💡Okay, so I have this problem here about shifting a sine function and finding the center of symmetry. Let me try to figure it out step by step.First, the original function is y = 2 sin(2x). I know that the general form of a sine function is y = A sin(Bx + C) + D, where A is the amplitude, B affects the period, C is the phase shift, and D is the vertical shift. In this case, A is 2, B is 2, and there's no C or D mentioned, so it's just y = 2 sin(2x).The problem says that this graph is shifted to the left by π/12 units to get f(x). Shifting left by π/12 means we're adding π/12 to the x inside the sine function. So, the new function f(x) should be y = 2 sin(2(x + π/12)). Let me write that out:f(x) = 2 sin(2(x + π/12)).Simplifying that, it becomes f(x) = 2 sin(2x + π/6). Okay, so that's the transformed function.Now, the question is about the center of symmetry of f(x). Hmm, centers of symmetry for sine functions... I remember that sine functions are symmetric about their midpoints, which are the points where the function crosses the x-axis. So, for the standard sine function y = sin(x), the centers of symmetry are at the points where it crosses the x-axis, like (kπ, 0) for integer k.But in this case, the function is transformed. So, I need to find the points where f(x) crosses the x-axis, which would be the centers of symmetry. Let's set f(x) equal to zero and solve for x:2 sin(2x + π/6) = 0.Dividing both sides by 2:sin(2x + π/6) = 0.The sine function is zero at integer multiples of π, so:2x + π/6 = kπ, where k is any integer.Let me solve for x:2x = kπ - π/6x = (kπ - π/6)/2x = (kπ)/2 - π/12.So, the x-coordinates where the function crosses the x-axis are at x = (kπ)/2 - π/12. These are the centers of symmetry. Now, I need to see which of the given options matches one of these points.Looking at the options:A: (π/12, 0)B: (5π/12, 0)C: (-5π/12, 0)D: (π/6, 0)Let me plug in some integer values for k to see which of these points come up.Starting with k = 0:x = (0)/2 - π/12 = -π/12. Hmm, that's not one of the options except for C, which is (-5π/12, 0). Wait, that's not the same as -π/12. Maybe k = 1?k = 1:x = (π)/2 - π/12 = (6π/12 - π/12) = 5π/12. That's option B.k = 2:x = (2π)/2 - π/12 = π - π/12 = 11π/12. Not an option.k = -1:x = (-π)/2 - π/12 = (-6π/12 - π/12) = -7π/12. Not an option.So, for k = 1, we get x = 5π/12, which is option B. That seems to be one of the centers of symmetry.Wait, but why is it only considering k = 1? Is there another point? Let me check k = 0 again. It gives x = -π/12, which is not an option, but maybe the question is asking for the principal center or the one closest to the origin? Or perhaps the first positive center?Looking at the options, B is 5π/12, which is positive, and C is -5π/12, which is negative. Since the question doesn't specify, but the original function is shifted to the left, maybe the center is shifted as well. Let me think.The original function y = 2 sin(2x) has centers of symmetry at x = kπ/2. After shifting left by π/12, the centers would shift left by π/12 as well. So, the new centers would be at x = kπ/2 - π/12.So, for k = 1, we have x = π/2 - π/12 = 5π/12, which is option B. That makes sense because shifting left by π/12 from π/2 (which is 6π/12) would give 5π/12.Alternatively, if we take k = 0, we get x = -π/12, which is option C, but it's negative. The question doesn't specify whether it wants the positive or negative one, but since the shift is to the left, maybe the center is shifted to the left, but still, 5π/12 is to the right of the origin.Wait, actually, shifting left by π/12 would mean the center moves left, but the center at k = 1 is moving from π/2 to 5π/12, which is actually moving left by π/12. Because π/2 is 6π/12, subtract π/12 gives 5π/12. So, yes, that's consistent.Therefore, the center of symmetry is at (5π/12, 0), which is option B.I think that's the correct answer. Let me just double-check.Original function: y = 2 sin(2x). Its centers are at x = kπ/2. After shifting left by π/12, the centers become x = kπ/2 - π/12. For k = 1, x = π/2 - π/12 = 5π/12. Yep, that's correct.So, the answer should be B.

💡Let ( x ) be the total number of pages in the book.1. **Reading on the first day:** Hui reads ( frac{1}{6}x + 10 ). Pages remaining after the first day: [ x - left(frac{1}{6}x + 10right) = frac{5}{6}x - 10 ]2. **Reading on the second day:** Hui reads ( frac{1}{5}left(frac{5}{6}x - 10right) + 20 ). Pages remaining after the second day: [ left(frac{5}{6}x - 10right) - left(frac{1}{5}left(frac{5}{6}x - 10right) + 20right) = frac{4}{5}left(frac{5}{6}x - 10right) - 20 = frac{2x}{3} - 28 ]3. **Reading on the third day:** Hui reads ( frac{1}{4}left(frac{2x}{3} - 28right) + 25 ). The remaining pages after the third day are: [ left(frac{2x}{3} - 28right) - left(frac{1}{4}left(frac{2x}{3} - 28right) + 25right) = frac{3}{4}left(frac{2x}{3} - 28right) - 25 = frac{1}{2}x - 46 ]4. **Final remaining pages:** It is given that after the third day, there are 74 pages left. [ frac{1}{2}x - 46 = 74 ]5. **Solving for ( x ):** [ frac{1}{2}x = 74 + 46 = 120 implies x = 120 times 2 = 240 ]Therefore, there are 240 pages in the book.

💡To determine the highest value of 'a' such that the number 365a16 is divisible by 8, we focus on the last three digits of the number, which are 'a16'. Divisibility by 8 depends on these last three digits.We systematically check each possible value of 'a' from 0 to 9:- For a = 0, the last three digits are 016, which is divisible by 8.- For a = 1, the last three digits are 116, which is not divisible by 8.- For a = 2, the last three digits are 216, which is not divisible by 8.- For a = 3, the last three digits are 316, which is not divisible by 8.- For a = 4, the last three digits are 416, which is divisible by 8.- For a = 5, the last three digits are 516, which is not divisible by 8.- For a = 6, the last three digits are 616, which is not divisible by 8.- For a = 7, the last three digits are 716, which is not divisible by 8.- For a = 8, the last three digits are 816, which is divisible by 8.- For a = 9, the last three digits are 916, which is not divisible by 8.Among these, the highest value of 'a' that satisfies the condition is 8.

💡First, I need to determine the value of ( a ) in the linear regression equation ( hat{y} = bx + a ). I know that the regression line passes through the sample center point ( (4, 50) ), so I can substitute these values into the equation to solve for ( a ).Substituting ( x = 4 ) and ( y = 50 ) into the equation:[50 = -2 times 4 + a][50 = -8 + a][a = 58]Now that I have the complete regression equation ( hat{y} = -2x + 58 ), I can use it to predict the forecast variable when the explanatory variable ( x ) is 6.Substituting ( x = 6 ) into the equation:[hat{y} = -2 times 6 + 58][hat{y} = -12 + 58][hat{y} = 46]Therefore, when the explanatory variable is 6, the forecast variable is approximately 46.

💡First, I need to determine the total value of all the coins in paise. Since the total sum is Rs. 70, I'll convert this to paise:70 Rupees = 70 * 100 = 7000 paiseNext, I'll calculate the total value contributed by the 220 coins of 20 paise each:220 coins * 20 paise/coin = 4400 paiseNow, I'll subtract this value from the total value to find the value of the remaining coins:7000 paise (total) - 4400 paise (20 paise coins) = 2600 paiseI know there are 324 coins in total, and 220 of them are 20 paise coins. So, the number of remaining coins is:324 coins (total) - 220 coins (20 paise coins) = 104 coinsFinally, to find the denomination of the remaining coins, I'll divide the total value of these coins by the number of coins:2600 paise / 104 coins = 25 paise/coinTherefore, the denomination of the remaining coins is 25 paise.

💡Okay, so I have this problem here, and I need to figure out which of the four functions has a minimum positive period of π and is decreasing on the interval (π/2, π). The options are:A: y = 2|sin x|B: y = cos xC: y = sin 2xD: y = |cos x|Alright, let's break this down step by step. First, I need to understand what it means for a function to have a minimum positive period of π. The period of a function is the smallest positive number T for which the function repeats its values every T units. So, for example, sin x has a period of 2π, meaning it repeats every 2π units. If I have a function like sin 2x, the period is π because it completes a full cycle twice as fast.Next, I need to check which of these functions have a period of π. Let's go through each option:**Option A: y = 2|sin x|**The basic sin x function has a period of 2π. However, when we take the absolute value of sin x, |sin x|, the graph is reflected above the x-axis, effectively making the function repeat every π instead of 2π. So, the period becomes π. Multiplying by 2 just vertically stretches the graph but doesn't affect the period. So, the period of y = 2|sin x| is π.**Option B: y = cos x**The cosine function, cos x, has a period of 2π, just like sin x. So, this doesn't satisfy the condition of having a period of π.**Option C: y = sin 2x**As I thought earlier, sin 2x has a period of π because the coefficient 2 in front of x compresses the graph horizontally, making it complete a full cycle in half the length. So, the period here is π.**Option D: y = |cos x|**Similar to |sin x|, taking the absolute value of cos x reflects the negative parts of the cosine graph above the x-axis. This causes the function to repeat every π instead of 2π. So, the period of y = |cos x| is π.Okay, so from the period analysis, options A, C, and D have a period of π, while option B has a period of 2π. So, we can eliminate option B.Now, the next part is to check which of these functions is decreasing on the interval (π/2, π). Let's analyze each remaining option.**Option A: y = 2|sin x|**First, let's consider the behavior of sin x on the interval (π/2, π). Sin x is positive in this interval because it's in the second quadrant, where sine values are positive. Also, sin x is decreasing on this interval because it starts at sin(π/2) = 1 and decreases to sin(π) = 0. Since we're taking the absolute value, which doesn't change anything in this interval because sin x is already positive, y = 2|sin x| will also be decreasing on (π/2, π). Multiplying by 2 just makes the function steeper but doesn't affect whether it's increasing or decreasing.**Option C: y = sin 2x**Let's look at sin 2x on the interval (π/2, π). First, let's find the corresponding interval for 2x. If x is between π/2 and π, then 2x is between π and 2π. Now, sin θ is decreasing on (π/2, 3π/2) and increasing on (3π/2, 2π). So, in the interval (π, 2π), sin θ is decreasing from π to 3π/2 and increasing from 3π/2 to 2π. Therefore, sin 2x is decreasing on (π/2, 3π/4) and increasing on (3π/4, π). So, it's not decreasing throughout the entire interval (π/2, π); it changes behavior in the middle. Therefore, option C doesn't satisfy the condition of being decreasing on the entire interval.**Option D: y = |cos x|**Let's analyze |cos x| on the interval (π/2, π). Cos x is negative in this interval because it's in the second quadrant, where cosine values are negative. Taking the absolute value reflects these negative values above the x-axis, making y = |cos x| positive. Now, let's see if it's decreasing. Cos x is decreasing on (0, π), but since we're taking the absolute value, the behavior changes. Specifically, as x increases from π/2 to π, cos x goes from 0 to -1, so |cos x| goes from 0 to 1. Therefore, |cos x| is increasing on this interval, not decreasing. So, option D is increasing on (π/2, π), which doesn't satisfy the condition.So, summarizing:- Option A: Decreasing on (π/2, π)- Option C: Not decreasing throughout (π/2, π)- Option D: Increasing on (π/2, π)Therefore, the only function that satisfies both conditions is option A: y = 2|sin x|.**Final Answer**boxed{A}