💡Okay, so I have this complex number z given by the expression z = [(1 + i)^2 + 3(1 - i)] / (2 + i). I need to find the modulus of z, which is |z|, and then find the real numbers a and b such that z² + a z + b = 1 + i. Hmm, okay, let's take this step by step.First, I need to simplify z. The expression for z is a fraction where both the numerator and the denominator are complex numbers. So, maybe I should start by simplifying the numerator and then deal with the denominator.The numerator is (1 + i)^2 + 3(1 - i). Let me compute each part separately. Starting with (1 + i)^2. I remember that (a + b)^2 = a² + 2ab + b², so applying that here:(1 + i)^2 = 1² + 2*1*i + i² = 1 + 2i + i².But wait, i² is equal to -1, right? So substituting that in:1 + 2i + (-1) = (1 - 1) + 2i = 0 + 2i = 2i.Okay, so (1 + i)^2 simplifies to 2i.Now, the other part of the numerator is 3(1 - i). Let's compute that:3(1 - i) = 3*1 - 3*i = 3 - 3i.So, the numerator is (2i) + (3 - 3i). Let's combine like terms:2i + 3 - 3i = 3 + (2i - 3i) = 3 - i.So, the numerator simplifies to 3 - i.Now, the denominator is 2 + i. So, z is (3 - i)/(2 + i). Hmm, to simplify this complex fraction, I think I need to multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of 2 + i is 2 - i. So, let's do that.Multiply numerator and denominator by (2 - i):Numerator: (3 - i)(2 - i)Denominator: (2 + i)(2 - i)Let me compute the numerator first. Using the distributive property (FOIL):(3 - i)(2 - i) = 3*2 + 3*(-i) + (-i)*2 + (-i)*(-i)= 6 - 3i - 2i + i²Again, i² is -1, so:6 - 3i - 2i + (-1) = (6 - 1) + (-3i - 2i) = 5 - 5i.Okay, so the numerator becomes 5 - 5i.Now, the denominator: (2 + i)(2 - i). That's a difference of squares, so it should be 2² - (i)².Calculating that:2² - i² = 4 - (-1) = 4 + 1 = 5.So, the denominator is 5.Putting it all together, z = (5 - 5i)/5. Let's simplify that:Divide both terms in the numerator by 5:5/5 - (5i)/5 = 1 - i.So, z simplifies to 1 - i. That was a bit involved, but I think that's correct.Now, part (I) asks for the modulus of z, which is |z|. Since z is 1 - i, the modulus is the square root of the sum of the squares of the real and imaginary parts.In other words, |z| = sqrt((1)^2 + (-1)^2) = sqrt(1 + 1) = sqrt(2). So, |z| is sqrt(2). That seems straightforward.Moving on to part (II), I need to find real numbers a and b such that z² + a z + b = 1 + i. Since z is 1 - i, let's compute z² first.Calculating z²: (1 - i)^2.Again, using the formula (a - b)^2 = a² - 2ab + b²:(1 - i)^2 = 1² - 2*1*i + i² = 1 - 2i + (-1) = (1 - 1) - 2i = 0 - 2i = -2i.So, z² is -2i.Now, let's plug z and z² into the equation:z² + a z + b = (-2i) + a*(1 - i) + b.Let me expand this:-2i + a*(1 - i) + b = (-2i) + a - a i + b.Now, combine like terms. The real parts are a and b, and the imaginary parts are -2i and -a i.So, grouping them:Real part: a + bImaginary part: (-2 - a)iSo, the equation becomes:(a + b) + (-2 - a)i = 1 + i.Since two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal, we can set up the following system of equations:1. a + b = 1 (equating the real parts)2. -2 - a = 1 (equating the imaginary parts)Let me solve equation 2 first:-2 - a = 1Adding 2 to both sides:-a = 1 + 2-a = 3Multiplying both sides by -1:a = -3Now, substitute a = -3 into equation 1:-3 + b = 1Adding 3 to both sides:b = 1 + 3b = 4So, the values are a = -3 and b = 4.Let me double-check my calculations to make sure I didn't make any mistakes.Starting with z = (1 + i)^2 + 3(1 - i) all over (2 + i). We simplified the numerator to 3 - i and the denominator to 2 + i, then multiplied numerator and denominator by the conjugate 2 - i, resulting in (5 - 5i)/5 = 1 - i. That seems correct.Modulus of z is sqrt(1^2 + (-1)^2) = sqrt(2). Correct.Then, z squared is (1 - i)^2 = -2i. Correct.Plugging into the equation: z² + a z + b = -2i + a(1 - i) + b. Expanding gives a + b + (-2 - a)i. Setting equal to 1 + i, we get a + b = 1 and -2 - a = 1. Solving gives a = -3, b = 4. That seems correct.I think I did everything right. So, the modulus of z is sqrt(2), and the values of a and b are -3 and 4, respectively.**Final Answer**(I) The modulus of ( z ) is (boxed{sqrt{2}}).(II) The values of ( a ) and ( b ) are (boxed{-3}) and (boxed{4}), respectively.

💡Okay, so I have this function f(x) = sqrt(3) sin(ωx) + cos(ωx), and ω is greater than 0. The problem says that for some x1 and x2, the absolute difference |f(x1) - f(x2)| is 4, and the minimum value of |x1 - x2| is π/2. I need to figure out which of the options A, B, C, D are correct.First, I remember that functions like a sinθ + b cosθ can be rewritten as a single sine or cosine function with a phase shift. Maybe that will help simplify things. Let me try that.So, f(x) = sqrt(3) sin(ωx) + cos(ωx). To combine these into a single sine function, I can use the identity A sinθ + B cosθ = C sin(θ + φ), where C = sqrt(A² + B²) and tanφ = B/A.Calculating C: sqrt((sqrt(3))² + (1)²) = sqrt(3 + 1) = sqrt(4) = 2.So, f(x) can be written as 2 sin(ωx + φ). Now, to find φ, tanφ = B/A = 1/sqrt(3), which means φ = π/6. So, f(x) = 2 sin(ωx + π/6).Alright, that simplifies the function. Now, the problem says that |f(x1) - f(x2)| = 4. Since the maximum value of sin is 1 and the minimum is -1, the maximum difference between f(x1) and f(x2) would be 2*1 - 2*(-1) = 4. So, this occurs when one is at the maximum and the other at the minimum.Therefore, x1 and x2 must be points where f(x) is at its maximum and minimum. The minimum distance between such points would be half the period of the function because the sine function reaches max and min every half period.The period T of f(x) is 2π/ω. So, half the period is π/ω. According to the problem, the minimum |x1 - x2| is π/2. Therefore, π/ω = π/2, which implies ω = 2.So, ω is 2. That means option A is correct.Moving on to option B: If g(x) = f(x + φ) is an even function, then φ = π/6.An even function satisfies g(x) = g(-x). So, f(x + φ) = f(-x + φ). Let's write f(x + φ) = 2 sin(2(x + φ) + π/6) = 2 sin(2x + 2φ + π/6).For this to be even, it must satisfy sin(2x + 2φ + π/6) = sin(-2x + 2φ + π/6). Hmm, how does that work?Wait, sin(A) = sin(B) implies A = B + 2πk or A = π - B + 2πk for some integer k. So, for all x, sin(2x + 2φ + π/6) = sin(-2x + 2φ + π/6). Let's set them equal:2x + 2φ + π/6 = -2x + 2φ + π/6 + 2πk or 2x + 2φ + π/6 = π - (-2x + 2φ + π/6) + 2πk.Simplify the first case:2x + 2φ + π/6 = -2x + 2φ + π/6 + 2πkSubtract 2φ + π/6 from both sides:2x = -2x + 2πk4x = 2πkx = πk/2But this must hold for all x, which is only possible if k = 0, but then x = 0, which isn't for all x. So, this case doesn't work.Now, the second case:2x + 2φ + π/6 = π - (-2x + 2φ + π/6) + 2πkSimplify the right side:π - (-2x + 2φ + π/6) = π + 2x - 2φ - π/6 = (π - π/6) + 2x - 2φ = (5π/6) + 2x - 2φSo, the equation becomes:2x + 2φ + π/6 = 5π/6 + 2x - 2φ + 2πkSubtract 2x from both sides:2φ + π/6 = 5π/6 - 2φ + 2πkBring 2φ to the left and π/6 to the right:4φ = 5π/6 - π/6 + 2πk4φ = 4π/6 + 2πkSimplify 4π/6 to 2π/3:4φ = 2π/3 + 2πkDivide both sides by 4:φ = π/6 + πk/2Since 0 < φ < π/2, let's find k such that φ is in this interval.For k = 0: φ = π/6, which is within (0, π/2).For k = 1: φ = π/6 + π/2 = 2π/3, which is greater than π/2, so not acceptable.For k = -1: φ = π/6 - π/2 = -π/3, which is negative, so not acceptable.Therefore, the only solution is φ = π/6. So, option B is correct.Moving on to option C: The equation f(x) = sqrt(3) has 4 distinct real roots in the interval x ∈ (π/16, 2π).Given f(x) = 2 sin(2x + π/6). So, set 2 sin(2x + π/6) = sqrt(3). Then sin(2x + π/6) = sqrt(3)/2.The solutions to sinθ = sqrt(3)/2 are θ = π/3 + 2πk and θ = 2π/3 + 2πk for integer k.So, 2x + π/6 = π/3 + 2πk or 2x + π/6 = 2π/3 + 2πk.Solving for x:Case 1: 2x = π/3 - π/6 + 2πk = π/6 + 2πk => x = π/12 + πkCase 2: 2x = 2π/3 - π/6 + 2πk = π/2 + 2πk => x = π/4 + πkSo, the solutions are x = π/12 + πk and x = π/4 + πk.Now, let's find all solutions in x ∈ (π/16, 2π).First, let's list the solutions:For x = π/12 + πk:k=0: π/12 ≈ 0.2618k=1: π/12 + π = 13π/12 ≈ 3.400k=2: π/12 + 2π = 25π/12 ≈ 6.544k=3: π/12 + 3π = 37π/12 ≈ 9.688But 2π ≈ 6.283, so 25π/12 ≈ 6.544 is just above 2π, so k=2 is beyond the interval.Similarly, for x = π/4 + πk:k=0: π/4 ≈ 0.7854k=1: π/4 + π = 5π/4 ≈ 3.927k=2: π/4 + 2π = 9π/4 ≈ 7.069k=3: π/4 + 3π = 13π/4 ≈ 10.210Again, 9π/4 ≈ 7.069 is beyond 2π, so k=2 is beyond.Now, let's check which of these solutions are within (π/16, 2π):π/16 ≈ 0.1963So, for x = π/12 ≈ 0.2618: This is greater than π/16, so it's included.x = π/4 ≈ 0.7854: Included.x = 13π/12 ≈ 3.400: Included.x = 5π/4 ≈ 3.927: Included.x = 25π/12 ≈ 6.544: This is greater than 2π ≈ 6.283, so excluded.x = 9π/4 ≈ 7.069: Excluded.So, the solutions within (π/16, 2π) are:π/12, π/4, 13π/12, 5π/4.That's four distinct roots. So, option C is correct.Finally, option D: If the minimum value of f(x) on [-π/12, m] is -2, then m ∈ [2π/3, +∞).We know that f(x) = 2 sin(2x + π/6). The minimum value of sin is -1, so the minimum of f(x) is -2.We need to find the smallest m such that f(x) reaches -2 on [-π/12, m]. The function reaches -2 when sin(2x + π/6) = -1, which occurs when 2x + π/6 = 3π/2 + 2πk.Solving for x:2x = 3π/2 - π/6 + 2πk = (9π/6 - π/6) + 2πk = 8π/6 + 2πk = 4π/3 + 2πkx = 2π/3 + πkWe need the smallest m ≥ -π/12 such that x = 2π/3 + πk is within [-π/12, m].The smallest such x is when k=0: x = 2π/3 ≈ 2.094.Check if 2π/3 is within [-π/12, m]. Since m must be at least 2π/3 to include this point. So, m must be in [2π/3, +∞). Therefore, option D is correct.So, all options A, B, C, D are correct.**Final Answer**boxed{ABCD}

💡Okay, so I have this problem here: for any real number t, the parabola given by the equation y = 4x² + 2tx - 3t passes through a fixed point. I need to find that fixed point. Hmm, interesting. Let me try to figure this out step by step.First, I know that a fixed point means that no matter what value t takes, the point (x, y) will always satisfy the equation y = 4x² + 2tx - 3t. So, essentially, the equation should hold true for all values of t, which suggests that the terms involving t must somehow cancel out or not affect the equation's validity.Let me write down the equation again:y = 4x² + 2tx - 3tI can see that there are two terms involving t: 2tx and -3t. Maybe I can factor out t from these terms to make it easier to analyze. Let's try that:y = 4x² + t(2x - 3)So now, the equation is y = 4x² + t(2x - 3). Since this equation must hold true for all values of t, the coefficient of t must be zero; otherwise, y would change as t changes, which contradicts the idea of a fixed point. That makes sense because if the coefficient of t isn't zero, then y would depend on t, and the point wouldn't be fixed.So, to eliminate the dependence on t, the coefficient of t, which is (2x - 3), must be zero. Let me set that equal to zero and solve for x:2x - 3 = 0Adding 3 to both sides:2x = 3Dividing both sides by 2:x = 3/2Okay, so x must be 3/2. Now, let's plug this value back into the original equation to find the corresponding y-coordinate. But wait, since we're looking for a fixed point, y should be the same regardless of t. Let me substitute x = 3/2 into the equation:y = 4*(3/2)² + 2t*(3/2) - 3tFirst, calculate (3/2)²:(3/2)² = 9/4So, 4*(9/4) = 9.Next, calculate 2t*(3/2):2t*(3/2) = 3tAnd then subtract 3t:3t - 3t = 0So, putting it all together:y = 9 + 0 = 9Therefore, when x = 3/2, y = 9, regardless of the value of t. That means the fixed point is (3/2, 9).Wait, let me double-check my calculations to make sure I didn't make a mistake. So, starting with x = 3/2:y = 4*(3/2)² + 2t*(3/2) - 3tCalculating each term:4*(9/4) = 92t*(3/2) = 3tSo, y = 9 + 3t - 3t = 9 + 0 = 9Yep, that seems correct. So, regardless of the value of t, when x is 3/2, y is always 9. Therefore, the fixed point is indeed (3/2, 9).I think that's the solution. Let me just recap to ensure I didn't skip any steps or make any errors. I started by recognizing that for the point to be fixed, the equation must hold for all t, meaning the coefficients of t must cancel out. By factoring out t, I found that 2x - 3 must equal zero, leading to x = 3/2. Substituting x back into the equation gave me y = 9, confirming that (3/2, 9) is the fixed point.I don't see any mistakes in my reasoning, so I'm confident that (3/2, 9) is the correct fixed point through which all such parabolas pass.

💡First, I identify the given information: the principal amount is Rs. 2,400, the time period is 3 years, and the additional interest earned with the higher rate is Rs. 72.Next, I set up the equation for simple interest for both the original rate and the higher rate. The difference between these two interests is Rs. 72.I then simplify the equation to solve for the difference in the interest rates. By canceling out common factors and solving the equation, I find that the difference in the interest rate is 1%.

💡Alright, I'm trying to solve this problem about the function ( f(x) = ax - 2ln x + 2 ). There are two parts: first, finding the value of ( a ) such that ( f(x) ) has an extremum at ( x = 1 ), and second, determining the range of ( a ) for which ( f(x) ) has two zeros.Starting with part (1). I remember that to find extrema, I need to take the derivative of the function and set it equal to zero at the given point. So, let's compute ( f'(x) ).The function is ( f(x) = ax - 2ln x + 2 ). The derivative of ( ax ) with respect to ( x ) is ( a ). The derivative of ( -2ln x ) is ( -2 times frac{1}{x} = -frac{2}{x} ). The derivative of the constant 2 is 0. So, putting it all together, ( f'(x) = a - frac{2}{x} ).Now, since there's an extremum at ( x = 1 ), we set ( f'(1) = 0 ). Plugging in ( x = 1 ), we get:( f'(1) = a - frac{2}{1} = a - 2 = 0 ).Solving for ( a ), we find ( a = 2 ). That seems straightforward.Moving on to part (2). We need to find the range of ( a ) such that ( f(x) ) has two zeros. That means the equation ( f(x) = 0 ) has two solutions. To analyze this, I think I should look at the behavior of the function and its critical points.First, let's recall that for a function to have two zeros, it must cross the x-axis twice. This typically happens when the function has a minimum that dips below the x-axis and then rises again, or a maximum that peaks above the x-axis and then falls. Since we're dealing with a function that involves a logarithm, which is only defined for ( x > 0 ), we'll consider ( x > 0 ).From part (1), we already have the derivative ( f'(x) = a - frac{2}{x} ). Setting this equal to zero gives the critical points:( a - frac{2}{x} = 0 ) ( frac{2}{x} = a ) ( x = frac{2}{a} ).So, the function has a critical point at ( x = frac{2}{a} ). Now, the nature of this critical point (whether it's a minimum or maximum) depends on the second derivative or the sign changes of the first derivative.Let me compute the second derivative to check concavity. The first derivative is ( f'(x) = a - frac{2}{x} ), so the second derivative ( f''(x) ) is the derivative of ( f'(x) ):( f''(x) = frac{d}{dx}left(a - frac{2}{x}right) = 0 - left(-frac{2}{x^2}right) = frac{2}{x^2} ).Since ( x > 0 ), ( f''(x) = frac{2}{x^2} > 0 ). This means the function is concave upward at the critical point ( x = frac{2}{a} ), so this critical point is a local minimum.Therefore, the function has a single minimum at ( x = frac{2}{a} ). For the function to have two zeros, this minimum must lie below the x-axis. So, we need ( fleft(frac{2}{a}right) < 0 ).Let's compute ( fleft(frac{2}{a}right) ):( fleft(frac{2}{a}right) = a times frac{2}{a} - 2lnleft(frac{2}{a}right) + 2 ).Simplifying term by term:1. ( a times frac{2}{a} = 2 ).2. ( -2lnleft(frac{2}{a}right) ) remains as is.3. The constant term is +2.So, putting it all together:( fleft(frac{2}{a}right) = 2 - 2lnleft(frac{2}{a}right) + 2 = 4 - 2lnleft(frac{2}{a}right) ).We need this to be less than zero:( 4 - 2lnleft(frac{2}{a}right) < 0 ).Let's solve this inequality step by step.First, subtract 4 from both sides:( -2lnleft(frac{2}{a}right) < -4 ).Divide both sides by -2, remembering that dividing by a negative number reverses the inequality:( lnleft(frac{2}{a}right) > 2 ).Now, exponentiate both sides to eliminate the natural logarithm:( frac{2}{a} > e^2 ).Solving for ( a ):( a < frac{2}{e^2} ).But we also need to consider the domain of ( a ). Since ( x = frac{2}{a} ) must be positive, ( a ) must be positive. If ( a ) were zero or negative, the critical point would be undefined or negative, which isn't in our domain ( x > 0 ). So, ( a > 0 ).Therefore, combining both conditions, ( a ) must satisfy:( 0 < a < frac{2}{e^2} ).To double-check, let's consider the behavior of ( f(x) ) as ( x ) approaches 0 and infinity.As ( x to 0^+ ), ( ln x to -infty ), so ( -2ln x to +infty ). Therefore, ( f(x) to +infty ).As ( x to infty ), the term ( ax ) dominates if ( a > 0 ), so ( f(x) to +infty ).Since the function tends to infinity at both ends and has a single minimum in between, if the minimum is below the x-axis, the function must cross the x-axis twice. Hence, our condition ( 0 < a < frac{2}{e^2} ) is correct.I think that's solid. I don't see any mistakes in the reasoning. The key steps were taking the derivative, finding the critical point, determining it's a minimum, evaluating the function at that point, setting it less than zero, and solving for ( a ). Also, considering the domain restrictions on ( a ) was important.**Final Answer**(1) The value of ( a ) is boxed{2}.(2) The range of values for ( a ) is boxed{left(0, dfrac{2}{e^2}right)}.

💡Okay, so I have this equation here: (R/2)xy = 21x + 20y - 13. Hmm, I need to find the value of T, which is xy, right? But wait, the problem doesn't specify what R is. Is R a given constant or do I need to figure it out? Hmm, maybe R is a typo or maybe it's supposed to be something else. Let me think.Wait, in the problem statement, it just says R/2. Maybe R is a known value? Or perhaps it's a misprint and should be another number. Let me check the original problem again. It says, "Given that x and y are integers satisfying the equation (R/2)xy = 21x + 20y - 13." Hmm, no, it doesn't specify R. Maybe I need to assume R is a specific value? Or perhaps it's supposed to be another variable?Wait, maybe R is a constant that I can solve for? But then, if R is a constant, I have two variables, x and y, so I can't solve for both. Hmm, this is confusing. Maybe I need to look at the problem differently.Wait, perhaps R is a known value, and I just didn't notice it. Let me see. Is there any other information given? No, just that x and y are integers. Maybe R is supposed to be 30? Because 30 divided by 2 is 15, which is a nice number. Let me try assuming R is 30.So, if R is 30, then the equation becomes (30/2)xy = 21x + 20y - 13, which simplifies to 15xy = 21x + 20y - 13. Okay, that seems manageable. Now, I need to solve for integers x and y such that 15xy = 21x + 20y - 13.Let me rearrange this equation to make it easier to handle. Let's move all terms to one side:15xy - 21x - 20y + 13 = 0.Hmm, this looks like a quadratic in terms of x and y. Maybe I can factor this equation. Let me try to factor it. I'll look for factors of the form (ax + b)(cy + d) = e, where a, b, c, d, and e are integers.Looking at the coefficients, 15xy suggests that the factors might be (3x + something)(5y + something else). Let me try that.Let me assume it factors as (3x + m)(5y + n) = p, where m and n are constants to be determined, and p is some integer.Expanding this, I get 15xy + 3x*n + 5y*m + m*n = p.Comparing this to the original equation 15xy -21x -20y +13 = 0, I can set up the following equations:3n = -21 (coefficient of x)5m = -20 (coefficient of y)m*n = 13 (constant term)Wait, let's see:From 3n = -21, so n = -7.From 5m = -20, so m = -4.Now, m*n = (-4)*(-7) = 28. But in the original equation, the constant term is +13, not +28. Hmm, that doesn't match. So, my assumption that it factors as (3x + m)(5y + n) = p is incorrect because the constant term doesn't match.Maybe I need to adjust the factors. Let me try a different approach. Perhaps I can rearrange the equation to group x and y terms.Starting from 15xy -21x -20y +13 = 0.Let me group the x terms and y terms:x(15y -21) -20y +13 = 0.Hmm, not sure if that helps. Maybe I can factor out something else.Alternatively, let's try to write the equation in terms of x:15xy -21x = 20y -13.Factor x out:x(15y -21) = 20y -13.Then, x = (20y -13)/(15y -21).Since x must be an integer, (20y -13) must be divisible by (15y -21). Let's denote k = 15y -21, so x = (20y -13)/k.But k = 15y -21, so let's express 20y -13 in terms of k:20y -13 = (20y -13) = (20y -13) = (20y -13).Wait, maybe I can express 20y in terms of k:From k = 15y -21, we can solve for y:15y = k +21 => y = (k +21)/15.Then, 20y = 20*(k +21)/15 = (4/3)*(k +21).So, 20y -13 = (4/3)*(k +21) -13.Let me compute that:(4/3)k + 28 -13 = (4/3)k +15.So, x = (4/3 k +15)/k = (4/3) + 15/k.Since x must be an integer, 15/k must be an integer minus 4/3. Wait, this is getting complicated.Alternatively, maybe I can write x = (20y -13)/(15y -21) and look for integer values of y that make x an integer.Let me denote y as an integer and compute x for various y.Let me try y = 2:x = (20*2 -13)/(15*2 -21) = (40 -13)/(30 -21) = 27/9 = 3. So x=3, y=2. That works.Let me try y=1:x=(20 -13)/(15 -21)=7/(-6). Not integer.y=3:x=(60 -13)/(45 -21)=47/24. Not integer.y=4:x=(80 -13)/(60 -21)=67/39. Not integer.y=5:x=(100 -13)/(75 -21)=87/54=29/18. Not integer.y=6:x=(120 -13)/(90 -21)=107/69. Not integer.y=7:x=(140 -13)/(105 -21)=127/84. Not integer.y=8:x=(160 -13)/(120 -21)=147/99=49/33. Not integer.y=9:x=(180 -13)/(135 -21)=167/114. Not integer.y=10:x=(200 -13)/(150 -21)=187/129. Not integer.Hmm, seems like y=2 is the only small integer that works. Let me try negative y values.y=0:x=(0 -13)/(0 -21)=(-13)/(-21)=13/21. Not integer.y=-1:x=(-20 -13)/(-15 -21)=(-33)/(-36)=11/12. Not integer.y=-2:x=(-40 -13)/(-30 -21)=(-53)/(-51)=53/51. Not integer.y=-3:x=(-60 -13)/(-45 -21)=(-73)/(-66)=73/66. Not integer.y=-4:x=(-80 -13)/(-60 -21)=(-93)/(-81)=31/27. Not integer.y=-5:x=(-100 -13)/(-75 -21)=(-113)/(-96)=113/96. Not integer.Hmm, so y=2 is the only integer that gives an integer x=3. So, x=3, y=2.Therefore, T=xy=3*2=6.Wait, but I assumed R=30 earlier. Is that correct? Because the original equation had R/2. If R is 30, then it works. But if R is different, the solution might be different. But since the problem didn't specify R, maybe R is supposed to be 30? Or maybe I misread the problem.Wait, looking back, the problem says: "Given that x and y are integers satisfying the equation (R/2)xy = 21x + 20y - 13." It doesn't specify R, so maybe R is a known value, or perhaps it's a misprint. But since I got a valid solution with R=30, maybe that's the intended value.Alternatively, maybe R is supposed to be another value, but without more information, I can't determine it. So, assuming R=30, the solution is x=3, y=2, so T=6.I think that's the answer.