💡Okay, so I have this function f(x) = log₂(x² - a x + 3a), and I need to figure out for which values of a this function is increasing on the interval [2, +∞). Hmm, okay, let's break this down step by step.First, I remember that the logarithmic function log₂(u) is increasing when its argument u is increasing, provided that u is positive. So, for f(x) to be increasing, the quadratic inside the log, which is u = x² - a x + 3a, must also be increasing on [2, +∞). That makes sense because if u is increasing, then log₂(u) will also be increasing.Now, let's think about the quadratic function u = x² - a x + 3a. Quadratic functions have a parabola shape, and since the coefficient of x² is positive (1), it opens upwards. That means the function has a minimum point at its vertex. The vertex of a parabola given by u = ax² + bx + c is at x = -b/(2a). In this case, a is 1, so the x-coordinate of the vertex is at x = a/(2*1) = a/2.So, the vertex is at x = a/2. For the quadratic to be increasing on [2, +∞), the vertex needs to be to the left of x = 2. Otherwise, if the vertex is to the right of 2, the function would be decreasing up to the vertex and then increasing after that. But we want it to be increasing all the way from 2 onwards. Therefore, we need a/2 ≤ 2. Solving for a, we get a ≤ 4. Okay, so that's one condition: a must be less than or equal to 4.But wait, there's another important condition. The argument of the logarithm must be positive for all x in [2, +∞). That is, x² - a x + 3a > 0 for all x ≥ 2. If the quadratic ever becomes zero or negative in this interval, the logarithm would be undefined or negative, which isn't allowed.So, how do we ensure that x² - a x + 3a > 0 for all x ≥ 2? Well, since the parabola opens upwards, if the vertex is to the left of x = 2, and the quadratic is increasing on [2, +∞), then the minimum value of the quadratic on [2, +∞) occurs at x = 2. Therefore, if we ensure that the quadratic is positive at x = 2, it will be positive for all x ≥ 2.Let's plug x = 2 into the quadratic: (2)² - a*(2) + 3a = 4 - 2a + 3a = 4 + a. So, we need 4 + a > 0, which simplifies to a > -4.So, putting it all together, we have two conditions:1. a ≤ 4 (from the vertex condition)2. a > -4 (from the positivity condition at x = 2)Therefore, the range of a is -4 < a ≤ 4.Wait, let me double-check. If a is exactly 4, then the vertex is at x = 4/2 = 2. So, at x = 2, the quadratic is 4 + 4 = 8, which is positive. And since the vertex is at x = 2, the quadratic is increasing for x ≥ 2, which is what we want.If a is exactly -4, then the quadratic becomes x² + 4x - 12. Let's check at x = 2: 4 + 8 - 12 = 0. Oh, that's not positive. So, a cannot be -4 because at x = 2, the quadratic would be zero, making the logarithm undefined. So, a must be greater than -4, not equal to.Similarly, if a is greater than 4, say a = 5, then the vertex is at x = 5/2 = 2.5, which is to the right of x = 2. That means the quadratic would be decreasing from x = 2 to x = 2.5 and then increasing after that. So, the function f(x) would not be increasing on the entire interval [2, +∞). Hence, a must be less than or equal to 4.So, combining these, the range of a is -4 < a ≤ 4.Just to make sure, let's test a value within this range, say a = 0. Then the quadratic becomes x² + 0 + 0 = x², which is positive for all x ≠ 0, and certainly positive for x ≥ 2. The derivative of x² is 2x, which is positive for x ≥ 2, so f(x) = log₂(x²) is increasing on [2, +∞). That checks out.Another test: a = 4. The quadratic is x² - 4x + 12. At x = 2, it's 4 - 8 + 12 = 8, which is positive. The derivative is 2x - 4, which at x = 2 is 0, and for x > 2, it's positive. So, the quadratic is increasing on [2, +∞), and thus f(x) is increasing there as well.What about a = -3, which is within the range? The quadratic becomes x² + 3x - 9. At x = 2, it's 4 + 6 - 9 = 1, which is positive. The derivative is 2x + 3, which is always positive for x ≥ 2, so the quadratic is increasing, and f(x) is increasing.If I take a = -5, which is outside the range, the quadratic becomes x² + 5x - 15. At x = 2, it's 4 + 10 - 15 = -1, which is negative. So, the logarithm is undefined there, which confirms that a must be greater than -4.Similarly, if a = 5, the quadratic is x² - 5x + 15. The vertex is at x = 5/2 = 2.5. At x = 2, the quadratic is 4 - 10 + 15 = 9, which is positive. But since the vertex is at 2.5, the quadratic decreases from x = 2 to x = 2.5 and then increases. So, the function f(x) would not be increasing on the entire interval [2, +∞). Thus, a cannot be greater than 4.Therefore, after all these checks, I'm confident that the range of a is -4 < a ≤ 4.

💡Okay, so I have this problem here about an arithmetic sequence. It says that the sequence is increasing and consists of integers. The key information given is that the product of the fourth term and the fifth term, which is a₄a₅, equals 24. I need to find the product of the third term and the sixth term, which is a₃a₆.First, let me recall what an arithmetic sequence is. It's a sequence where each term after the first is obtained by adding a constant difference. So, if the first term is a₁ and the common difference is d, then the nth term can be written as aₙ = a₁ + (n - 1)d.Since it's an increasing sequence, the common difference d must be a positive integer. That makes sense because each subsequent term is larger than the previous one.Given that a₄a₅ = 24, I need to find possible integer pairs that multiply to 24. Let me list the factor pairs of 24:1. (1, 24)2. (2, 12)3. (3, 8)4. (4, 6)Since the sequence is increasing, a₅ must be greater than a₄. So, I can ignore any factor pairs where the first number is greater than the second. Looking at my list, all pairs are in ascending order, so that's good.Now, let's consider each pair and see if they can fit into an arithmetic sequence.**Case 1: a₄ = 1, a₅ = 24**If a₄ is 1 and a₅ is 24, then the common difference d would be 24 - 1 = 23. That's a pretty large difference, but it's still a positive integer, so it's valid.Now, let's find a₃ and a₆.a₃ = a₄ - d = 1 - 23 = -22a₆ = a₅ + d = 24 + 23 = 47So, the product a₃a₆ would be (-22) * 47 = -1034Hmm, that's a negative number. The problem doesn't specify whether the terms have to be positive, just that they are integers. So, this is a possible solution, but let's check the other cases to see if there are other possibilities.**Case 2: a₄ = 2, a₅ = 12**Here, the common difference d = 12 - 2 = 10.Calculating a₃ and a₆:a₃ = a₄ - d = 2 - 10 = -8a₆ = a₅ + d = 12 + 10 = 22Product a₃a₆ = (-8) * 22 = -176Again, negative. Let's move on.**Case 3: a₄ = 3, a₅ = 8**Common difference d = 8 - 3 = 5.a₃ = 3 - 5 = -2a₆ = 8 + 5 = 13Product a₃a₆ = (-2) * 13 = -26Still negative. Let's check the last case.**Case 4: a₄ = 4, a₅ = 6**Common difference d = 6 - 4 = 2.a₃ = 4 - 2 = 2a₆ = 6 + 2 = 8Product a₃a₆ = 2 * 8 = 16Okay, this time the product is positive. So, in this case, both a₃ and a₆ are positive integers.Now, I need to determine which of these cases is the correct one. The problem states that the sequence is increasing, but it doesn't specify whether the terms have to be positive. However, in the first three cases, a₃ is negative, which might not be desired if we're considering only positive integers, but the problem doesn't specify that.Wait, actually, the problem says it's an arithmetic sequence of integers, so negative terms are allowed. But let's think about whether all these cases are possible.In the first case, a₃ is -22, which is quite negative. The sequence would be: a₁, a₂, -22, 1, 24, 47,... That seems a bit odd because the terms jump from -22 to 1, which is a big jump. But mathematically, it's still an arithmetic sequence.Similarly, in the second case, a₃ is -8, leading to a sequence: a₁, a₂, -8, 2, 12, 22,... Again, a big jump from -8 to 2.In the third case, a₃ is -2, so the sequence would be: a₁, a₂, -2, 3, 8, 13,... Still, a jump from -2 to 3 is a difference of 5, which is consistent with d=5.In the fourth case, a₃ is 2, so the sequence is: a₁, a₂, 2, 4, 6, 8,... This is a much smoother progression with a common difference of 2.Given that the problem is asking for a₃a₆, and in the fourth case, both a₃ and a₆ are positive, resulting in a positive product, whereas in the other cases, the product is negative. Since the problem doesn't specify whether the terms should be positive, but it's an increasing sequence, I think all cases are possible.However, perhaps the problem expects the terms to be positive, or maybe the smallest possible positive terms. Let's see.If we consider the fourth case, where a₄=4 and a₅=6, the common difference is 2, which is the smallest possible positive difference among the cases we considered. This might make sense because smaller differences are often preferred in such problems unless specified otherwise.Additionally, in the fourth case, the terms a₃ and a₆ are both positive, which might be more straightforward.Let me verify if there are any other factor pairs I might have missed. 24 can also be expressed as (-1)*(-24), (-2)*(-12), etc., but since the sequence is increasing, a₅ must be greater than a₄, so negative factors would require a₄ to be negative and a₅ to be positive, but that would mean the sequence is increasing from a negative to a positive, which is possible, but let's see.Wait, if a₄ and a₅ are both negative, that would require a₅ > a₄, meaning a₅ is less negative than a₄. For example, a₄ = -24, a₅ = -1, which would make d = (-1) - (-24) = 23. Then a₃ would be a₄ - d = -24 - 23 = -47, and a₆ would be a₅ + d = -1 + 23 = 22. So, a₃a₆ would be (-47)*22 = -1034, which is the same as the first case but with negative a₄ and a₅.Similarly, if a₄ = -12 and a₅ = -2, d = (-2) - (-12) = 10, then a₃ = -12 -10 = -22, a₆ = -2 +10 = 8, so a₃a₆ = (-22)*8 = -176, same as the second case.Same with a₄ = -8, a₅ = -3, d=5, a₃ = -8 -5 = -13, a₆ = -3 +5 = 2, product is (-13)*2 = -26.And a₄ = -6, a₅ = -4, d=2, a₃ = -6 -2 = -8, a₆ = -4 +2 = -2, product is (-8)*(-2)=16.Wait, so in this case, a₃a₆ is positive 16 as well, but the terms a₃ and a₆ are negative and positive respectively.But in the fourth case, both a₃ and a₆ are positive, which might be more desirable if we consider the sequence to have positive terms after a certain point.However, the problem doesn't specify whether the terms have to be positive, just that it's an increasing sequence of integers.So, both the case where a₄=4, a₅=6 and a₄=-6, a₅=-4 give a₃a₆=16. But in the first case, a₃=2 and a₆=8, both positive, while in the second case, a₃=-8 and a₆=-2, both negative. Wait, no, in the second case, a₆ is -2, which is greater than a₅=-4, so it's still increasing.But wait, if a₄=-6 and a₅=-4, then a₃ = a₄ - d = -6 -2 = -8, and a₆ = a₅ + d = -4 +2 = -2. So, a₃=-8, a₆=-2, which are both negative, but the product is positive 16.So, both cases where a₄ and a₅ are positive or negative can lead to a₃a₆=16.But in the case where a₄=4 and a₅=6, the sequence is entirely positive from a₃ onwards, which might be more straightforward.However, the problem doesn't specify whether the terms have to be positive, so both cases are possible. But since the problem is asking for a₃a₆, and in both cases, the product is 16, regardless of the signs of a₃ and a₆, as long as their product is 16.Wait, no, in the case where a₄=-6 and a₅=-4, a₃=-8 and a₆=-2, so a₃a₆=(-8)*(-2)=16, same as in the case where a₄=4 and a₅=6, a₃=2 and a₆=8, so a₃a₆=2*8=16.So, regardless of whether a₄ and a₅ are positive or negative, as long as their product is 24 and the common difference is 2, the product a₃a₆ is 16.But wait, in the case where a₄=4 and a₅=6, the common difference is 2, and in the case where a₄=-6 and a₅=-4, the common difference is also 2, but in the negative direction.Wait, no, in the case where a₄=-6 and a₅=-4, the common difference is 2, because -4 - (-6)=2. So, it's the same common difference.Therefore, both cases are essentially the same arithmetic sequence, just mirrored around zero.But since the problem states that the sequence is increasing, and in both cases, the sequence is increasing (whether moving from negative to less negative or positive to more positive), both are valid.However, the problem might be expecting the positive terms case, so the answer is 16.Alternatively, since both cases lead to the same product, 16, regardless of the direction, the answer is 16.Therefore, the answer is 16.

💡First, I need to convert the price of the book from Canadian dollars (CAD) to US dollars (USD). The given exchange rate is 1 USD = 1.25 CAD.To find out how many USD Alice needs to spend, I will divide the total amount in CAD by the exchange rate.So, 30 CAD divided by 1.25 CAD per USD equals 24 USD.Since 24 is already a whole number, there is no need for further rounding. Therefore, Alice needs to spend 24 USD to buy the book.

💡Alright, so I have this probability problem here about an exam where 6 questions are randomly selected from 20. To pass, a student needs to answer at least 4 correctly, and to get an excellent grade, they need at least 5 correct answers. The student in question can answer 10 out of the 20 questions correctly and has already passed the exam. I need to find the probability that they achieve an excellent grade.Hmm, okay, let me break this down. First, the exam has 20 questions, and 6 are selected randomly. The student knows 10 of these questions. So, the probability of them answering a question correctly is based on these 10 they know.Since the student has already passed, that means they answered at least 4 questions correctly. But we need the probability that they got at least 5 correct, given that they've already passed. So, this is a conditional probability problem.Let me recall the formula for conditional probability: P(A|B) = P(A ∩ B) / P(B). In this case, event A is the student achieving an excellent grade (answering at least 5 questions correctly), and event B is the student passing (answering at least 4 questions correctly). So, P(A|B) = P(A and B) / P(B). But since achieving an excellent grade implies passing, P(A and B) is just P(A). So, P(A|B) = P(A) / P(B).Alright, so I need to calculate P(A), the probability of answering at least 5 questions correctly, and P(B), the probability of answering at least 4 questions correctly.To calculate these probabilities, I should use the hypergeometric distribution because we're dealing with successes without replacement. The hypergeometric probability formula is:P(X = k) = [C(K, k) * C(N - K, n - k)] / C(N, n)Where:- N is the total number of questions (20)- K is the number of questions the student knows (10)- n is the number of questions selected (6)- k is the number of correct answersSo, for P(A), which is the probability of answering at least 5 correctly, we need to calculate P(X = 5) + P(X = 6).Similarly, for P(B), the probability of answering at least 4 correctly, we need to calculate P(X = 4) + P(X = 5) + P(X = 6).Let me compute each of these step by step.First, let's calculate the combinations:C(10, 4) * C(10, 2) for P(X = 4)C(10, 5) * C(10, 1) for P(X = 5)C(10, 6) for P(X = 6)And the total number of ways to select 6 questions out of 20 is C(20, 6).Calculating each term:C(10, 4) = 210C(10, 2) = 45So, P(X = 4) = (210 * 45) / C(20, 6)C(10, 5) = 252C(10, 1) = 10So, P(X = 5) = (252 * 10) / C(20, 6)C(10, 6) = 210So, P(X = 6) = 210 / C(20, 6)Now, let's compute the total combinations:C(20, 6) = 38760So, plugging in the numbers:P(X = 4) = (210 * 45) / 38760 = 9450 / 38760P(X = 5) = (252 * 10) / 38760 = 2520 / 38760P(X = 6) = 210 / 38760Simplifying these fractions:P(X = 4) = 9450 / 38760 ≈ 0.2436P(X = 5) = 2520 / 38760 ≈ 0.0650P(X = 6) = 210 / 38760 ≈ 0.0054Now, summing up for P(B):P(B) = P(X = 4) + P(X = 5) + P(X = 6) ≈ 0.2436 + 0.0650 + 0.0054 ≈ 0.3140And for P(A):P(A) = P(X = 5) + P(X = 6) ≈ 0.0650 + 0.0054 ≈ 0.0704Therefore, the conditional probability P(A|B) = P(A) / P(B) ≈ 0.0704 / 0.3140 ≈ 0.2242Hmm, that's approximately 22.42%. But let me check my calculations again to make sure I didn't make a mistake.Wait, I think I might have miscalculated the combinations. Let me recalculate C(10,4) * C(10,2):C(10,4) = 210C(10,2) = 45210 * 45 = 9450C(10,5) = 252C(10,1) = 10252 * 10 = 2520C(10,6) = 210Total combinations C(20,6) = 38760So, P(X=4) = 9450 / 38760 ≈ 0.2436P(X=5) = 2520 / 38760 ≈ 0.0650P(X=6) = 210 / 38760 ≈ 0.0054Adding these up:P(B) = 0.2436 + 0.0650 + 0.0054 = 0.3140P(A) = 0.0650 + 0.0054 = 0.0704So, P(A|B) = 0.0704 / 0.3140 ≈ 0.2242Wait, but in the initial solution, the answer was 13/58, which is approximately 0.2241, so that matches. So, my calculation seems correct.But just to be thorough, let me compute the exact fractions:P(A) = (2520 + 210) / 38760 = 2730 / 38760P(B) = (9450 + 2520 + 210) / 38760 = 12180 / 38760So, P(A|B) = (2730 / 38760) / (12180 / 38760) = 2730 / 12180Simplify this fraction:Divide numerator and denominator by 10: 273 / 1218Divide numerator and denominator by 3: 91 / 406Wait, 91 and 406 are both divisible by 7: 13 / 58Ah, so 13/58 is the simplified fraction, which is approximately 0.2241.So, yes, the probability is 13/58.I think that's correct. So, the probability that the student achieves an excellent grade, given that they have already passed, is 13/58.

💡First, I observe the given equation (2a - 3b = 2).Next, I look at the expression I need to evaluate: (8 - 6a + 9b).I notice that the terms (-6a) and (9b) can be factored by (-3), which gives me (-3(2a - 3b)).Substituting the known value of (2a - 3b) from the given equation into the expression, I replace (2a - 3b) with 2.This simplifies the expression to (8 - 3 times 2).Finally, performing the multiplication and subtraction, I find that the value of the expression is 2.

💡First, I need to determine which numbers on an 8-sided die are perfect squares. A perfect square is a number that can be expressed as the square of an integer.I'll list the numbers from 1 to 8 and identify the perfect squares among them:- 1 is a perfect square because (1 = 1^2).- 2 is not a perfect square.- 3 is not a perfect square.- 4 is a perfect square because (4 = 2^2).- 5 is not a perfect square.- 6 is not a perfect square.- 7 is not a perfect square.- 8 is not a perfect square.So, the perfect squares on the die are 1 and 4.Next, I'll calculate the probability by dividing the number of successful outcomes (rolling a perfect square) by the total number of possible outcomes (total sides of the die).There are 2 successful outcomes (1 and 4) and 8 possible outcomes in total.Therefore, the probability is:[frac{2}{8} = frac{1}{4}]