:[4n^3 - n = 3,000,000]Yes, that's correct.Testing (n = 90) and (n = 91) shows that the function crosses 3,000,000 between these two integers, but since (n) must be an integer, there is no solution.Therefore, my conclusion is correct: it's impossible for the sum of the areas to equal (100) m².

answer：Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. The problem has two parts. Let me try to tackle them one by one.Starting with part (1): I need to find the value of sin C given that sin C + cos C = 1 - sin(C/2). Hmm, that seems a bit tricky, but maybe I can manipulate the equation using some trigonometric identities.First, I remember that sin C can be written as 2 sin(C/2) cos(C/2). Similarly, cos C can be expressed using the double-angle identity as 1 - 2 sin²(C/2). Let me try substituting these into the equation.So, replacing sin C and cos C:2 sin(C/2) cos(C/2) + [1 - 2 sin²(C/2)] = 1 - sin(C/2)Let me simplify the left side:2 sin(C/2) cos(C/2) + 1 - 2 sin²(C/2) = 1 - sin(C/2)Subtracting 1 from both sides:2 sin(C/2) cos(C/2) - 2 sin²(C/2) = - sin(C/2)Hmm, let's factor out 2 sin(C/2) from the left side:2 sin(C/2) [cos(C/2) - sin(C/2)] = - sin(C/2)Now, I can divide both sides by sin(C/2), assuming sin(C/2) ≠ 0. Since C is an angle in a triangle, it must be between 0 and π, so sin(C/2) can't be zero. So, dividing both sides:2 [cos(C/2) - sin(C/2)] = -1Divide both sides by 2:cos(C/2) - sin(C/2) = -1/2Wait, that's interesting. Let me write that as:sin(C/2) - cos(C/2) = 1/2Because I multiplied both sides by -1. So, sin(C/2) - cos(C/2) = 1/2.Now, I can square both sides to eliminate the trigonometric functions. Let's do that:[sin(C/2) - cos(C/2)]² = (1/2)²Expanding the left side:sin²(C/2) - 2 sin(C/2) cos(C/2) + cos²(C/2) = 1/4I know that sin²θ + cos²θ = 1, so sin²(C/2) + cos²(C/2) = 1. Therefore, substituting:1 - 2 sin(C/2) cos(C/2) = 1/4Simplify:1 - 1/4 = 2 sin(C/2) cos(C/2)So, 3/4 = 2 sin(C/2) cos(C/2)But 2 sin(C/2) cos(C/2) is equal to sin C, right? So:sin C = 3/4Okay, that seems to be the answer for part (1). Let me just double-check my steps to make sure I didn't make a mistake.Starting from sin C + cos C = 1 - sin(C/2), I used the double-angle identities correctly. Then, after substituting, I simplified and ended up with sin(C/2) - cos(C/2) = 1/2. Squaring both sides gave me 1 - sin C = 1/4, so sin C = 3/4. That seems consistent.Moving on to part (2): Given that a² + b² = 4(a + b) - 8, find the value of side c.First, let's try to simplify the given equation:a² + b² = 4a + 4b - 8Let me rearrange the terms:a² - 4a + b² - 4b = -8Hmm, this looks like it can be written as two separate quadratic expressions. Let me complete the square for both a and b.For a: a² - 4a. To complete the square, take half of -4, which is -2, square it to get 4. So, a² - 4a + 4 = (a - 2)².Similarly, for b: b² - 4b. Half of -4 is -2, square it to get 4. So, b² - 4b + 4 = (b - 2)².But since I added 4 for a and 4 for b, I need to add them to the right side as well. Let's do that:(a² - 4a + 4) + (b² - 4b + 4) = -8 + 4 + 4Simplify:(a - 2)² + (b - 2)² = 0Wait, the sum of two squares equals zero. That means each square must be zero individually because squares are non-negative.So, (a - 2)² = 0 and (b - 2)² = 0.Therefore, a = 2 and b = 2.So, sides a and b are both 2 units long.Now, I need to find side c. Since I know angles A, B, and C are related to sides a, b, c via the Law of Cosines or Law of Sines. But since I know sin C from part (1), maybe I can use the Law of Cosines.Law of Cosines states that c² = a² + b² - 2ab cos C.I know a = 2, b = 2, so let's compute that.First, compute a² + b²: 2² + 2² = 4 + 4 = 8.Now, compute 2ab: 2 * 2 * 2 = 8.So, c² = 8 - 8 cos C.But I need cos C. From part (1), I know sin C = 3/4. Since sin² C + cos² C = 1, we can find cos C.So, sin C = 3/4, so sin² C = 9/16.Therefore, cos² C = 1 - 9/16 = 7/16.Thus, cos C = ±√(7/16) = ±√7 / 4.But in a triangle, angle C is between 0 and π. From part (1), when we had sin(C/2) - cos(C/2) = 1/2, we can infer something about angle C.Wait, when we had sin(C/2) - cos(C/2) = 1/2, we squared it and found sin C = 3/4. But let me think about the sign of cos C.If angle C is acute (less than π/2), then cos C is positive. If it's obtuse (greater than π/2), cos C is negative.From the equation sin(C/2) - cos(C/2) = 1/2, let's see if we can determine whether C is acute or obtuse.Let me denote θ = C/2. So, θ = C/2, which means C = 2θ. Since C is between 0 and π, θ is between 0 and π/2.So, the equation becomes sin θ - cos θ = 1/2.Let me analyze this equation. sin θ - cos θ = 1/2.I can write this as √2 sin(θ - 45°) = 1/2, but maybe that's overcomplicating.Alternatively, let's square both sides:(sin θ - cos θ)² = (1/2)²Which gives sin² θ - 2 sin θ cos θ + cos² θ = 1/4Again, sin² θ + cos² θ = 1, so:1 - 2 sin θ cos θ = 1/4Thus, 2 sin θ cos θ = 1 - 1/4 = 3/4So, sin(2θ) = 3/4But 2θ = C, so sin C = 3/4, which matches part (1).But wait, since θ is between 0 and π/2, 2θ is between 0 and π. So, sin C = 3/4, which is positive, so C is between 0 and π, which is consistent.But does that tell us whether C is acute or obtuse? Let's see.If sin C = 3/4, then C could be in the first or second quadrant. So, C could be acute or obtuse.But from the equation sin θ - cos θ = 1/2, let's see if θ is greater than 45° or less.If θ = 45°, sin θ - cos θ = 0. So, sin θ - cos θ = 1/2 is positive, which implies that sin θ > cos θ.Since sin θ > cos θ, θ must be greater than 45°, because in the first quadrant, sin θ increases and cos θ decreases as θ increases beyond 45°.So, θ > 45°, which means C = 2θ > 90°, so angle C is obtuse.Therefore, cos C is negative.So, cos C = -√7 / 4.Therefore, going back to the Law of Cosines:c² = a² + b² - 2ab cos CWe have a = 2, b = 2, so:c² = 4 + 4 - 2*2*2*(-√7 / 4)Simplify:c² = 8 - 8*(-√7 / 4)Compute the second term:-8*(-√7 / 4) = 8*(√7 / 4) = 2√7Therefore, c² = 8 + 2√7So, c = √(8 + 2√7)Hmm, can this be simplified? Let me see.Let me factor out a 2 from inside the square root:√(2*(4 + √7)) = √2 * √(4 + √7)Not sure if that helps. Alternatively, maybe express it as √( (sqrt7 + 1)^2 )?Wait, let me compute (sqrt7 + 1)^2:(sqrt7 + 1)^2 = 7 + 2 sqrt7 + 1 = 8 + 2 sqrt7Yes! So, 8 + 2 sqrt7 is equal to (sqrt7 + 1)^2.Therefore, c = sqrt( (sqrt7 + 1)^2 ) = sqrt7 + 1.Since c is a length, it must be positive, so c = 1 + sqrt7.Let me just recap:From part (1), sin C = 3/4, which led us to determine that angle C is obtuse, so cos C = -sqrt7 / 4.Then, using the Law of Cosines with a = b = 2, we found c² = 8 + 2 sqrt7, which simplifies to (1 + sqrt7)^2, so c = 1 + sqrt7.That seems consistent. Let me double-check the calculations:1. Starting with a² + b² = 4(a + b) - 8.2. Completing the square gives (a - 2)^2 + (b - 2)^2 = 0, so a = 2, b = 2.3. Using Law of Cosines: c² = 2² + 2² - 2*2*2*cos C.4. From part (1), sin C = 3/4, so cos C = -sqrt7 / 4.5. Plugging in: c² = 4 + 4 - 8*(-sqrt7 /4) = 8 + 2 sqrt7.6. Which is (1 + sqrt7)^2, so c = 1 + sqrt7.Everything checks out. So, I think that's the correct answer.**Final Answer**(1) boxed{dfrac{3}{4}}(2) boxed{1 + sqrt{7}}