💡Alright, so I have this problem here: For how many positive integers ( m ) is ( frac{990}{m^2 - 2} ) a positive integer? The options are from A to E, with E being more than four. Hmm, okay, let me try to figure this out step by step.First, I need to understand what the problem is asking. It says that ( frac{990}{m^2 - 2} ) should be a positive integer. That means that ( m^2 - 2 ) has to be a divisor of 990 because if you divide 990 by something and get an integer, that something must be a factor of 990.So, my first thought is to factorize 990 to find all its divisors. Let me do that. 990 is an even number, so it's divisible by 2. Dividing 990 by 2 gives me 495. Now, 495 ends with a 5, so it's divisible by 5. Dividing 495 by 5 gives me 99. 99 is 9 times 11, and 9 is 3 squared. So, putting it all together, the prime factorization of 990 is ( 2 times 3^2 times 5 times 11 ).Now, to find all the divisors of 990, I can use the prime factors. The exponents for each prime factor are 1 for 2, 2 for 3, 1 for 5, and 1 for 11. The number of divisors is calculated by adding 1 to each of the exponents and then multiplying them together. So, that would be ( (1+1)(2+1)(1+1)(1+1) = 2 times 3 times 2 times 2 = 24 ) divisors. Okay, so there are 24 divisors of 990.But I don't need all the divisors; I need the ones that can be expressed as ( m^2 - 2 ) where ( m ) is a positive integer. So, let me think about this. If ( m^2 - 2 ) is a divisor of 990, then ( m^2 ) must be equal to that divisor plus 2. So, ( m^2 = d + 2 ), where ( d ) is a positive divisor of 990.Therefore, I need to find all the divisors ( d ) of 990 such that ( d + 2 ) is a perfect square. That makes sense because ( m ) has to be an integer, so ( m^2 ) must be a perfect square.Alright, so let me list all the positive divisors of 990. Since 990 has 24 divisors, that's a bit lengthy, but let me try to list them systematically.Starting from the smallest:1, 2, 3, 5, 6, 9, 10, 11, 15, 18, 22, 30, 33, 45, 55, 90, 99, 110, 165, 198, 330, 495, 990.Wait, that's 23 divisors. Did I miss one? Let me recount. Hmm, maybe I missed 135? Wait, no, 135 isn't a divisor of 990 because 990 divided by 135 is not an integer. Let me check again.Wait, 990 divided by 2 is 495, divided by 3 is 330, divided by 5 is 198, divided by 6 is 165, divided by 9 is 110, divided by 10 is 99, divided by 11 is 90, divided by 15 is 66, divided by 18 is 55, divided by 22 is 45, divided by 30 is 33, divided by 33 is 30, divided by 45 is 22, divided by 55 is 18, divided by 90 is 11, divided by 99 is 10, divided by 110 is 9, divided by 165 is 6, divided by 198 is 5, divided by 330 is 3, divided by 495 is 2, and divided by 990 is 1. So, actually, I think I have all 24 divisors listed. Wait, no, in my initial list, I have 23. Hmm, maybe I missed one. Let me see.Wait, 990 divided by 15 is 66, which I didn't include earlier. So, adding 66 to the list. So, the complete list is:1, 2, 3, 5, 6, 9, 10, 11, 15, 18, 22, 30, 33, 45, 55, 66, 90, 99, 110, 165, 198, 330, 495, 990.Yes, that's 24 divisors. Okay, so now I have all the divisors.Now, for each divisor ( d ), I need to check if ( d + 2 ) is a perfect square. So, let me go through each divisor and compute ( d + 2 ), then see if it's a perfect square.Starting with ( d = 1 ): ( 1 + 2 = 3 ). Is 3 a perfect square? No, because 1^2 is 1 and 2^2 is 4, so 3 is not a perfect square.Next, ( d = 2 ): ( 2 + 2 = 4 ). 4 is a perfect square, it's 2 squared. So, that's one value: ( m = 2 ).Next, ( d = 3 ): ( 3 + 2 = 5 ). 5 is not a perfect square.( d = 5 ): ( 5 + 2 = 7 ). Not a perfect square.( d = 6 ): ( 6 + 2 = 8 ). Not a perfect square.( d = 9 ): ( 9 + 2 = 11 ). Not a perfect square.( d = 10 ): ( 10 + 2 = 12 ). Not a perfect square.( d = 11 ): ( 11 + 2 = 13 ). Not a perfect square.( d = 15 ): ( 15 + 2 = 17 ). Not a perfect square.( d = 18 ): ( 18 + 2 = 20 ). Not a perfect square.( d = 22 ): ( 22 + 2 = 24 ). Not a perfect square.( d = 30 ): ( 30 + 2 = 32 ). Not a perfect square.( d = 33 ): ( 33 + 2 = 35 ). Not a perfect square.( d = 45 ): ( 45 + 2 = 47 ). Not a perfect square.( d = 55 ): ( 55 + 2 = 57 ). Not a perfect square.( d = 66 ): ( 66 + 2 = 68 ). Not a perfect square.( d = 90 ): ( 90 + 2 = 92 ). Not a perfect square.( d = 99 ): ( 99 + 2 = 101 ). Not a perfect square.( d = 110 ): ( 110 + 2 = 112 ). Not a perfect square.( d = 165 ): ( 165 + 2 = 167 ). Not a perfect square.( d = 198 ): ( 198 + 2 = 200 ). 200 is not a perfect square because 14^2 is 196 and 15^2 is 225, so 200 is in between.( d = 330 ): ( 330 + 2 = 332 ). Not a perfect square.( d = 495 ): ( 495 + 2 = 497 ). Not a perfect square.( d = 990 ): ( 990 + 2 = 992 ). Not a perfect square.Wait a second, so the only divisor where ( d + 2 ) is a perfect square is when ( d = 2 ), giving ( m = 2 ). But hold on, earlier I thought of ( m = 3 ) as well. Let me check that.Wait, when ( d = 7 ), ( d + 2 = 9 ), which is 3 squared. But 7 is not a divisor of 990. Wait, is 7 a divisor of 990? Let me check. 990 divided by 7 is approximately 141.428..., which is not an integer. So, 7 is not a divisor of 990. Therefore, ( d = 7 ) is not in our list.But wait, I thought earlier that ( m = 3 ) would work because ( 3^2 - 2 = 9 - 2 = 7 ), but 7 is not a divisor of 990. So, that means ( m = 3 ) is not a solution because 7 doesn't divide 990. Hmm, so maybe I was wrong earlier.Wait, let me check ( m = 3 ): ( m^2 - 2 = 9 - 2 = 7 ). 990 divided by 7 is approximately 141.428..., which is not an integer. So, ( m = 3 ) doesn't work. So, only ( m = 2 ) works.But wait, let me double-check my list of divisors. Did I miss any divisor that when added to 2 gives a perfect square?Looking back at the list:1: 3 (not square)2: 4 (square, m=2)3:5 (not)5:7 (not)6:8 (not)9:11 (not)10:12 (not)11:13 (not)15:17 (not)18:20 (not)22:24 (not)30:32 (not)33:35 (not)45:47 (not)55:57 (not)66:68 (not)90:92 (not)99:101 (not)110:112 (not)165:167 (not)198:200 (not)330:332 (not)495:497 (not)990:992 (not)So, indeed, only ( d = 2 ) gives ( m = 2 ). So, that would mean there's only one value of ( m ), which is 2.But wait, the answer choices are A) one, B) two, C) three, D) four, E) more than four. So, according to my calculation, the answer is A) one.But hold on, I initially thought ( m = 3 ) might work, but it doesn't because 7 doesn't divide 990. But let me think again: is there any other way?Wait, maybe I missed some divisors. Let me check the list again. Did I list all 24 divisors correctly?Starting from 1:1, 2, 3, 5, 6, 9, 10, 11, 15, 18, 22, 30, 33, 45, 55, 66, 90, 99, 110, 165, 198, 330, 495, 990.Yes, that's 24 divisors. So, I think I have all of them.Wait, but let me think differently. Maybe I can approach this problem by considering the equation ( m^2 - 2 ) divides 990, so ( m^2 - 2 ) must be a positive divisor of 990. Therefore, ( m^2 - 2 ) must be less than or equal to 990. So, ( m^2 leq 992 ), which means ( m leq sqrt{992} approx 31.5 ). So, ( m ) can be at most 31.So, ( m ) can be from 1 to 31. Let me check for each ( m ) from 1 to 31 whether ( m^2 - 2 ) divides 990.Starting with ( m = 1 ): ( 1 - 2 = -1 ). But we need positive integers, so discard.( m = 2 ): ( 4 - 2 = 2 ). 2 divides 990, yes.( m = 3 ): ( 9 - 2 = 7 ). 7 doesn't divide 990, as we saw earlier.( m = 4 ): ( 16 - 2 = 14 ). 14 divides 990? Let's check: 990 divided by 14 is 70.714..., which is not an integer. So, no.( m = 5 ): ( 25 - 2 = 23 ). 23 doesn't divide 990 because 990 divided by 23 is approximately 43.04, not integer.( m = 6 ): ( 36 - 2 = 34 ). 34 divides 990? 990 divided by 34 is approximately 29.117, not integer.( m = 7 ): ( 49 - 2 = 47 ). 47 doesn't divide 990.( m = 8 ): ( 64 - 2 = 62 ). 62 divides 990? 990 divided by 62 is approximately 15.967, not integer.( m = 9 ): ( 81 - 2 = 79 ). 79 doesn't divide 990.( m = 10 ): ( 100 - 2 = 98 ). 98 divides 990? 990 divided by 98 is approximately 10.102, not integer.( m = 11 ): ( 121 - 2 = 119 ). 119 doesn't divide 990.( m = 12 ): ( 144 - 2 = 142 ). 142 divides 990? 990 divided by 142 is approximately 6.971, not integer.( m = 13 ): ( 169 - 2 = 167 ). 167 doesn't divide 990.( m = 14 ): ( 196 - 2 = 194 ). 194 divides 990? 990 divided by 194 is approximately 5.103, not integer.( m = 15 ): ( 225 - 2 = 223 ). 223 doesn't divide 990.( m = 16 ): ( 256 - 2 = 254 ). 254 divides 990? 990 divided by 254 is approximately 3.90, not integer.( m = 17 ): ( 289 - 2 = 287 ). 287 doesn't divide 990.( m = 18 ): ( 324 - 2 = 322 ). 322 divides 990? 990 divided by 322 is approximately 3.074, not integer.( m = 19 ): ( 361 - 2 = 359 ). 359 doesn't divide 990.( m = 20 ): ( 400 - 2 = 398 ). 398 divides 990? 990 divided by 398 is approximately 2.487, not integer.( m = 21 ): ( 441 - 2 = 439 ). 439 doesn't divide 990.( m = 22 ): ( 484 - 2 = 482 ). 482 divides 990? 990 divided by 482 is approximately 2.054, not integer.( m = 23 ): ( 529 - 2 = 527 ). 527 doesn't divide 990.( m = 24 ): ( 576 - 2 = 574 ). 574 divides 990? 990 divided by 574 is approximately 1.725, not integer.( m = 25 ): ( 625 - 2 = 623 ). 623 doesn't divide 990.( m = 26 ): ( 676 - 2 = 674 ). 674 divides 990? 990 divided by 674 is approximately 1.47, not integer.( m = 27 ): ( 729 - 2 = 727 ). 727 doesn't divide 990.( m = 28 ): ( 784 - 2 = 782 ). 782 divides 990? 990 divided by 782 is approximately 1.266, not integer.( m = 29 ): ( 841 - 2 = 839 ). 839 doesn't divide 990.( m = 30 ): ( 900 - 2 = 898 ). 898 divides 990? 990 divided by 898 is approximately 1.102, not integer.( m = 31 ): ( 961 - 2 = 959 ). 959 doesn't divide 990.So, after checking all ( m ) from 1 to 31, the only value that works is ( m = 2 ). Therefore, there is only one positive integer ( m ) that satisfies the condition.Wait, but the answer choices include up to four, and the option A is one. So, according to my calculations, the answer should be A) one.But hold on, I initially thought ( m = 3 ) might work, but it doesn't because 7 doesn't divide 990. So, only ( m = 2 ) works. Therefore, the correct answer is A) one.But wait, let me think again. Maybe I missed something. Is there a way that ( m^2 - 2 ) could be a negative divisor? But the problem specifies positive integers ( m ), and ( m^2 - 2 ) must be positive because 990 is positive, and the fraction is positive. So, ( m^2 - 2 ) must be positive, meaning ( m^2 > 2 ), so ( m geq 2 ). So, ( m = 1 ) is invalid because ( 1 - 2 = -1 ), which is negative, and we need positive integers.Therefore, yes, only ( m = 2 ) works. So, the answer is A) one.Wait, but the initial problem statement says "positive integers ( m )", so ( m ) must be positive, and ( m^2 - 2 ) must be positive, so ( m geq 2 ). So, yeah, only ( m = 2 ) works.But wait, let me check ( m = 0 ). Wait, ( m ) is a positive integer, so ( m = 0 ) is not considered.Wait, another thought: maybe ( m^2 - 2 ) could be a negative divisor of 990, but since the fraction is positive, both numerator and denominator must be positive or both negative. But since 990 is positive, ( m^2 - 2 ) must also be positive. So, ( m^2 - 2 > 0 ), hence ( m geq 2 ).Therefore, only ( m = 2 ) works. So, the answer is A) one.But wait, I'm a bit confused because I thought earlier that ( m = 3 ) might work, but it doesn't. So, maybe I was overcomplicating it. Let me just confirm once more.If ( m = 2 ), then ( m^2 - 2 = 4 - 2 = 2 ), and 990 divided by 2 is 495, which is an integer. So, that works.If ( m = 3 ), ( m^2 - 2 = 9 - 2 = 7 ), and 990 divided by 7 is not an integer, as 7 times 141 is 987, and 990 - 987 is 3, so it's not divisible.Therefore, only ( m = 2 ) works. So, the answer is A) one.Wait, but the answer choices are A) one, B) two, etc. So, I think I'm confident now that the answer is A) one.But hold on, let me think again. Maybe I missed a divisor. Let me check the list of divisors again:1, 2, 3, 5, 6, 9, 10, 11, 15, 18, 22, 30, 33, 45, 55, 66, 90, 99, 110, 165, 198, 330, 495, 990.So, for each of these, ( d + 2 ) must be a perfect square. So, let's compute ( d + 2 ) for each:1 + 2 = 3 (not square)2 + 2 = 4 (square, m=2)3 + 2 = 5 (not)5 + 2 = 7 (not)6 + 2 = 8 (not)9 + 2 = 11 (not)10 + 2 = 12 (not)11 + 2 = 13 (not)15 + 2 = 17 (not)18 + 2 = 20 (not)22 + 2 = 24 (not)30 + 2 = 32 (not)33 + 2 = 35 (not)45 + 2 = 47 (not)55 + 2 = 57 (not)66 + 2 = 68 (not)90 + 2 = 92 (not)99 + 2 = 101 (not)110 + 2 = 112 (not)165 + 2 = 167 (not)198 + 2 = 200 (not)330 + 2 = 332 (not)495 + 2 = 497 (not)990 + 2 = 992 (not)So, indeed, only ( d = 2 ) gives a perfect square when added to 2. Therefore, only ( m = 2 ) works.Wait, but I just thought of something else. Maybe ( m^2 - 2 ) could be equal to 990 itself. Let's check: ( m^2 - 2 = 990 ) implies ( m^2 = 992 ). Is 992 a perfect square? Let's see, 31^2 is 961, 32^2 is 1024, so 992 is between them, not a perfect square. So, no.Similarly, ( m^2 - 2 = 495 ) would imply ( m^2 = 497 ), not a square.( m^2 - 2 = 330 ) implies ( m^2 = 332 ), not a square.Same with others. So, no other ( m ) beyond 2 works.Therefore, I'm confident that the answer is A) one.

💡First, I need to expand the expression (2(x + 3)(x + 8)). To do this, I'll start by expanding the product of the two binomials ((x + 3)) and ((x + 8)).I'll use the distributive property, also known as the FOIL method, to multiply the two binomials:- Multiply the First terms: (x cdot x = x^2)- Multiply the Outer terms: (x cdot 8 = 8x)- Multiply the Inner terms: (3 cdot x = 3x)- Multiply the Last terms: (3 cdot 8 = 24)Adding these products together gives:[x^2 + 8x + 3x + 24]Next, I'll combine the like terms (8x) and (3x) to simplify the expression:[x^2 + 11x + 24]Now, I'll distribute the coefficient 2 across each term in the simplified expression:[2 cdot x^2 + 2 cdot 11x + 2 cdot 24 = 2x^2 + 22x + 48]Therefore, the expanded and simplified form of (2(x + 3)(x + 8)) is (2x^2 + 22x + 48).

💡First, identify the amount of pepper Brennan used and the amount he has left.He used 0.16 grams of pepper and has 0.09 grams remaining.To find the initial amount of pepper, add the amount used to the amount left.0.16 grams + 0.09 grams equals 0.25 grams.Therefore, Brennan initially had 0.25 grams of pepper.

💡Okay, so I'm trying to figure out whether a function ( f : mathbb{Q} rightarrow mathbb{R} ) with a certain property must be bounded on some interval. The property is that for every ( h in mathbb{Q} ) and every ( x_0 in mathbb{R} ), the difference ( f(x + h) - f(x) ) approaches 0 as ( x ) approaches ( x_0 ) through rational numbers. First, I need to understand what this property really means. It says that if I pick any rational number ( h ) and any real number ( x_0 ), then as ( x ) gets closer and closer to ( x_0 ) (but staying within the rationals), the difference between ( f(x + h) ) and ( f(x) ) becomes negligible. So, in some sense, the function doesn't change much when you shift the input by a small rational number, especially near any real number.Now, the question is whether such a function ( f ) must be bounded on some interval. That is, does there exist an interval ( [a, b] ) such that ( f(x) ) doesn't go to infinity or negative infinity as ( x ) varies within ( [a, b] cap mathbb{Q} )?I remember that in real analysis, functions with certain continuity properties are often bounded on compact intervals. But here, ( f ) is defined only on the rationals, which are dense in the reals but not a compact space. So, maybe the usual theorems don't apply directly.I also recall that the rationals are countable, so maybe we can construct a function that is unbounded on every interval but still satisfies the given property. That might be a way to show that the answer is no, such a function doesn't have to be bounded on any interval.Let me think about how to construct such a function. Maybe using some enumeration of the rationals or primes? I remember that primes can be used to create sequences with specific properties.Wait, the user mentioned something about primes and defining ( q_n ) as the product of the first ( n ) primes raised to the power ( n ). So, ( q_n = (p_1 p_2 ldots p_n)^n ). Then, they defined sets ( A_n ) as ( { m / q_n mid m in mathbb{Z} } ). These sets are getting finer as ( n ) increases because the denominators are getting larger, so the spacing between consecutive elements in ( A_n ) is getting smaller.Moreover, ( A_n subset A_{n+1} ) because each subsequent set includes all the previous ones with more points. The union of all ( A_n ) covers ( mathbb{Q} ), which makes sense because every rational number can be expressed as a fraction with some denominator, and eventually, that denominator will be a product of primes raised to some power.Next, they defined ( B_n ) as ( A_n setminus A_{n-1} ). So each ( B_n ) contains the new points added in ( A_n ) that weren't in the previous sets. This way, every rational number is in exactly one ( B_n ), and the union of all ( B_n ) is ( mathbb{Q} ).The key property here is that for any fixed ( h in mathbb{Q} ), if ( x in B_n ), then ( x + h ) is also in ( B_n ) for sufficiently large ( n ). This is because the spacing in ( A_n ) is so fine that adding a fixed ( h ) doesn't take you out of the set ( A_n ), especially since ( h ) is rational and can be expressed with a denominator that's a product of primes, which is accounted for in ( q_n ) for large enough ( n ).Now, defining the function ( f ) as ( f(x) = n ) if ( x in B_n ). So, each ( B_n ) is assigned a value ( n ), which increases without bound as ( n ) increases. Therefore, ( f ) is unbounded on any interval because, no matter how small an interval you take, it will contain elements from infinitely many ( B_n ), each contributing a larger and larger value.But does this function satisfy the given property? Let's check. For any ( h in mathbb{Q} ) and ( x_0 in mathbb{R} ), we need to show that ( f(x + h) - f(x) ) approaches 0 as ( x ) approaches ( x_0 ) through ( mathbb{Q} ).Since ( x_0 ) is a real number, and ( x ) approaches ( x_0 ) through rationals, we can choose a neighborhood around ( x_0 ) that doesn't include any elements from ( B_1, B_2, ldots, B_N ) for some ( N ). This is possible because each ( B_n ) is countable and discrete in some sense. So, beyond a certain ( N ), all ( x ) in this neighborhood will be in ( B_n ) for ( n > N ), and adding ( h ) will keep them in the same ( B_n ). Therefore, ( f(x + h) = f(x) ) for these ( x ), making the difference zero.Hence, this function ( f ) satisfies the given property but is unbounded on every interval. Therefore, the answer is no; such a function doesn't have to be bounded on any interval.I think this makes sense. The function is constructed in a way that it's unbounded because it takes on larger and larger values on sets that are getting finer and finer. But the property about the difference ( f(x + h) - f(x) ) going to zero is satisfied because, near any real number, you can find a neighborhood where the function is constant (or doesn't change much) due to the construction of the sets ( B_n ).So, in conclusion, the function doesn't have to be bounded on any interval. The key idea is constructing a function that's unbounded but still meets the given condition by carefully controlling the behavior around each real number using the structure of the rationals.

💡Okay, so I have this problem where I need to find the analytic expression of a function f(x) and then determine its extreme values. The function is given as f(x) = ax - b/x + c, where a, b, c are real numbers. It also says that the function passes through the point (1, 0), and the tangent line at x=2 is y = -x + 3. Alright, let me break this down. First, I need to find the values of a, b, and c. Since the function passes through (1, 0), that means when x=1, f(x)=0. So I can plug in x=1 into the function and set it equal to 0. That gives me one equation.Next, the tangent line at x=2 is given as y = -x + 3. I know that the tangent line at a point has two properties: it passes through the point (2, f(2)) and its slope is equal to the derivative of f at x=2. So I can use this information to set up two more equations.First, let me write down the function and its derivative:f(x) = ax - b/x + cf'(x) = derivative of ax is a, derivative of -b/x is b/x², and derivative of c is 0. So f'(x) = a + b/x².Now, let's use the given information step by step.1. The function passes through (1, 0). So f(1) = 0.Plugging x=1 into f(x):f(1) = a(1) - b/1 + c = a - b + c = 0So equation 1: a - b + c = 02. The tangent line at x=2 is y = -x + 3. First, the point (2, f(2)) lies on the tangent line, so f(2) must equal the value of the tangent line at x=2.Calculating the tangent line at x=2:y = -2 + 3 = 1So f(2) = 1Plugging x=2 into f(x):f(2) = a(2) - b/2 + c = 2a - b/2 + c = 1So equation 2: 2a - (b/2) + c = 13. The slope of the tangent line at x=2 is equal to f'(2). The given tangent line is y = -x + 3, so its slope is -1.Therefore, f'(2) = -1Plugging x=2 into f'(x):f'(2) = a + b/(2)^2 = a + b/4 = -1So equation 3: a + b/4 = -1Now, I have three equations:1. a - b + c = 02. 2a - (b/2) + c = 13. a + (b/4) = -1I need to solve this system of equations for a, b, c.Let me write them again:Equation 1: a - b + c = 0Equation 2: 2a - (b/2) + c = 1Equation 3: a + (b/4) = -1Let me try to solve equations 1 and 3 first to express a and c in terms of b, or something like that.From equation 3: a + (b/4) = -1Let me solve for a:a = -1 - (b/4)So a is expressed in terms of b.Now, let's plug this into equation 1.Equation 1: a - b + c = 0Substitute a:(-1 - b/4) - b + c = 0Simplify:-1 - b/4 - b + c = 0Combine like terms:-1 - (b/4 + b) + c = 0Convert b to quarters:-1 - (b/4 + 4b/4) + c = 0-1 - (5b/4) + c = 0So, c = 1 + (5b/4)So now, c is expressed in terms of b.Now, let's plug a and c into equation 2.Equation 2: 2a - (b/2) + c = 1We have a = -1 - b/4 and c = 1 + 5b/4Substitute:2*(-1 - b/4) - (b/2) + (1 + 5b/4) = 1Let me compute each term:2*(-1) = -22*(-b/4) = -b/2So first part: -2 - b/2Second term: -b/2Third term: 1 + 5b/4So putting it all together:(-2 - b/2) - b/2 + (1 + 5b/4) = 1Combine like terms:Constants: -2 + 1 = -1b terms: -b/2 - b/2 + 5b/4Let me convert all to quarters:-b/2 = -2b/4So:-2b/4 - 2b/4 + 5b/4 = (-2 -2 +5)b/4 = 1b/4So overall:-1 + (b/4) = 1So:b/4 = 1 + 1 = 2Therefore, b = 8Now, knowing that b=8, let's find a and c.From equation 3: a = -1 - b/4 = -1 - 8/4 = -1 - 2 = -3From c = 1 + 5b/4 = 1 + 5*8/4 = 1 + 10 = 11So a = -3, b=8, c=11Therefore, the function is f(x) = -3x - 8/x + 11Wait, let me double-check.f(1) = -3(1) -8/1 +11 = -3 -8 +11 = 0, which is correct.f(2) = -3*2 -8/2 +11 = -6 -4 +11 = 1, which is correct.f'(x) = -3 + 8/x²f'(2) = -3 + 8/4 = -3 + 2 = -1, which matches the slope of the tangent line.Great, so part I is done.Now, part II: find the extreme values of f(x).Extreme values occur where the derivative is zero or undefined, but since f(x) is defined for x ≠ 0, and the derivative f'(x) = -3 + 8/x².Set f'(x) = 0:-3 + 8/x² = 0So 8/x² = 3Multiply both sides by x²:8 = 3x²So x² = 8/3Therefore, x = sqrt(8/3) or x = -sqrt(8/3)Simplify sqrt(8/3):sqrt(8)/sqrt(3) = 2*sqrt(2)/sqrt(3) = 2*sqrt(6)/3So critical points at x = 2√6/3 and x = -2√6/3Now, to determine if these are maxima or minima, we can use the second derivative test or analyze the sign of the first derivative around these points.Let me compute the second derivative.f'(x) = -3 + 8x^{-2}f''(x) = 0 + (-16)x^{-3} = -16/x³So f''(x) = -16/x³Evaluate f''(x) at x = 2√6/3:f''(2√6/3) = -16/( (2√6/3)^3 )Compute denominator:(2√6/3)^3 = 8*(6√6)/27 = 48√6/27 = 16√6/9So f''(2√6/3) = -16 / (16√6/9) = -16 * (9)/(16√6) = -9/√6Which is negative, so the function is concave down at this point, meaning it's a local maximum.Similarly, at x = -2√6/3:f''(-2√6/3) = -16/( (-2√6/3)^3 )Compute denominator:(-2√6/3)^3 = -8*(6√6)/27 = -48√6/27 = -16√6/9So f''(-2√6/3) = -16 / (-16√6/9) = -16 * (-9)/(16√6) = 9/√6Which is positive, so the function is concave up at this point, meaning it's a local minimum.Therefore, x = 2√6/3 is a local maximum, and x = -2√6/3 is a local minimum.Now, let's find the function values at these points.First, at x = 2√6/3:f(x) = -3x -8/x +11Compute each term:-3x = -3*(2√6/3) = -2√6-8/x = -8/(2√6/3) = -8*(3)/(2√6) = -12/√6 = -12√6/6 = -2√6So f(x) = -2√6 -2√6 +11 = (-4√6) +11So f(2√6/3) = 11 -4√6Similarly, at x = -2√6/3:f(x) = -3x -8/x +11Compute each term:-3x = -3*(-2√6/3) = 2√6-8/x = -8/(-2√6/3) = 8*(3)/(2√6) = 12/√6 = 12√6/6 = 2√6So f(x) = 2√6 + 2√6 +11 = 4√6 +11So f(-2√6/3) = 11 +4√6Therefore, the function has a local maximum at x = 2√6/3 with value 11 -4√6, and a local minimum at x = -2√6/3 with value 11 +4√6.Wait, let me double-check the calculations for f(x) at x = 2√6/3:f(x) = -3x -8/x +11x = 2√6/3-3x = -3*(2√6/3) = -2√6-8/x = -8/(2√6/3) = -8*(3)/(2√6) = -12/√6 = -12√6/6 = -2√6So f(x) = -2√6 -2√6 +11 = -4√6 +11, which is 11 -4√6. Correct.Similarly, at x = -2√6/3:-3x = -3*(-2√6/3) = 2√6-8/x = -8/(-2√6/3) = 8*(3)/(2√6) = 12/√6 = 2√6So f(x) = 2√6 +2√6 +11 = 4√6 +11. Correct.So that's the conclusion.**Final Answer**Ⅰ. The analytic expression of ( f(x) ) is (boxed{-3x - frac{8}{x} + 11}).Ⅱ. The extreme values of ( f(x) ) are a local minimum of (boxed{11 + 4sqrt{6}}) and a local maximum of (boxed{11 - 4sqrt{6}}).

💡Okay, so I have this problem: "At most how many common tangents can two ellipses have if they share one of their foci?" Hmm, interesting. I remember that ellipses have some properties related to their foci, and tangents... Let me try to recall.First, an ellipse is defined as the set of points where the sum of the distances from two fixed points (the foci) is constant. So, if two ellipses share a focus, they each have another focus somewhere else. Let me visualize this: imagine two ellipses, both having a common focus at point F, but each has another focus, say F1 and F2, at different locations.Now, the question is about common tangents. A tangent to an ellipse is a line that touches the ellipse at exactly one point. For two ellipses, a common tangent is a line that is tangent to both. So, how many such lines can exist?I remember that two circles can have up to four common tangents: two direct and two transverse. But ellipses are more complicated because they are not circles. However, maybe some similar logic applies.Wait, but these ellipses share a focus. That might impose some restrictions or perhaps symmetries that could affect the number of common tangents. Let me think about the properties of ellipses related to their foci and tangents.I recall that for an ellipse, the tangent at any point makes equal angles with the lines joining the point to each of the foci. So, if I have a tangent line, and I pick a point on the tangent, the angles between the tangent and the lines to each focus are equal. That's a key property.Given that, if two ellipses share a focus, say F, then for a common tangent, the reflection property must hold for both ellipses. That is, for the tangent line, the reflection of the other focus over the tangent line must lie on the line connecting the shared focus and the point of tangency.Wait, that might be a bit abstract. Let me try to formalize it.Suppose we have two ellipses, E1 and E2. Both share a focus at F. E1 has another focus at F1, and E2 has another focus at F2. For a line to be tangent to both E1 and E2, it must satisfy the reflection property for both ellipses.So, if L is a common tangent, then for E1, the reflection of F1 over L must lie on the line connecting F and the point of tangency on E1. Similarly, for E2, the reflection of F2 over L must lie on the line connecting F and the point of tangency on E2.But since L is the same tangent line for both ellipses, these two reflections must somehow be consistent. Hmm, this seems complicated. Maybe there's another way to think about it.I remember that for an ellipse, the set of all possible reflections of one focus over all possible tangent lines forms a circle centered at the other focus with radius equal to the major axis length. Is that right?Let me verify. If I take all tangent lines to an ellipse and reflect one focus over each tangent, the set of these reflections should form a circle. The radius of this circle would be equal to the length of the major axis of the ellipse. So, if the ellipse has major axis length 2a, then the radius of this circle is 2a.Yes, that makes sense because the reflection property ensures that the distance from the other focus to the reflection point is constant, equal to the major axis length.So, applying this to our problem, for ellipse E1, reflecting focus F1 over all possible tangent lines gives a circle centered at F with radius 2a1, where 2a1 is the major axis length of E1. Similarly, for ellipse E2, reflecting focus F2 over all possible tangent lines gives a circle centered at F with radius 2a2, where 2a2 is the major axis length of E2.Now, the common tangents to both ellipses correspond to the common tangent lines that are also tangent to both circles centered at F with radii 2a1 and 2a2. Wait, is that correct?No, actually, the reflections of F1 and F2 over the common tangent line must lie on both circles. So, the reflection points must lie on the intersection of the two circles: one circle centered at F with radius 2a1 and another centered at F with radius 2a2.But if both circles are centered at the same point F, then they are concentric circles. The number of intersection points between two concentric circles depends on their radii. If the radii are different, they don't intersect. If the radii are equal, they are the same circle.Wait, that can't be right because if the radii are different, the circles don't intersect, meaning there are no common reflections, hence no common tangents. But that contradicts my earlier thought that two ellipses sharing a focus can have common tangents.Hmm, maybe I made a mistake in the reasoning. Let me go back.The reflection of F1 over the tangent line lies on the circle centered at F with radius 2a1, and the reflection of F2 over the same tangent line lies on the circle centered at F with radius 2a2. So, for a common tangent, the reflection points must lie on both circles.But since both reflections are over the same tangent line, the two reflection points are related. Specifically, the tangent line is the perpendicular bisector of the segment joining F1 and its reflection, and also the perpendicular bisector of the segment joining F2 and its reflection.Therefore, the tangent line must be the perpendicular bisector of both F1F1' and F2F2', where F1' and F2' are the reflections of F1 and F2 over the tangent line.This implies that the tangent line is equidistant from F1 and F1', and also from F2 and F2'. Therefore, F1' must lie on the circle centered at F with radius 2a1, and F2' must lie on the circle centered at F with radius 2a2.But since the tangent line is the same for both, the reflections F1' and F2' must lie on the same line, which is the tangent line. Wait, no, F1' and F2' are points, not lines.Wait, maybe I need to think differently. The tangent line is the external bisector of the angle formed by F1E and F2E, where E is the point of tangency. So, for each ellipse, the tangent line has this property.But since both ellipses share a focus F, maybe we can relate the reflections of the other foci over the tangent line.Let me try to formalize this.For ellipse E1 with foci F and F1, the tangent line at point E satisfies the condition that the reflection of F1 over the tangent line lies on the line FE. Similarly, for ellipse E2 with foci F and F2, the reflection of F2 over the tangent line lies on the line FE.Therefore, for a common tangent, the reflection of F1 and F2 over the tangent line must both lie on the same line FE.But FE is a line from F to E, which is a point on both ellipses. Wait, but E is a point on both ellipses? No, E is a point on the tangent line, but not necessarily on both ellipses.Wait, no, E is the point of tangency on each ellipse. So, for E1, E is a point on E1, and for E2, E is a point on E2. But since the tangent line is common, E must be the same point for both ellipses? No, that can't be, because the tangent line touches each ellipse at their own points.Wait, maybe I'm confusing the points of tangency. Let me clarify.Suppose the common tangent line touches E1 at point E1 and E2 at point E2. Then, for E1, the reflection of F1 over the tangent line lies on the line FE1. Similarly, for E2, the reflection of F2 over the tangent line lies on the line FE2.But since the tangent line is the same, the reflections F1' and F2' must lie on the same line, which is the tangent line. Wait, no, F1' lies on FE1, and F2' lies on FE2.Hmm, this is getting a bit tangled. Maybe another approach is needed.I remember that the number of common tangents between two conic sections can be up to four, but in this case, since they share a focus, maybe the number is limited.Wait, let me think about specific cases. If two ellipses share a focus and are otherwise identical, they would coincide, and thus have infinitely many common tangents. But the problem says "at most," so we need to consider when they are distinct.If the ellipses are different but share a focus, how many common tangents can they have? Maybe two?Wait, I think the maximum number is two. Let me try to reason why.If we consider the two ellipses, each has a focus at F, and another focus at F1 and F2 respectively. The common tangents must satisfy the reflection property for both ellipses.As I thought earlier, reflecting the other foci over the tangent line gives points on circles centered at F with radii equal to the major axes of the ellipses.So, if we have two circles centered at F with different radii, the number of common tangents between these two circles would be the number of lines that are tangent to both circles.Wait, but in this case, the reflections must lie on both circles. So, the number of common tangents corresponds to the number of common tangent lines to these two circles.But two circles can have up to four common tangents: two direct and two transverse. However, in this case, the circles are both centered at F, so they are concentric.Wait, concentric circles have no common tangents because all their tangent lines are parallel, but they don't coincide. So, if the circles are concentric, they don't have any common tangents.But that contradicts the idea that the ellipses can have common tangents. So, maybe my earlier reasoning is flawed.Wait, no, because the reflections F1' and F2' are not necessarily on the same tangent line. Each reflection is on a different tangent line, but for a common tangent, both reflections must lie on the same tangent line.Wait, no, for a common tangent line, the reflection of F1 over that line must lie on FE1, and the reflection of F2 over that same line must lie on FE2.But since FE1 and FE2 are different lines (unless E1 and E2 coincide, which they don't), the reflections F1' and F2' must lie on different lines.Hmm, this is confusing. Maybe another approach is needed.Let me think about the director circle of an ellipse. The director circle is the locus of all points from which the ellipse is seen at a right angle, and it's related to the common tangents.But I'm not sure if that helps here.Wait, another idea: the number of common tangents between two ellipses can be found by solving their equations simultaneously and ensuring that the discriminant is zero for tangency.But that might be too algebraic and complicated.Alternatively, I remember that two ellipses can intersect at up to four points, but common tangents are different.Wait, but common tangents don't necessarily intersect the ellipses at the same points.Hmm.Wait, going back to the reflection property. For a common tangent, the reflection of F1 over the tangent line must lie on the line FE1, and the reflection of F2 over the same tangent line must lie on FE2.But since FE1 and FE2 are different lines (unless E1 and E2 are colinear with F, which is not necessarily the case), the reflections F1' and F2' must lie on different lines.But the tangent line is the same for both, so the reflections must be consistent with that.Wait, maybe the key is that the tangent line must be such that it reflects F1 to a point on FE1 and F2 to a point on FE2.So, for each ellipse, the tangent line is determined by the reflection of its other focus over the tangent line lying on FE.Therefore, for a common tangent, the tangent line must satisfy both conditions simultaneously.So, how many such lines can exist?I think it's related to the number of solutions to the system where the reflection of F1 over L lies on FE1 and the reflection of F2 over L lies on FE2.This seems like a system of equations that could have multiple solutions.But I'm not sure exactly how many.Wait, maybe considering the geometric transformations involved.Reflecting F1 over L gives a point on FE1, and reflecting F2 over L gives a point on FE2.So, L must be such that it reflects F1 to FE1 and F2 to FE2.This seems like a problem of finding lines L such that L is the perpendicular bisector of F1F1' and F2F2', where F1' is on FE1 and F2' is on FE2.But since F1' and F2' are variable points on FE1 and FE2, respectively, the number of such lines L depends on how FE1 and FE2 are arranged.But since FE1 and FE2 are lines from F to points E1 and E2 on the ellipses, which are different, the number of common tangents could be limited.Wait, maybe the maximum number is two.I think I've heard before that two ellipses sharing a focus can have at most two common tangents.But I need to verify this.Alternatively, think about the case where the two ellipses are such that their other foci are on opposite sides of the shared focus. Then, they might have two common tangents on either side.Yes, that makes sense.So, in the best case, they can have two common tangents.Therefore, the maximum number of common tangents two ellipses can have if they share one of their foci is two.**Final Answer**The maximum number of common tangents is boxed{2}.