💡Okay, so I've got this problem here about a pedestrian and a cyclist. Let me try to figure it out step by step. First, the pedestrian starts walking at 9:00 AM. An hour later, at 10:00 AM, the cyclist starts from the same point. Then, at 10:30 AM, the cyclist catches up with the pedestrian. After that, the cyclist continues ahead but then the bike breaks down. After some time repairing, the cyclist starts again and catches up with the pedestrian at 1:00 PM. I need to find out how long the repair took.Alright, let's break this down. From 9:00 AM to 10:30 AM, the pedestrian has been walking for 1.5 hours. The cyclist started at 10:00 AM, so by 10:30 AM, the cyclist has been cycling for 0.5 hours. At that point, they meet, which means they've covered the same distance from the starting point.Let me denote the pedestrian's speed as ( v_p ) and the cyclist's speed as ( v_c ). So, the distance the pedestrian has walked by 10:30 AM is ( 1.5 times v_p ). The distance the cyclist has cycled by 10:30 AM is ( 0.5 times v_c ). Since they meet at this point, these distances are equal:[ 1.5 v_p = 0.5 v_c ]Solving for ( v_c ):[ v_c = frac{1.5}{0.5} v_p = 3 v_p ]So, the cyclist is three times faster than the pedestrian.Now, after 10:30 AM, the cyclist continues ahead but then the bike breaks down. Let's say the breakdown happens some time after 10:30 AM. The cyclist repairs the bike and then starts again, eventually catching up with the pedestrian at 1:00 PM.Let me figure out the total time each has been traveling by 1:00 PM.The pedestrian started at 9:00 AM, so by 1:00 PM, that's 4 hours. So, the pedestrian has walked for 4 hours, covering a distance of ( 4 v_p ).The cyclist started at 10:00 AM, so by 1:00 PM, that's 3 hours. However, the cyclist had a breakdown and repair, so the actual cycling time is less than 3 hours.Let me denote the time the cyclist spent cycling before the breakdown as ( t_1 ), and the time after the repair as ( t_2 ). The total cycling time is ( t_1 + t_2 ), and the repair time is ( 3 - (t_1 + t_2) ).But wait, actually, the cyclist catches up at 1:00 PM, which is 4 hours after the pedestrian started. So, the distance covered by the cyclist is also ( 4 v_p ). But the cyclist's speed is ( 3 v_p ), so the time the cyclist spent cycling is:[ frac{4 v_p}{3 v_p} = frac{4}{3} text{ hours} = 1 frac{1}{3} text{ hours} = 80 text{ minutes} ]So, the cyclist should have spent 80 minutes cycling if there was no breakdown. But the cyclist actually had 3 hours from 10:00 AM to 1:00 PM, which is 180 minutes. Therefore, the repair time is the difference between the actual time and the cycling time:[ 180 text{ minutes} - 80 text{ minutes} = 100 text{ minutes} ]Wait, that seems straightforward, but let me double-check.From 10:00 AM to 10:30 AM, the cyclist catches up, which took 0.5 hours. Then, after that, the cyclist continues ahead, breaks down, repairs, and then catches up again at 1:00 PM.So, from 10:30 AM to 1:00 PM is 2.5 hours. During this time, the cyclist was either cycling or repairing.But the total distance the cyclist needs to cover from 10:30 AM to 1:00 PM is the same as the pedestrian's distance in that time.Wait, no. The pedestrian is walking continuously, so from 10:30 AM to 1:00 PM, the pedestrian walks for 2.5 hours, covering ( 2.5 v_p ).But the cyclist, after 10:30 AM, had already covered ( 1.5 v_p ) (same as the pedestrian). Then, the cyclist continues ahead, breaks down, repairs, and then catches up again.Wait, maybe I need to model this differently.Let me denote the time from 10:30 AM until the breakdown as ( t ), and the repair time as ( r ). Then, after repair, the cyclist cycles again until 1:00 PM, which is another ( t' ) time.But this might complicate things. Maybe another approach.Since the cyclist's speed is 3 times the pedestrian's, the relative speed is ( 3 v_p - v_p = 2 v_p ). So, the cyclist is gaining on the pedestrian at 2 v_p per hour.But wait, after 10:30 AM, the cyclist is ahead, then breaks down, repairs, and then catches up again.So, perhaps the distance between them when the cyclist breaks down is some distance, and then during the repair time, the pedestrian continues walking, and then the cyclist starts again and catches up.Let me think.At 10:30 AM, both are at the same point. Then, the cyclist continues ahead. Let's say the cyclist cycles for ( t ) hours until the breakdown. During this time, the cyclist covers ( 3 v_p times t ). The pedestrian, in the same time, walks ( v_p times t ). So, the distance between them when the cyclist breaks down is ( 3 v_p t - v_p t = 2 v_p t ).Then, the cyclist stops for repair time ( r ). During this repair time, the pedestrian continues walking, covering ( v_p times r ). So, the distance between them when the cyclist starts again is ( 2 v_p t - v_p r ).Then, the cyclist starts again and needs to cover this distance to catch up. The relative speed is still ( 2 v_p ), so the time to catch up is ( frac{2 v_p t - v_p r}{2 v_p} = t - frac{r}{2} ).But the total time from 10:30 AM to 1:00 PM is 2.5 hours. So, the sum of ( t ) (cycling before breakdown), ( r ) (repair), and ( t - frac{r}{2} ) (cycling after repair) should equal 2.5 hours.So:[ t + r + left( t - frac{r}{2} right) = 2.5 ]Simplify:[ 2 t + frac{r}{2} = 2.5 ]Multiply both sides by 2:[ 4 t + r = 5 ]But we also know that the total distance the cyclist covers from 10:30 AM to 1:00 PM is the same as the pedestrian's distance in that time.Wait, no. The pedestrian's distance from 10:30 AM to 1:00 PM is ( 2.5 v_p ). The cyclist's distance is ( 3 v_p times (t + t - frac{r}{2}) ) because the cyclist cycles for ( t ) before breakdown and ( t - frac{r}{2} ) after repair.So:[ 3 v_p times left( t + t - frac{r}{2} right) = 2.5 v_p ]Simplify:[ 3 times left( 2 t - frac{r}{2} right) = 2.5 ][ 6 t - frac{3 r}{2} = 2.5 ]Now we have two equations:1. ( 4 t + r = 5 )2. ( 6 t - frac{3 r}{2} = 2.5 )Let me solve equation 1 for ( r ):[ r = 5 - 4 t ]Substitute into equation 2:[ 6 t - frac{3 (5 - 4 t)}{2} = 2.5 ]Multiply through by 2 to eliminate the fraction:[ 12 t - 3 (5 - 4 t) = 5 ][ 12 t - 15 + 12 t = 5 ][ 24 t - 15 = 5 ][ 24 t = 20 ][ t = frac{20}{24} = frac{5}{6} text{ hours} approx 50 text{ minutes} ]Now, substitute ( t = frac{5}{6} ) into equation 1:[ 4 times frac{5}{6} + r = 5 ][ frac{20}{6} + r = 5 ][ frac{10}{3} + r = 5 ][ r = 5 - frac{10}{3} = frac{15}{3} - frac{10}{3} = frac{5}{3} text{ hours} approx 100 text{ minutes} ]So, the repair time is ( frac{5}{3} ) hours, which is 100 minutes.Wait, that matches my initial calculation. So, the repair took 100 minutes.But let me just verify again.From 10:30 AM to 1:00 PM is 2.5 hours. The cyclist spent ( t = frac{5}{6} ) hours cycling before breakdown, then repaired for ( r = frac{5}{3} ) hours, and then cycled again for ( t - frac{r}{2} = frac{5}{6} - frac{5}{6} = 0 ) hours? Wait, that can't be right.Wait, no. The time after repair is ( t - frac{r}{2} ). Let's compute that:( t = frac{5}{6} ) hours, ( r = frac{5}{3} ) hours.So, ( t - frac{r}{2} = frac{5}{6} - frac{5}{6} = 0 ). That doesn't make sense. It implies the cyclist didn't cycle after repair, which contradicts the problem statement.Hmm, maybe I made a mistake in setting up the equations.Let me go back.After the breakdown, the distance between the cyclist and pedestrian is ( 2 v_p t ). During repair time ( r ), the pedestrian walks ( v_p r ), so the distance becomes ( 2 v_p t - v_p r ). Then, the cyclist starts again and needs to cover this distance at a relative speed of ( 2 v_p ), so time taken is ( frac{2 v_p t - v_p r}{2 v_p} = t - frac{r}{2} ).But the total time from 10:30 AM to 1:00 PM is 2.5 hours, which should equal ( t + r + (t - frac{r}{2}) ).So:[ t + r + t - frac{r}{2} = 2.5 ][ 2 t + frac{r}{2} = 2.5 ]Multiply by 2:[ 4 t + r = 5 ]That's correct.And the total distance the cyclist covers is ( 3 v_p times (t + t - frac{r}{2}) ), which should equal the pedestrian's distance ( 2.5 v_p ).So:[ 3 (2 t - frac{r}{2}) = 2.5 ][ 6 t - frac{3 r}{2} = 2.5 ]That's correct.Solving these, we get ( t = frac{5}{6} ) hours and ( r = frac{5}{3} ) hours.But then, the time after repair is ( t - frac{r}{2} = frac{5}{6} - frac{5}{6} = 0 ). That suggests the cyclist didn't cycle after repair, which is impossible because the problem states that the cyclist did catch up again.This inconsistency means there's a mistake in my setup.Perhaps the issue is in how I'm calculating the distance after repair.Wait, when the cyclist breaks down, the distance between them is ( 2 v_p t ). During repair time ( r ), the pedestrian walks ( v_p r ), so the distance becomes ( 2 v_p t - v_p r ). Then, the cyclist starts again and needs to cover this distance at a relative speed of ( 2 v_p ), so time taken is ( frac{2 v_p t - v_p r}{2 v_p} = t - frac{r}{2} ).But if ( t - frac{r}{2} ) is negative, that would mean the cyclist doesn't need to cycle because the pedestrian has already caught up, which contradicts the problem.So, perhaps my initial assumption about the distance after repair is incorrect.Wait, actually, when the cyclist breaks down, the pedestrian is behind. So, during the repair time, the pedestrian continues walking, reducing the distance between them. Then, when the cyclist starts again, he needs to cover the remaining distance.But if the repair time is too long, the pedestrian might have already passed the cyclist's breakdown point.Wait, but in this case, the cyclist catches up again at 1:00 PM, so the repair time must have been such that the cyclist could still catch up.Alternatively, maybe I should consider the total distance each has traveled by 1:00 PM.Pedestrian: 4 hours walking, distance ( 4 v_p ).Cyclist: Started at 10:00 AM, so 3 hours total. But with a breakdown. So, the cyclist's total cycling time is less than 3 hours.But the distance the cyclist covers is also ( 4 v_p ), so:[ 3 v_p times text{cycling time} = 4 v_p ]Thus, cycling time is ( frac{4}{3} ) hours, which is 80 minutes.So, the cyclist spent 80 minutes cycling and the rest was repair.Total time from 10:00 AM to 1:00 PM is 180 minutes.Thus, repair time is ( 180 - 80 = 100 ) minutes.This seems straightforward and avoids the earlier inconsistency.So, the repair took 100 minutes.

💡Alright, so I have this problem here: I need to prove that in the sequence formed by the integers a, 2a, 3a, ..., ba, the number of terms divisible by b is equal to the greatest common divisor of a and b. Hmm, okay. Let me try to wrap my head around this.First, let me make sure I understand the problem correctly. We have a sequence where each term is a multiple of a, starting from a and going up to ba. So, the sequence is a, 2a, 3a, ..., up to ba. Now, among these terms, some of them will be divisible by b. The claim is that the number of such terms is exactly equal to the greatest common divisor (gcd) of a and b.Alright, so my goal is to show that if I count how many terms in this sequence are divisible by b, that count will be the same as gcd(a, b). Let me think about how to approach this.Maybe I should start by recalling what gcd(a, b) represents. The greatest common divisor of a and b is the largest integer that divides both a and b without leaving a remainder. So, if d = gcd(a, b), then d divides both a and b, and any other common divisor of a and b also divides d.Given that, perhaps I can express a and b in terms of their gcd. Let me denote d = gcd(a, b). Then, I can write a as d times some integer, say, m, and b as d times another integer, say, n. So, a = d*m and b = d*n. Since d is the gcd, m and n must be coprime, meaning their gcd is 1.Okay, so now, substituting back into the sequence, each term is k*a, where k ranges from 1 to b. So, the sequence becomes d*m, 2*d*m, 3*d*m, ..., b*d*m. But since b = d*n, the last term is d*n*d*m = d^2*n*m. Hmm, that seems a bit complicated. Maybe I should think differently.Wait, perhaps instead of substituting a and b, I can think about the divisibility condition. A term k*a is divisible by b if and only if b divides k*a. In other words, b | k*a. Since b = d*n and a = d*m, this condition becomes d*n | k*d*m. Simplifying, we can cancel out the d, so we get n | k*m.Now, since m and n are coprime (because a = d*m and b = d*n, and d is the gcd), this means that n must divide k. Because if n divides k*m and n and m are coprime, then n must divide k. That's a property of coprime numbers.So, n divides k, which means k must be a multiple of n. Therefore, the values of k for which k*a is divisible by b are exactly the multiples of n in the range from 1 to b.Now, how many multiples of n are there between 1 and b? Well, since b = d*n, the multiples of n up to b are n, 2n, 3n, ..., d*n. So, there are exactly d multiples of n in this range.Therefore, the number of terms in the sequence a, 2a, 3a, ..., ba that are divisible by b is equal to d, which is the gcd of a and b.Wait, let me double-check this reasoning. So, I expressed a and b in terms of their gcd, which is d, and then showed that the divisibility condition reduces to n dividing k, where n = b/d and m = a/d are coprime. Then, since n and m are coprime, n must divide k, leading to k being multiples of n. The number of such k's is d, which is the gcd.This seems solid, but let me think if there's another way to approach this, maybe using modular arithmetic or something else.Alternatively, I could think about the sequence modulo b. Each term k*a modulo b. We want to find how many times this is congruent to 0 modulo b. So, k*a ≡ 0 mod b. Which is the same as saying b divides k*a.Again, since d = gcd(a, b), we can write a = d*m and b = d*n with gcd(m, n) = 1. Then, k*a = k*d*m. For b = d*n to divide k*d*m, we need n to divide k*m. Since m and n are coprime, n must divide k. So, k must be a multiple of n, and as before, there are d such multiples between 1 and b.Hmm, same conclusion. Maybe I can think of an example to test this.Let's take a = 6 and b = 9. Then, gcd(6, 9) = 3. The sequence is 6, 12, 18, 24, 30, 36, 42, 48, 54. Now, let's see which terms are divisible by 9.6 ÷ 9 = 0.666... → Not divisible12 ÷ 9 = 1.333... → Not divisible18 ÷ 9 = 2 → Divisible24 ÷ 9 = 2.666... → Not divisible30 ÷ 9 ≈ 3.333... → Not divisible36 ÷ 9 = 4 → Divisible42 ÷ 9 ≈ 4.666... → Not divisible48 ÷ 9 ≈ 5.333... → Not divisible54 ÷ 9 = 6 → DivisibleSo, the terms divisible by 9 are 18, 36, and 54. That's 3 terms, which is equal to gcd(6, 9) = 3. Okay, that checks out.Another example: a = 8, b = 12. gcd(8, 12) = 4. The sequence is 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96. Let's see which are divisible by 12.8 ÷ 12 ≈ 0.666... → No16 ÷ 12 ≈ 1.333... → No24 ÷ 12 = 2 → Yes32 ÷ 12 ≈ 2.666... → No40 ÷ 12 ≈ 3.333... → No48 ÷ 12 = 4 → Yes56 ÷ 12 ≈ 4.666... → No64 ÷ 12 ≈ 5.333... → No72 ÷ 12 = 6 → Yes80 ÷ 12 ≈ 6.666... → No88 ÷ 12 ≈ 7.333... → No96 ÷ 12 = 8 → YesSo, the terms divisible by 12 are 24, 48, 72, and 96. That's 4 terms, which is equal to gcd(8, 12) = 4. Good, another confirmation.Wait, what if a and b are coprime? Let's say a = 3, b = 4. gcd(3, 4) = 1. The sequence is 3, 6, 9, 12. Which terms are divisible by 4?3 ÷ 4 = 0.75 → No6 ÷ 4 = 1.5 → No9 ÷ 4 = 2.25 → No12 ÷ 4 = 3 → YesOnly one term, 12, which is divisible by 4. That's equal to gcd(3, 4) = 1. Perfect.What if a = b? Let's say a = 5, b = 5. gcd(5, 5) = 5. The sequence is 5, 10, 15, 20, 25. How many terms are divisible by 5? All of them, right? 5, 10, 15, 20, 25. That's 5 terms, which is equal to gcd(5, 5) = 5. Makes sense.Another edge case: a = 1, b = 1. gcd(1, 1) = 1. The sequence is just 1. Is 1 divisible by 1? Yes. So, one term, which matches the gcd.Alright, so through examples and reasoning, it seems that the number of terms in the sequence a, 2a, 3a, ..., ba that are divisible by b is indeed equal to gcd(a, b). The key idea is that by expressing a and b in terms of their gcd, we can reduce the problem to counting multiples of a certain number, which gives us the gcd as the count.I think I've got a good grasp on this now. The initial step of expressing a and b in terms of their gcd simplifies the problem, and using properties of coprime numbers helps in determining the divisibility condition. The examples also helped solidify the understanding.

💡Alright, so I've got this problem about a function and some conditions. Let me try to figure it out step by step.First, the function is given as ( f(x) = frac{a}{3}x^3 - frac{3}{2}x^2 + (a+1)x + 1 ), where ( a ) is a real number. There are two parts to this problem.**Part 1: Finding the value of ( a ) such that ( f(x) ) has an extremum at ( x = 1 ).**Okay, so I remember that extrema occur where the first derivative is zero. So, I need to find the derivative of ( f(x) ) and set it equal to zero at ( x = 1 ).Let's compute ( f'(x) ):( f'(x) = frac{d}{dx} left( frac{a}{3}x^3 - frac{3}{2}x^2 + (a+1)x + 1 right) )Calculating term by term:- The derivative of ( frac{a}{3}x^3 ) is ( a x^2 ).- The derivative of ( -frac{3}{2}x^2 ) is ( -3x ).- The derivative of ( (a+1)x ) is ( a + 1 ).- The derivative of the constant 1 is 0.So, putting it all together:( f'(x) = a x^2 - 3x + (a + 1) )Now, since there's an extremum at ( x = 1 ), we set ( f'(1) = 0 ):( f'(1) = a(1)^2 - 3(1) + (a + 1) = a - 3 + a + 1 = 2a - 2 )Set this equal to zero:( 2a - 2 = 0 )Solving for ( a ):( 2a = 2 )( a = 1 )Okay, so ( a = 1 ). That seems straightforward.**Part 2: Finding the range of real numbers ( x ) such that ( f'(x) > x^2 - x - a + 1 ) holds for any ( a in (0, +infty) ).**Hmm, this seems a bit more involved. Let's see.First, let's write down the inequality:( f'(x) > x^2 - x - a + 1 )We already have ( f'(x) = a x^2 - 3x + (a + 1) ). Let's substitute that in:( a x^2 - 3x + (a + 1) > x^2 - x - a + 1 )Now, let's bring all terms to one side to simplify:( a x^2 - 3x + a + 1 - x^2 + x + a - 1 > 0 )Wait, let me double-check that:Subtract ( x^2 - x - a + 1 ) from both sides:( a x^2 - 3x + a + 1 - (x^2 - x - a + 1) > 0 )Which becomes:( a x^2 - 3x + a + 1 - x^2 + x + a - 1 > 0 )Now, combine like terms:- ( a x^2 - x^2 = (a - 1) x^2 )- ( -3x + x = -2x )- ( a + a = 2a )- ( 1 - 1 = 0 )So, the inequality simplifies to:( (a - 1) x^2 - 2x + 2a > 0 )Okay, so we have:( (a - 1) x^2 - 2x + 2a > 0 )And this needs to hold for any ( a > 0 ).Hmm, so this is a quadratic inequality in terms of ( x ), but the coefficients depend on ( a ). We need this inequality to be true for all positive ( a ). That seems tricky because as ( a ) varies, the quadratic can change its behavior.Let me think about how to approach this. Maybe rearrange the inequality to group terms with ( a ) together.Let's rewrite the inequality:( (a - 1) x^2 - 2x + 2a > 0 )Let me factor out ( a ) from the terms that have it:( a x^2 - x^2 - 2x + 2a > 0 )Grouping the ( a ) terms:( a(x^2 + 2) - x^2 - 2x > 0 )So, we have:( a(x^2 + 2) > x^2 + 2x )Since ( a > 0 ), we can divide both sides by ( a ):( x^2 + 2 > frac{x^2 + 2x}{a} )Wait, but ( a ) is any positive real number. So, as ( a ) approaches zero, the right-hand side ( frac{x^2 + 2x}{a} ) can become very large if ( x^2 + 2x ) is positive, or very negative if ( x^2 + 2x ) is negative.But since the inequality must hold for all ( a > 0 ), including as ( a ) approaches zero, we need the right-hand side to not exceed the left-hand side even in the limit as ( a to 0 ).So, let's analyze the inequality:( x^2 + 2 > frac{x^2 + 2x}{a} )As ( a to 0^+ ), the right-hand side ( frac{x^2 + 2x}{a} ) tends to ( +infty ) if ( x^2 + 2x > 0 ), and to ( -infty ) if ( x^2 + 2x < 0 ).But the left-hand side ( x^2 + 2 ) is always positive because ( x^2 geq 0 ) and we add 2. So, if ( x^2 + 2x > 0 ), then as ( a to 0^+ ), the right-hand side becomes larger than the left-hand side, which would violate the inequality. Therefore, to satisfy the inequality for all ( a > 0 ), we must have ( x^2 + 2x leq 0 ).So, ( x^2 + 2x leq 0 )Let's solve this inequality:( x^2 + 2x leq 0 )Factor:( x(x + 2) leq 0 )This is a quadratic inequality. The roots are at ( x = 0 ) and ( x = -2 ). The parabola opens upwards, so the inequality ( x(x + 2) leq 0 ) holds between the roots.Therefore, the solution is ( -2 leq x leq 0 ).So, the range of real numbers ( x ) is ( [-2, 0] ).But wait, let me verify this because I might have made a mistake in the reasoning.We had:( a(x^2 + 2) > x^2 + 2x )Which rearranged to:( a > frac{x^2 + 2x}{x^2 + 2} )Since this must hold for all ( a > 0 ), the maximum value of ( frac{x^2 + 2x}{x^2 + 2} ) must be less than or equal to zero because ( a ) can be made arbitrarily small (approaching zero). Therefore, ( frac{x^2 + 2x}{x^2 + 2} leq 0 ).So, ( frac{x^2 + 2x}{x^2 + 2} leq 0 )Since ( x^2 + 2 > 0 ) for all real ( x ), the sign of the fraction is determined by the numerator ( x^2 + 2x ). So, ( x^2 + 2x leq 0 ), which as before, gives ( -2 leq x leq 0 ).Therefore, the range of ( x ) is indeed ( [-2, 0] ).**Final Answer**1. The value of ( a ) is (boxed{1}).2. The range of real numbers ( x ) is (boxed{[-2, 0]}).

💡Okay, so I have this problem where I need to prove that for non-negative numbers ( x ) and ( y ) satisfying ( x + y leq 1 ), the inequality ( 12xy leq 4x(1 - y) + 9y(1 - x) ) holds. Hmm, let me try to figure this out step by step.First, I should understand what the given condition ( x + y leq 1 ) implies. Since both ( x ) and ( y ) are non-negative, this means neither ( x ) nor ( y ) can exceed 1. Also, if ( x ) increases, ( y ) must decrease to keep their sum less than or equal to 1, and vice versa.Now, looking at the inequality I need to prove: ( 12xy leq 4x(1 - y) + 9y(1 - x) ). Let me try to simplify the right-hand side to see if I can make it look more manageable.Expanding the right-hand side:[ 4x(1 - y) = 4x - 4xy ][ 9y(1 - x) = 9y - 9xy ]So, adding these together:[ 4x - 4xy + 9y - 9xy = 4x + 9y - 13xy ]So, the inequality becomes:[ 12xy leq 4x + 9y - 13xy ]Hmm, let me bring all the terms to one side to see what I get:[ 12xy + 13xy leq 4x + 9y ][ 25xy leq 4x + 9y ]Okay, so now I have ( 25xy leq 4x + 9y ). I need to show that this holds given ( x + y leq 1 ).I wonder if I can use some inequality like AM-GM here. The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Maybe I can apply it to ( 4x ) and ( 9y ).Let me try that. Applying AM-GM to ( 4x ) and ( 9y ):[ frac{4x + 9y}{2} geq sqrt{4x cdot 9y} ][ frac{4x + 9y}{2} geq sqrt{36xy} ][ frac{4x + 9y}{2} geq 6sqrt{xy} ]Multiplying both sides by 2:[ 4x + 9y geq 12sqrt{xy} ]Hmm, so ( 4x + 9y ) is greater than or equal to ( 12sqrt{xy} ). But I need to relate this to ( 25xy ). I know that ( sqrt{xy} leq frac{x + y}{2} ) from AM-GM as well. Since ( x + y leq 1 ), this would mean ( sqrt{xy} leq frac{1}{2} ).Wait, but I'm not sure if that directly helps. Let me think differently. Maybe I can express ( xy ) in terms of ( x ) and ( y ) using the given condition ( x + y leq 1 ).Since ( x + y leq 1 ), I can write ( y leq 1 - x ) and ( x leq 1 - y ). Maybe substituting ( y ) with ( 1 - x ) would help, but I'm not sure.Alternatively, perhaps I can consider the ratio ( frac{4x + 9y}{xy} ) and see if it's greater than or equal to 25. But that might complicate things.Wait, another approach: since ( x + y leq 1 ), maybe I can set ( x = a ) and ( y = b ) where ( a + b leq 1 ), and then try to maximize ( 25ab ) subject to ( a + b leq 1 ). If I can show that the maximum of ( 25ab ) is less than or equal to ( 4a + 9b ), then the inequality would hold.To maximize ( ab ) under ( a + b leq 1 ), we know from AM-GM that the maximum occurs when ( a = b ), but since the coefficients in ( 4a + 9b ) are different, maybe the maximum of ( 25ab ) isn't straightforward.Alternatively, perhaps I can use substitution. Let me set ( y = 1 - x - z ) where ( z geq 0 ) to satisfy ( x + y leq 1 ). Then, substitute ( y ) into the inequality and see if it simplifies.But that might get messy. Let me go back to the inequality ( 25xy leq 4x + 9y ). Maybe I can divide both sides by ( xy ) (assuming ( x ) and ( y ) are positive, which they are since they are non-negative and the inequality involves ( xy )).So, dividing both sides by ( xy ):[ 25 leq frac{4}{y} + frac{9}{x} ]Hmm, now I have ( frac{4}{y} + frac{9}{x} geq 25 ). I need to show that this holds given ( x + y leq 1 ).This looks like a form where I can apply the Cauchy-Schwarz inequality. The Cauchy-Schwarz inequality states that for positive real numbers ( a_1, a_2, ..., a_n ) and ( b_1, b_2, ..., b_n ):[ (a_1b_1 + a_2b_2 + ... + a_nb_n)^2 leq (a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + b_2^2 + ... + b_n^2) ]Maybe I can set ( a_1 = 2 ), ( a_2 = 3 ), ( b_1 = sqrt{frac{1}{y}} ), and ( b_2 = sqrt{frac{1}{x}} ). Let me try that.Applying Cauchy-Schwarz:[ (2 cdot sqrt{frac{1}{y}} + 3 cdot sqrt{frac{1}{x}})^2 leq (2^2 + 3^2)left(frac{1}{y} + frac{1}{x}right) ][ left(2sqrt{frac{1}{y}} + 3sqrt{frac{1}{x}}right)^2 leq (4 + 9)left(frac{1}{y} + frac{1}{x}right) ][ left(2sqrt{frac{1}{y}} + 3sqrt{frac{1}{x}}right)^2 leq 13left(frac{1}{y} + frac{1}{x}right) ]But I'm not sure if this helps me get to ( frac{4}{y} + frac{9}{x} geq 25 ). Maybe I need a different approach.Let me consider using Lagrange multipliers to maximize ( 25xy ) subject to ( x + y leq 1 ). But that might be overcomplicating things for this problem.Wait, another idea: since ( x + y leq 1 ), perhaps I can express ( y ) as ( y = 1 - x - t ) where ( t geq 0 ). Then, substitute this into the inequality and see if it holds.But I'm not sure if that's the best way. Maybe I should try specific values to see if the inequality holds, which might give me some insight.For example, let me set ( x = 0 ). Then, ( y leq 1 ). The inequality becomes:[ 12 cdot 0 cdot y leq 4 cdot 0 cdot (1 - y) + 9y(1 - 0) ][ 0 leq 0 + 9y ]Which is true since ( y geq 0 ).Similarly, if ( y = 0 ), then ( x leq 1 ). The inequality becomes:[ 12x cdot 0 leq 4x(1 - 0) + 9 cdot 0 cdot (1 - x) ][ 0 leq 4x + 0 ]Which is also true since ( x geq 0 ).Now, let me try ( x = y = 0.5 ). Then, ( x + y = 1 ), which satisfies the condition. Plugging into the inequality:[ 12 cdot 0.5 cdot 0.5 = 12 cdot 0.25 = 3 ][ 4 cdot 0.5 cdot (1 - 0.5) + 9 cdot 0.5 cdot (1 - 0.5) = 4 cdot 0.5 cdot 0.5 + 9 cdot 0.5 cdot 0.5 = 1 + 2.25 = 3.25 ]So, ( 3 leq 3.25 ), which holds.Another test case: ( x = 0.2 ), ( y = 0.3 ). Then, ( x + y = 0.5 leq 1 ). Compute both sides:Left-hand side: ( 12 cdot 0.2 cdot 0.3 = 12 cdot 0.06 = 0.72 )Right-hand side: ( 4 cdot 0.2 cdot (1 - 0.3) + 9 cdot 0.3 cdot (1 - 0.2) = 4 cdot 0.2 cdot 0.7 + 9 cdot 0.3 cdot 0.8 = 0.56 + 2.16 = 2.72 )So, ( 0.72 leq 2.72 ), which is true.Hmm, these test cases seem to support the inequality. But I need a general proof, not just specific examples.Going back to the inequality ( 25xy leq 4x + 9y ), maybe I can rearrange it to:[ 4x + 9y - 25xy geq 0 ]Let me factor this expression. Maybe factor out ( x ) and ( y ):[ x(4 - 25y) + y(9) geq 0 ]Not sure if that helps. Alternatively, perhaps I can write it as:[ 4x + 9y geq 25xy ]Divide both sides by ( x ) (assuming ( x > 0 )):[ 4 + 9frac{y}{x} geq 25y ]But I'm not sure if that's helpful.Wait, another approach: since ( x + y leq 1 ), let me express ( y ) as ( y = 1 - x - t ) where ( t geq 0 ). Then, substitute into the inequality:[ 25x(1 - x - t) leq 4x + 9(1 - x - t) ]But this might complicate things further.Alternatively, maybe I can consider the function ( f(x, y) = 4x + 9y - 25xy ) and show that it's always non-negative under the constraint ( x + y leq 1 ).To do this, I can analyze the function ( f(x, y) ) over the domain ( x geq 0 ), ( y geq 0 ), and ( x + y leq 1 ).First, let's check the boundaries:1. When ( x = 0 ): ( f(0, y) = 0 + 9y - 0 = 9y geq 0 ).2. When ( y = 0 ): ( f(x, 0) = 4x + 0 - 0 = 4x geq 0 ).3. When ( x + y = 1 ): Substitute ( y = 1 - x ) into ( f(x, y) ):[ f(x, 1 - x) = 4x + 9(1 - x) - 25x(1 - x) ][ = 4x + 9 - 9x - 25x + 25x^2 ][ = 9 - 20x + 25x^2 ]This is a quadratic in ( x ): ( 25x^2 - 20x + 9 ). Let's check its discriminant:[ D = (-20)^2 - 4 cdot 25 cdot 9 = 400 - 900 = -500 ]Since the discriminant is negative, the quadratic is always positive. So, ( f(x, 1 - x) > 0 ) for all ( x ).Since ( f(x, y) ) is non-negative on all boundaries and the function is linear in ( x ) and ( y ) within the domain, it should be non-negative throughout the domain ( x + y leq 1 ).Therefore, ( 4x + 9y - 25xy geq 0 ), which implies ( 25xy leq 4x + 9y ), and thus ( 12xy leq 4x(1 - y) + 9y(1 - x) ).So, the inequality holds for all non-negative ( x ) and ( y ) satisfying ( x + y leq 1 ).

💡First, I need to substitute the given expressions for ( y_1 ) and ( y_2 ) into the equation for ( y ).So, ( y = y_1 + xy_2 ) becomes:[y = (x^2 - 7x + 6) + x(7x - 3)]Next, I'll expand the terms:[y = x^2 - 7x + 6 + 7x^2 - 3x]Now, I'll combine like terms:[y = (x^2 + 7x^2) + (-7x - 3x) + 6 = 8x^2 - 10x + 6]Finally, I'll substitute ( x = 2 ) into the simplified equation:[y = 8(2)^2 - 10(2) + 6 = 32 - 20 + 6 = 18]

💡Okay, so I have this problem here: I need to find the real number ( m ) such that the power function ( f(x) = (m^2 - 9m + 19)x^{m - 4} ) is monotonically increasing on the interval ( (0, +infty) ). Hmm, let me think about how to approach this.First, I remember that a function is monotonically increasing if its derivative is non-negative throughout the interval. Since this is a power function, its derivative should be straightforward to compute. But before jumping into calculus, maybe I can recall some properties of power functions.A power function has the form ( f(x) = kx^n ), where ( k ) is a constant and ( n ) is the exponent. The behavior of the function depends on both ( k ) and ( n ). For the function to be monotonically increasing on ( (0, +infty) ), two conditions must be satisfied:1. The coefficient ( k ) must be positive. If ( k ) were negative, the function would be decreasing if the exponent is positive, or increasing if the exponent is negative, but since we want it to be increasing, ( k ) must be positive.2. The exponent ( n ) must be positive. If ( n ) is positive, then as ( x ) increases, ( x^n ) also increases, making the function monotonically increasing. If ( n ) is negative, the function would decrease as ( x ) increases.So, applying this to our function ( f(x) = (m^2 - 9m + 19)x^{m - 4} ), we need both the coefficient ( (m^2 - 9m + 19) ) to be positive and the exponent ( (m - 4) ) to be positive.Let me write down these two conditions:1. ( m^2 - 9m + 19 > 0 )2. ( m - 4 > 0 )Starting with the second condition because it seems simpler:( m - 4 > 0 ) implies ( m > 4 ). So, ( m ) must be greater than 4.Now, the first condition is a quadratic inequality: ( m^2 - 9m + 19 > 0 ). To solve this, I should find the roots of the quadratic equation ( m^2 - 9m + 19 = 0 ) and then determine where the quadratic is positive.Using the quadratic formula:( m = frac{9 pm sqrt{81 - 76}}{2} = frac{9 pm sqrt{5}}{2} )So, the roots are ( m = frac{9 + sqrt{5}}{2} ) and ( m = frac{9 - sqrt{5}}{2} ). Let me approximate these to understand the intervals:( sqrt{5} ) is approximately 2.236.So,( m = frac{9 + 2.236}{2} = frac{11.236}{2} approx 5.618 )and( m = frac{9 - 2.236}{2} = frac{6.764}{2} approx 3.382 )So, the quadratic ( m^2 - 9m + 19 ) factors as ( (m - 5.618)(m - 3.382) ). Since the coefficient of ( m^2 ) is positive, the parabola opens upwards. Therefore, the quadratic is positive outside the interval ( (3.382, 5.618) ) and negative inside this interval.So, ( m^2 - 9m + 19 > 0 ) when ( m < 3.382 ) or ( m > 5.618 ).But from the second condition, we already have ( m > 4 ). So, combining both conditions, ( m ) must satisfy both ( m > 4 ) and ( m > 5.618 ). Therefore, the more restrictive condition is ( m > 5.618 ).But wait, the problem asks for the real number ( m ). It doesn't specify that ( m ) has to be an integer, but given the context, it's possible that ( m ) is an integer. Let me check if ( m ) is an integer.Looking back at the quadratic equation ( m^2 - 9m + 19 = 1 ), which was set to 1 in the initial solution. Wait, why was it set to 1? Hmm, maybe that was a mistake in the initial thought process.Wait, no, actually, in the initial solution, the user set ( m^2 - 9m + 19 = 1 ) to simplify, but that might not be the right approach. Let me think again.The coefficient ( m^2 - 9m + 19 ) just needs to be positive, not necessarily equal to 1. So, the initial approach of setting it to 1 might have been incorrect. Instead, we should just ensure that ( m^2 - 9m + 19 > 0 ) and ( m - 4 > 0 ).So, as I found earlier, ( m > 5.618 ) approximately. But the problem is asking for a specific real number ( m ). That suggests that there might be a unique solution, which contradicts my previous conclusion that ( m ) can be any number greater than approximately 5.618.Wait, maybe I missed something. Let me go back to the problem statement.It says, "the power function ( f(x) = (m^2 - 9m + 19)x^{m - 4} ) is monotonically increasing on ( (0, +infty) )." So, it's a power function, which generally has the form ( kx^n ). For it to be monotonically increasing on ( (0, +infty) ), as I thought earlier, ( k > 0 ) and ( n > 0 ).So, ( m^2 - 9m + 19 > 0 ) and ( m - 4 > 0 ).But if the problem is asking for a specific real number ( m ), perhaps there is a unique solution where both conditions are satisfied, but maybe there's an additional constraint.Wait, perhaps the function is not just any power function, but it's a specific one where the coefficient is 1? Because in the initial solution, the user set ( m^2 - 9m + 19 = 1 ). Maybe that's a hint.Let me try that. If ( m^2 - 9m + 19 = 1 ), then:( m^2 - 9m + 18 = 0 )Factoring:( (m - 3)(m - 6) = 0 )So, ( m = 3 ) or ( m = 6 ).Now, checking the exponent for each:For ( m = 3 ), exponent is ( 3 - 4 = -1 ). So, ( f(x) = 1 cdot x^{-1} = frac{1}{x} ), which is decreasing on ( (0, +infty) ).For ( m = 6 ), exponent is ( 6 - 4 = 2 ). So, ( f(x) = 1 cdot x^{2} ), which is increasing on ( (0, +infty) ).Therefore, ( m = 6 ) is the solution.But wait, why did we set ( m^2 - 9m + 19 = 1 )? Is that a necessary condition?In the problem statement, it's just given as ( f(x) = (m^2 - 9m + 19)x^{m - 4} ). There's no indication that the coefficient has to be 1. So, perhaps the initial approach was incorrect.Let me think again. The function is monotonically increasing if both the coefficient is positive and the exponent is positive.So, ( m^2 - 9m + 19 > 0 ) and ( m - 4 > 0 ).From earlier, ( m > 5.618 ). So, any ( m ) greater than approximately 5.618 would satisfy both conditions.But the problem is asking for "the real number ( m )", implying a unique solution. So, perhaps there's a specific value of ( m ) that makes the function not just increasing, but maybe also satisfies another condition, like being a monomial with integer exponents or something else.Alternatively, maybe the function is supposed to be a monomial with a specific exponent, but the problem doesn't specify that.Wait, perhaps I'm overcomplicating. Let me check the initial solution again.In the initial solution, the user set ( m^2 - 9m + 19 = 1 ), leading to ( m = 3 ) or ( m = 6 ). Then, checked which one makes the function increasing. Since ( m = 6 ) gives a positive exponent, that's the answer.But why set the coefficient to 1? Maybe because the function is a power function, and sometimes power functions are considered with leading coefficient 1, but that's not necessarily the case.Alternatively, maybe the problem is designed so that the coefficient is positive and the exponent is positive, and the only integer solution is ( m = 6 ).Given that, perhaps the answer is ( m = 6 ).But to be thorough, let me consider if there are other possible values of ( m ) that satisfy both ( m^2 - 9m + 19 > 0 ) and ( m - 4 > 0 ).As I found earlier, ( m > 5.618 ). So, any ( m ) greater than approximately 5.618 would work. But since the problem asks for "the real number ( m )", it's likely that there's a specific solution, probably an integer.Given that, ( m = 6 ) is the smallest integer greater than 5.618, so that's probably the intended answer.Therefore, the real number ( m ) is 6.