💡Alright, so I have this problem here: I need to determine the value of |3h - 2k| given that two factors of the polynomial 3x³ - hx - k are (x - 3) and (x + 1). The options are A) 24, B) 25, C) 26, D) 27, E) 28.Hmm, okay. Let me think about how to approach this. Since (x - 3) and (x + 1) are factors of the polynomial, that means when I substitute x = 3 and x = -1 into the polynomial, the result should be zero. That's from the Factor Theorem, right? So, I can set up two equations by plugging in these values and then solve for h and k.Let me write that down. First, substituting x = 3 into the polynomial:3*(3)³ - h*(3) - k = 0Calculating 3³ is 27, so 3*27 is 81. So, the equation becomes:81 - 3h - k = 0Okay, that's my first equation: 81 - 3h - k = 0.Now, substituting x = -1 into the polynomial:3*(-1)³ - h*(-1) - k = 0Calculating (-1)³ is -1, so 3*(-1) is -3. The equation becomes:-3 + h - k = 0So, my second equation is: -3 + h - k = 0.Now, I have a system of two equations:1) 81 - 3h - k = 02) -3 + h - k = 0I need to solve for h and k. Let me rearrange the second equation to express one variable in terms of the other. Let's solve for k.From equation 2: -3 + h - k = 0So, h - k = 3Which means, k = h - 3Okay, so k is h minus 3. Now, I can substitute this expression for k into equation 1.Substituting k = h - 3 into equation 1:81 - 3h - (h - 3) = 0Let me simplify that:81 - 3h - h + 3 = 0Combine like terms:81 + 3 - 3h - h = 084 - 4h = 0Now, solve for h:-4h = -84h = (-84)/(-4)h = 21Alright, so h is 21. Now, let's find k using k = h - 3:k = 21 - 3k = 18So, h is 21 and k is 18. Now, the problem asks for |3h - 2k|.Let me compute 3h - 2k first:3h = 3*21 = 632k = 2*18 = 36So, 3h - 2k = 63 - 36 = 27Taking the absolute value, |27| is just 27.Looking back at the options, D is 27. So, the answer should be D) 27.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, substituting x = 3:3*(3)^3 = 3*27 = 8181 - 3h - k = 0With h = 21 and k = 18:81 - 63 - 18 = 081 - 81 = 0, which is correct.Now, substituting x = -1:3*(-1)^3 = 3*(-1) = -3-3 - h*(-1) - k = 0Which is -3 + h - k = 0With h = 21 and k = 18:-3 + 21 - 18 = 0-3 + 3 = 0, which is also correct.So, both substitutions check out. Therefore, h = 21 and k = 18 are correct. Then, |3h - 2k| = |63 - 36| = |27| = 27.Yep, that seems solid. I think I did everything correctly. So, the answer is 27, which is option D.**Final Answer**boxed{D}

💡Okay, so I have this problem here: For the function f(x) = -x² + 2x, where x is between -1 and 3, I need to find the probability that if I randomly pick a point x₀ in that interval, f(x₀) is greater than or equal to zero. The options are A: 1/6, B: 1/3, C: 1/2, D: 2/3.Alright, let me break this down. First, I know that probability in this context is about the length of the interval where f(x) is non-negative divided by the total length of the interval from -1 to 3. So, I need to figure out where f(x) is greater than or equal to zero, and then see what fraction that is of the entire interval.So, let's start by solving the inequality f(x) ≥ 0. That means:- x² + 2x ≥ 0Hmm, okay. Let me rearrange this to make it easier to solve. I can factor out a negative sign, but maybe it's better to bring all terms to one side. Alternatively, I can factor the quadratic expression.Wait, let me write it as:-x² + 2x ≥ 0I can factor out an x, but it might be easier to factor the quadratic. Let me rewrite it:-x² + 2x = -(x² - 2x)So, that's - (x² - 2x). Hmm, can I factor x² - 2x? Yes, x(x - 2). So, putting it all together:- x² + 2x = -x(x - 2)So, the inequality becomes:- x(x - 2) ≥ 0Alternatively, I can write this as:x(x - 2) ≤ 0Because if I multiply both sides by -1, the inequality flips. So, x(x - 2) ≤ 0.Now, I need to find the values of x where this product is less than or equal to zero. Let's find the critical points where the expression equals zero. That happens when x = 0 or x - 2 = 0, so x = 0 or x = 2.These points divide the real line into intervals. Let's test each interval to see where the product is negative or zero.The critical points are at x = 0 and x = 2, so the intervals are:1. x < 02. 0 < x < 23. x > 2Let's pick test points in each interval.1. For x < 0, let's choose x = -1: (-1)(-1 - 2) = (-1)(-3) = 3, which is positive. So, the product is positive here.2. For 0 < x < 2, let's choose x = 1: (1)(1 - 2) = (1)(-1) = -1, which is negative. So, the product is negative here.3. For x > 2, let's choose x = 3: (3)(3 - 2) = (3)(1) = 3, which is positive. So, the product is positive here.So, the product x(x - 2) is less than or equal to zero in the interval [0, 2]. Therefore, the inequality x(x - 2) ≤ 0 holds when x is between 0 and 2, inclusive.But wait, our original function is f(x) = -x² + 2x, and we set f(x) ≥ 0, which led us to x(x - 2) ≤ 0, which is true for x in [0, 2].So, the values of x where f(x) is non-negative are from 0 to 2.Now, the total interval we're considering is from -1 to 3. So, the length of the interval where f(x) is non-negative is 2 - 0 = 2 units. The total length of the interval from -1 to 3 is 3 - (-1) = 4 units.Therefore, the probability is the ratio of these two lengths: 2 / 4 = 1/2.Wait, but let me double-check. Is the interval where f(x) is non-negative exactly from 0 to 2? Let me plug in some numbers to make sure.At x = -1: f(-1) = -(-1)^2 + 2*(-1) = -1 - 2 = -3, which is negative.At x = 0: f(0) = -0 + 0 = 0.At x = 1: f(1) = -1 + 2 = 1, which is positive.At x = 2: f(2) = -4 + 4 = 0.At x = 3: f(3) = -9 + 6 = -3, which is negative.So, yes, f(x) is zero at x = 0 and x = 2, positive in between, and negative outside of that interval within our domain of [-1, 3].Therefore, the length where f(x) is non-negative is indeed from 0 to 2, which is 2 units. The total interval is from -1 to 3, which is 4 units. So, the probability is 2/4, which simplifies to 1/2.Looking back at the options, C is 1/2, so that should be the answer.But just to make sure, let me visualize the function f(x) = -x² + 2x. It's a downward opening parabola because the coefficient of x² is negative. The vertex of the parabola is at x = -b/(2a) = -2/(2*(-1)) = 1. So, the vertex is at x = 1, and plugging that in, f(1) = -1 + 2 = 1. So, the vertex is at (1, 1), which is the maximum point.The parabola crosses the x-axis at x = 0 and x = 2, which we found earlier. So, between 0 and 2, the function is above the x-axis, and outside of that, it's below. Since our domain is from -1 to 3, the portion where the function is non-negative is indeed from 0 to 2.Calculating the lengths again: from -1 to 3 is 4 units. From 0 to 2 is 2 units. So, 2/4 = 1/2. Yep, that seems correct.I don't think I made any mistakes here. The key steps were solving the inequality correctly, identifying the interval where the function is non-negative, and then computing the probability as the ratio of the lengths. Everything checks out.**Final Answer**boxed{C}

💡Okay, so I have this problem here: Let T be the sum of all the real coefficients of the expansion of (1 + ix)^1012. I need to find log base 2 of T. Hmm, okay. Let me try to figure this out step by step.First, I remember that when you expand a binomial like (a + b)^n, you can use the Binomial Theorem. So, in this case, (1 + ix)^1012 can be expanded as the sum from k=0 to 1012 of (1012 choose k) * (ix)^k. That makes sense.Now, the problem is asking for the sum of all the real coefficients. Hmm, so in the expansion, each term will have a coefficient, and some of them will be real numbers, while others will have an imaginary component because of the 'i' in (ix)^k. So, I need to figure out which terms are real and sum up their coefficients.Wait, when is (ix)^k real? Let me think. The term (ix)^k is equal to i^k * x^k. So, i^k cycles through i, -1, -i, 1, and so on. So, when k is even, i^k is either 1 or -1, which are real numbers. When k is odd, i^k is either i or -i, which are imaginary. Therefore, the real coefficients come from the even powers of k, right? So, the real terms are when k is even, and their coefficients are (1012 choose k) * (i^k). But since i^k is either 1 or -1, the coefficients are just (1012 choose k) multiplied by 1 or -1.Wait, but the problem says "the sum of all the real coefficients." So, does that mean I just need to sum up all the coefficients where the term is real? So, that would be the sum over even k of (1012 choose k) * (i^k). But since i^k is either 1 or -1, depending on whether k is a multiple of 4 or 2 mod 4. Hmm, this might complicate things.Alternatively, maybe I can think of another approach. I remember that for a binomial expansion, the sum of all coefficients is obtained by setting x=1. So, if I set x=1 in (1 + ix)^1012, I get the sum of all coefficients, but that includes both real and imaginary parts. But I only want the real coefficients. Hmm.Wait, another idea: if I consider both (1 + ix)^1012 and (1 - ix)^1012, and add them together, the imaginary parts will cancel out, leaving me with twice the sum of the real terms. Let me check that.So, (1 + ix)^1012 + (1 - ix)^1012. Expanding both using the Binomial Theorem, we get:Sum from k=0 to 1012 of (1012 choose k) * (ix)^k + Sum from k=0 to 1012 of (1012 choose k) * (-ix)^k.When we add these two, the terms where k is odd will cancel out because (ix)^k + (-ix)^k = 0 when k is odd. The terms where k is even will add up because (ix)^k + (-ix)^k = 2*(ix)^k when k is even. So, the sum becomes 2 times the sum from k even of (1012 choose k) * (ix)^k.But wait, I want the sum of the real coefficients, so maybe I should set x=1 after adding them? Let me try that.So, if I set x=1 in the sum above, I get:2 * Sum from k even of (1012 choose k) * (i)^k.But since k is even, i^k is either 1 or -1. Specifically, if k is a multiple of 4, i^k is 1, and if k is 2 mod 4, i^k is -1. Hmm, so the sum becomes 2 times the sum over even k of (1012 choose k) * (-1)^{k/2}.Wait, that might complicate things because now I have alternating signs. Is there a better way to compute this?Alternatively, maybe I can use a different approach. I remember that the sum of the coefficients of even powers in a binomial expansion can be found using (f(1) + f(-1))/2, where f(x) is the polynomial. Let me see if that applies here.In this case, f(x) = (1 + ix)^1012. So, f(1) = (1 + i)^1012 and f(-1) = (1 - i)^1012. Then, (f(1) + f(-1))/2 would give me the sum of the coefficients of the even powers, which is exactly what I need because the even powers correspond to the real terms.So, let me compute f(1) + f(-1):f(1) = (1 + i)^1012f(-1) = (1 - i)^1012So, f(1) + f(-1) = (1 + i)^1012 + (1 - i)^1012Now, I need to compute this sum. Let me recall that (1 + i) and (1 - i) can be expressed in polar form. The modulus of (1 + i) is sqrt(1^2 + 1^2) = sqrt(2), and its argument is pi/4. Similarly, (1 - i) has modulus sqrt(2) and argument -pi/4.Therefore, (1 + i)^1012 = (sqrt(2))^1012 * e^{i * 1012 * pi/4}Similarly, (1 - i)^1012 = (sqrt(2))^1012 * e^{-i * 1012 * pi/4}Let me compute (sqrt(2))^1012 first. Since sqrt(2) is 2^{1/2}, so (2^{1/2})^1012 = 2^{1012/2} = 2^{506}.Now, let's compute the exponents:1012 * pi/4 = (1012/4) * pi = 253 * piSimilarly, -1012 * pi/4 = -253 * piSo, e^{i * 253 * pi} and e^{-i * 253 * pi}But 253 is an odd number, so 253 * pi is equivalent to pi modulo 2pi, because 253 = 2*126 + 1, so 253 * pi = 126*2pi + pi, which is the same as pi.Therefore, e^{i * 253 * pi} = e^{i * pi} = -1Similarly, e^{-i * 253 * pi} = e^{-i * pi} = -1So, putting it all together:(1 + i)^1012 + (1 - i)^1012 = 2^{506}*(-1) + 2^{506}*(-1) = -2^{506} - 2^{506} = -2^{507}Wait, but earlier I thought that the sum of the real coefficients would be positive. Hmm, maybe I made a mistake here.Wait, let's double-check the exponents. 1012 * pi/4 is indeed 253 * pi, which is an odd multiple of pi, so e^{i * 253 * pi} is -1, correct. So, both terms are -2^{506}, so their sum is -2^{507}.But wait, the sum of the real coefficients should be a positive number, right? Because coefficients are positive or negative, but when we sum them, depending on the terms, it could be positive or negative. Hmm, but in this case, the sum is negative. That seems odd.Wait, maybe I made a mistake in interpreting the real coefficients. Let me think again. The real coefficients are the coefficients of the real terms, which are the even powers of x. But in the expansion of (1 + ix)^1012, the coefficients alternate in sign depending on the power of i.Wait, when k is even, i^k is either 1 or -1. So, the coefficients are (1012 choose k) * i^k, which is either (1012 choose k) or -(1012 choose k). So, the sum of the real coefficients is actually the sum over even k of (1012 choose k) * (-1)^{k/2}.Wait, that might be the case. So, when k is a multiple of 4, i^k is 1, and when k is 2 mod 4, i^k is -1. So, the sum would be Sum_{k even} (1012 choose k) * (-1)^{k/2}.But how does that relate to f(1) + f(-1)? Let me see. Earlier, I thought that f(1) + f(-1) would give me twice the sum of the real coefficients, but now I see that it's actually giving me twice the sum of the real terms evaluated at x=1, which includes the signs from i^k.Wait, so maybe I need to adjust my approach. Let me think again.If I set x=1 in (1 + ix)^1012 + (1 - ix)^1012, I get 2 * Sum_{k even} (1012 choose k) * (i)^k. But since i^k is either 1 or -1, this sum is 2 * [Sum_{k even} (1012 choose k) * (-1)^{k/2}]. So, that's the sum of the real terms, but with alternating signs.But the problem says "the sum of all the real coefficients." So, does that mean I need to sum the absolute values of the coefficients, or just the coefficients as they are, including their signs?Wait, the problem says "the sum of all the real coefficients." So, I think it means the sum of the coefficients, considering their signs. So, in that case, the sum would be Sum_{k even} (1012 choose k) * (-1)^{k/2}, which is equal to [f(1) + f(-1)] / 2, where f(x) = (1 + ix)^1012.But earlier, I computed f(1) + f(-1) as -2^{507}, so dividing by 2 gives -2^{506}. But that would mean T is negative, which seems odd because coefficients can be negative, but the sum being negative? Hmm.Wait, maybe I made a mistake in the calculation. Let me double-check.(1 + i)^1012: modulus is sqrt(2), argument pi/4. So, (sqrt(2))^1012 = 2^{506}, and the angle is 1012 * pi/4. Let's compute 1012 divided by 4: 1012 / 4 = 253. So, 253 * pi. Since 253 is odd, 253 * pi = pi + 252 * pi, which is pi modulo 2pi. So, e^{i * 253 * pi} = e^{i * pi} = -1. So, (1 + i)^1012 = 2^{506} * (-1) = -2^{506}.Similarly, (1 - i)^1012: modulus sqrt(2), argument -pi/4. So, (sqrt(2))^1012 = 2^{506}, and the angle is -1012 * pi/4 = -253 * pi. Again, -253 * pi is equivalent to -pi modulo 2pi, so e^{-i * 253 * pi} = e^{-i * pi} = -1. So, (1 - i)^1012 = 2^{506} * (-1) = -2^{506}.Therefore, f(1) + f(-1) = -2^{506} + (-2^{506}) = -2^{507}.So, [f(1) + f(-1)] / 2 = (-2^{507}) / 2 = -2^{506}.Wait, but that would mean T = -2^{506}, but T is supposed to be the sum of the real coefficients. But coefficients can be negative, so maybe that's okay. But then log base 2 of T would be log2(-2^{506}), which is undefined because log of a negative number is not real. Hmm, that can't be right.So, I must have made a mistake somewhere. Let me think again.Wait, maybe I misapplied the formula. The sum of the coefficients of even powers is [f(1) + f(-1)] / 2, but in this case, f(x) = (1 + ix)^1012. So, f(1) = (1 + i)^1012 and f(-1) = (1 - i)^1012.But when I compute f(1) + f(-1), I get -2^{507}, so [f(1) + f(-1)] / 2 = -2^{506}. But that would imply that the sum of the real coefficients is negative, which might not make sense because the coefficients themselves can be positive or negative, but their sum being negative is possible.Wait, but let's think about the expansion. The expansion of (1 + ix)^1012 will have coefficients that alternate in sign for the real terms because i^k alternates between 1 and -1 for even k. So, the sum of the real coefficients could indeed be negative.But then, if T is -2^{506}, then log2(T) would be log2(-2^{506}), which is not a real number. That can't be right because the problem is asking for log2(T), implying that T is positive.So, I must have made a mistake in interpreting the problem. Let me read it again."Let T be the sum of all the real coefficients of the expansion of (1 + ix)^1012. What is log_{2}(T)?"Wait, maybe I misinterpreted what "real coefficients" means. Perhaps it refers to the coefficients of the real terms, i.e., the terms where the power of x is even, and those coefficients are real numbers. So, in that case, the coefficients themselves are real, but their sum could be negative.But then, log2(T) would be undefined if T is negative. So, perhaps I need to take the absolute value of T? Or maybe I made a mistake in the calculation.Wait, let's go back to the beginning. The expansion of (1 + ix)^1012 is Sum_{k=0}^{1012} (1012 choose k) (ix)^k. The real terms are when k is even, so the coefficients are (1012 choose k) * i^k, which is (1012 choose k) * (-1)^{k/2} because i^k = (i^2)^{k/2} = (-1)^{k/2}.So, the sum of the real coefficients is Sum_{k even} (1012 choose k) * (-1)^{k/2}.Alternatively, this can be written as Sum_{m=0}^{506} (1012 choose 2m) * (-1)^m.Hmm, that's an alternating sum. Maybe there's a way to compute this sum.Alternatively, perhaps I can use generating functions or some identity to compute this sum.Wait, another idea: consider that (1 + ix)^1012 + (1 - ix)^1012 = 2 * Sum_{k even} (1012 choose k) (ix)^k. So, if I set x=1, I get 2 * Sum_{k even} (1012 choose k) i^k = 2 * [Sum_{k even} (1012 choose k) (-1)^{k/2}]. So, that's equal to 2T, where T is the sum of the real coefficients.But earlier, I computed this as -2^{507}, so 2T = -2^{507}, so T = -2^{506}.But again, log2(T) would be log2(-2^{506}), which is undefined. So, maybe I need to take the absolute value? Or perhaps I made a mistake in the sign somewhere.Wait, let's think about the expansion again. When k is even, i^k is (-1)^{k/2}. So, for k=0, i^0=1; k=2, i^2=-1; k=4, i^4=1; and so on. So, the coefficients alternate between positive and negative.So, the sum of the real coefficients is indeed Sum_{k even} (1012 choose k) * (-1)^{k/2}, which is equal to [f(1) + f(-1)] / 2, where f(x) = (1 + ix)^1012.But as we saw, f(1) + f(-1) = -2^{507}, so T = -2^{506}.But since log2(T) is undefined for negative T, perhaps the problem is referring to the absolute value of T? Or maybe I made a mistake in interpreting the problem.Wait, perhaps the problem is asking for the sum of the absolute values of the real coefficients. That would make sense because then T would be positive, and log2(T) would be defined.But the problem says "the sum of all the real coefficients," not the sum of the absolute values. So, I think it's referring to the algebraic sum, which could be negative.But then, log2(T) would be undefined. Hmm, that's a problem.Wait, maybe I made a mistake in the calculation of f(1) + f(-1). Let me double-check.f(1) = (1 + i)^1012. Let's compute (1 + i)^1012.We know that (1 + i) = sqrt(2) * e^{i pi/4}, so (1 + i)^1012 = (sqrt(2))^1012 * e^{i * 1012 * pi/4}.(sqrt(2))^1012 = (2^{1/2})^1012 = 2^{506}.1012 * pi/4 = 253 * pi.So, e^{i * 253 * pi} = e^{i * pi} = -1 because 253 is odd.So, (1 + i)^1012 = 2^{506} * (-1) = -2^{506}.Similarly, (1 - i)^1012 = (sqrt(2))^1012 * e^{-i * 253 * pi} = 2^{506} * (-1) = -2^{506}.So, f(1) + f(-1) = -2^{506} + (-2^{506}) = -2^{507}.Therefore, [f(1) + f(-1)] / 2 = -2^{506}.So, T = -2^{506}.But then, log2(T) is log2(-2^{506}), which is not a real number. That can't be right because the problem is asking for log2(T), implying that T is positive.Wait, maybe I need to take the absolute value of T? So, |T| = 2^{506}, and then log2(|T|) = 506.But the problem didn't specify taking the absolute value. Hmm.Alternatively, maybe I made a mistake in interpreting the real coefficients. Perhaps the problem is referring to the coefficients of the real terms, which are the even-powered terms, but without considering the sign from i^k. So, maybe the coefficients are just (1012 choose k), and the sum is Sum_{k even} (1012 choose k).Wait, that would make more sense because then the sum would be positive, and log2(T) would be defined.Let me think about that. If I consider the real terms as the even-powered terms, regardless of the sign from i^k, then the coefficients are (1012 choose k), and their sum is Sum_{k even} (1012 choose k).In that case, how do I compute that sum?I remember that the sum of the coefficients of even powers in a binomial expansion is equal to (f(1) + f(-1))/2, where f(x) is the polynomial. So, in this case, f(x) = (1 + ix)^1012.So, f(1) = (1 + i)^1012 and f(-1) = (1 - i)^1012.Wait, but earlier, I computed f(1) + f(-1) as -2^{507}, which would make the sum of the coefficients of even powers equal to -2^{506}, which is negative. But that contradicts the idea that the sum of the coefficients of even powers is positive.Wait, maybe I'm confusing the sum of the coefficients with the sum of the terms. Let me clarify.When I set x=1, f(1) gives me the sum of all coefficients, including the imaginary parts. Similarly, f(-1) gives me the sum of all coefficients with x=-1. But when I add f(1) and f(-1), the imaginary parts cancel out, and I get twice the sum of the real coefficients (the even-powered terms). But in this case, the sum is negative.But if I consider the sum of the coefficients of the even-powered terms without considering the sign from i^k, then I need a different approach.Wait, perhaps I can use the fact that (1 + ix)^1012 + (1 - ix)^1012 = 2 * Sum_{k even} (1012 choose k) (ix)^k. So, if I set x=1, I get 2 * Sum_{k even} (1012 choose k) i^k. But i^k is either 1 or -1, so the sum is 2 * [Sum_{k even} (1012 choose k) (-1)^{k/2}].But that's the same as before, leading to T = -2^{506}.Alternatively, maybe I should consider that the real coefficients are the coefficients of the real terms, which are the even-powered terms, but without the sign from i^k. So, the coefficients are just (1012 choose k), and their sum is Sum_{k even} (1012 choose k).In that case, how do I compute that sum?I recall that the sum of the coefficients of even powers in (1 + x)^n is equal to 2^{n-1}. Wait, is that correct?Wait, let's test it with a small n. For example, n=2: (1 + x)^2 = 1 + 2x + x^2. The sum of the coefficients of even powers is 1 + 1 = 2, which is 2^{2-1} = 2^1 = 2. Correct.Similarly, for n=3: (1 + x)^3 = 1 + 3x + 3x^2 + x^3. Sum of even coefficients: 1 + 3 = 4, which is 2^{3-1} = 4. Correct.So, in general, the sum of the coefficients of even powers in (1 + x)^n is 2^{n-1}.But in our case, the polynomial is (1 + ix)^1012, not (1 + x)^1012. So, does the same logic apply?Wait, no, because in (1 + ix)^1012, the coefficients of the even powers are multiplied by i^k, which can be 1 or -1. So, the sum of the coefficients of the even powers is not simply 2^{1011}.Wait, but earlier, I thought that the sum of the coefficients of even powers is [f(1) + f(-1)] / 2, which in this case is (-2^{507}) / 2 = -2^{506}. But that contradicts the idea that it should be 2^{1011}.Wait, maybe I'm confusing two different things. Let me clarify.In the standard binomial expansion (1 + x)^n, the sum of the coefficients of even powers is indeed 2^{n-1}. But in our case, the polynomial is (1 + ix)^n, so the coefficients of the even powers are (n choose k) * i^k, which can be positive or negative.Therefore, the sum of the coefficients of the even powers is not 2^{n-1}, but rather [f(1) + f(-1)] / 2, which in our case is -2^{506}.But that would mean that the sum of the real coefficients is negative, which seems odd. However, mathematically, it's correct because the coefficients alternate in sign.But the problem is asking for log2(T), and if T is negative, log2(T) is undefined. So, perhaps the problem is referring to the sum of the absolute values of the real coefficients. In that case, T would be 2^{506}, and log2(T) would be 506.But the problem didn't specify absolute values, so I'm not sure. Alternatively, maybe I made a mistake in interpreting the problem.Wait, another approach: perhaps the problem is referring to the sum of the real parts of the coefficients, treating each coefficient as a complex number. But in this case, the coefficients are either real or purely imaginary. So, the real coefficients are the ones with even k, and their sum is T.But as we saw, T = -2^{506}, which is negative. So, log2(T) is undefined. Hmm.Wait, maybe I need to consider that the sum of the real coefficients is actually 2^{1011}, regardless of the sign. Let me think about that.In the standard binomial expansion, the sum of the coefficients of even powers is 2^{n-1}. So, for (1 + x)^1012, the sum of the coefficients of even powers is 2^{1011}. But in our case, the polynomial is (1 + ix)^1012, so the coefficients of the even powers are multiplied by i^k, which can be 1 or -1. So, the sum is not 2^{1011}, but rather [f(1) + f(-1)] / 2, which is -2^{506}.But that's conflicting with the standard result. So, perhaps the problem is referring to the sum of the absolute values of the real coefficients, which would be 2^{1011}.Wait, let's compute the sum of the absolute values of the real coefficients. That would be Sum_{k even} |(1012 choose k) * (-1)^{k/2}| = Sum_{k even} (1012 choose k). Because the absolute value of (-1)^{k/2} is 1.So, Sum_{k even} (1012 choose k) is equal to 2^{1011}, as per the standard binomial theorem.Therefore, if the problem is referring to the sum of the absolute values of the real coefficients, then T = 2^{1011}, and log2(T) = 1011.But the problem says "the sum of all the real coefficients," which usually means the algebraic sum, not the sum of absolute values. However, in this case, the algebraic sum is negative, leading to an undefined log2(T). Therefore, it's more likely that the problem is referring to the sum of the absolute values of the real coefficients, making T = 2^{1011}, and log2(T) = 1011.Alternatively, perhaps the problem is referring to the sum of the real parts of the coefficients, treating each coefficient as a complex number. But in this case, the coefficients are either real or purely imaginary, so the real parts are just the real coefficients, which we've already determined sum to -2^{506}. But again, log2 of a negative number is undefined.Wait, maybe I made a mistake in the calculation of f(1) + f(-1). Let me try a different approach.Consider that (1 + ix)^1012 + (1 - ix)^1012 = 2 * Sum_{k even} (1012 choose k) (ix)^k.If I set x=1, I get 2 * Sum_{k even} (1012 choose k) i^k.But i^k = (i^2)^{k/2} = (-1)^{k/2}.So, Sum_{k even} (1012 choose k) i^k = Sum_{k even} (1012 choose k) (-1)^{k/2}.Therefore, 2 * Sum_{k even} (1012 choose k) (-1)^{k/2} = (1 + i)^1012 + (1 - i)^1012.As we computed earlier, this equals -2^{507}.So, Sum_{k even} (1012 choose k) (-1)^{k/2} = -2^{506}.Therefore, T = -2^{506}.But again, log2(T) is undefined. So, perhaps the problem is referring to the sum of the absolute values of the real coefficients, which is 2^{1011}.Wait, let's compute Sum_{k even} (1012 choose k). This is equal to 2^{1011}, as per the standard binomial theorem.Because in (1 + x)^1012, the sum of the coefficients of even powers is 2^{1011}.But in our case, the coefficients are multiplied by (-1)^{k/2}, so the sum is -2^{506}.But if we take the absolute values, it's 2^{1011}.Therefore, perhaps the problem is referring to the sum of the absolute values of the real coefficients, making T = 2^{1011}, and log2(T) = 1011.Alternatively, maybe the problem is referring to the sum of the real parts of the coefficients, which in this case, since the coefficients are real or purely imaginary, the real parts are just the real coefficients. But as we saw, their sum is negative, leading to an undefined log2(T).Given that the problem is asking for log2(T), and log2 of a negative number is undefined, it's more likely that the problem is referring to the sum of the absolute values of the real coefficients, making T = 2^{1011}, and log2(T) = 1011.Therefore, the answer is 1011.

💡Alright, so I have this problem here: "In what time will a train 120 meters long cross an electric pole, if its speed is 160 kilometers per hour?" Okay, let's break this down step by step. First, I need to understand what it means for a train to cross an electric pole. I think it means that the entire length of the train has to pass by the pole. So, the train isn't just moving past the pole; the entire 120-meter length needs to go by.Now, the speed is given in kilometers per hour, but the length is in meters. I remember that to make calculations easier, it's better to have the units consistent. So, I should probably convert the speed from km/hr to m/s because the length is in meters. How do I do that? I think there's a conversion factor. Let me recall... Oh yeah, 1 kilometer is 1000 meters, and 1 hour is 3600 seconds. So, to convert km/hr to m/s, I multiply by (1000/3600), which simplifies to 5/18.Okay, so let's do that. The speed is 160 km/hr. Multiplying that by 5/18 should give me the speed in m/s. Let me calculate that: 160 * (5/18). Hmm, 160 divided by 18 is approximately 8.888..., and then multiplied by 5 gives me around 44.444... m/s. So, the train is moving at roughly 44.44 meters per second.Now, I need to find out how long it takes for the entire train to pass the pole. Since the train is 120 meters long, and it's moving at 44.44 m/s, the time it takes to pass the pole should be the time it takes to cover 120 meters at that speed. I remember the formula for time is distance divided by speed. So, time equals distance divided by speed.Plugging in the numbers: time = 120 meters / 44.44 m/s. Let me do that division. 120 divided by 44.44 is approximately 2.7 seconds. So, it should take about 2.7 seconds for the train to completely pass the electric pole.Wait a minute, does that make sense? Let me double-check my calculations. 160 km/hr converted to m/s is indeed 44.44 m/s. And 120 meters divided by 44.44 m/s is roughly 2.7 seconds. Yeah, that seems right. I don't think I made any mistakes there.But just to be thorough, let me think about it another way. If the train is moving at 44.44 meters per second, how far does it travel in one second? It travels 44.44 meters in one second. So, to cover 120 meters, it would take a little over two seconds, which aligns with my previous calculation of 2.7 seconds.Okay, I think I'm confident with this answer. The train will cross the electric pole in approximately 2.7 seconds.

💡First, I recognize that if ( x = -1 ) is a root of the quadratic equation ( x^2 + 3x + k = 0 ), it must satisfy the equation when substituted in place of ( x ).Substituting ( x = -1 ) into the equation:[(-1)^2 + 3(-1) + k = 0]Simplifying the equation:[1 - 3 + k = 0]Combining like terms:[-2 + k = 0]Solving for ( k ):[k = 2]Therefore, the value of ( k ) is 2.

💡First, I need to evaluate the expression (log_3 (9^3)).I know that (9) can be expressed as a power of (3), specifically (9 = 3^2). Therefore, (9^3) can be rewritten as ((3^2)^3).Using the power of a power property, ((3^2)^3 = 3^{2 times 3} = 3^6).Now, applying the logarithmic identity (log_b(a^n) = n log_b(a)), I have:[log_3(3^6) = 6 log_3(3)]Since (log_3(3) = 1) (because (3^1 = 3)), the expression simplifies to:[log_3(3^6) = 6 times 1 = 6]Hence, (log_3 (9^3) = boxed{6}).