💡Alright, so I have this problem here with two parts. Let me try to tackle them one by one. Starting with part (1): When a is less than 0, what is a divided by the absolute value of a? Hmm, okay. I remember that the absolute value of a number is always positive, right? So if a is negative, then |a| would be -a because that would make it positive. So, if I take a and divide it by |a|, which is -a, then a divided by -a should be -1. That seems straightforward. Let me double-check: if a is negative, say a = -2, then |a| is 2, and -2 divided by 2 is indeed -1. Okay, so part (1) seems to be -1.Now, moving on to part (2). This one looks a bit more complicated. I need to find the minimum value of the expression: (ab)/|ab| + |bc|/bc + (ca)/|ca| + |abc|/abc. All right, let's break this down.First, I notice that each term in the expression is a fraction where the numerator and denominator are related by absolute value. So, for any term like x/|x|, if x is positive, this fraction is 1, and if x is negative, it's -1. That's because the absolute value of x is always positive, so if x is negative, dividing by its absolute value flips the sign.So, each of these terms can only be either 1 or -1. Therefore, the entire expression is the sum of four terms, each of which is either 1 or -1. So, the maximum possible value of this expression would be 4 (if all terms are 1), and the minimum possible value would be -4 (if all terms are -1). But wait, is that actually possible? Let me think.If all terms are 1, that would mean that ab, bc, ca, and abc are all positive. But if ab is positive, that means a and b have the same sign. Similarly, bc positive means b and c have the same sign, and ca positive means c and a have the same sign. So, if a and b have the same sign, and b and c have the same sign, then a and c must also have the same sign. Therefore, all three variables a, b, and c must have the same sign. If they are all positive, then ab, bc, ca, and abc are all positive, so each term is 1, and the sum is 4. That makes sense.Now, what about the minimum value? If all terms are -1, that would mean ab, bc, ca, and abc are all negative. Let's see if that's possible. If ab is negative, then a and b have opposite signs. Similarly, bc negative means b and c have opposite signs, and ca negative means c and a have opposite signs. So, if a and b have opposite signs, and b and c have opposite signs, then a and c must have the same sign. But then, if a and c have the same sign, then ca would be positive, which contradicts the requirement that ca is negative. Therefore, it's impossible for all four terms to be -1 simultaneously.So, the minimum value can't be -4. Let's see what's the next possible minimum. If three terms are -1 and one is 1, the total would be -2. Is that achievable? Let's check.Suppose ab is negative, bc is negative, ca is negative, and abc is positive. Wait, but if ab is negative, that means a and b have opposite signs. If bc is negative, b and c have opposite signs. So, a and c must have the same sign. Then, ca would be positive, which contradicts ca being negative. So, that doesn't work.Alternatively, maybe ab is negative, bc is negative, ca is negative, and abc is negative. But as we saw earlier, if ab, bc, and ca are negative, then a and c must have the same sign, making ca positive, which is a contradiction.Hmm, maybe I need to approach this differently. Let's consider different cases based on the number of negative variables among a, b, and c.Case 1: All three variables are positive. Then, ab, bc, ca, and abc are all positive, so each term is 1, and the sum is 4.Case 2: Exactly one variable is negative. Let's say a is negative, and b and c are positive. Then, ab is negative, bc is positive, ca is negative, and abc is negative. So, (ab)/|ab| = -1, |bc|/bc = 1, (ca)/|ca| = -1, and |abc|/abc = -1. Adding these up: -1 + 1 -1 -1 = -2.Case 3: Exactly two variables are negative. Let's say a and b are negative, and c is positive. Then, ab is positive, bc is negative, ca is negative, and abc is positive. So, (ab)/|ab| = 1, |bc|/bc = -1, (ca)/|ca| = -1, and |abc|/abc = 1. Adding these up: 1 -1 -1 +1 = 0.Case 4: All three variables are negative. Then, ab is positive, bc is positive, ca is positive, and abc is negative. So, (ab)/|ab| = 1, |bc|/bc = 1, (ca)/|ca| = 1, and |abc|/abc = -1. Adding these up: 1 +1 +1 -1 = 2.So, from these cases, the possible sums are 4, -2, 0, and 2. Therefore, the minimum value is -2.Wait, let me verify this. In Case 2, when exactly one variable is negative, the sum is -2. Is there a case where the sum is less than -2? From the earlier analysis, it seems not, because having two variables negative gives a sum of 0, and having all three negative gives a sum of 2. So, the minimum is indeed -2.Therefore, the answers are:(1) When a < 0, a/|a| = -1.(2) The minimum value of the expression is -2.

💡First, I need to plot the vertices of the trapezoid to understand its shape and orientation in the coordinate system.Next, I'll identify the two parallel sides (bases) of the trapezoid. Looking at the coordinates, sides ( AB ) and ( CD ) are both vertical lines, which means they are parallel.To find the lengths of these bases, I'll calculate the distance between the ( y )-coordinates of points ( A ) and ( B ) for base ( AB ), and between points ( C ) and ( D ) for base ( CD ).Once I have the lengths of both bases, I'll determine the height of the trapezoid, which is the horizontal distance between the two vertical sides. This is found by calculating the difference between the ( x )-coordinates of any corresponding points from the two bases.Finally, I'll use the formula for the area of a trapezoid, which is ( frac{1}{2} times (b_1 + b_2) times h ), where ( b_1 ) and ( b_2 ) are the lengths of the two bases, and ( h ) is the height. Plugging in the values I've calculated, I'll compute the area of the trapezoid.

💡First, I need to determine the relative speed between the train and the man. Since they are moving in opposite directions, I add their speeds to find the relative speed.The train's speed is 60 km/h, and the man's speed is 6 km/h.Relative speed = Speed of train + Speed of manRelative speed = 60 km/h + 6 km/hRelative speed = 66 km/hNext, I convert the relative speed from kilometers per hour to meters per second to match the length of the train, which is given in meters.1 km/h = 1000 meters / 3600 seconds66 km/h = 66 * (1000 / 3600) m/s66 km/h = 66 * (5 / 18) m/s66 km/h = 18.333... m/s (approximately)Now that I have the relative speed in meters per second, I can calculate the time it takes for the train to pass the man completely.Time = Distance / SpeedTime = Length of train / Relative speedTime = 385 meters / 18.333... m/sTime ≈ 21.005 secondsTherefore, it will take approximately 21.005 seconds for the train to pass the man.

💡Okay, so I have this problem here where I need to figure out how many times the graphs of ( y = 3log{x} ) and ( y = log{x^2} + 1 ) intersect. The options are from A to E, with E being that they coincide entirely. Hmm, let's break this down step by step.First, I remember that to find the intersection points of two graphs, I need to set their equations equal to each other and solve for ( x ). So, I'll start by setting ( 3log{x} = log{x^2} + 1 ).Now, I recall that ( log{x^2} ) can be simplified using logarithmic properties. Specifically, ( log{x^2} = 2log{|x|} ). But since we're dealing with logarithms, ( x ) has to be positive anyway because you can't take the log of a negative number or zero. So, I can safely say ( log{x^2} = 2log{x} ).Substituting that back into the equation, it becomes ( 3log{x} = 2log{x} + 1 ). Hmm, okay, let's subtract ( 2log{x} ) from both sides to get ( log{x} = 1 ).Now, solving for ( x ), I remember that ( log{x} = 1 ) means ( x = 10 ) because ( log{10} = 1 ). So, it looks like ( x = 10 ) is a solution. But wait, is that the only solution?Let me think about the behavior of these functions. The function ( y = 3log{x} ) is a logarithmic function stretched vertically by a factor of 3, while ( y = log{x^2} + 1 ) simplifies to ( y = 2log{x} + 1 ), which is a logarithmic function stretched by 2 and shifted up by 1.Since both functions are logarithmic, they are defined for ( x > 0 ). Let's analyze their growth rates. The function ( 3log{x} ) grows faster than ( 2log{x} + 1 ) as ( x ) increases because 3 is greater than 2. Conversely, for values of ( x ) less than 10, ( 3log{x} ) is less than ( 2log{x} + 1 ) because the coefficient 3 makes it steeper but starting from a lower point.So, at ( x = 10 ), both functions meet. For ( x > 10 ), ( 3log{x} ) will be above ( 2log{x} + 1 ), and for ( x < 10 ), ( 3log{x} ) will be below ( 2log{x} + 1 ). This suggests that they intersect only once at ( x = 10 ).Wait, but just to be thorough, could there be another intersection point somewhere else? Let me consider the possibility of multiple solutions. If I set ( 3log{x} = 2log{x} + 1 ), simplifying gives ( log{x} = 1 ), which only has one solution at ( x = 10 ). There's no quadratic or higher-degree equation here, so only one solution exists.Therefore, the graphs intersect at exactly one point. So, the answer should be option B: They intersect at 1 point only.**Final Answer**boxed{B}

💡Okay, so I have this problem here: "The coefficient of ( x^{7} ) in the expansion of ( (x + a)^{10} ) is 15. Find the value of ( a )." Hmm, let me try to figure this out step by step.First, I remember that when you expand a binomial like ( (x + a)^{n} ), you can use the binomial theorem. The binomial theorem says that each term in the expansion is of the form ( binom{n}{k} x^{n - k} a^{k} ), where ( binom{n}{k} ) is the binomial coefficient, also known as "n choose k."In this problem, ( n ) is 10 because the exponent on ( (x + a) ) is 10. We're interested in the coefficient of ( x^{7} ). So, I need to find the term where the power of ( x ) is 7. Let me write down the general term from the binomial expansion:[ T_{k+1} = binom{10}{k} x^{10 - k} a^{k} ]We want the term where the exponent of ( x ) is 7. So, I set up the equation:[ 10 - k = 7 ]Solving for ( k ), I subtract 7 from both sides:[ 10 - 7 = k ][ k = 3 ]Okay, so the term we're interested in is when ( k = 3 ). Plugging this back into the general term:[ T_{4} = binom{10}{3} x^{7} a^{3} ]Now, the coefficient of ( x^{7} ) is given as 15. That means:[ binom{10}{3} a^{3} = 15 ]I need to compute ( binom{10}{3} ). The formula for combinations is:[ binom{n}{k} = frac{n!}{k!(n - k)!} ]So,[ binom{10}{3} = frac{10!}{3! cdot 7!} ]Calculating this:[ frac{10 times 9 times 8 times 7!}{3 times 2 times 1 times 7!} ]The ( 7! ) cancels out from numerator and denominator:[ frac{10 times 9 times 8}{3 times 2 times 1} ]Calculating the numerator:[ 10 times 9 = 90 ][ 90 times 8 = 720 ]Denominator:[ 3 times 2 = 6 ][ 6 times 1 = 6 ]So,[ frac{720}{6} = 120 ]Therefore, ( binom{10}{3} = 120 ).Now, going back to the equation:[ 120 cdot a^{3} = 15 ]To solve for ( a ), I divide both sides by 120:[ a^{3} = frac{15}{120} ]Simplify the fraction:[ frac{15}{120} = frac{1}{8} ]So,[ a^{3} = frac{1}{8} ]To find ( a ), take the cube root of both sides:[ a = sqrt[3]{frac{1}{8}} ]Since ( sqrt[3]{frac{1}{8}} = frac{1}{2} ), because ( left( frac{1}{2} right)^{3} = frac{1}{8} ).Therefore, the value of ( a ) is ( frac{1}{2} ).Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the binomial expansion, I identified the correct term by setting ( 10 - k = 7 ), which gives ( k = 3 ). Then, I correctly calculated the binomial coefficient ( binom{10}{3} = 120 ). Plugging that into the equation ( 120a^{3} = 15 ), I solved for ( a^{3} = frac{1}{8} ), leading to ( a = frac{1}{2} ). That seems right.I think I've got it. The value of ( a ) is ( frac{1}{2} ).**Final Answer**The value of ( a ) is boxed{dfrac{1}{2}}.

💡Okay, so I've got this problem here where I need to figure out how many integers n from 1 to 100 inclusive will eventually reach 1 when I apply this function f repeatedly. The function f is defined differently depending on whether n is odd or even. If n is odd, f(n) is n squared plus 1, and if n is even, f(n) is n over 2 plus 3. Alright, let me start by understanding what the function does. If I have an odd number, squaring it and adding 1 will give me an even number because any odd number squared is odd, and adding 1 makes it even. On the other hand, if I have an even number, dividing it by 2 and adding 3 will give me another number, which could be either odd or even depending on the original number.So, the function alternates between these two operations based on whether the current number is odd or even. My goal is to find all starting numbers n between 1 and 100 such that after some number of applications of f, I end up at 1.Let me think about what happens when we apply f repeatedly. If I start with n = 1, which is odd, then f(1) = 1^2 + 1 = 2. Now, 2 is even, so f(2) = 2/2 + 3 = 1 + 3 = 4. Then, f(4) = 4/2 + 3 = 2 + 3 = 5. 5 is odd, so f(5) = 5^2 + 1 = 25 + 1 = 26. 26 is even, so f(26) = 26/2 + 3 = 13 + 3 = 16. 16 is even, so f(16) = 16/2 + 3 = 8 + 3 = 11. 11 is odd, so f(11) = 11^2 + 1 = 121 + 1 = 122. 122 is even, so f(122) = 122/2 + 3 = 61 + 3 = 64. 64 is even, so f(64) = 64/2 + 3 = 32 + 3 = 35. 35 is odd, so f(35) = 35^2 + 1 = 1225 + 1 = 1226. 1226 is even, so f(1226) = 1226/2 + 3 = 613 + 3 = 616. 616 is even, so f(616) = 616/2 + 3 = 308 + 3 = 311. 311 is odd, so f(311) = 311^2 + 1 = 96721 + 1 = 96722. 96722 is even, so f(96722) = 96722/2 + 3 = 48361 + 3 = 48364. 48364 is even, so f(48364) = 48364/2 + 3 = 24182 + 3 = 24185. 24185 is odd, so f(24185) = 24185^2 + 1. That's a huge number, and it seems like we're just getting larger and larger without ever coming back down to 1.Wait, so starting from n=1, we go to 2, then to 4, 5, 26, 16, 11, 122, 64, 35, 1226, 616, 311, 96722, 48364, 24185, and so on. It doesn't seem like we're ever getting back to 1. In fact, we're just increasing the number each time, especially when we hit an odd number because squaring it makes it much larger.Let me try another starting number. How about n=2? f(2) = 2/2 + 3 = 1 + 3 = 4. Then f(4) = 4/2 + 3 = 2 + 3 = 5. Then f(5) = 25 + 1 = 26, and we're back to the same sequence as before. So n=2 also doesn't lead us to 1.What about n=3? f(3) = 3^2 + 1 = 9 + 1 = 10. 10 is even, so f(10) = 10/2 + 3 = 5 + 3 = 8. 8 is even, so f(8) = 8/2 + 3 = 4 + 3 = 7. 7 is odd, so f(7) = 49 + 1 = 50. 50 is even, so f(50) = 50/2 + 3 = 25 + 3 = 28. 28 is even, so f(28) = 28/2 + 3 = 14 + 3 = 17. 17 is odd, so f(17) = 289 + 1 = 290. 290 is even, so f(290) = 290/2 + 3 = 145 + 3 = 148. 148 is even, so f(148) = 148/2 + 3 = 74 + 3 = 77. 77 is odd, so f(77) = 5929 + 1 = 5930. 5930 is even, so f(5930) = 5930/2 + 3 = 2965 + 3 = 2968. 2968 is even, so f(2968) = 2968/2 + 3 = 1484 + 3 = 1487. 1487 is odd, so f(1487) = 1487^2 + 1, which is a very large number. Again, we're just increasing and not getting back to 1.Hmm, maybe I should try a smaller number. Let's try n=4. f(4) = 4/2 + 3 = 2 + 3 = 5. Then f(5) = 25 + 1 = 26, and we know from before that 26 leads to 16, 11, 122, etc., so it's not going to reach 1.What about n=5? f(5) = 25 + 1 = 26, same as above.n=6: f(6) = 6/2 + 3 = 3 + 3 = 6. Wait, so f(6) = 6. That means 6 is a fixed point. So starting from 6, we just stay at 6 forever. So 6 doesn't lead to 1.n=7: f(7) = 49 + 1 = 50, which we saw earlier leads to 28, 17, 290, etc., not reaching 1.n=8: f(8) = 8/2 + 3 = 4 + 3 = 7, which we saw leads to 50, 28, etc.n=9: f(9) = 81 + 1 = 82. 82 is even, so f(82) = 82/2 + 3 = 41 + 3 = 44. 44 is even, so f(44) = 44/2 + 3 = 22 + 3 = 25. 25 is odd, so f(25) = 625 + 1 = 626. 626 is even, so f(626) = 626/2 + 3 = 313 + 3 = 316. 316 is even, so f(316) = 316/2 + 3 = 158 + 3 = 161. 161 is odd, so f(161) = 161^2 + 1 = 25921 + 1 = 25922. 25922 is even, so f(25922) = 25922/2 + 3 = 12961 + 3 = 12964. 12964 is even, so f(12964) = 12964/2 + 3 = 6482 + 3 = 6485. 6485 is odd, so f(6485) = 6485^2 + 1, which is huge. So again, we're just increasing.n=10: f(10) = 5 + 3 = 8, which leads to 7, 50, etc.n=11: f(11) = 121 + 1 = 122, which we've seen before.n=12: f(12) = 12/2 + 3 = 6 + 3 = 9. Then f(9) = 81 + 1 = 82, which leads to 44, 25, etc.n=13: f(13) = 169 + 1 = 170. 170 is even, so f(170) = 170/2 + 3 = 85 + 3 = 88. 88 is even, so f(88) = 88/2 + 3 = 44 + 3 = 47. 47 is odd, so f(47) = 2209 + 1 = 2210. 2210 is even, so f(2210) = 2210/2 + 3 = 1105 + 3 = 1108. 1108 is even, so f(1108) = 1108/2 + 3 = 554 + 3 = 557. 557 is odd, so f(557) = 557^2 + 1, which is a large number.n=14: f(14) = 14/2 + 3 = 7 + 3 = 10, which leads to 8, 7, etc.n=15: f(15) = 225 + 1 = 226. 226 is even, so f(226) = 226/2 + 3 = 113 + 3 = 116. 116 is even, so f(116) = 116/2 + 3 = 58 + 3 = 61. 61 is odd, so f(61) = 3721 + 1 = 3722. 3722 is even, so f(3722) = 3722/2 + 3 = 1861 + 3 = 1864. 1864 is even, so f(1864) = 1864/2 + 3 = 932 + 3 = 935. 935 is odd, so f(935) = 935^2 + 1, which is huge.n=16: f(16) = 16/2 + 3 = 8 + 3 = 11, which we've seen leads to 122, etc.n=17: f(17) = 289 + 1 = 290, which leads to 145, 77, etc.n=18: f(18) = 18/2 + 3 = 9 + 3 = 12, which leads to 6, a fixed point.n=19: f(19) = 361 + 1 = 362. 362 is even, so f(362) = 362/2 + 3 = 181 + 3 = 184. 184 is even, so f(184) = 184/2 + 3 = 92 + 3 = 95. 95 is odd, so f(95) = 9025 + 1 = 9026. 9026 is even, so f(9026) = 9026/2 + 3 = 4513 + 3 = 4516. 4516 is even, so f(4516) = 4516/2 + 3 = 2258 + 3 = 2261. 2261 is odd, so f(2261) = 2261^2 + 1, which is a huge number.n=20: f(20) = 20/2 + 3 = 10 + 3 = 13, which leads to 82, etc.Hmm, I'm noticing a pattern here. Whenever I start with an odd number, applying f once gives me an even number, which then gets divided by 2 and added 3. If that result is even again, it continues, but if it becomes odd, then squaring it makes it much larger, and the cycle continues without ever reaching 1.Even numbers, when divided by 2 and added 3, can sometimes lead to odd numbers, which then get squared, making them even larger. It seems like once we hit an odd number, the sequence tends to grow without bound, never coming back down to 1.Let me check if there's any number that can lead to 1. Suppose there exists some n such that f(n) = 1. Then, n must satisfy either n^2 + 1 = 1 (if n is odd) or n/2 + 3 = 1 (if n is even).For n odd: n^2 + 1 = 1 => n^2 = 0 => n = 0. But 0 is not in our range of 1 to 100.For n even: n/2 + 3 = 1 => n/2 = -2 => n = -4. Again, not in our range.So, there is no n in 1 to 100 such that f(n) = 1. Therefore, it's impossible for any number in this range to reach 1 in one step.But maybe it can reach 1 in more than one step? Let's see. Suppose there's a number m such that f(m) = k, and f(k) = 1. But as we saw, there's no k such that f(k) = 1, because the only possible k would have to be 0 or -4, which aren't in our range. So, even if we could reach such a k, it's not possible.Therefore, it seems like there are no numbers n from 1 to 100 for which repeated applications of f(n) will ever reach 1. Every starting number either leads to a fixed point (like 6) or enters a cycle that grows indefinitely without ever reaching 1.Wait, let me double-check. Maybe there's a cycle that includes 1? If 1 were part of a cycle, then starting from 1, we would loop back to 1 after some steps. But as we saw earlier, starting from 1 leads to 2, 4, 5, 26, 16, 11, 122, 64, 35, 1226, 616, 311, 96722, etc., which doesn't include 1 again. So, 1 is not part of a cycle.Alternatively, maybe there's a number that eventually maps to 1 through some other path. But as we saw, the only way to get to 1 is if f(n) = 1, which isn't possible for any n in 1 to 100. So, even if we could reach 1 in multiple steps, the intermediate steps would have to somehow lead to a number that maps to 1, which doesn't exist in our range.Therefore, after considering all possibilities, I conclude that there are no integers n from 1 to 100 for which repeated applications of f(n) will result in 1.