💡Okay, so I have this integral to solve: the integral from 0 to π/3 of (1 - 2 sin²(θ/2)) dθ. Hmm, let me think about how to approach this. I remember that there are some trigonometric identities that might simplify this expression. First, I recall that sin²(x) can be expressed using a double angle formula. Specifically, sin²(x) = (1 - cos(2x))/2. Maybe I can use that here. Let me try substituting x = θ/2. So, sin²(θ/2) would be (1 - cos(θ))/2. Let me write that down:1 - 2 sin²(θ/2) = 1 - 2 * [(1 - cosθ)/2] = 1 - (1 - cosθ) = cosθ.Oh, that's nice! So the integrand simplifies to cosθ. That makes the integral much easier to handle. So now, the integral becomes the integral from 0 to π/3 of cosθ dθ.I remember that the integral of cosθ is sinθ. So, applying the fundamental theorem of calculus, I can evaluate this as sinθ evaluated from 0 to π/3. Let me compute that:sin(π/3) - sin(0).I know that sin(π/3) is √3/2 and sin(0) is 0. So, subtracting these gives √3/2 - 0 = √3/2.Wait, let me double-check that I didn't make any mistakes. Did I apply the identity correctly? Yes, sin²(θ/2) is indeed (1 - cosθ)/2, so multiplying by 2 gives 1 - cosθ, and subtracting that from 1 gives cosθ. That seems right.And the integral of cosθ is definitely sinθ. Evaluating from 0 to π/3, sin(π/3) is √3/2, and sin(0) is 0. So, yes, the result is √3/2.Looking back at the options given:A: -√3/2B: -1/2C: 1/2D: √3/2So, the correct answer should be D: √3/2. That matches my calculation. I don't see any errors in my steps, so I'm confident with this answer.

💡Okay, so I have this problem about triangle ABC where angle B is twice angle A. There are also two vectors given, m and n, which are orthogonal to each other. I need to find the size of angle B. Let me try to figure this out step by step.First, since it's a triangle, the sum of the internal angles should be 180 degrees, or π radians. So, A + B + C = π. We know that B = 2A, so I can substitute that in: A + 2A + C = π, which simplifies to 3A + C = π. Therefore, C = π - 3A. Hmm, that might be useful later.Now, the vectors m and n are orthogonal, which means their dot product is zero. Vector m is (cos A, -sin B) and vector n is (cos B, sin A). So, the dot product of m and n is:cos A * cos B + (-sin B) * sin A = 0Wait, actually, the dot product is (cos A)(cos B) + (-sin B)(sin A). So that's cos A cos B - sin B sin A. Hmm, that looks familiar. Isn't that the formula for cos(A + B)? Let me check:Yes, cos(A + B) = cos A cos B - sin A sin B. So, in this case, it's cos A cos B - sin B sin A, which is the same as cos(A + B). So, the dot product is cos(A + B) = 0.Therefore, cos(A + B) = 0. When is cosine zero? At π/2, 3π/2, etc. But since A and B are angles in a triangle, they must be between 0 and π, so A + B must be π/2 or 3π/2. But 3π/2 is more than π, which isn't possible because A + B + C = π. So, A + B must be π/2.Wait, hold on. If A + B = π/2, then C = π - (A + B) = π - π/2 = π/2. So, angle C is a right angle. That's interesting.But wait, earlier I had C = π - 3A. So, if C is π/2, then π - 3A = π/2. Solving for A: π - π/2 = 3A => π/2 = 3A => A = π/6. So, angle A is 30 degrees, angle B is 2A, which is 60 degrees, and angle C is 90 degrees. That makes sense because 30 + 60 + 90 = 180.So, angle B is 60 degrees, or π/3 radians. That should be the answer to part (1).Now, moving on to part (2). We have a function f(x) = cos(ωx - B/2) + sin ωx, where ω > 0. The smallest positive period of f(x) is π. We need to find the monotonically increasing interval of f(x) and the maximum value of f(x) on [0, π/2].First, let's note that B is π/3, so B/2 is π/6. Therefore, f(x) = cos(ωx - π/6) + sin ωx.I think I can rewrite this function using trigonometric identities to make it easier to analyze. Let me try to combine the terms.We have cos(ωx - π/6) which can be expanded using the cosine of a difference identity:cos(ωx - π/6) = cos ωx cos π/6 + sin ωx sin π/6We know that cos π/6 = √3/2 and sin π/6 = 1/2. So, substituting these in:cos(ωx - π/6) = (√3/2) cos ωx + (1/2) sin ωxTherefore, f(x) becomes:f(x) = (√3/2) cos ωx + (1/2) sin ωx + sin ωxCombine the sin ωx terms:(√3/2) cos ωx + (1/2 + 1) sin ωx = (√3/2) cos ωx + (3/2) sin ωxHmm, this looks like a linear combination of sine and cosine functions with the same argument ωx. I can write this as a single sine or cosine function using the amplitude-phase form.Recall that a cos θ + b sin θ = R cos(θ - φ), where R = √(a² + b²) and tan φ = b/a.Alternatively, it can also be written as R sin(θ + φ), depending on the phase shift. Let me choose to write it as a sine function because the coefficients are positive.So, let's compute R:R = √[(√3/2)² + (3/2)²] = √[(3/4) + (9/4)] = √[12/4] = √3Okay, so R is √3. Now, let's find φ such that:sin φ = (√3/2) / R = (√3/2) / √3 = 1/2cos φ = (3/2) / R = (3/2) / √3 = (3)/(2√3) = √3/2So, sin φ = 1/2 and cos φ = √3/2. That means φ = π/6.Therefore, f(x) can be written as:f(x) = √3 sin(ωx + π/6)Wait, let me check that. If I have a cos θ + b sin θ = R sin(θ + φ), then:R sin(θ + φ) = R sin θ cos φ + R cos θ sin φComparing with a cos θ + b sin θ, we have:a = R sin φb = R cos φSo, in our case, a = √3/2 and b = 3/2.So, R sin φ = √3/2R cos φ = 3/2We found R = √3, so:sin φ = (√3/2) / √3 = 1/2cos φ = (3/2) / √3 = √3/2Which again gives φ = π/6.Therefore, f(x) = √3 sin(ωx + π/6)Okay, that seems correct.Now, the function f(x) is a sine function with amplitude √3, angular frequency ω, and phase shift π/6.The period of f(x) is 2π / ω. The problem states that the smallest positive period is π. Therefore:2π / ω = π => ω = 2So, ω is 2. Therefore, f(x) = √3 sin(2x + π/6)Now, we need to find the monotonically increasing interval of f(x). Since f(x) is a sine function, its derivative will be a cosine function, and we can find where the derivative is positive.Let's compute the derivative f’(x):f’(x) = √3 * 2 cos(2x + π/6) = 2√3 cos(2x + π/6)We need to find where f’(x) > 0, which is where cos(2x + π/6) > 0.The cosine function is positive in the intervals (-π/2 + 2πk, π/2 + 2πk) for integers k.So, let's solve:-π/2 + 2πk < 2x + π/6 < π/2 + 2πkSubtract π/6:-π/2 - π/6 + 2πk < 2x < π/2 - π/6 + 2πkSimplify the angles:-π/2 - π/6 = - (3π/6 + π/6) = -4π/6 = -2π/3π/2 - π/6 = (3π/6 - π/6) = 2π/6 = π/3So, we have:-2π/3 + 2πk < 2x < π/3 + 2πkDivide by 2:-π/3 + πk < x < π/6 + πkTherefore, the function f(x) is increasing on intervals (-π/3 + πk, π/6 + πk) for integers k.But since we're talking about the function's behavior over all real numbers, these intervals repeat every π. So, the monotonically increasing intervals are (-π/3 + πk, π/6 + πk) for all integers k.But the problem doesn't specify a particular interval, so I think that's the general answer.Now, the second part is to find the maximum value of f(x) on [0, π/2].We have f(x) = √3 sin(2x + π/6). The maximum value of sine is 1, so the maximum value of f(x) is √3 * 1 = √3.But we need to check if this maximum is achieved within the interval [0, π/2].Let's find the critical points by setting the derivative equal to zero:f’(x) = 2√3 cos(2x + π/6) = 0So, cos(2x + π/6) = 0Which implies 2x + π/6 = π/2 + πkSolving for x:2x = π/2 - π/6 + πk = (3π/6 - π/6) + πk = (2π/6) + πk = π/3 + πkThus, x = π/6 + (π/2)kNow, let's find the critical points in [0, π/2]:For k = 0: x = π/6 ≈ 0.523, which is within [0, π/2].For k = 1: x = π/6 + π/2 = 2π/3 ≈ 2.094, which is greater than π/2 ≈ 1.571, so it's outside the interval.For k = -1: x = π/6 - π/2 = -π/3, which is negative, so also outside.Therefore, the only critical point in [0, π/2] is x = π/6.Now, let's evaluate f(x) at the endpoints and at the critical point:At x = 0:f(0) = √3 sin(0 + π/6) = √3 sin(π/6) = √3 * 1/2 = √3/2 ≈ 0.866At x = π/6:f(π/6) = √3 sin(2*(π/6) + π/6) = √3 sin(π/3 + π/6) = √3 sin(π/2) = √3 * 1 = √3 ≈ 1.732At x = π/2:f(π/2) = √3 sin(2*(π/2) + π/6) = √3 sin(π + π/6) = √3 sin(7π/6) = √3 * (-1/2) = -√3/2 ≈ -0.866So, the maximum value on [0, π/2] is √3, achieved at x = π/6.Therefore, the maximum value is √3.Wait, but let me double-check the calculation at x = π/2:2x + π/6 = 2*(π/2) + π/6 = π + π/6 = 7π/6. The sine of 7π/6 is indeed -1/2, so f(π/2) is -√3/2. That's correct.So, yes, the maximum is √3 at x = π/6.Therefore, summarizing:(1) Angle B is π/3 radians or 60 degrees.(2) The function f(x) has monotonically increasing intervals of (-π/3 + πk, π/6 + πk) for all integers k, and the maximum value on [0, π/2] is √3.**Final Answer**(1) The size of angle ( B ) is (boxed{dfrac{pi}{3}}).(2) The monotonically increasing interval of ( f(x) ) is (left( -dfrac{pi}{3} + kpi, dfrac{pi}{6} + kpi right)) for all integers ( k ), and the maximum value of ( f(x) ) on ([0, dfrac{pi}{2}]) is (boxed{sqrt{3}}).

💡Okay, so I've got these two equations: (a² - b²)² = 100 and (a³ * b³) = 512. I need to find a⁴ + b⁴. Hmm, let's see how to approach this.First, looking at the first equation, (a² - b²)² = 100. If I take the square root of both sides, that should simplify things a bit. So, sqrt[(a² - b²)²] = sqrt(100), which means a² - b² = ±10. Okay, so a² - b² could be either 10 or -10. I'll keep that in mind.Next, the second equation is (a³ * b³) = 512. I think I can rewrite this as (a * b)³ = 512. Taking the cube root of both sides, that would give me a * b = cube root of 512. Since 8 * 8 * 8 = 512, the cube root of 512 is 8. So, a * b = 8. That's helpful because it gives me a relationship between a and b.Now, I need to find a⁴ + b⁴. I remember that a⁴ + b⁴ can be expressed in terms of (a² + b²)² - 2a²b². So, if I can find a² + b², I can plug it into this formula.But how do I find a² + b²? I know a² - b² is either 10 or -10, and I also know that a * b = 8. Maybe I can use these to find a² + b².I recall that (a² + b²)² = (a² - b²)² + 4a²b². Let me verify that. Expanding (a² + b²)² gives a⁴ + 2a²b² + b⁴, and (a² - b²)² gives a⁴ - 2a²b² + b⁴. So, if I add 4a²b² to (a² - b²)², I get a⁴ - 2a²b² + b⁴ + 4a²b² = a⁴ + 2a²b² + b⁴, which is indeed (a² + b²)². Okay, that makes sense.So, using that identity, (a² + b²)² = (a² - b²)² + 4a²b². I already know (a² - b²)² is 100, and a * b is 8, so a²b² is (a * b)² = 8² = 64. Therefore, 4a²b² = 4 * 64 = 256.Plugging these into the equation: (a² + b²)² = 100 + 256 = 356. So, (a² + b²)² = 356. Taking the square root of both sides, a² + b² = sqrt(356). Hmm, sqrt(356) is approximately 18.867, but I'll keep it as sqrt(356) for exactness.Now, going back to the expression for a⁴ + b⁴, which is (a² + b²)² - 2a²b². We already have (a² + b²)² = 356 and a²b² = 64, so 2a²b² = 128.Therefore, a⁴ + b⁴ = 356 - 128 = 228. So, the value of a⁴ + b⁴ is 228.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the beginning:1. (a² - b²)² = 100 => a² - b² = ±10.2. (a³ * b³) = 512 => (a * b)³ = 512 => a * b = 8.3. a⁴ + b⁴ = (a² + b²)² - 2a²b².4. (a² + b²)² = (a² - b²)² + 4a²b² = 100 + 256 = 356.5. So, a² + b² = sqrt(356).6. Then, a⁴ + b⁴ = 356 - 128 = 228.Everything seems to check out. I think I'm confident with this answer.

💡Okay, so I have this problem about a right triangle ABC, which is scalene. That means all sides are different lengths, right? The right angle is at B, I assume, since P is on the hypotenuse AC. There's a point P on AC such that angle ABP is 45 degrees. They also tell me that AP is 2 units and PC is 3 units. I need to find the area of triangle ABC.First, let me visualize the triangle. ABC is a right triangle with the right angle at B. So, sides AB and BC are the legs, and AC is the hypotenuse. Point P is somewhere on AC, closer to A since AP is 2 and PC is 3, so AC must be 5 units long.Now, angle ABP is 45 degrees. Hmm, so BP is a line from B to P on AC, making a 45-degree angle with AB. That seems important. Maybe I can use some trigonometric relationships here.I remember that in a right triangle, if an angle is 45 degrees, it's an isosceles right triangle. But here, ABP is 45 degrees, but ABP isn't necessarily a right triangle. Wait, no, ABC is a right triangle, and BP is just a line from B to P. So, maybe I can use the Angle Bisector Theorem or something like that.Wait, angle ABP is 45 degrees. Since ABC is a right triangle, angle ABC is 90 degrees. So, if BP splits angle ABC into two angles, one of which is 45 degrees, then the other angle would also be 45 degrees. So, BP is actually the angle bisector of the right angle at B. That makes sense.So, if BP is the angle bisector, then by the Angle Bisector Theorem, the ratio of the adjacent sides is equal to the ratio of the segments that the bisector creates on the opposite side. In this case, the opposite side is AC, which is divided by P into AP = 2 and PC = 3. So, the ratio of AB to BC should be equal to the ratio of AP to PC, which is 2:3.Let me write that down. So, AB / BC = AP / PC = 2 / 3. Therefore, I can let AB = 2x and BC = 3x for some positive real number x.Since ABC is a right triangle, by the Pythagorean theorem, AB² + BC² = AC². We know AC is 5, so:(2x)² + (3x)² = 5² 4x² + 9x² = 25 13x² = 25 x² = 25 / 13 x = 5 / √13Okay, so AB = 2x = 10 / √13 and BC = 3x = 15 / √13. But I need the area of triangle ABC, which is (AB * BC) / 2.Let me compute that:Area = (AB * BC) / 2 = (10 / √13 * 15 / √13) / 2 = (150 / 13) / 2 = 75 / 13So, the area is 75/13.Wait, let me double-check if I applied the Angle Bisector Theorem correctly. The theorem states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. So, AP / PC = AB / BC. Yes, that's correct. So, 2 / 3 = AB / BC, which led me to set AB = 2x and BC = 3x. Then, using Pythagoras, I found x² = 25 / 13, so x = 5 / √13.Calculating AB and BC, then the area, seems right. So, I think my answer is correct.**Final Answer**The area of triangle (ABC) is (boxed{dfrac{75}{13}}).

💡Okay, so I have this physics problem here where a body moves in a straight line from x = 5 meters to x = 10 meters, and it's under the influence of a force given by F(x) = 3x² - 2x + 5 Newtons. I need to find the work done by this force. The options are A: 925 J, B: 850 J, C: 825 J, and D: 800 J.Hmm, work done by a force... I remember that work is related to force and displacement. But since the force here is variable—it changes with position x—I can't just use the simple formula Work = Force × Distance. Instead, I think I need to use calculus, specifically integration, to calculate the work done.So, work done by a variable force is given by the integral of the force over the displacement. The formula should be something like:W = ∫ F(x) dx from the initial position to the final position.In this case, the initial position is x = 5 m, and the final position is x = 10 m. So, plugging in the values, the work done W should be:W = ∫ from 5 to 10 of (3x² - 2x + 5) dx.Alright, now I need to compute this integral. Let me recall how to integrate polynomials. The integral of x^n is (x^(n+1))/(n+1), right? So, applying that to each term in the force function.First term: 3x². The integral of 3x² is 3*(x³/3) = x³.Second term: -2x. The integral of -2x is -2*(x²/2) = -x².Third term: 5. The integral of 5 is 5x.So, putting it all together, the integral of F(x) dx is:x³ - x² + 5x.Now, I need to evaluate this from x = 5 to x = 10. That means I'll calculate the antiderivative at x = 10 and subtract the antiderivative at x = 5.Let me compute each part step by step.First, evaluate at x = 10:10³ - 10² + 5*10.Calculating each term:10³ = 1000,10² = 100,5*10 = 50.So, plugging these in:1000 - 100 + 50 = 1000 - 100 is 900, plus 50 is 950.So, the antiderivative at x = 10 is 950.Now, evaluate at x = 5:5³ - 5² + 5*5.Calculating each term:5³ = 125,5² = 25,5*5 = 25.So, plugging these in:125 - 25 + 25.125 - 25 is 100, plus 25 is 125.So, the antiderivative at x = 5 is 125.Now, subtract the value at x = 5 from the value at x = 10 to get the work done:950 - 125 = 825.So, the work done is 825 Joules.Looking back at the options, that's option C: 825 J.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, integrating F(x):∫(3x² - 2x + 5) dx = x³ - x² + 5x + C. That seems correct.Evaluating at x = 10:10³ = 1000,10² = 100,5*10 = 50.1000 - 100 + 50 = 950. That's correct.Evaluating at x = 5:5³ = 125,5² = 25,5*5 = 25.125 - 25 + 25 = 125. That's correct.Subtracting: 950 - 125 = 825. Yep, that's right.So, the work done is indeed 825 J, which is option C.I think that's solid. I don't see any errors in my steps. The integral was straightforward, and the arithmetic checks out. So, I feel confident that option C is the correct answer.**Final Answer**boxed{C}

💡Okay, so I have this problem about an operation defined on two real numbers a and b, where ab ≠ 1. The operation is given by a ∗ b = (a + b - 2ab)/(1 - ab). We start with a list of n real numbers, each between 0 and 1 (not including 1). The task is to repeatedly pick any two numbers, apply the ∗ operation, and replace them with the result until only one number remains. Part a asks to prove that this final number is the same regardless of the choices made at each step. Part b considers the case where the numbers can be up to 1, including 1, and specifically asks what happens if there's exactly one 1 in the list.Alright, let's tackle part a first. I need to show that no matter how I pair the numbers, the final result is the same. This kind of reminds me of operations that are associative and commutative, where the order and grouping don't matter. Maybe I can show that this ∗ operation is both commutative and associative, which would imply that the order of operations doesn't affect the result.First, let's check commutativity. Is a ∗ b equal to b ∗ a? Let's compute both:a ∗ b = (a + b - 2ab)/(1 - ab)b ∗ a = (b + a - 2ba)/(1 - ba)Since addition and multiplication are commutative, these two expressions are equal. So, yes, the operation is commutative.Next, let's check associativity. Is (a ∗ b) ∗ c equal to a ∗ (b ∗ c)? This might be more involved. Let's compute both sides.First, compute (a ∗ b) ∗ c:Let’s denote a ∗ b as d, where d = (a + b - 2ab)/(1 - ab). Then, d ∗ c = (d + c - 2dc)/(1 - dc).Substituting d:d ∗ c = [( (a + b - 2ab)/(1 - ab) ) + c - 2*( (a + b - 2ab)/(1 - ab) )*c ] / [1 - ( (a + b - 2ab)/(1 - ab) )*c ]This looks complicated. Let me try to simplify the numerator and denominator step by step.Numerator:[(a + b - 2ab)/(1 - ab) + c - 2c*(a + b - 2ab)/(1 - ab)]Let me factor out 1/(1 - ab):= [ (a + b - 2ab) + c*(1 - ab) - 2c*(a + b - 2ab) ] / (1 - ab)Similarly, the denominator:1 - [ (a + b - 2ab)/(1 - ab) * c ] = [ (1 - ab) - c*(a + b - 2ab) ] / (1 - ab)So, overall, d ∗ c becomes:[ (a + b - 2ab + c - abc - 2ac - 2bc + 4abc ) ] / [ (1 - ab - ac - bc + 2abc ) ]Wait, let me double-check that expansion:Numerator:(a + b - 2ab) + c*(1 - ab) - 2c*(a + b - 2ab)= a + b - 2ab + c - abc - 2ac - 2bc + 4abcCombine like terms:a + b + c - 2ab - 2ac - 2bc + ( - abc + 4abc )= a + b + c - 2ab - 2ac - 2bc + 3abcDenominator:(1 - ab) - c*(a + b - 2ab)= 1 - ab - ac - bc + 2abcSo, numerator is a + b + c - 2ab - 2ac - 2bc + 3abc, denominator is 1 - ab - ac - bc + 2abc.Now, let's compute a ∗ (b ∗ c). Let’s denote b ∗ c as e, where e = (b + c - 2bc)/(1 - bc). Then, a ∗ e = (a + e - 2ae)/(1 - ae).Substituting e:a ∗ e = [a + (b + c - 2bc)/(1 - bc) - 2a*(b + c - 2bc)/(1 - bc) ] / [1 - a*(b + c - 2bc)/(1 - bc) ]Again, let's simplify numerator and denominator.Numerator:a + (b + c - 2bc)/(1 - bc) - 2a*(b + c - 2bc)/(1 - bc)Factor out 1/(1 - bc):= [ a*(1 - bc) + (b + c - 2bc) - 2a*(b + c - 2bc) ] / (1 - bc)Denominator:1 - [ a*(b + c - 2bc)/(1 - bc) ] = [ (1 - bc) - a*(b + c - 2bc) ] / (1 - bc)So, numerator:a*(1 - bc) + b + c - 2bc - 2a*b - 2a*c + 4a*bc= a - abc + b + c - 2bc - 2ab - 2ac + 4abcCombine like terms:a + b + c - 2ab - 2ac - 2bc + ( - abc + 4abc )= a + b + c - 2ab - 2ac - 2bc + 3abcDenominator:(1 - bc) - a*b - a*c + 2a*bc= 1 - ab - ac - bc + 2abcSo, a ∗ (b ∗ c) is equal to (a + b + c - 2ab - 2ac - 2bc + 3abc)/(1 - ab - ac - bc + 2abc), which is the same as (a ∗ b) ∗ c.Therefore, the operation ∗ is associative.Since ∗ is both commutative and associative, the way we group the operations doesn't matter. Therefore, regardless of the order in which we perform the operations, the final result will be the same. This proves part a.Now, moving on to part b. The condition is weakened to 0 < x ≤ 1, and we have exactly one 1 in the list. What happens?Let me think. If one of the numbers is 1, and all others are between 0 and 1, how does the operation ∗ behave when one of the operands is 1?Let’s compute 1 ∗ b, where b is between 0 and 1.1 ∗ b = (1 + b - 2*1*b)/(1 - 1*b) = (1 + b - 2b)/(1 - b) = (1 - b)/(1 - b) = 1.So, 1 ∗ b = 1, regardless of the value of b (as long as b ≠ 1, but in our case, b is less than or equal to 1, and since we have exactly one 1, the others are less than 1).This means that whenever we perform an operation involving 1, the result is always 1. So, if we have a list with one 1 and other numbers less than 1, every time we combine 1 with another number, we just get 1. Therefore, the 1 will persist through all operations, and eventually, the final number will be 1.To confirm, let's take a small example. Suppose we have three numbers: 1, a, b, where 0 < a, b < 1.First, combine 1 and a: 1 ∗ a = 1. Now, the list is [1, b]. Combine them: 1 ∗ b = 1. So, the final number is 1.Alternatively, first combine a and b: a ∗ b = (a + b - 2ab)/(1 - ab). Since a and b are less than 1, 1 - ab is positive, so the result is a number less than 1. Then, combine this result with 1: 1 ∗ (a ∗ b) = 1. So, again, the final number is 1.Therefore, regardless of the order, the presence of exactly one 1 in the list ensures that the final number is 1.**Final Answer**a. boxed{ text{The single number is unique regardless of the operations' order.} }b. boxed{1}