💡Okay, so I have this problem about an isosceles triangle. Let me try to understand what it's asking. An isosceles triangle has two sides that are equal in length, and the angles opposite those sides are also equal. The problem says that one of the angles opposite an equal side is 30 degrees. I need to find the measure of the triangle's largest angle.Alright, let's break this down. First, in an isosceles triangle, if two sides are equal, then the angles opposite those sides are equal. So, if one of those angles is 30 degrees, the other one must also be 30 degrees. That makes sense because the triangle has symmetry.Now, I remember that the sum of all angles in any triangle is always 180 degrees. So, if I have two angles that are each 30 degrees, I can add those together and then subtract from 180 to find the third angle. Let me write that out:30 degrees + 30 degrees = 60 degrees.So, the two equal angles add up to 60 degrees. Now, subtracting that from 180 degrees gives me the measure of the third angle:180 degrees - 60 degrees = 120 degrees.Hmm, so the third angle is 120 degrees. Since 120 is larger than 30, that must be the largest angle in the triangle. Therefore, the measure of the triangle's largest angle is 120 degrees.Wait, let me double-check to make sure I didn't make a mistake. If the two equal angles are 30 degrees each, adding them gives 60 degrees. Subtracting from 180 gives 120 degrees for the third angle. Yes, that seems correct. Also, in an isosceles triangle, the largest angle is always opposite the longest side, which in this case would be the side that's not equal to the other two. So, that makes sense.I think I've got it right. The largest angle is 120 degrees.

💡Alright, so I have this problem: "How many even positive integers less than 1000 can be written as the sum of two positive perfect cubes?" Hmm, okay. Let me try to figure this out step by step.First, I know that a perfect cube is a number multiplied by itself three times, like 1³=1, 2³=8, 3³=27, and so on. The problem is asking for even numbers less than 1000 that can be written as the sum of two such cubes. So, I need to find all pairs of cubes (a³ + b³) where a and b are positive integers, their sum is even, and the total is less than 1000.Wait, the sum has to be even. So, when is a sum of two numbers even? I remember that an even plus an even is even, and an odd plus an odd is also even. But an even plus an odd is odd. So, for the sum a³ + b³ to be even, both a and b must be even, or both must be odd. That makes sense.Now, I need to figure out the range of a and b. Since we're dealing with numbers less than 1000, let's see what the maximum cube we need to consider is. Let's calculate 10³, which is 1000. But since we need numbers less than 1000, the largest cube we can have is 9³=729. So, a and b can be from 1 to 9.Okay, so a and b can be 1, 2, 3, 4, 5, 6, 7, 8, or 9. Now, I need to list all possible sums a³ + b³ where both a and b are in this range, and the sum is even and less than 1000.But listing all possible sums manually would take a lot of time. Maybe I can find a smarter way. Let me think. Since a and b can range from 1 to 9, I can create a table where one axis is a and the other is b, and each cell is the sum a³ + b³. Then, I can go through the table, pick out the even numbers less than 1000, and count them, making sure not to count duplicates.Wait, that sounds like a plan. Let me try to visualize this table. The rows can represent a from 1 to 9, and the columns can represent b from 1 to 9. Each cell (i,j) will contain i³ + j³. But since addition is commutative, the table will be symmetric along the diagonal. So, I can just consider the upper triangle or the lower triangle to avoid duplicates.But actually, since I'm only interested in the sums, not the specific pairs (a,b), maybe I can just compute all possible sums and then filter them based on the criteria: even and less than 1000. Then, I can collect these sums into a set to automatically handle duplicates, and the size of the set will be the answer.Yes, that sounds efficient. So, let me start by listing all possible a³ where a is from 1 to 9:1³ = 12³ = 83³ = 274³ = 645³ = 1256³ = 2167³ = 3438³ = 5129³ = 729Okay, so these are the cubes. Now, I need to compute all possible sums a³ + b³ where a and b are between 1 and 9, and the sum is even and less than 1000.Since I need the sum to be even, as I thought earlier, both a and b must be even or both must be odd. So, maybe I can separate the cubes into even and odd and then compute the sums accordingly.Let me separate the cubes:Odd cubes (from odd a):1³ = 13³ = 275³ = 1257³ = 3439³ = 729Even cubes (from even a):2³ = 84³ = 646³ = 2168³ = 512So, now, to get even sums, I can add two odd cubes or two even cubes.Let me compute all possible sums of two odd cubes:1 + 1 = 21 + 27 = 281 + 125 = 1261 + 343 = 3441 + 729 = 73027 + 27 = 5427 + 125 = 15227 + 343 = 37027 + 729 = 756125 + 125 = 250125 + 343 = 468125 + 729 = 854343 + 343 = 686343 + 729 = 1072 (this is over 1000, so we can ignore this)729 + 729 = 1458 (also over 1000, ignore)Okay, so from the odd cubes, the sums less than 1000 are:2, 28, 126, 344, 730, 54, 152, 370, 756, 250, 468, 854, 686Now, let me compute all possible sums of two even cubes:8 + 8 = 168 + 64 = 728 + 216 = 2248 + 512 = 52064 + 64 = 12864 + 216 = 28064 + 512 = 576216 + 216 = 432216 + 512 = 728512 + 512 = 1024 (over 1000, ignore)So, from the even cubes, the sums less than 1000 are:16, 72, 224, 520, 128, 280, 576, 432, 728Now, let me compile all these sums together:From odd cubes:2, 28, 126, 344, 730, 54, 152, 370, 756, 250, 468, 854, 686From even cubes:16, 72, 224, 520, 128, 280, 576, 432, 728Now, let me list them all out and make sure there are no duplicates:2, 16, 28, 54, 72, 126, 128, 152, 224, 250, 280, 344, 370, 432, 468, 520, 576, 686, 728, 730, 756, 854Wait, let me count them:1. 22. 163. 284. 545. 726. 1267. 1288. 1529. 22410. 25011. 28012. 34413. 37014. 43215. 46816. 52017. 57618. 68619. 72820. 73021. 75622. 854Hmm, that's 22 numbers. Wait, but earlier I thought there were 25. Did I miss some?Let me check my earlier lists again.From odd cubes, I had:2, 28, 126, 344, 730, 54, 152, 370, 756, 250, 468, 854, 686That's 13 numbers.From even cubes:16, 72, 224, 520, 128, 280, 576, 432, 728That's 9 numbers.Total should be 22, but I thought the answer was 25. Maybe I missed some sums?Wait, let me check if I computed all possible sums correctly.From odd cubes:1 + 1 = 21 + 27 = 281 + 125 = 1261 + 343 = 3441 + 729 = 73027 + 27 = 5427 + 125 = 15227 + 343 = 37027 + 729 = 756125 + 125 = 250125 + 343 = 468125 + 729 = 854343 + 343 = 686343 + 729 = 1072 (over 1000)729 + 729 = 1458 (over 1000)So, that's 13 sums.From even cubes:8 + 8 = 168 + 64 = 728 + 216 = 2248 + 512 = 52064 + 64 = 12864 + 216 = 28064 + 512 = 576216 + 216 = 432216 + 512 = 728512 + 512 = 1024 (over 1000)So, that's 9 sums.Total 22 sums. But the initial thought process mentioned 25. Maybe I missed some?Wait, perhaps I didn't consider all possible pairs. Let me think. For example, when a and b are different, like a=1 and b=2, but wait, 1 is odd and 2 is even, so their sum would be odd, which we don't want. So, only pairs where both are odd or both are even.But maybe I missed some combinations where a and b are different but both odd or both even.Wait, let me check the odd cubes again. Did I compute all possible pairs?1³ + 1³ = 21³ + 3³ = 281³ + 5³ = 1261³ + 7³ = 3441³ + 9³ = 7303³ + 3³ = 543³ + 5³ = 1523³ + 7³ = 3703³ + 9³ = 7565³ + 5³ = 2505³ + 7³ = 4685³ + 9³ = 8547³ + 7³ = 6867³ + 9³ = 1072 (over)9³ + 9³ = 1458 (over)So, that's 13 sums.For even cubes:2³ + 2³ = 162³ + 4³ = 722³ + 6³ = 2242³ + 8³ = 5204³ + 4³ = 1284³ + 6³ = 2804³ + 8³ = 5766³ + 6³ = 4326³ + 8³ = 7288³ + 8³ = 1024 (over)So, that's 9 sums.Total 22. Hmm.Wait, maybe I missed some sums where a ≠ b but both are odd or both are even. Let me check.For example, 1³ + 3³ = 28, which I have.But what about 3³ + 1³? That's the same as 1³ + 3³, so it's already counted.Similarly, 1³ + 5³ = 126, and 5³ + 1³ is the same.So, I think I have all unique sums.Wait, but in the initial thought process, the user had 25 numbers. Maybe I missed some.Let me check the initial thought process again.They listed:2, 16, 28, 54, 72, 128, 152, 224, 250, 280, 342, 344, 370, 432, 468, 520, 539, 576, 637, 686, 728, 793, 854, 855, 945.Wait, that's 25 numbers. But in my list, I have 22. So, I must have missed some.Wait, let me see. Where did they get 342, 539, 637, 793, 855, 945?Wait, those numbers are not in my list. So, perhaps I missed some sums.Wait, how did they get 342? 342 is even. Let me see: 342 = a³ + b³.What cubes add up to 342? Let's see:342 - 1 = 341, which is not a cube.342 - 8 = 334, not a cube.342 - 27 = 315, not a cube.342 - 64 = 278, not a cube.342 - 125 = 217, not a cube.342 - 216 = 126, which is 5³ + 1³ = 126. Wait, 126 is already in my list. So, 342 = 126 + 216? Wait, no, 126 + 216 = 342. But 126 is 1³ + 5³, and 216 is 6³. So, 342 = (1³ + 5³) + 6³? Wait, no, that's three cubes. The problem is about the sum of two cubes. So, 342 must be a sum of two cubes.Wait, let me check: 342 = a³ + b³.Let me try a=7: 343 is too big, so 342 - 343 is negative.a=6: 216. 342 - 216 = 126. 126 is 5³ + 1³, but that's two cubes, so 342 = 6³ + (5³ + 1³). Wait, that's three cubes. Hmm.Wait, maybe 342 is 3³ + something. 3³=27. 342 -27=315, not a cube.4³=64. 342-64=278, not a cube.5³=125. 342-125=217, not a cube.7³=343, too big.Wait, maybe 342 is not a sum of two cubes. So, perhaps the initial thought process had a mistake.Similarly, 539: let's see if 539 is a sum of two cubes.539 - 1=538, not a cube.539 -8=531, not a cube.539 -27=512, which is 8³. So, 539 = 27 + 512 = 3³ + 8³. So, that's a valid sum. But in my earlier list, I had 3³ + 8³ = 539. Wait, did I include that?Wait, in my odd cubes sums, I had 3³ + 8³? Wait, no, because 8 is even, so 3 is odd and 8 is even, so their sum is odd, which we don't want. So, 539 is odd, so it shouldn't be in our list. But the initial thought process included it. So, that's a mistake.Similarly, 637: let's see if it's a sum of two cubes.637 -1=636, not a cube.637 -8=629, not a cube.637 -27=610, not a cube.637 -64=573, not a cube.637 -125=512, which is 8³. So, 637 = 125 + 512 = 5³ + 8³. Again, 5 is odd, 8 is even, so their sum is odd, which we don't want. So, 637 shouldn't be in our list.Similarly, 793: 793 - 1=792, not a cube.793 -8=785, not a cube.793 -27=766, not a cube.793 -64=729, which is 9³. So, 793 = 64 + 729 = 4³ + 9³. 4 is even, 9 is odd, so their sum is odd. So, 793 shouldn't be in our list.855: 855 -1=854, which is 854=9³ + 5³=729 + 125. So, 855=854 +1=9³ +5³ +1³. That's three cubes. Alternatively, 855= a³ + b³.Let me check: 855 - 1=854, which is 9³ +5³.855 -8=847, not a cube.855 -27=828, not a cube.855 -64=791, not a cube.855 -125=730, which is 9³ +1³=729 +1=730. So, 855=730 +125=9³ +1³ +5³. Again, three cubes.Wait, maybe 855 is not a sum of two cubes. Let me check 855= a³ + b³.a=9: 729. 855-729=126=5³ +1³. So, again, three cubes.a=8: 512. 855-512=343=7³. So, 855=512 +343=8³ +7³. 8 is even, 7 is odd, so their sum is odd. So, 855 is odd, shouldn't be in our list.Similarly, 945: 945 -1=944, not a cube.945 -8=937, not a cube.945 -27=918, not a cube.945 -64=881, not a cube.945 -125=820, not a cube.945 -216=729=9³. So, 945=216 +729=6³ +9³. 6 is even, 9 is odd, so their sum is odd. So, 945 shouldn't be in our list.So, the initial thought process had some errors, including odd sums and sums that are actually three cubes.Therefore, my initial count of 22 seems correct. But wait, let me double-check.Wait, in my list, I have:From odd cubes:2, 28, 126, 344, 730, 54, 152, 370, 756, 250, 468, 854, 686From even cubes:16, 72, 224, 520, 128, 280, 576, 432, 728Total 22 numbers.But let me check if I missed any sums where a and b are different but both odd or both even.For example, 1³ + 3³=28, which is in the list.1³ + 5³=126, in the list.1³ +7³=344, in the list.1³ +9³=730, in the list.3³ +5³=152, in the list.3³ +7³=370, in the list.3³ +9³=756, in the list.5³ +7³=468, in the list.5³ +9³=854, in the list.7³ +9³=1072, over 1000.So, that's all for odd cubes.For even cubes:2³ +4³=72, in the list.2³ +6³=224, in the list.2³ +8³=520, in the list.4³ +6³=280, in the list.4³ +8³=576, in the list.6³ +8³=728, in the list.Also, 2³ +2³=16, 4³ +4³=128, 6³ +6³=432, 8³ +8³=1024 (over).So, that's all.Wait, but in my list, I have 22 numbers, but the initial thought process had 25, including some odd numbers and some sums that are actually three cubes.So, perhaps the correct answer is 22.But let me check another way. Maybe I can list all possible even sums less than 1000 and count them.Let me list all possible a³ + b³ where a and b are from 1 to 9, both odd or both even, and the sum is less than 1000.Starting with a=1 (odd):1 +1=21 +3=281 +5=1261 +7=3441 +9=730a=3 (odd):3 +3=543 +5=1523 +7=3703 +9=756a=5 (odd):5 +5=2505 +7=4685 +9=854a=7 (odd):7 +7=6867 +9=1072 (over)a=9 (odd):9 +9=1458 (over)Now, even a:a=2:2 +2=162 +4=722 +6=2242 +8=520a=4:4 +4=1284 +6=2804 +8=576a=6:6 +6=4326 +8=728a=8:8 +8=1024 (over)So, compiling all these:From a=1: 2,28,126,344,730From a=3:54,152,370,756From a=5:250,468,854From a=7:686From a=2:16,72,224,520From a=4:128,280,576From a=6:432,728From a=8: noneNow, let's list them all:2,28,126,344,730,54,152,370,756,250,468,854,686,16,72,224,520,128,280,576,432,728Now, let's sort them to check for duplicates:2,16,28,54,72,126,128,152,224,250,280,344,370,432,468,520,576,686,728,730,756,854That's 22 numbers.So, I think the correct answer is 22.Wait, but in the initial thought process, the user had 25, including some odd numbers and some sums that are actually three cubes. So, perhaps the correct answer is 22.But let me double-check if I missed any sums.Wait, what about 342? Is 342 a sum of two cubes? Let me check:342= a³ + b³.Let me try a=7: 343 is too big.a=6:216. 342-216=126=5³ +1³. So, 342=6³ +5³ +1³. That's three cubes.Alternatively, 342= a³ + b³.Is there a pair where a³ + b³=342?Let me check:a=5:125. 342-125=217, not a cube.a=4:64. 342-64=278, not a cube.a=3:27. 342-27=315, not a cube.a=2:8. 342-8=334, not a cube.a=1:1. 342-1=341, not a cube.So, 342 cannot be expressed as the sum of two cubes. So, it shouldn't be in the list.Similarly, 539=3³ +8³=27+512=539. But 3 is odd, 8 is even, so their sum is odd. So, 539 is odd, shouldn't be in our list.Same with 637=5³ +8³=125+512=637, which is odd.793=4³ +9³=64+729=793, which is odd.855=8³ +7³=512+343=855, which is odd.945=6³ +9³=216+729=945, which is odd.So, all these numbers are odd and shouldn't be in our list.Therefore, the correct count is 22.Wait, but in the initial thought process, the user had 25, including these odd numbers. So, perhaps the correct answer is 22.But let me check another source or way to confirm.Alternatively, I can list all possible even sums less than 1000 and count them.Let me list them again:From odd cubes:2,28,126,344,730,54,152,370,756,250,468,854,686From even cubes:16,72,224,520,128,280,576,432,728Total 22 numbers.So, I think the correct answer is 22.Wait, but the initial thought process had 25. Maybe the user included some sums that are actually three cubes or included odd sums by mistake.Therefore, I think the correct answer is 22.But wait, let me check if I missed any sums.Wait, what about 342? As I checked earlier, it's not a sum of two cubes.Similarly, 539,637,793,855,945 are all odd and shouldn't be included.So, I think 22 is correct.But wait, let me check another way. Maybe I can use a program or a calculator to list all possible sums.But since I'm doing this manually, let me try to see if I missed any.Wait, from a=1, I have 2,28,126,344,730.From a=3, I have 54,152,370,756.From a=5, I have 250,468,854.From a=7, I have 686.From a=2, I have16,72,224,520.From a=4, I have128,280,576.From a=6, I have432,728.From a=8, none.So, that's 5+4+3+1+4+3+2=22.Yes, 22 numbers.Therefore, the correct answer is 22.

💡First, identify the individual probabilities of each brother being selected:- Probability of Ram being selected: 2/7- Probability of Ravi being selected: 1/5- Probability of Raj being selected: 3/8Next, multiply these probabilities together to find the probability that all three are selected:(2/7) * (1/5) * (3/8) = (2 * 1 * 3) / (7 * 5 * 8) = 6/280Finally, simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2:6 ÷ 2 = 3280 ÷ 2 = 140The simplified probability that all three brothers are selected is 3/140.

💡Okay, so I have this problem where I need to find the equation of a curve that satisfies two different sets of conditions. Let me try to break it down step by step.First, the problem is divided into two parts. The first part is about finding the equation of a line, and the second part is about finding the equation of a circle. I need to tackle each part separately and then present both solutions.Starting with the first part:1. **Finding the equation of the line:** - The line passes through the intersection point of two given lines: (2x + y - 8 = 0) and (x - 2y + 1 = 0). - It is also perpendicular to the line (6x - 8y + 3 = 0).Alright, so I need to find the intersection point of the two lines first. To do that, I can solve the system of equations:[begin{cases}2x + y - 8 = 0 x - 2y + 1 = 0end{cases}]Let me solve this system. Maybe I can use substitution or elimination. Let's try elimination.From the first equation: (2x + y = 8). Let me solve for y: (y = 8 - 2x).Now, substitute this into the second equation:(x - 2(8 - 2x) + 1 = 0)Simplify:(x - 16 + 4x + 1 = 0)Combine like terms:(5x - 15 = 0)So, (5x = 15), which gives (x = 3).Now, substitute (x = 3) back into (y = 8 - 2x):(y = 8 - 6 = 2).So, the intersection point is (3, 2). Got that.Next, the line we're looking for is perpendicular to (6x - 8y + 3 = 0). I need to find the slope of this line first.Let me rewrite (6x - 8y + 3 = 0) in slope-intercept form (y = mx + b):Subtract 6x and 3: (-8y = -6x - 3)Divide by -8: (y = (6/8)x + 3/8), which simplifies to (y = (3/4)x + 3/8).So, the slope of this line is (3/4). Since our desired line is perpendicular to this, its slope should be the negative reciprocal. The negative reciprocal of (3/4) is (-4/3).Therefore, the slope of our line is (-4/3).Now, we have a point (3, 2) and a slope (-4/3). Let me write the equation of the line using point-slope form:(y - y_1 = m(x - x_1))Plugging in the values:(y - 2 = (-4/3)(x - 3))Let me simplify this:Multiply both sides by 3 to eliminate the fraction:(3(y - 2) = -4(x - 3))Expand both sides:(3y - 6 = -4x + 12)Bring all terms to one side:(4x + 3y - 18 = 0)So, the equation of the line is (4x + 3y - 18 = 0). That seems straightforward.Moving on to the second part:2. **Finding the equation of the circle:** - The circle passes through points C(-1, 1) and D(1, 3). - Its center is on the x-axis.Alright, so the center of the circle is on the x-axis, which means its coordinates are (h, 0) for some h. The circle passes through points C and D, so the distance from the center to both points should be equal (since both are radii).Let me denote the center as (h, 0). The distance from (h, 0) to C(-1, 1) should equal the distance from (h, 0) to D(1, 3).Using the distance formula, the distance squared from center to C is:[(h + 1)^2 + (0 - 1)^2 = (h + 1)^2 + 1]Similarly, the distance squared from center to D is:[(h - 1)^2 + (0 - 3)^2 = (h - 1)^2 + 9]Since these distances are equal:[(h + 1)^2 + 1 = (h - 1)^2 + 9]Let me expand both sides:Left side: (h^2 + 2h + 1 + 1 = h^2 + 2h + 2)Right side: (h^2 - 2h + 1 + 9 = h^2 - 2h + 10)Set them equal:(h^2 + 2h + 2 = h^2 - 2h + 10)Subtract (h^2) from both sides:(2h + 2 = -2h + 10)Bring all terms to one side:(2h + 2 + 2h - 10 = 0)Combine like terms:(4h - 8 = 0)So, (4h = 8), which gives (h = 2).Therefore, the center is at (2, 0). Now, let's find the radius. We can use either point C or D. Let's use point C(-1, 1):Radius squared is:[(2 + 1)^2 + (0 - 1)^2 = 3^2 + (-1)^2 = 9 + 1 = 10]So, the radius is (sqrt{10}).Thus, the equation of the circle is:[(x - 2)^2 + (y - 0)^2 = 10]Simplifying, it's:[(x - 2)^2 + y^2 = 10]Let me just verify this with point D(1, 3):[(1 - 2)^2 + (3)^2 = (-1)^2 + 9 = 1 + 9 = 10]Yes, that works. So, the equation is correct.So, summarizing both parts:1. The line has equation (4x + 3y - 18 = 0).2. The circle has equation ((x - 2)^2 + y^2 = 10).I think that's it. I didn't encounter any issues, but let me just double-check my calculations.For the line:- Intersection point: (3, 2). Correct.- Slope of given line: 3/4, so perpendicular slope is -4/3. Correct.- Equation using point-slope: Correct steps, ended up with 4x + 3y - 18 = 0. Seems good.For the circle:- Center on x-axis: (h, 0). Correct.- Equate distances from center to C and D: Correct setup.- Solved for h: Got h = 2. Correct.- Radius squared: 10. Correct.- Equation: (x - 2)^2 + y^2 = 10. Correct.Everything seems to check out.**Final Answer**The equation of the line is boxed{4x + 3y - 18 = 0} and the equation of the circle is boxed{(x - 2)^2 + y^2 = 10}.

💡To determine how many of the given statements imply the truth of ((p wedge q) rightarrow r), we evaluate each statement by checking the truth value of ((p wedge q) rightarrow r) under the conditions specified by each statement.1. **Statement 1: (p wedge q wedge sim r)** - Here, (p) and (q) are true, and (r) is false. - (p wedge q) is true. - Therefore, ((p wedge q) rightarrow r) is equivalent to ( text{true} rightarrow text{false} ), which is false.2. **Statement 2: (sim p wedge q wedge r)** - Here, (p) is false, (q) is true, and (r) is true. - (p wedge q) is false because (p) is false. - Therefore, ((p wedge q) rightarrow r) is equivalent to ( text{false} rightarrow text{true} ), which is true.3. **Statement 3: (p wedge sim q wedge sim r)** - Here, (p) is true, (q) is false, and (r) is false. - (p wedge q) is false because (q) is false. - Therefore, ((p wedge q) rightarrow r) is equivalent to ( text{false} rightarrow text{false} ), which is true.4. **Statement 4: (sim p wedge sim q wedge sim r)** - Here, (p), (q), and (r) are all false. - (p wedge q) is false because both are false. - Therefore, ((p wedge q) rightarrow r) is equivalent to ( text{false} rightarrow text{false} ), which is true.From the above evaluations, Statements 2, 3, and 4 imply the truth of ((p wedge q) rightarrow r). Thus, the number of statements that imply the truth of the given implication is 3.

💡Okay, so I have this problem here: I need to find the coefficient of ( x^4 ) in the expansion of the function ( f(x) = left( x^2 + frac{1}{x} right)^n ), where ( n ) is given by ( 5 int_{0}^{frac{pi}{2}} cos x , dx ). Hmm, let me break this down step by step.First, I need to figure out what ( n ) is. The problem says ( n = 5 times ) the integral of ( cos x ) from 0 to ( frac{pi}{2} ). I remember that the integral of ( cos x ) is ( sin x ), so let me compute that.Calculating the integral:[int_{0}^{frac{pi}{2}} cos x , dx = sin x Big|_{0}^{frac{pi}{2}} = sinleft( frac{pi}{2} right) - sin(0) = 1 - 0 = 1]So, the integral is 1. Then, multiplying by 5 gives ( n = 5 times 1 = 5 ). Got it, so ( n = 5 ).Now, the function becomes ( f(x) = left( x^2 + frac{1}{x} right)^5 ). I need to expand this and find the coefficient of ( x^4 ). I think the binomial theorem is the way to go here. The binomial theorem states that:[(a + b)^k = sum_{m=0}^{k} binom{k}{m} a^{k - m} b^m]In this case, ( a = x^2 ), ( b = frac{1}{x} ), and ( k = 5 ). So, each term in the expansion will be:[binom{5}{m} (x^2)^{5 - m} left( frac{1}{x} right)^m]Simplifying each term:[binom{5}{m} x^{2(5 - m)} times x^{-m} = binom{5}{m} x^{10 - 2m - m} = binom{5}{m} x^{10 - 3m}]So, each term has an exponent of ( 10 - 3m ). I need the exponent to be 4, so I set up the equation:[10 - 3m = 4]Solving for ( m ):[10 - 4 = 3m implies 6 = 3m implies m = 2]So, the term where ( m = 2 ) will have the exponent ( x^4 ). Now, let's find the coefficient for this term:[binom{5}{2} = frac{5!}{2!(5 - 2)!} = frac{120}{2 times 6} = frac{120}{12} = 10]Therefore, the coefficient of ( x^4 ) is 10.Wait, let me double-check my steps to make sure I didn't make a mistake. First, calculating ( n ): integral of ( cos x ) from 0 to ( pi/2 ) is indeed 1, so ( n = 5 ). Then, using the binomial theorem on ( (x^2 + 1/x)^5 ), each term is ( binom{5}{m} x^{10 - 3m} ). Setting ( 10 - 3m = 4 ) gives ( m = 2 ). Calculating ( binom{5}{2} ) gives 10. That seems correct.I don't see any errors in my calculations. So, I think the coefficient is indeed 10.**Final Answer**The coefficient of ( x^4 ) is boxed{10}.