💡Okay, so I have this problem where I need to find the maximum value of the function ( f(x) = frac{x^4 - x^2}{x^6 + 2x^3 - 1} ) for real numbers ( x > 1 ). Hmm, that looks a bit complicated, but maybe I can simplify it somehow.First, let me write down the function again to make sure I have it right:[ f(x) = frac{x^4 - x^2}{x^6 + 2x^3 - 1} ]I notice that both the numerator and the denominator are polynomials in ( x ). Maybe I can factor them or find a substitution that makes the expression easier to handle.Looking at the numerator: ( x^4 - x^2 ). I can factor out an ( x^2 ):[ x^4 - x^2 = x^2(x^2 - 1) = x^2(x - 1)(x + 1) ]Okay, that's factored. Now, the denominator: ( x^6 + 2x^3 - 1 ). Hmm, that's a bit trickier. Maybe I can factor it as a quadratic in terms of ( x^3 ). Let me set ( y = x^3 ), then the denominator becomes:[ y^2 + 2y - 1 ]That's a quadratic equation. Let me see if I can factor it or find its roots. The quadratic formula gives:[ y = frac{-2 pm sqrt{4 + 4}}{2} = frac{-2 pm sqrt{8}}{2} = frac{-2 pm 2sqrt{2}}{2} = -1 pm sqrt{2} ]So, the denominator factors as:[ (x^3 + 1 + sqrt{2})(x^3 + 1 - sqrt{2}) ]Hmm, not sure if that helps directly, but maybe I can use substitution or another approach.Wait, maybe I can rewrite the original function in terms of ( t = x - frac{1}{x} ). Let me try that substitution because sometimes such substitutions simplify expressions involving ( x ) and ( frac{1}{x} ).Let ( t = x - frac{1}{x} ). Then, ( t ) is positive because ( x > 1 ), so ( x - frac{1}{x} > 0 ).Let me express the numerator and denominator in terms of ( t ).First, the numerator:[ x^4 - x^2 = x^2(x^2 - 1) = x^2(x - 1)(x + 1) ]But I'm not sure how to express this in terms of ( t ). Maybe I need a different approach.Alternatively, let me consider dividing both numerator and denominator by ( x^3 ) to simplify the expression:[ f(x) = frac{frac{x^4 - x^2}{x^3}}{frac{x^6 + 2x^3 - 1}{x^3}} = frac{x - frac{1}{x}}{x^3 + 2 - frac{1}{x^3}} ]Okay, that looks a bit better. Now, let me set ( t = x - frac{1}{x} ) as before. Then, I can express ( x^3 - frac{1}{x^3} ) in terms of ( t ).Recall that:[ (x - frac{1}{x})^3 = x^3 - 3x + frac{3}{x} - frac{1}{x^3} ]So,[ t^3 = x^3 - frac{1}{x^3} - 3(x - frac{1}{x}) ]Therefore,[ x^3 - frac{1}{x^3} = t^3 + 3t ]Great, so substituting back into the denominator:[ x^3 + 2 - frac{1}{x^3} = (x^3 - frac{1}{x^3}) + 2 = t^3 + 3t + 2 ]So now, the function ( f(x) ) becomes:[ f(x) = frac{t}{t^3 + 3t + 2} ]Now, I need to find the maximum value of ( frac{t}{t^3 + 3t + 2} ) where ( t = x - frac{1}{x} ) and ( x > 1 ). Since ( x > 1 ), ( t ) will be positive because ( x - frac{1}{x} > 0 ) when ( x > 1 ).So, let me consider ( t > 0 ) and find the maximum of ( f(t) = frac{t}{t^3 + 3t + 2} ).To find the maximum, I can take the derivative of ( f(t) ) with respect to ( t ) and set it equal to zero.First, compute the derivative ( f'(t) ):Using the quotient rule:[ f'(t) = frac{(1)(t^3 + 3t + 2) - t(3t^2 + 3)}{(t^3 + 3t + 2)^2} ]Simplify the numerator:[ (t^3 + 3t + 2) - t(3t^2 + 3) = t^3 + 3t + 2 - 3t^3 - 3t = -2t^3 + 0t + 2 ]So,[ f'(t) = frac{-2t^3 + 2}{(t^3 + 3t + 2)^2} ]Set the derivative equal to zero to find critical points:[ -2t^3 + 2 = 0 ][ -2t^3 = -2 ][ t^3 = 1 ][ t = 1 ]So, the critical point is at ( t = 1 ). Now, I need to check if this is a maximum.Let me test values around ( t = 1 ). For ( t < 1 ), say ( t = 0.5 ):[ f'(0.5) = frac{-2(0.5)^3 + 2}{(0.5^3 + 3*0.5 + 2)^2} = frac{-2*(0.125) + 2}{(0.125 + 1.5 + 2)^2} = frac{-0.25 + 2}{(3.625)^2} = frac{1.75}{13.140625} > 0 ]So, the function is increasing before ( t = 1 ).For ( t > 1 ), say ( t = 2 ):[ f'(2) = frac{-2*(8) + 2}{(8 + 6 + 2)^2} = frac{-16 + 2}{16^2} = frac{-14}{256} < 0 ]So, the function is decreasing after ( t = 1 ). Therefore, ( t = 1 ) is indeed a maximum.Now, compute ( f(1) ):[ f(1) = frac{1}{1^3 + 3*1 + 2} = frac{1}{1 + 3 + 2} = frac{1}{6} ]Wait, that's ( frac{1}{6} ), but I thought earlier it might be ( frac{1}{5} ). Did I make a mistake?Wait, let me double-check the substitution. Earlier, I set ( t = x - frac{1}{x} ), and at ( t = 1 ), what is ( x )?Let me solve for ( x ) when ( t = 1 ):[ x - frac{1}{x} = 1 ]Multiply both sides by ( x ):[ x^2 - 1 = x ][ x^2 - x - 1 = 0 ]Solving this quadratic equation:[ x = frac{1 pm sqrt{1 + 4}}{2} = frac{1 pm sqrt{5}}{2} ]Since ( x > 1 ), we take the positive root:[ x = frac{1 + sqrt{5}}{2} approx 1.618 ]Okay, so ( x ) is the golden ratio. Now, let me compute ( f(x) ) at this ( x ):First, compute ( x^4 - x^2 ):Let me compute ( x^2 ):[ x = frac{1 + sqrt{5}}{2} ][ x^2 = left(frac{1 + sqrt{5}}{2}right)^2 = frac{1 + 2sqrt{5} + 5}{4} = frac{6 + 2sqrt{5}}{4} = frac{3 + sqrt{5}}{2} ]Then, ( x^4 ):[ x^4 = (x^2)^2 = left(frac{3 + sqrt{5}}{2}right)^2 = frac{9 + 6sqrt{5} + 5}{4} = frac{14 + 6sqrt{5}}{4} = frac{7 + 3sqrt{5}}{2} ]So,[ x^4 - x^2 = frac{7 + 3sqrt{5}}{2} - frac{3 + sqrt{5}}{2} = frac{4 + 2sqrt{5}}{2} = 2 + sqrt{5} ]Now, the denominator ( x^6 + 2x^3 - 1 ). Let me compute ( x^3 ) first.Since ( x = frac{1 + sqrt{5}}{2} ), and we know that ( x ) satisfies the equation ( x^2 = x + 1 ). Let me use this to compute higher powers.Compute ( x^3 ):[ x^3 = x * x^2 = x*(x + 1) = x^2 + x = (x + 1) + x = 2x + 1 ]So,[ x^3 = 2x + 1 ]Then, ( x^6 = (x^3)^2 = (2x + 1)^2 = 4x^2 + 4x + 1 )But ( x^2 = x + 1 ), so:[ x^6 = 4(x + 1) + 4x + 1 = 4x + 4 + 4x + 1 = 8x + 5 ]Therefore, the denominator:[ x^6 + 2x^3 - 1 = (8x + 5) + 2*(2x + 1) - 1 = 8x + 5 + 4x + 2 - 1 = 12x + 6 ]So, putting it all together:[ f(x) = frac{2 + sqrt{5}}{12x + 6} ]But ( x = frac{1 + sqrt{5}}{2} ), so:[ 12x + 6 = 12*frac{1 + sqrt{5}}{2} + 6 = 6(1 + sqrt{5}) + 6 = 6 + 6sqrt{5} + 6 = 12 + 6sqrt{5} ]Therefore,[ f(x) = frac{2 + sqrt{5}}{12 + 6sqrt{5}} ]Simplify this fraction by rationalizing the denominator:Multiply numerator and denominator by the conjugate of the denominator's denominator, which is ( 12 - 6sqrt{5} ):[ f(x) = frac{(2 + sqrt{5})(12 - 6sqrt{5})}{(12 + 6sqrt{5})(12 - 6sqrt{5})} ]Compute the denominator:[ (12)^2 - (6sqrt{5})^2 = 144 - 36*5 = 144 - 180 = -36 ]Compute the numerator:[ (2)(12) + (2)(-6sqrt{5}) + (sqrt{5})(12) + (sqrt{5})(-6sqrt{5}) ][ = 24 - 12sqrt{5} + 12sqrt{5} - 6*5 ][ = 24 - 12sqrt{5} + 12sqrt{5} - 30 ][ = (24 - 30) + (-12sqrt{5} + 12sqrt{5}) ][ = -6 + 0 ][ = -6 ]So,[ f(x) = frac{-6}{-36} = frac{1}{6} ]Wait, that's the same result as before. So, the maximum value is ( frac{1}{6} ). But earlier, I thought it might be ( frac{1}{5} ). Did I make a mistake in the substitution?Wait, let me go back to the substitution step. I set ( t = x - frac{1}{x} ), and then expressed the function as ( frac{t}{t^3 + 3t + 2} ). Then, I found that the maximum occurs at ( t = 1 ), giving ( frac{1}{6} ).But when I computed ( f(x) ) directly at ( x = frac{1 + sqrt{5}}{2} ), I also got ( frac{1}{6} ). So, it seems consistent.Wait, but in the initial problem statement, the user mentioned a previous thought process where they thought the maximum was ( frac{1}{5} ). Maybe they made a mistake there.Alternatively, perhaps I made a mistake in the substitution. Let me double-check.Wait, when I set ( t = x - frac{1}{x} ), then ( x^3 - frac{1}{x^3} = t^3 + 3t ). So, the denominator becomes ( x^3 + 2 - frac{1}{x^3} = (x^3 - frac{1}{x^3}) + 2 = t^3 + 3t + 2 ). That seems correct.Then, the function becomes ( frac{t}{t^3 + 3t + 2} ). Taking derivative, got critical point at ( t = 1 ), leading to ( f(t) = frac{1}{6} ). So, I think that's correct.But let me check another approach. Maybe using calculus directly on the original function.Let me compute the derivative of ( f(x) = frac{x^4 - x^2}{x^6 + 2x^3 - 1} ).Using the quotient rule:[ f'(x) = frac{(4x^3 - 2x)(x^6 + 2x^3 - 1) - (x^4 - x^2)(6x^5 + 6x^2)}{(x^6 + 2x^3 - 1)^2} ]That's a bit messy, but let's try to simplify the numerator.First, expand the terms:Numerator:[ (4x^3 - 2x)(x^6 + 2x^3 - 1) - (x^4 - x^2)(6x^5 + 6x^2) ]Let me compute each part separately.Compute ( (4x^3 - 2x)(x^6 + 2x^3 - 1) ):Multiply term by term:- ( 4x^3 * x^6 = 4x^9 )- ( 4x^3 * 2x^3 = 8x^6 )- ( 4x^3 * (-1) = -4x^3 )- ( -2x * x^6 = -2x^7 )- ( -2x * 2x^3 = -4x^4 )- ( -2x * (-1) = 2x )So, combining these:[ 4x^9 + 8x^6 - 4x^3 - 2x^7 - 4x^4 + 2x ]Now, compute ( (x^4 - x^2)(6x^5 + 6x^2) ):Multiply term by term:- ( x^4 * 6x^5 = 6x^9 )- ( x^4 * 6x^2 = 6x^6 )- ( -x^2 * 6x^5 = -6x^7 )- ( -x^2 * 6x^2 = -6x^4 )So, combining these:[ 6x^9 + 6x^6 - 6x^7 - 6x^4 ]Now, subtract the second product from the first:Numerator:[ (4x^9 + 8x^6 - 4x^3 - 2x^7 - 4x^4 + 2x) - (6x^9 + 6x^6 - 6x^7 - 6x^4) ]Distribute the negative sign:[ 4x^9 + 8x^6 - 4x^3 - 2x^7 - 4x^4 + 2x - 6x^9 - 6x^6 + 6x^7 + 6x^4 ]Combine like terms:- ( x^9: 4x^9 - 6x^9 = -2x^9 )- ( x^7: -2x^7 + 6x^7 = 4x^7 )- ( x^6: 8x^6 - 6x^6 = 2x^6 )- ( x^4: -4x^4 + 6x^4 = 2x^4 )- ( x^3: -4x^3 )- ( x: 2x )So, the numerator simplifies to:[ -2x^9 + 4x^7 + 2x^6 + 2x^4 - 4x^3 + 2x ]Now, set the numerator equal to zero to find critical points:[ -2x^9 + 4x^7 + 2x^6 + 2x^4 - 4x^3 + 2x = 0 ]Factor out a common factor of ( -2x ):[ -2x(x^8 - 2x^6 - x^5 - x^3 + 2x^2 - 1) = 0 ]Since ( x > 1 ), ( x neq 0 ), so we can divide both sides by ( -2x ):[ x^8 - 2x^6 - x^5 - x^3 + 2x^2 - 1 = 0 ]This is an eighth-degree polynomial equation, which is quite complicated. Maybe I can factor it or find rational roots.Let me try possible rational roots using the Rational Root Theorem. Possible roots are ( pm1 ).Test ( x = 1 ):[ 1 - 2 - 1 - 1 + 2 - 1 = -2 neq 0 ]Test ( x = -1 ):[ 1 - 2 + 1 + 1 + 2 - 1 = 1 neq 0 ]So, no rational roots. Maybe I can factor by grouping.Looking at the polynomial:[ x^8 - 2x^6 - x^5 - x^3 + 2x^2 - 1 ]Let me group terms:- ( x^8 - 2x^6 )- ( -x^5 )- ( -x^3 + 2x^2 )- ( -1 )Hmm, not obvious. Maybe factor out ( x^6 ) from the first two terms:[ x^6(x^2 - 2) - x^5 - x^3 + 2x^2 - 1 ]Still not helpful. Alternatively, maybe factor out ( x^5 ):[ x^5(x^3 - 2x - 1) - x^3 + 2x^2 - 1 ]Not helpful either. Maybe try to factor as a product of lower-degree polynomials.Alternatively, perhaps use substitution. Let me set ( y = x^3 ). Then, ( x^8 = x^2 * x^6 = x^2 y^2 ), but that might not help.Alternatively, set ( z = x^2 ). Then, ( x^8 = z^4 ), ( x^6 = z^3 ), ( x^5 = x*z^2 ), ( x^3 = z^{1.5} ), which complicates things.Alternatively, perhaps notice that ( x = frac{1 + sqrt{5}}{2} ) is a root, as found earlier.Let me test ( x = frac{1 + sqrt{5}}{2} ) in the polynomial:Compute ( x^8 - 2x^6 - x^5 - x^3 + 2x^2 - 1 ).But this would be time-consuming. Alternatively, since we already found that ( x = frac{1 + sqrt{5}}{2} ) is a critical point from the substitution method, and it gives ( f(x) = frac{1}{6} ), which seems to be the maximum, perhaps this is the only critical point in ( x > 1 ).Alternatively, maybe there are other critical points, but since the function tends to zero as ( x ) approaches infinity, and it's positive for ( x > 1 ), the maximum is likely at ( x = frac{1 + sqrt{5}}{2} ).Therefore, the maximum value is ( frac{1}{6} ).Wait, but earlier I thought the maximum was ( frac{1}{5} ). Maybe I confused it with another problem. Let me double-check.Wait, in the substitution method, I got ( f(t) = frac{1}{6} ) at ( t = 1 ). When I computed directly, I also got ( frac{1}{6} ). So, it seems consistent.Therefore, the maximum value is ( frac{1}{6} ).But wait, let me check the behavior of the function as ( x ) approaches 1 from the right.Compute ( lim_{x to 1^+} f(x) ):Numerator: ( 1 - 1 = 0 )Denominator: ( 1 + 2 - 1 = 2 )So, ( f(x) ) approaches 0 as ( x ) approaches 1.As ( x ) approaches infinity, the leading terms dominate:Numerator: ( x^4 )Denominator: ( x^6 )So, ( f(x) ) behaves like ( frac{x^4}{x^6} = frac{1}{x^2} ), which approaches 0.Therefore, the function starts at 0 when ( x = 1 ), increases to a maximum, and then decreases back to 0 as ( x ) approaches infinity. So, the maximum is indeed at ( x = frac{1 + sqrt{5}}{2} ), giving ( f(x) = frac{1}{6} ).Wait, but in the initial problem statement, the user mentioned a previous thought process where they thought the maximum was ( frac{1}{5} ). Maybe they made a mistake there.Alternatively, perhaps I made a mistake in the substitution. Let me go back.Wait, when I set ( t = x - frac{1}{x} ), then ( t = 1 ) corresponds to ( x = frac{1 + sqrt{5}}{2} ), and the function value is ( frac{1}{6} ). So, I think that's correct.Therefore, the maximum value is ( frac{1}{6} ).

💡First, I need to determine the number of peach trees planted. The family planted 3 times as many peach trees as apricot trees. Given that there are 58 apricot trees, I can calculate the number of peach trees by multiplying 58 by 3.Next, I need to find out how many cherry trees were planted. The family planted 5 times as many cherry trees as peach trees. Using the number of peach trees calculated in the previous step, I can determine the number of cherry trees by multiplying the peach trees by 5.Finally, to find the total number of trees planted, I will add the number of apricot trees, peach trees, and cherry trees together.

💡First, express both 16 and 64 as powers of 2:- 16 is (2^4)- 64 is (2^6)Rewrite the original expression using these bases:[frac{(2^4)^{24}}{(2^6)^8}]Apply the power of a power rule ((a^m)^n = a^{m times n}):[frac{2^{96}}{2^{48}}]Subtract the exponents when dividing like bases:[2^{96 - 48} = 2^{48}]Convert (2^{48}) back to a power of 16:[2^{48} = (2^4)^{12} = 16^{12}]

💡First, I recognize that the total amount to be divided is Rs. 1210 among three persons: p, q, and r.Given that the ratio of p to q is 5:4, I can represent the amounts that p and q receive as 5x and 4x respectively, where x is a common multiplier.It's also given that r receives Rs. 400.To find the value of x, I set up the equation based on the total amount:5x + 4x + 400 = 1210Combining like terms:9x + 400 = 1210Subtracting 400 from both sides:9x = 810Dividing both sides by 9:x = 90Now, I can determine the amount that q receives:q = 4x = 4 * 90 = 360Finally, to find the ratio of q to r:q : r = 360 : 400Simplifying the ratio by dividing both numbers by their greatest common divisor, which is 40:q : r = 9 : 10

💡Okay, so I have this problem about a tennis tournament with 128 players. The strongest 36 players get a bye in the first round, which means they automatically move to the next round without playing. The remaining 92 players have to compete in the first round. They are paired off into 46 matches because 92 divided by 2 is 46. Alright, so in the first round, there are 46 matches. Each match has a winner and a loser, and since it's single elimination, the losers are out of the tournament. That means after the first round, we'll have 46 winners from these matches. Now, these 46 winners will join the 36 players who had a bye. So, in the next round, there will be 46 + 36 = 82 players. Hmm, 82 players. But wait, in a single-elimination tournament, each round should have an even number of players because they pair up to play matches. 82 is an even number, so that's fine. But wait, 82 is not a power of 2, which is usually how tournaments are structured. Normally, you have 128 players, which is 2^7, so the tournament would have 7 rounds. But in this case, because of the byes, the numbers don't align perfectly. So, moving on. Each subsequent round will have half the number of players as the previous round, right? Because each match eliminates one player, so the number of players halves each time. But since 82 isn't a power of 2, the tournament might have some byes in the next rounds as well to make the numbers work out. But wait, the question is asking for the total number of matches played in the entire tournament. Maybe I don't need to worry about the structure beyond the first round. Let me think. In any single-elimination tournament, the total number of matches is always one less than the number of players. Because each match eliminates one player, and you need to eliminate all but one player to determine the champion. So, if there are 128 players, you would normally have 127 matches. But in this case, the first round only has 46 matches because 92 players are competing, and 36 are getting a bye. So, does that change the total number of matches? Or is it still 127 because regardless of the byes, you still need to eliminate 127 players to have one champion? Let me check. If there are 128 players, you need 127 losses to determine the champion. Each match results in one loss, so you need 127 matches. The byes just mean that some players don't have to play in the first round, but the total number of matches remains the same because the number of losses needed is still 127. So, even though the first round has 46 matches, and the subsequent rounds will have more matches, the total number of matches in the entire tournament is still 127. Wait, but let me make sure. If I have 92 players in the first round, they play 46 matches, resulting in 46 winners. Then, these 46 join the 36 who had a bye, making 82 players. Then, in the next round, 82 players would play 41 matches, resulting in 41 winners. Then, 41 players would need to play, but 41 is odd, so one person would get a bye again. This continues until we have a final match. But adding up the matches: 46 (first round) + 41 (second round) + 21 (third round, since 41 becomes 21 after byes) + 11 (fourth round) + 6 (fifth round) + 3 (sixth round) + 2 (seventh round) + 1 (eighth round) = Let me add these up. 46 + 41 is 87. 87 + 21 is 108. 108 + 11 is 119. 119 + 6 is 125. 125 + 3 is 128. 128 + 2 is 130. 130 + 1 is 131. Wait, that's more than 127. Hmm, maybe my approach is wrong. Because if I count the matches round by round, I get a different number. But the general rule is that in single elimination, the total number of matches is one less than the number of players. So, 128 players should have 127 matches. But why does my round-by-round count give me 131? Maybe I'm overcounting because of the byes. Each bye doesn't count as a match, but it affects the number of players in the next round. Wait, let's think differently. The total number of matches is equal to the total number of losses, which is 127. Because each match results in one loss, and you need 127 losses to determine the champion. So, regardless of how the byes are structured, the total number of matches is 127. Therefore, the answer should be 127 matches.

💡Okay, so I have this problem here: "One evening, a restaurant served a total of 0.2 loaf of wheat bread and 0.4 loaf of white bread. How many loaves were served in all?" Alright, let's see. I need to find the total number of loaves served that evening. First, I need to understand what's being asked. It says they served 0.2 loaf of wheat bread and 0.4 loaf of white bread. So, they're talking about two types of bread: wheat and white. Each is measured in loaves, but the amounts are given in decimals. Hmm, decimals can sometimes be tricky, but I think I can handle this.So, the question is asking for the total number of loaves served. That means I need to add the amount of wheat bread to the amount of white bread. That makes sense because if you have some of one thing and some of another, to find the total, you add them together.Alright, so let's break it down. The restaurant served 0.2 loaf of wheat bread. That's like saying they served 20% of a loaf. And they also served 0.4 loaf of white bread, which is 40% of a loaf. So, if I add those two together, I should get the total number of loaves served.Let me write that out: 0.2 (wheat) + 0.4 (white) = ?Okay, adding decimals. I remember that when adding decimals, you line up the decimal points and add each column just like you would with whole numbers. So, 0.2 plus 0.4. Let's see, 2 plus 4 is 6, and since both numbers have one decimal place, the result should also have one decimal place. So, 0.6.Wait, that seems straightforward. So, 0.2 plus 0.4 equals 0.6. That means the restaurant served a total of 0.6 loaves of bread that evening.But let me double-check to make sure I didn't make a mistake. Maybe I should think about it in fractions to confirm. 0.2 is the same as 1/5, and 0.4 is the same as 2/5. So, adding 1/5 and 2/5 together gives me 3/5, which is 0.6 in decimal form. Okay, that matches what I got before. So, that seems correct.Another way to think about it is in terms of money. If I have 0.20 and I get another 0.40, how much do I have in total? That's 0.60. So, it's the same concept. Adding the two amounts together gives me the total.Wait, but the question says "how many loaves were served in all." So, is 0.6 the total number of loaves? Yes, because they served 0.2 of one type and 0.4 of another, and adding them together gives the total loaves served.I think I'm confident with this answer. It seems pretty straightforward, but it's always good to double-check, especially when dealing with decimals and fractions. Sometimes, it's easy to mix up the places or miscount, but in this case, everything lines up.So, to summarize, the restaurant served 0.2 loaf of wheat bread and 0.4 loaf of white bread. Adding those together gives a total of 0.6 loaves served that evening.