🧠:[<think>Okay, so I need to figure out how many solutions there are to the equation ( x^{2} - left[x^{2}right] = (x - [x])^{2} ) where ( 1 leq x leq n ). Hmm, let's start by understanding the notation here. The brackets [x] usually denote the floor function, which means the greatest integer less than or equal to x. So, [x] is the integer part of x, and x - [x] would be the fractional part, right? Let me confirm that. Yes, if x is a real number, then x = [x] + {x}, where {x} is the fractional part, between 0 (inclusive) and 1 (exclusive). So, rewriting the equation with this in mind: the left-hand side is ( x^2 - lfloor x^2 rfloor ), which is the fractional part of ( x^2 ). The right-hand side is ( (x - lfloor x rfloor)^2 ), which is the square of the fractional part of x. So the equation is saying that the fractional part of ( x^2 ) is equal to the square of the fractional part of x. Interesting. Let me denote {x} as the fractional part of x, so {x} = x - [x]. Then the equation becomes:( { x^2 } = ( { x } )^2 )So, we need to find all x in [1, n] such that the fractional part of x squared is equal to the square of the fractional part of x.First, let's consider x as an integer. If x is an integer, then {x} = 0, so the right-hand side is 0. The left-hand side is {x²} which is also 0 because x² is an integer. So for all integer x, the equation holds. That means all integers from 1 to n are solutions. So there are n integer solutions. But we also need to consider non-integer solutions, right? Because the problem says "the number of solutions" without specifying they have to be integers. So x can be any real number between 1 and n, including non-integers. So we need to find all real numbers x in [1, n] that satisfy the equation, and then count them. But wait, how do we count solutions when there might be infinitely many? But the problem says "the number of solutions", so maybe it's countable? Wait, the equation may have solutions in intervals or something. Let me think again.Wait, no, the equation is for x in [1, n]. Since x is a real variable, unless restricted, there could be infinitely many solutions. But the problem is given a natural number n, determine the number of solutions. So perhaps the number is finite? Maybe the solutions are only integers? Wait, but let's check with a specific example.Suppose x is not an integer. Let x = k + f, where k is an integer (k >=1 since x >=1) and 0 < f <1. Then, {x} = f, so the right-hand side is f². The left-hand side is {x²} = x² - [x²]. Let's compute x²:x² = (k + f)² = k² + 2k f + f². Therefore, [x²] = k² + [2k f + f²]. Since 0 < f <1, 2k f + f² is between 0 and 2k + 1. But 2k f + f² can be written as f(2k + f). Since 0 < f <1, then 2k + f < 2k +1. So [x²] = k² + m, where m is the integer part of 2k f + f². Therefore, the fractional part {x²} = x² - [x²] = (k² + 2k f + f²) - (k² + m) = 2k f + f² - m. So the equation becomes:2k f + f² - m = f² ⇒ 2k f - m = 0 ⇒ m = 2k f.But m is an integer, since it's the floor of 2k f + f². Wait, m is the integer part of 2k f + f², which is equal to floor(2k f + f²). But since m = 2k f, we have that 2k f must be an integer? Because m is integer, so if m = 2k f, then 2k f must be integer. But f is between 0 and 1, so 2k f is between 0 and 2k. So 2k f is integer. Let me write m = 2k f, where m is integer and 0 ≤ m < 2k +1 (since 2k f + f² < 2k*1 +1² = 2k +1). But m is the floor of 2k f + f². Wait, but we set m = floor(2k f + f²). But according to the equation, m = 2k f. So 2k f must be an integer, let's denote m = 2k f. Then, since m is integer, f = m/(2k). But f must be in (0,1), so m must be an integer with 1 ≤ m ≤ 2k -1 (since m = 0 would make f = 0, which is excluded, and m = 2k would make f =1, which is also excluded). Wait, but m is floor(2k f + f²), but in our case, m = 2k f. Let's check if this holds.Suppose m = 2k f. Then 2k f must be integer. So f = m/(2k), where m is integer from 1 to 2k -1. Then, substituting back into the expression for floor(2k f + f²):2k f + f² = m + (m/(2k))². Since m is integer, the integer part of this is m + floor((m/(2k))²). But (m/(2k))² is less than 1, because m < 2k, so (m/(2k))² < (2k/(2k))² =1. Therefore, floor(2k f + f²) = m + 0 = m. Therefore, our equation reduces to m = 2k f, which is satisfied. Therefore, for each integer m from 1 to 2k -1, f = m/(2k) will satisfy the equation, provided that when we compute 2k f + f², its integer part is m. Wait, but we just saw that with f = m/(2k), 2k f = m, so 2k f + f² = m + (m/(2k))². Since (m/(2k))² is positive but less than 1, the floor of that is m. Therefore, floor(2k f + f²) = m. Therefore, yes, for each m from 1 to 2k -1, f = m/(2k) gives a solution.But wait, let me check with a specific example. Let's take k =1. Then x is in [1,2), so k =1. Then m can be from 1 to 2*1 -1 =1. So m=1. Then f=1/(2*1)=0.5. Therefore, x =1 +0.5=1.5. Let's check if this is a solution.Left-hand side: x²=2.25, so fractional part is 0.25. Right-hand side: fractional part of x is 0.5, square is 0.25. So yes, 0.25=0.25. So x=1.5 is a solution. So in the interval [1,2), there is one non-integer solution at x=1.5.Similarly, let's check k=2. Then m can be from 1 to 3 (since 2k -1=3). So m=1,2,3. Then f=1/4, 2/4=1/2, 3/4. Let's check x=2 +1/4=2.25.Left-hand side: x²=5.0625, fractional part is 0.0625. Right-hand side: fractional part of x is 0.25, square is 0.0625. So yes, equal. Similarly for m=2: x=2.5. x²=6.25, fractional part 0.25. Fractional part of x is 0.5, square 0.25. Equal. For m=3: x=2.75. x²=7.5625, fractional part 0.5625. Fractional part of x is 0.75, square is 0.5625. Equal. So in [2,3), there are three non-integer solutions at 2.25, 2.5, 2.75.Wait, so for each integer k, in the interval [k, k+1), there are (2k -1) non-integer solutions? Wait, when k=1, we had 1 solution, which is 2k -1=1. For k=2, 3 solutions, which is 2k -1=3. So yes, it seems that in each interval [k, k+1), there are 2k -1 non-integer solutions. Therefore, for each integer k from 1 to n-1, the interval [k, k+1) has 2k -1 non-integer solutions, and in the interval [n, n+1), but since x is up to n, we only consider up to x=n. Wait, x is in [1, n], so if n is an integer, then the last interval is [n, n], which is just the integer n. So for k from 1 to n-1, each interval [k, k+1) contributes 2k -1 non-integer solutions plus the integer solution at k+1. Wait, no, the integer solutions are at the integers themselves. Wait, actually, in the interval [k, k+1), the integer k is included, but x is from 1 to n. Wait, maybe I need to separate the analysis.Wait, the total solutions would be:- All integer solutions from 1 to n: that's n solutions.- Plus all non-integer solutions in each interval [k, k+1) for k from 1 to n-1. Because the interval [n, n+1) is beyond n, except for x=n, which is already counted as an integer solution.So, for each k from 1 to n-1, in the interval [k, k+1), there are 2k -1 non-integer solutions. Therefore, total non-integer solutions would be sum_{k=1}^{n-1} (2k -1). Then total solutions would be n + sum_{k=1}^{n-1} (2k -1).Wait, let's compute the sum sum_{k=1}^{n-1} (2k -1). That's equal to 2*sum_{k=1}^{n-1}k - sum_{k=1}^{n-1}1 = 2*( (n-1)n)/2 ) - (n-1) = (n-1)n - (n-1) = (n-1)(n -1) = (n-1)^2.Therefore, total solutions would be n + (n-1)^2 = n + n^2 -2n +1 = n^2 -n +1.Wait, let me check with n=1. If n=1, then the interval is [1,1], which only includes x=1, which is an integer, so solutions=1. According to the formula n^2 -n +1, when n=1, 1 -1 +1=1. Correct.For n=2, total solutions would be 4 -2 +1=3. Let's check manually. For x between 1 and 2, we have integer solutions at 1 and 2. But wait, in the interval [1,2), the integer 1 is included, but x=2 is in the interval [2,3). Wait, maybe my initial analysis was slightly off.Wait, when we have k from 1 to n-1, each interval [k, k+1) has 2k -1 non-integer solutions. For n=2, k=1, so non-integer solutions in [1,2) are 1. So total solutions would be integer solutions (1 and 2) plus non-integer solution (1.5). So total 3 solutions. According to the formula, n=2: 4 -2 +1=3. Correct.For n=3, formula gives 9 -3 +1=7. Let's check. Integer solutions at 1,2,3. Non-integer solutions:In [1,2): 1.5In [2,3): 2.25, 2.5, 2.75Total non-integer solutions: 1 +3=4. Total solutions:3 +4=7. Correct. So the formula seems to work.Wait, but how did we get sum_{k=1}^{n-1} (2k -1) = (n-1)^2. Let's verify:For n=2: sum_{k=1}^1 (2*1 -1)=1, which is (2-1)^2=1. Correct.For n=3: sum_{k=1}^2 (2k -1)= (2*1 -1)+(2*2 -1)=1 +3=4=(3-1)^2=4. Correct.Yes, the formula holds. Therefore, total solutions are n + (n-1)^2 = n^2 -n +1.Therefore, the answer should be n² -n +1. Wait, but let me check with another example. Take n=4. According to formula, 16 -4 +1=13.Integer solutions:1,2,3,4.Non-integer solutions:[1,2):1.5[2,3):2.25,2.5,2.75[3,4):3.25,3.5,3.75,3.9375? Wait, no. Wait, for k=3, in [3,4), the non-integer solutions are at x=3 + m/(2*3)=3 + m/6, where m=1,2,3,4,5. Wait, hold on. Wait, earlier for k=1, solutions were at 1 +1/2=1.5.For k=2, solutions at 2 +1/4,2 +2/4,2 +3/4=2.25,2.5,2.75.Wait, but according to earlier logic, for each k, m ranges from 1 to 2k -1, so for k=3, m=1 to 5, since 2*3 -1=5. Therefore, f=1/6,2/6,3/6,4/6,5/6. So x=3 +1/6≈3.1667, 3 +1/3≈3.3333, 3.5, 3 +2/3≈3.6667, 3 +5/6≈3.8333.Wait, but when k=3, m=1 to 5. Then x=3 + m/(2*3)=3 +m/6. So that's 6 solutions? Wait, no, m is from 1 to 5, so five solutions. So in [3,4), there are 5 non-integer solutions.But according to the formula sum_{k=1}^{n-1} (2k -1) when n=4, sum is sum_{k=1}^3 (2k -1)=1 +3 +5=9. Then total solutions would be 4 +9=13. But according to actual count:Integer solutions:4.Non-integer solutions: [1,2):1, [2,3):3, [3,4):5. Total non-integer solutions:1+3+5=9. Total solutions:4+9=13. Correct. So yes, for k=3, 5 non-integer solutions.Wait, but in the problem, the equation is for x in [1,n]. So when n is an integer, the interval [n, n+1) is not included, except x=n. So if n=4, the interval [4,5) is not included. So the formula works.Therefore, the total number of solutions is n^2 -n +1.But let me confirm once more with n=1,2,3,4.n=1:1 solution (only x=1). 1² -1 +1=1. Correct.n=2:3 solutions. 2² -2 +1=3. Correct.n=3:7 solutions. 9 -3 +1=7. Correct.n=4:13 solutions. 16 -4 +1=13. Correct.Therefore, the formula holds. Hence, the answer is n² -n +1.But let me check the case when x is an integer. For example, x=2. Then fractional part of x is 0, so RHS=0. LHS=x² - floor(x²)=4 -4=0. So 0=0. Correct. So integers are indeed solutions.For non-integer x, like x=1.5. As checked earlier, works. x=2.25: x=2.25, fractional part 0.25. (0.25)^2=0.0625. x²=5.0625, fractional part 0.0625. Equal. Correct.x=3.5: fractional part 0.5. Square is 0.25. x²=12.25, fractional part 0.25. Equal. Correct.x=3.1667 (≈3 +1/6): fractional part≈0.1667. Square≈0.0278. x²≈10.0278, fractional part≈0.0278. Correct.So all these work. Therefore, the formula seems solid.Thus, the number of solutions is n squared minus n plus one, which is boxed{n^2 - n + 1}.Final AnswerThe number of solutions is boxed{n^2 - n + 1}.</think>Given a natural number ( n ), we need to determine the number of solutions to the equation[x^{2} - leftlfloor x^{2} rightrfloor = (x - leftlfloor x rightrfloor)^{2}]for which ( 1 leq x leq n ).First, we recognize that the equation involves the fractional parts of ( x ) and ( x^2 ). Let ( { x } ) denote the fractional part of ( x ). The equation simplifies to:[{ x^2 } = ( { x } )^2]For integer values of ( x ), both sides of the equation are zero, so all integers ( x ) in the range ( [1, n] ) are solutions. There are ( n ) such integer solutions.Next, we consider non-integer solutions. Let ( x = k + f ) where ( k ) is an integer and ( 0 < f < 1 ). The equation becomes:[{ (k + f)^2 } = f^2]Expanding ( (k + f)^2 ) gives:[k^2 + 2k f + f^2]The fractional part of this expression is ( 2k f + f^2 - leftlfloor 2k f + f^2 rightrfloor ). For this to equal ( f^2 ), we need:[2k f = leftlfloor 2k f + f^2 rightrfloor]This implies ( 2k f ) must be an integer. Let ( m = 2k f ), where ( m ) is an integer. Then ( f = frac{m}{2k} ) with ( 1 leq m leq 2k - 1 ). This gives ( 2k - 1 ) non-integer solutions in each interval ( [k, k+1) ).For each integer ( k ) from 1 to ( n-1 ), there are ( 2k - 1 ) non-integer solutions. Summing these solutions:[sum_{k=1}^{n-1} (2k - 1) = (n-1)^2]Adding the ( n ) integer solutions, the total number of solutions is:[n + (n-1)^2 = n^2 - n + 1]Thus, the number of solutions is (boxed{n^2 - n + 1}).,<think>Okay, so I need to figure out how many solutions there are to the equation ( x^{2} - left[x^{2}right] = (x - [x])^{2} ) where ( x ) is between 1 and ( n ), inclusive. Hmm, let's start by understanding what this equation means. First, the notation [x] is the floor function, right? So [x] is the greatest integer less than or equal to x. That means ( x - [x] ) is the fractional part of x, often written as ( {x} ). So the right side of the equation is just the square of the fractional part of x. The left side is ( x^{2} - left[x^{2}right] ), which is the fractional part of ( x^{2} ). So the equation simplifies to ( {x^{2}} = ({x})^{2} ). That's an interesting equation. So we need all x in [1, n] such that the fractional part of x squared is equal to the square of the fractional part of x.Let me write x as ( k + f ), where ( k = [x] ) (so k is an integer) and ( f = {x} ) (so ( 0 leq f < 1 )). Then x = k + f, where k is an integer from 1 to n-1 (since x is at least 1 and less than n+1, but x can be up to n. Wait, actually, when x is in [1, n], k can be from 1 up to n-1, and when x is exactly an integer, like n, then f=0. So maybe k ranges from 1 to n, but if x is exactly n, then f=0. Hmm, need to be careful with the upper bound.But let's proceed. If x = k + f, then ( x^2 = (k + f)^2 = k^2 + 2kf + f^2 ). Therefore, the integer part of x^2 is [x^2] = k^2 + [2kf + f^2], because k^2 is integer and 2kf + f^2 is the remaining part. Then the fractional part of x^2 is ( {x^2} = 2kf + f^2 - [2kf + f^2] ).But according to the equation, this fractional part must equal ( ({x})^2 = f^2 ). Therefore, we have:( 2kf + f^2 - [2kf + f^2] = f^2 )Simplify this equation. Subtract ( f^2 ) from both sides:( 2kf - [2kf + f^2] = 0 )Which implies:( [2kf + f^2] = 2kf )But since the floor of ( 2kf + f^2 ) is equal to ( 2kf ), this means that ( 2kf + f^2 ) must be an integer. Wait, no. Wait, the floor function of a number is equal to that number minus its fractional part. So, [2kf + f^2] = 2kf + f^2 - {2kf + f^2}. But according to the equation, [2kf + f^2] = 2kf. Therefore,2kf + f^2 - {2kf + f^2} = 2kfSubtract 2kf from both sides:f^2 - {2kf + f^2} = 0Which implies:{2kf + f^2} = f^2So the fractional part of ( 2kf + f^2 ) is equal to ( f^2 ). Therefore, the equation simplifies to:( 2kf + f^2 = m + f^2 ), where m is an integer such that ( m leq 2kf + f^2 < m + 1 )Subtracting ( f^2 ) from both sides:( 2kf = m )Therefore, ( m = 2kf ). But m must be an integer, so 2kf is an integer. Therefore, 2kf is integer. Since k is a positive integer (from 1 to n-1 or n, depending on x) and f is in [0,1), we have that 2kf must be an integer.But f is in [0,1), so 2kf is in [0, 2k). Since m = 2kf must be an integer, m can be 0, 1, 2, ..., 2k - 1. However, since f < 1, 2kf < 2k. So m can be 0, 1, ..., 2k - 1. But m = 2kf, which gives f = m/(2k). But f must also satisfy that when you compute 2kf + f^2, its fractional part is f^2. Wait, but we already derived that m must be 2kf. Let's check if this is sufficient.If f = m/(2k), then substitute back into the original equation. Let's check the fractional part of 2kf + f^2. Since f = m/(2k), then:2kf + f^2 = 2k*(m/(2k)) + (m/(2k))^2 = m + m^2/(4k^2)So the integer part of this is m, and the fractional part is m^2/(4k^2). But according to the equation, the fractional part must be f^2 = (m/(2k))^2 = m^2/(4k^2). Therefore, this holds true for any integer m. Therefore, if f = m/(2k), where m is an integer such that 0 ≤ m < 2k (since f < 1 implies m/(2k) < 1 ⇒ m < 2k), then the equation holds. Therefore, for each integer k (from 1 to n), the fractional part f must be of the form m/(2k), where m is an integer from 0 to 2k - 1. But since x is in [1, n], k can be from 1 to n - 1 if x is not an integer. Wait, but x can be an integer. Let's see.Wait, x is in [1, n]. So x can be any real number between 1 and n. So k, the integer part [x], can be from 1 up to n. When k = n, then x is in [n, n+1), but since x ≤ n, when k = n, x must be exactly n, so f = 0. Let's check if x = n is a solution.When x = n, which is an integer, then {x} = 0, so the right-hand side is 0. The left-hand side is x² - [x²] = n² - [n²] = 0. So yes, x = n is a solution. Similarly, any integer x in [1, n] would satisfy the equation because both sides would be zero. So integers are solutions.But in our analysis above, when f = 0, which corresponds to x being integer, m must be 0. So m = 0 gives f = 0, which is valid. Therefore, integers are included in the solutions where f = m/(2k) with m = 0.So for each k from 1 to n, x = k + f, where f = m/(2k) and m is an integer from 0 to 2k - 1. However, we need to ensure that x is within [1, n]. But when k = n, f must be 0, as x cannot exceed n. For k < n, f can be up to (2k - 1)/(2k) which is less than 1. So x = k + m/(2k) where k is from 1 to n and m is from 0 to 2k - 1. But when k = n, m can only be 0, since x = n + m/(2n) must be ≤ n. So for k = n, m must be 0. For k from 1 to n - 1, m can be from 0 to 2k - 1. Therefore, the number of solutions would be the sum over each k from 1 to n of the number of valid m's.Wait, but x has to be in [1, n]. For each k from 1 to n - 1, x = k + m/(2k) where m is 0, 1, ..., 2k - 1. Then x ranges from k to k + (2k - 1)/(2k) = k + 1 - 1/(2k) < k + 1. So x is in [k, k + 1 - 1/(2k)), which is within [1, n) since k ≤ n - 1. For k = n, m can only be 0, so x = n + 0 = n, which is within [1, n].Therefore, the total number of solutions is for each k from 1 to n, the number of m's is 2k when k < n, and 1 when k = n. Wait, no. Wait, for k from 1 to n - 1, m can be from 0 to 2k - 1, which is 2k values. For k = n, m can only be 0, so 1 value. So total solutions would be sum_{k=1}^{n-1} 2k + 1. But wait, is that correct?Wait, but each x is a unique solution. For each k from 1 to n - 1, there are 2k values of m (from 0 to 2k - 1), each corresponding to a different x. For k = n, only x = n is a solution. Therefore, total solutions would be sum_{k=1}^{n-1} 2k + 1.But sum_{k=1}^{n-1} 2k is 2 * sum_{k=1}^{n-1} k = 2 * (n-1)n/2 = (n-1)n. Then adding 1 gives (n-1)n + 1 = n² - n + 1. But wait, let me verify.Wait, for k = 1: m from 0 to 1 (2 values)k = 2: m from 0 to 3 (4 values)...k = n-1: m from 0 to 2(n-1)-1 = 2n - 3 (2n - 2 values)Therefore, for each k from 1 to n-1, the number of m's is 2k. So the sum is sum_{k=1}^{n-1} 2k = 2 * sum_{k=1}^{n-1} k = 2*(n-1)n/2 = n(n - 1). Then for k = n, m = 0, so 1 solution. Therefore, total solutions is n(n - 1) + 1 = n² - n + 1. But let's test this with small n. Let's take n = 1. Then the interval is [1, 1]. So x = 1. Check if it's a solution. Left-hand side: 1² - [1²] = 1 - 1 = 0. Right-hand side: (1 - [1])² = 0² = 0. So x = 1 is a solution. According to the formula, n² - n + 1 = 1 - 1 + 1 = 1, which matches.n = 2. Let's compute manually.x in [1, 2]. Let's check integer x = 1 and 2. Both will satisfy the equation as before. Now non-integer x between 1 and 2. Let x = 1 + f, where f is in [0,1). The equation is {x²} = f².x² = (1 + f)^2 = 1 + 2f + f². So fractional part is 2f + f² - [2f + f²]. According to the equation, this must equal f². Therefore, 2f + f² - [2f + f²] = f² ⇒ [2f + f²] = 2f. So 2f must be integer. Since f ∈ [0,1), 2f ∈ [0,2). So [2f + f²] = 2f ⇒ 2f is integer. Therefore, 2f is integer, so f = m/2 where m = 0,1. But f ∈ [0,1), so m = 0 or 1. But m =1 gives f = 0.5. Thus, possible f are 0 and 0.5. However, f = 0 corresponds to x =1, which is already counted as integer. So non-integer solutions are f = 0.5, x = 1.5.Check x =1.5: left-hand side is (1.5)^2 - [2.25] = 2.25 - 2 = 0.25. Right-hand side: (1.5 -1)^2 = 0.5^2 = 0.25. So yes, solution. Therefore, in n=2, solutions are x=1, x=1.5, x=2. Total 3 solutions. According to formula, n² -n +1 =4 -2 +1=3. Correct.Another test: n=3. According to formula, 9 -3 +1=7 solutions.Let's check:Integer solutions: x=1,2,3. That's 3 solutions.Non-integer solutions: For k=1 (x between 1 and 2): f = m/2, m=0,1. But m=0 gives x=1, which is integer. So m=1 gives x=1.5. Already counted in n=2 case. Wait, for n=3, we need to check x in [1,3]. So k=1,2.For k=1: x=1 + m/2, m=0,1. So x=1, 1.5. But x=1 is integer. So non-integer x=1.5. But x=1.5 is within [1,2], so for n=3, it's still valid.For k=2: x=2 + m/(4), since k=2, m=0,1,2,3. So m from 0 to 3. So x=2, 2.25, 2.5, 2.75. x=2 is integer. The others: 2.25, 2.5, 2.75. Check if these are solutions.Take x=2.25. {x²} = {5.0625} =0.0625. Right-hand side: (0.25)^2=0.0625. Equal. So yes.x=2.5: {6.25}=0.25. Right-hand side: (0.5)^2=0.25. Equal.x=2.75: {7.5625}=0.5625. Right-hand side: (0.75)^2=0.5625. Equal. So these three are solutions.Thus, for k=1: 1 non-integer solution (1.5). For k=2: 3 non-integer solutions (2.25, 2.5, 2.75). For k=3: x=3, which is integer. So total non-integer solutions: 1+3=4. Plus integer solutions 3: total 7. Which matches formula. So formula seems to hold.Therefore, the total number of solutions is n² - n +1. Wait, but let's check for n=3. The formula gives 9 -3 +1=7. Which matches. For n=2, 4 -2 +1=3. For n=1, 1 -1 +1=1. So seems correct.But wait, let me check for k=1 in n=3. The solutions for k=1 are x=1 and x=1.5, but x=1.5 is already included in the k=1, m=1 case. Then when k=2, m=0,1,2,3 gives x=2,2.25,2.5,2.75. So when n=3, for k=1, m=0 and 1 (but x=1.5 is allowed as x=1.5 ≤3). Wait, but in the formula earlier, for each k from 1 to n-1, we have 2k solutions. Wait, for n=3, n-1=2. So sum_{k=1}^{2} 2k = 2 +4=6. Then add 1 for k=3 (x=3), total 7. So that's consistent.But here, when k=1, the x's are 1, 1.5 (2 solutions). But 1 is included in k=1, m=0. Similarly, for k=2, m=0,1,2,3: x=2, 2.25, 2.5, 2.75 (4 solutions). So 2k solutions per k. So in general, for each k from 1 to n-1, there are 2k solutions (including the integer solution x=k). But wait, when m=0, x=k, which is integer, so these are overlapping with the integer solutions. However, in our count above, when we sum over all k from 1 to n-1 the 2k solutions, and then add 1 for x=n, we might be overcounting the integer solutions. Wait, no. Wait, in the formula, for each k from 1 to n-1, we have 2k solutions (including x=k as m=0). Then for k=n, m=0 gives x=n. But if we sum for k=1 to n-1, each k contributes 2k solutions (including x=k), and k=n contributes 1 solution (x=n). However, the integer solutions x=1,2,...,n are each counted once in the k=1 to n-1, m=0 cases, and x=n is counted once. Therefore, total integer solutions are n (from x=1 to x=n), but according to our formula, sum_{k=1}^{n-1} 2k +1 = n(n-1) +1. Wait, but n(n-1) +1 is not equal to n. For example, n=3, 3*2 +1=7, which includes 3 integers and 4 non-integers. So the formula counts both integer and non-integer solutions. So that's correct because in each k, the solutions are x=k + m/(2k), where m=0 gives x=k (integer) and m=1,...,2k-1 gives non-integers. Therefore, for each k, 2k solutions: 1 integer and 2k -1 non-integers. Therefore, total integer solutions are n (from k=1 to n, but k=n only contributes 1 solution, x=n). Wait, for k from 1 to n-1, each k has 2k solutions, of which 1 is integer (x=k) and 2k -1 are non-integers. Then for k=n, only x=n is a solution. Therefore, total integer solutions are (n-1) +1 =n, which is correct. Total non-integer solutions would be sum_{k=1}^{n-1} (2k -1). Sum_{k=1}^{n-1} 2k - sum_{k=1}^{n-1}1 = n(n-1) - (n-1) = (n-1)(n -1) = (n-1)^2. Then total solutions are n + (n-1)^2 = n + n² - 2n +1 = n² -n +1. Which matches the formula. Therefore, that formula is correct.Therefore, the number of solutions is n² -n +1. But wait, let me check once more for n=2. n² -n +1=4-2+1=3. Which is correct: 1,1.5,2. Similarly for n=3, 9-3+1=7, which was confirmed. So yes, the formula seems to hold.Therefore, the answer is boxed{n^2 - n + 1}.Final AnswerThe number of solutions is boxed{n^2 - n + 1}.</think>Given a natural number ( n ), we need to determine the number of solutions to the equation[x^{2} - leftlfloor x^{2} rightrfloor = left( x - leftlfloor x rightrfloor right)^{2}]for which ( 1 leq x leq n ).First, we recognize that the equation can be rewritten in terms of the fractional part of ( x ). Let ( x = k + f ), where ( k = lfloor x rfloor ) (an integer) and ( f = { x } ) (the fractional part of ( x ), such that ( 0 leq f < 1 )). The equation then becomes:[{ x^{2} } = f^{2}]Expanding ( x^2 ) as ( (k + f)^2 = k^2 + 2kf + f^2 ), the fractional part of ( x^2 ) is ( 2kf + f^2 - lfloor 2kf + f^2 rfloor ). For this to equal ( f^2 ), we must have:[2kf + f^2 - lfloor 2kf + f^2 rfloor = f^2]This simplifies to:[lfloor 2kf + f^2 rfloor = 2kf]This implies ( 2kf ) must be an integer. Let ( m = 2kf ), where ( m ) is an integer. Then ( f = frac{m}{2k} ) with ( 0 leq m < 2k ). For each integer ( k ) from 1 to ( n ), the values of ( m ) range from 0 to ( 2k - 1 ). However, for ( k = n ), ( x ) must be exactly ( n ), so ( m = 0 ).The total number of solutions is the sum of the valid ( m ) values for each ( k ):- For ( k ) from 1 to ( n-1 ), there are ( 2k ) solutions.- For ( k = n ), there is 1 solution.Summing these contributions:[sum_{k=1}^{n-1} 2k + 1 = 2 sum_{k=1}^{n-1} k + 1 = 2 cdot frac{(n-1)n}{2} + 1 = (n-1)n + 1 = n^2 - n + 1]Thus, the number of solutions is (boxed{n^2 - n + 1}).]

🔑:Solution. If we setx=m+alpha, m in{1,2, ldots, n-1}, 0 leq alpha<1the given equation becomesm^{2}+2 m alpha=left[m^{2}+2 m alpha+alpha^{2}right]According to this, the number m+alpha is a solution to the given equation if and only if 2 m alpha is an integer, i.e., if and only ifalpha inleft{0, frac{1}{2 m}, frac{2}{2 m}, ldots, frac{2 m-1}{2 m}right}which means that in the interval [m cdot m+1) the given equation has 2 m solutions. Since x=n is also a solution to the equation, we conclude that the number of solutions to this equation is1+2(1+2+ldots+(n-1))=1+n(n-1)=n^{2}-n+1

🧠:[<think>Okay, let's tackle this hyperbola problem step by step. First, let's parse the given information and figure out what we need to find.We have a hyperbola C defined by the equation x²/a² - y²/2 = 1, where a > 0. The foci of this hyperbola are F₁ and F₂, with F₁ being the left focus and F₂ the right focus. Then, there's a line l that passes through F₁ and is perpendicular to the x-axis. Since it's perpendicular to the x-axis, that means line l is vertical. This line intersects the hyperbola at points A and B. We're told that the area of triangle ABF₂ is 2√6, and we need to find the equation of the asymptotes of the hyperbola from the given options.First, let's recall some basic properties of hyperbolas. The standard form of a hyperbola centered at the origin opening horizontally is x²/a² - y²/b² = 1. Here, the given equation is x²/a² - y²/2 = 1, so we can see that b² = 2, hence b = √2.For hyperbolas, the distance to the foci from the center is c, where c² = a² + b². So, in this case, c² = a² + 2. Therefore, the foci are at (±c, 0), which are F₁ = (-c, 0) and F₂ = (c, 0).Next, we have a vertical line passing through F₁, which is at (-c, 0). A vertical line passing through this point has the equation x = -c. This line intersects the hyperbola at points A and B. Let's find the coordinates of points A and B by substituting x = -c into the hyperbola equation.Substituting x = -c into x²/a² - y²/2 = 1:(-c)² / a² - y² / 2 = 1c² / a² - y² / 2 = 1But we know that c² = a² + 2, so substituting that in:(a² + 2)/a² - y²/2 = 11 + 2/a² - y²/2 = 1Subtracting 1 from both sides:2/a² - y²/2 = 0So:2/a² = y²/2Multiplying both sides by 2:4/a² = y²Therefore, y = ±2/aSo the points A and B are at (-c, 2/a) and (-c, -2/a). Therefore, the coordinates are:A: (-c, 2/a)B: (-c, -2/a)Now, we need to find the area of triangle ABF₂, where F₂ is at (c, 0).First, let's recall that the area of a triangle given three vertices can be calculated using the determinant formula:Area = (1/2) |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|Alternatively, since AB is a vertical line segment (since both A and B have the same x-coordinate, -c), and F₂ is at (c, 0), maybe we can compute the base and height of the triangle.The length of AB is the distance between points A and B. Since they have the same x-coordinate, the distance is just the difference in y-coordinates:AB = |2/a - (-2/a)| = |4/a| = 4/aThe base of the triangle can be considered as AB, which is 4/a. The height of the triangle would be the horizontal distance from the line AB to the point F₂. Since AB is at x = -c and F₂ is at x = c, the horizontal distance is |c - (-c)| = 2c.But wait, in triangle area terms, if we take AB as the base, then the height would be the horizontal distance from F₂ to the line AB. However, since AB is vertical, the height isn't the horizontal distance; actually, the area can be calculated as (1/2) * base * height, where the base is AB and the height is the horizontal distance between the line AB and the point F₂. Wait, is that correct?Wait, maybe a better approach is to compute the area using coordinates. Let's list the coordinates of the three points:A: (-c, 2/a)B: (-c, -2/a)F₂: (c, 0)So, plotting these points, AB is a vertical line segment at x = -c, from (-c, 2/a) to (-c, -2/a). F₂ is at (c, 0). So the triangle ABF₂ is a triangle with vertices at these three points.To compute the area, perhaps we can use the formula for the area of a triangle with coordinates. Let's use the shoelace formula.Coordinates:A: (-c, 2/a)B: (-c, -2/a)F₂: (c, 0)Shoelace formula:Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|Plugging in:x₁ = -c, y₁ = 2/ax₂ = -c, y₂ = -2/ax₃ = c, y₃ = 0Compute each term:First term: x₁(y₂ - y₃) = (-c)(-2/a - 0) = (-c)(-2/a) = 2c/aSecond term: x₂(y₃ - y₁) = (-c)(0 - 2/a) = (-c)(-2/a) = 2c/aThird term: x₃(y₁ - y₂) = c(2/a - (-2/a)) = c(4/a) = 4c/aSum these terms:2c/a + 2c/a + 4c/a = 8c/aTake absolute value and multiply by 1/2:Area = (1/2)|8c/a| = (1/2)(8c/a) = 4c/aWe are told that the area is 2√6. Therefore:4c/a = 2√6Divide both sides by 2:2c/a = √6So:c/a = √6 / 2Therefore:c = (a√6)/2But we also know from the hyperbola that c² = a² + b², and since the hyperbola equation is x²/a² - y²/2 = 1, we have b² = 2. So:c² = a² + 2But we also have c = (a√6)/2. Let's square both sides:c² = (a² * 6)/4 = (6a²)/4 = (3a²)/2Therefore, set equal:(3a²)/2 = a² + 2Multiply both sides by 2 to eliminate the denominator:3a² = 2a² + 4Subtract 2a² from both sides:a² = 4Therefore, a = 2 (since a > 0)Now that we have a = 2, we can find the asymptotes of the hyperbola. The asymptotes of a hyperbola given by x²/a² - y²/b² = 1 are the lines y = ±(b/a)x.From the given equation, b² = 2, so b = √2. We found a = 2, so:Slope of asymptotes = ±(√2)/2Therefore, the equations of the asymptotes are:y = ±(√2/2)xLooking at the options provided:A: y=±√3xB: y=±(√3)/3xC: y=±√2xD: y=±(√2)/2xSo the correct answer should be option D.Wait, but let me double-check my steps to make sure I didn't make a mistake.First, foci at (-c, 0) and (c, 0). The line x = -c intersects the hyperbola at y = ±2/a. So points A and B are (-c, 2/a) and (-c, -2/a). Then, the area of triangle ABF₂ was calculated using the shoelace formula as 4c/a, set equal to 2√6, leading to 4c/a = 2√6 ⇒ 2c/a = √6 ⇒ c/a = √6/2.Then, using c² = a² + 2, and c = (a√6)/2, so c² = (6a²)/4 = (3a²)/2. Therefore, (3a²)/2 = a² + 2 ⇒ 3a² = 2a² + 4 ⇒ a² = 4 ⇒ a=2. Then, asymptotes are y = ±(b/a)x = ±(√2/2)x. So yes, that's option D.But let me check once again if the area calculation is correct.Alternatively, the base AB is vertical with length 4/a, and the horizontal distance from AB to F₂ is the horizontal distance between x = -c and x = c, which is 2c. But in the formula for the area of a triangle, if we take AB as the base, which is vertical, then the height would be the horizontal distance from F₂ to the line AB. However, when dealing with a vertical base, the height is indeed the horizontal distance from the third point to the line AB. Since AB is vertical at x = -c, the horizontal distance from F₂ (which is at x = c) to AB is |c - (-c)| = 2c. Therefore, area should be (1/2)*base*height = (1/2)*(4/a)*(2c) = (4c)/a. Which matches the earlier calculation. So that's correct. Therefore, 4c/a = 2√6, leading to c/a = √6/2. Then, c = (a√6)/2. Then c² = (6a²)/4 = (3a²)/2. But c² is also a² + 2, so 3a²/2 = a² + 2 ⇒ solving gives a²=4, so a=2, correct. Then asymptotes y=±(√2/2)x, which is option D. So answer D.I think that's solid. Let me check if the options are correctly labeled. The problem says:A: y=±√3xB: y=±(√3)/3xC: y=±√2xD: y=±(√2)/2xYes, so D is the correct answer. Therefore, the answer is D.Final Answerboxed{D}</think>Given the hyperbola ( C: frac{x^2}{a^2} - frac{y^2}{2} = 1 ) with foci ( F_1 ) and ( F_2 ), we need to find the equation of the asymptotes given that the area of triangle ( triangle ABF_2 ) is ( 2sqrt{6} ).1. Identify the foci: For the hyperbola ( frac{x^2}{a^2} - frac{y^2}{2} = 1 ), we have ( b^2 = 2 ). The distance to the foci ( c ) is given by ( c^2 = a^2 + b^2 ). Thus, ( c = sqrt{a^2 + 2} ). The foci are ( F_1 = (-c, 0) ) and ( F_2 = (c, 0) ).2. Find points A and B: The vertical line through ( F_1 ) is ( x = -c ). Substituting ( x = -c ) into the hyperbola equation: [ frac{(-c)^2}{a^2} - frac{y^2}{2} = 1 implies frac{c^2}{a^2} - frac{y^2}{2} = 1 ] Since ( c^2 = a^2 + 2 ), we get: [ frac{a^2 + 2}{a^2} - frac{y^2}{2} = 1 implies 1 + frac{2}{a^2} - frac{y^2}{2} = 1 implies frac{2}{a^2} = frac{y^2}{2} implies y = pm frac{2}{a} ] Thus, points ( A ) and ( B ) are ( (-c, frac{2}{a}) ) and ( (-c, -frac{2}{a}) ).3. Calculate the area of ( triangle ABF_2 ): - The length of ( AB ) is ( frac{4}{a} ). - The horizontal distance from ( F_2 ) to the line ( AB ) is ( 2c ). - The area is ( frac{1}{2} times frac{4}{a} times 2c = frac{4c}{a} ). - Given the area is ( 2sqrt{6} ), we solve ( frac{4c}{a} = 2sqrt{6} implies frac{2c}{a} = sqrt{6} implies frac{c}{a} = frac{sqrt{6}}{2} ).4. Relate ( c ) and ( a ): - From ( c = frac{asqrt{6}}{2} ), squaring both sides gives ( c^2 = frac{3a^2}{2} ). - Since ( c^2 = a^2 + 2 ), we equate ( frac{3a^2}{2} = a^2 + 2 implies 3a^2 = 2a^2 + 4 implies a^2 = 4 implies a = 2 ).5. Find the asymptotes: The asymptotes of the hyperbola are ( y = pm frac{b}{a}x ). With ( b^2 = 2 implies b = sqrt{2} ) and ( a = 2 ), the asymptotes are ( y = pm frac{sqrt{2}}{2}x ).Thus, the equation of the asymptotes is (boxed{D}).,<think>Okay, let's see. The problem is about a hyperbola, and we need to find the equation of its asymptotes based on the given conditions. The answer choices are A through D, each with different slopes. Let me start by recalling some basic properties of hyperbolas.The standard form of a hyperbola centered at the origin opening horizontally is (frac{x^2}{a^2} - frac{y^2}{b^2} = 1). The foci are located at ((pm c, 0)), where (c^2 = a^2 + b^2). The asymptotes of such a hyperbola are the lines (y = pm frac{b}{a}x). So, the answer should be in terms of (b/a).Given the hyperbola equation here is (frac{x^2}{a^2} - frac{y^2}{2} = 1). Comparing this with the standard form, (b^2 = 2), so (b = sqrt{2}). Therefore, the asymptotes should be (y = pm frac{sqrt{2}}{a}x). However, the answer choices don't have (a) in them, which means we need to find the value of (a) using the given conditions. Once we know (a), we can determine the slopes of the asymptotes.The problem mentions points (F_1) and (F_2) as the left and right foci. Since the hyperbola is centered at the origin, (F_1) is ((-c, 0)) and (F_2) is ((c, 0)). Remember (c^2 = a^2 + b^2), so (c = sqrt{a^2 + 2}).There's a line (l) passing through (F_1) and perpendicular to the x-axis. A line perpendicular to the x-axis is vertical, so its equation is (x = -c). This line intersects the hyperbola at points (A) and (B). Let's find the coordinates of (A) and (B) by plugging (x = -c) into the hyperbola equation.Substituting (x = -c) into (frac{x^2}{a^2} - frac{y^2}{2} = 1):[frac{(-c)^2}{a^2} - frac{y^2}{2} = 1 implies frac{c^2}{a^2} - frac{y^2}{2} = 1]Since (c^2 = a^2 + 2), substitute that in:[frac{a^2 + 2}{a^2} - frac{y^2}{2} = 1 implies 1 + frac{2}{a^2} - frac{y^2}{2} = 1]Subtracting 1 from both sides:[frac{2}{a^2} - frac{y^2}{2} = 0 implies frac{2}{a^2} = frac{y^2}{2}]Multiply both sides by 2:[frac{4}{a^2} = y^2 implies y = pm frac{2}{a}]Therefore, points (A) and (B) are ((-c, frac{2}{a})) and ((-c, -frac{2}{a})).Now, we need to find the area of triangle (ABF_2). Let's recall the coordinates:- (A = (-c, frac{2}{a}))- (B = (-c, -frac{2}{a}))- (F_2 = (c, 0))To compute the area of triangle (ABF_2), we can use the formula for the area of a triangle given three vertices. The formula is:[text{Area} = frac{1}{2} |x_A(y_B - y_{F_2}) + x_B(y_{F_2} - y_A) + x_{F_2}(y_A - y_B)|]But maybe there's a simpler way. Since points (A) and (B) are vertically aligned (same x-coordinate), the distance between (A) and (B) is vertical, which is (|frac{2}{a} - (-frac{2}{a})| = frac{4}{a}). The base of the triangle can be considered as this vertical segment (AB), and the height would be the horizontal distance from (F_2) to the line (AB). Since line (AB) is at (x = -c), and (F_2) is at (x = c), the horizontal distance is (c - (-c) = 2c). Wait, but in triangle area terms, if we take base as AB and height as the horizontal distance between F2 and the line AB, but actually, the height in the area formula should be the perpendicular distance from the third vertex to the base.Wait, maybe another approach. Let me sketch the scenario mentally. The line AB is vertical at x = -c, and F2 is at (c,0). So triangle ABF2 is a triangle with vertices at (-c, 2/a), (-c, -2/a), and (c,0). To compute the area, perhaps we can use coordinates.Alternatively, since AB is vertical, the base AB has length 4/a (from 2/a to -2/a), and the base is at x = -c. The third point is F2 at (c,0). The area can be calculated as (1/2)*base*height, where height is the horizontal distance between the line AB and the point F2. But actually, since AB is vertical, the height is the horizontal distance between x = -c and x = c, which is 2c. Wait, but in area terms, if we take the base as AB (length 4/a) and the height as the horizontal distance from F2 to AB (which is 2c), but actually, the area would then be (1/2)*base*height = (1/2)*(4/a)*(2c) = (4c)/a.But according to the problem, the area is 2√6. So:(4c)/a = 2√6 ⇒ (2c)/a = √6 ⇒ 2c = a√6 ⇒ c = (a√6)/2.But we know that c^2 = a^2 + 2. Let's substitute c:c = (a√6)/2 ⇒ c^2 = (a^2 * 6)/4 = (3a^2)/2So:(3a^2)/2 = a^2 + 2 ⇒ (3/2)a^2 - a^2 = 2 ⇒ (1/2)a^2 = 2 ⇒ a^2 = 4 ⇒ a = 2 (since a > 0)So, a = 2. Then, b^2 = 2, so b = √2. Therefore, the slopes of the asymptotes are ±b/a = ±√2/2. Looking at the answer choices:A: ±√3xB: ±√3/3 xC: ±√2xD: ±√2/2 xSo, the answer should be D: (y = pm frac{sqrt{2}}{2}x). Let me verify the steps again to make sure.First, foci at (-c,0) and (c,0), with c^2 = a^2 + 2. The vertical line through F1 is x = -c. Plugging into hyperbola gives y^2 = 4/a^2. So points A and B are (-c, 2/a) and (-c, -2/a). Then, triangle ABF2: coordinates are (-c, 2/a), (-c, -2/a), (c,0). Calculating the area.Alternatively, using coordinates, the area can be calculated using the shoelace formula. Let's try that.Coordinates:A: (-c, 2/a)B: (-c, -2/a)F2: (c,0)Shoelace formula:Area = 1/2 |sum of x_i(y_j - y_k)|So,1/2 | (-c)(-2/a - 0) + (-c)(0 - 2/a) + c(2/a - (-2/a)) |Compute each term:First term: (-c)(-2/a - 0) = (-c)(-2/a) = 2c/aSecond term: (-c)(0 - 2/a) = (-c)(-2/a) = 2c/aThird term: c(2/a - (-2/a)) = c(4/a) = 4c/aSum: 2c/a + 2c/a + 4c/a = 8c/aAbsolute value and multiply by 1/2: (1/2)*|8c/a| = 4c/aSo, area is 4c/a. Set equal to 2√6:4c/a = 2√6 ⇒ 2c/a = √6 ⇒ c = (a√6)/2Then, as before, c^2 = (a^2*6)/4 = (3a^2)/2But c^2 = a^2 + b^2 = a^2 + 2Therefore:3a^2/2 = a^2 + 2 ⇒ (3/2)a^2 - a^2 = (1/2)a^2 = 2 ⇒ a^2 = 4 ⇒ a = 2Thus, b^2 = 2, so b = √2Then, the asymptotes are y = ±(b/a)x = ±(√2/2)x, which is option D.Wait, but let me check why the area came out as 4c/a. Let me visualize the triangle. Points A and B are vertically aligned on x = -c, with y-coordinates ±2/a. Point F2 is at (c,0). So, the triangle is symmetric with respect to the x-axis. The base AB is vertical with length 4/a, and the third vertex is at (c,0). The horizontal distance between AB and F2 is 2c (from x = -c to x = c). So, if we consider AB as the base, the height is 2c, so area is (1/2)*base*height = (1/2)*(4/a)*(2c) = (4c)/a. Which matches the shoelace result. So, that's correct.Thus, solving for c in terms of a, and then relating to the hyperbola equation gives a = 2. Then, the asymptotes have slopes ±√2/2, which is option D. So, the answer should be D.Final Answerboxed{D}</think>Given the hyperbola (C: frac{x^2}{a^2} - frac{y^2}{2} = 1) with foci (F_1) and (F_2), we need to find the equation of its asymptotes given that the area of (triangle ABF_2) is (2sqrt{6}). 1. Identify the foci: The foci are located at ((pm c, 0)) where (c^2 = a^2 + 2). Thus, (F_1 = (-c, 0)) and (F_2 = (c, 0)).2. Equation of the vertical line through (F_1): This line is (x = -c).3. Find intersection points (A) and (B): Substitute (x = -c) into the hyperbola equation: [ frac{(-c)^2}{a^2} - frac{y^2}{2} = 1 implies frac{c^2}{a^2} - frac{y^2}{2} = 1 ] Using (c^2 = a^2 + 2): [ frac{a^2 + 2}{a^2} - frac{y^2}{2} = 1 implies 1 + frac{2}{a^2} - frac{y^2}{2} = 1 implies frac{2}{a^2} = frac{y^2}{2} implies y = pm frac{2}{a} ] Thus, points (A) and (B) are ((-c, frac{2}{a})) and ((-c, -frac{2}{a})).4. Calculate the area of (triangle ABF_2): - The base (AB) is vertical with length (frac{4}{a}). - The height from (F_2) to the line (AB) is the horizontal distance (2c). - Area formula: (frac{1}{2} times text{base} times text{height} = frac{1}{2} times frac{4}{a} times 2c = frac{4c}{a}). - Given area is (2sqrt{6}): [ frac{4c}{a} = 2sqrt{6} implies frac{2c}{a} = sqrt{6} implies c = frac{asqrt{6}}{2} ]5. Relate (c) to (a): - (c^2 = frac{3a^2}{2}) and (c^2 = a^2 + 2): [ frac{3a^2}{2} = a^2 + 2 implies frac{a^2}{2} = 2 implies a^2 = 4 implies a = 2 ]6. Determine the asymptotes: - The asymptotes of the hyperbola are (y = pm frac{b}{a}x) where (b^2 = 2). Thus, (b = sqrt{2}). - Slopes: (pm frac{sqrt{2}}{2}).The equation of the asymptotes is (y = pm frac{sqrt{2}}{2}x), which corresponds to option D.[boxed{D}]]

🔑:Let F_{1}(-c,0), A(-c,y_{0}), and c^{2}=a^{2}+2,Then frac {c^{2}}{a^{2}}- frac {y_{0}^{2}}{2}=1, so y_{0}^{2}=2cdot frac {c^{2}-a^{2}}{a^{2}}= frac {4}{a^{2}},Also, the area S_{triangle ABF_{2}}=2 sqrt {6},Which equals frac {1}{2}cdot 2ccdot |2y_{0}|= frac {4c}{a}=2 sqrt {6},Thus, frac {c}{a}= frac { sqrt {6}}{2}, then frac {b}{a}= sqrt { frac {c^{2}}{a^{2}}-1}= frac { sqrt {2}}{2},Therefore, the equation of the asymptote of this hyperbola is y=± frac { sqrt {2}}{2}x.Hence, the correct answer is boxed{D}.By setting F_{1}(-c,0), A(-c,y_{0}), and using the given c^{2}=a^{2}+2, we substitute point A into the equation of the hyperbola to solve for y_{0}. From the area formula of the triangle, we can derive the relationship between a and c, and subsequently between a and b, to find the equation of the asymptote.This problem examines the method of finding the equation of the asymptote of a hyperbola, the equation and application of hyperbolas, and computational skills. It is considered a medium-level question.

🧠:[<think>Okay, so I have this sequence where a₁ is 2, and then each next term is calculated using the formula aₙ₊₁ = (5aₙ - 13)/(3aₙ - 7). The question is asking me to determine the periodicity of the sequence. Hmm, periodicity means that the sequence repeats after a certain number of terms, right? So like, if it's periodic with period 3, then a₁ = a₄ = a₇, and so on. First, maybe I should just compute the first few terms and see if a pattern emerges. Let's start with a₁ = 2. Then, let's find a₂. Plugging into the formula: (5*2 - 13)/(3*2 - 7) = (10 - 13)/(6 - 7) = (-3)/(-1) = 3. So a₂ is 3.Next, a₃ would be (5*3 - 13)/(3*3 - 7) = (15 - 13)/(9 - 7) = 2/2 = 1. So a₃ is 1.Then a₄ is (5*1 - 13)/(3*1 - 7) = (5 - 13)/(3 - 7) = (-8)/(-4) = 2. Oh, wait, that's 2 again! So a₄ is 2, which is the same as a₁. Then a₅ should be the same as a₂, right? Let's check: a₄ is 2, so a₅ = (5*2 -13)/(3*2 -7) = (10 -13)/(6 -7) = (-3)/(-1) = 3. Yep, that's the same as a₂. Then a₆ would be 1 again, and so on. So the sequence is 2, 3, 1, 2, 3, 1, 2, 3, 1,... repeating every three terms. Therefore, the periodicity is 3. But wait, let me make sure I didn't make any calculation errors. Let me recompute a₄ again. a₃ is 1, so plug into the formula: (5*1 -13) is 5 -13 = -8, and denominator is 3*1 -7 = 3 -7 = -4. So -8/-4 = 2. Yep, correct. Then a₄ is 2, same as a₁, so it cycles. Therefore, the period is 3.Alternatively, maybe I can approach this problem algebraically. Let me consider the function f(x) = (5x -13)/(3x -7). If the sequence is periodic, then applying f three times should bring me back to the original number. So f(f(f(x))) = x. Let me verify this. Let's compute f(f(f(x))) and see if it simplifies to x.First, let's compute f(x) = (5x -13)/(3x -7).Then f(f(x)) would be f applied to (5x -13)/(3x -7). Let's compute that:f(f(x)) = [5*( (5x -13)/(3x -7) ) -13 ] / [3*( (5x -13)/(3x -7) ) -7 ]Let me compute numerator and denominator separately.Numerator:5*(5x -13)/(3x -7) -13 = [25x -65]/[3x -7] -13 = [25x -65 -13*(3x -7)]/[3x -7] = [25x -65 -39x +91]/[3x -7] = [ (25x -39x) + (-65 +91) ]/[3x -7] = [ -14x +26 ]/[3x -7]Denominator:3*(5x -13)/(3x -7) -7 = [15x -39]/[3x -7] -7 = [15x -39 -7*(3x -7)]/[3x -7] = [15x -39 -21x +49]/[3x -7] = [ (15x -21x) + (-39 +49) ]/[3x -7] = [ -6x +10 ]/[3x -7]So f(f(x)) = [ -14x +26 ] / [ -6x +10 ] = ( -14x +26 ) / ( -6x +10 )Simplify numerator and denominator by factoring out -2:Numerator: -2*(7x -13)Denominator: -2*(3x -5)So f(f(x)) = [ -2*(7x -13) ] / [ -2*(3x -5) ] = (7x -13)/(3x -5)Therefore, f(f(x)) simplifies to (7x -13)/(3x -5)Now compute f(f(f(x))) by applying f to the result of f(f(x)):f(f(f(x))) = f( (7x -13)/(3x -5) ) = [5*( (7x -13)/(3x -5) ) -13 ] / [3*( (7x -13)/(3x -5) ) -7 ]Again, compute numerator and denominator separately.Numerator:5*(7x -13)/(3x -5) -13 = [35x -65]/(3x -5) -13 = [35x -65 -13*(3x -5)]/(3x -5) = [35x -65 -39x +65]/(3x -5) = [ (35x -39x) + (-65 +65) ]/(3x -5) = [ -4x +0 ]/(3x -5) = -4x/(3x -5)Denominator:3*(7x -13)/(3x -5) -7 = [21x -39]/(3x -5) -7 = [21x -39 -7*(3x -5)]/(3x -5) = [21x -39 -21x +35]/(3x -5) = [ (21x -21x) + (-39 +35) ]/(3x -5) = [0x -4]/(3x -5) = -4/(3x -5)Therefore, f(f(f(x))) = [ -4x / (3x -5) ] / [ -4 / (3x -5) ] = (-4x)/(3x -5) * (3x -5)/(-4) = xSo, after three applications of f, we get back to x. Therefore, f^3(x) = x. That means that applying the function three times brings us back to the original value, which implies that the sequence is periodic with period 3, provided that none of the denominators in the iterations become zero, which would cause undefined terms. But in our initial calculations, starting from a₁ = 2, we didn't encounter any zero denominators. Let's check:a₁ = 2, then denominator for a₂ is 3*2 -7 = 6 -7 = -1 ≠ 0.a₂ = 3, denominator for a₃ is 3*3 -7 = 9 -7 = 2 ≠ 0.a₃ = 1, denominator for a₄ is 3*1 -7 = 3 -7 = -4 ≠ 0.So all denominators are non-zero, so the sequence is well-defined and periodic with period 3. Therefore, the periodicity is 3.Alternatively, to confirm, maybe I can check if applying f three times to any x gives back x. From the algebra above, f^3(x) = x for all x where the function is defined (i.e., where denominators aren't zero). So this is a general result, not just specific to the starting value a₁ = 2. Therefore, regardless of the starting value (as long as we don't hit a denominator zero), the sequence will be periodic with period 3. But in the given problem, the starting value is 2, which we've already confirmed leads to a cycle of 3 terms.Therefore, the periodicity is 3.Final AnswerThe periodicity of the sequence is boxed{3}.</think>Given the sequence left{a_{n}right} defined by a_{1}=2 and a_{n+1}=frac{5 a_{n}-13}{3 a_{n}-7} for n geqslant 1, we need to determine its periodicity.First, we compute the initial terms of the sequence:- a_1 = 2- a_2 = frac{5 cdot 2 - 13}{3 cdot 2 - 7} = frac{10 - 13}{6 - 7} = frac{-3}{-1} = 3- a_3 = frac{5 cdot 3 - 13}{3 cdot 3 - 7} = frac{15 - 13}{9 - 7} = frac{2}{2} = 1- a_4 = frac{5 cdot 1 - 13}{3 cdot 1 - 7} = frac{5 - 13}{3 - 7} = frac{-8}{-4} = 2We observe that a_4 = a_1, indicating the sequence repeats every 3 terms. To confirm this periodicity algebraically, we consider the function f(x) = frac{5x - 13}{3x - 7} and compute f(f(f(x))):1. Compute f(x) = frac{5x - 13}{3x - 7}.2. Compute f(f(x)): - Numerator: 5left(frac{5x - 13}{3x - 7}right) - 13 = frac{-14x + 26}{3x - 7} - Denominator: 3left(frac{5x - 13}{3x - 7}right) - 7 = frac{-6x + 10}{3x - 7} - Thus, f(f(x)) = frac{-14x + 26}{-6x + 10} = frac{7x - 13}{3x - 5}3. Compute f(f(f(x))): - Numerator: 5left(frac{7x - 13}{3x - 5}right) - 13 = frac{-4x}{3x - 5} - Denominator: 3left(frac{7x - 13}{3x - 5}right) - 7 = frac{-4}{3x - 5} - Thus, f(f(f(x))) = frac{-4x}{3x - 5} div frac{-4}{3x - 5} = xThis shows that applying f three times returns to the original value, confirming the periodicity. Therefore, the sequence is periodic with period 3.[boxed{3}],<think>Okay, so I need to figure out the periodicity of this sequence. Let me start by understanding what the problem is asking. The sequence is defined by a recursive formula: a₁ = 2, and each subsequent term is calculated using aₙ₊₁ = (5aₙ - 13)/(3aₙ - 7). They want to know if the sequence is periodic, and if so, what the period is.First, let me recall what periodicity means. A sequence is periodic if after a certain number of terms, it repeats itself. The smallest such number is called the period. For example, a sequence with period 2 would alternate between two values: a, b, a, b, etc.Given that this sequence is recursively defined, maybe if I compute the first few terms, I can see a pattern or repetition. Let me start calculating terms step by step.We know a₁ = 2. Let's find a₂:a₂ = (5a₁ - 13)/(3a₁ - 7) = (5*2 - 13)/(3*2 - 7) = (10 - 13)/(6 - 7) = (-3)/(-1) = 3. So a₂ = 3.Next, a₃ = (5a₂ - 13)/(3a₂ - 7) = (5*3 - 13)/(3*3 - 7) = (15 - 13)/(9 - 7) = 2/2 = 1. So a₃ = 1.Now, a₄ = (5a₃ - 13)/(3a₃ - 7) = (5*1 - 13)/(3*1 - 7) = (5 - 13)/(3 - 7) = (-8)/(-4) = 2. So a₄ = 2.Wait a minute, a₄ is 2, which is the same as a₁. Let me check if this continues. Let's compute a₅:a₅ = (5a₄ - 13)/(3a₄ - 7) = (5*2 - 13)/(3*2 - 7) = (10 - 13)/(6 - 7) = (-3)/(-1) = 3. So a₅ = 3, which is the same as a₂.Then a₆ = (5a₅ - 13)/(3a₅ - 7) = (15 - 13)/(9 - 7) = 2/2 = 1. So a₆ = 1, same as a₃.a₇ = (5*1 -13)/(3*1 -7) = (-8)/(-4) = 2. So a₇ = 2, same as a₄ and a₁.So the sequence is going 2, 3, 1, 2, 3, 1, 2, 3, 1,... repeating every three terms. Therefore, the period is 3.But wait, let me make sure that this pattern continues indefinitely. Since each term is determined by the previous one, if we have a cycle of three terms: 2 → 3 → 1 → 2 → ..., then it should keep repeating. Let me verify another cycle.Starting with a₇ = 2, so a₈ should be 3, and indeed:a₈ = (5*2 -13)/(3*2 -7) = (-3)/(-1) = 3. Then a₉ = (5*3 -13)/(9 -7) = 2/2 = 1. Then a₁₀ = 2 again. So yes, the cycle repeats every three terms. Therefore, the sequence is periodic with period 3.But let me also check if maybe there's a smaller period. For example, is it possible that the sequence repeats every 1 or 2 terms? Well, from the terms calculated: 2, 3, 1, 2, 3, 1, 2,... There's no repetition of a single term, so period 1 is out. What about period 2? Let's see: The terms go 2, 3, 1, 2, 3, 1,... so 2 is followed by 3, then 3 is followed by 1, 1 is followed by 2, etc. There's no two-term cycle here. So the minimal period is indeed 3.Alternatively, maybe there's a mathematical way to determine periodicity without computing terms? For example, considering the function f(x) = (5x -13)/(3x -7), which generates the sequence. If we compose this function with itself multiple times, we can see if after some compositions it becomes the identity function, which would imply periodicity.Let's try that. Let f(x) = (5x -13)/(3x -7). Let's compute f(f(x)):First, compute f(f(x)):Let me denote f(x) as (5x -13)/(3x -7). Then f(f(x)) would be:f(f(x)) = [5*( (5x -13)/(3x -7) ) -13 ] / [3*( (5x -13)/(3x -7) ) -7 ]Let me compute numerator and denominator separately.Numerator:5*(5x -13)/(3x -7) -13 = [25x -65 -13*(3x -7)] / (3x -7)= [25x -65 -39x +91] / (3x -7)= (-14x +26) / (3x -7)Denominator:3*(5x -13)/(3x -7) -7 = [15x -39 -7*(3x -7)] / (3x -7)= [15x -39 -21x +49] / (3x -7)= (-6x +10) / (3x -7)Therefore, f(f(x)) = [ (-14x +26) / (3x -7) ] / [ (-6x +10) / (3x -7) ] = (-14x +26)/(-6x +10) = (14x -26)/(6x -10) = simplifying numerator and denominator by 2: (7x -13)/(3x -5)So f(f(x)) = (7x -13)/(3x -5). Not the identity function, so f composed twice is not identity. Let's compute f(f(f(x))) to see if that gives us back x, which would mean that applying f three times brings us back to the original value, implying period 3.Compute f(f(f(x))) = f( (7x -13)/(3x -5) )Apply f to that: [5*( (7x -13)/(3x -5) ) -13 ] / [3*( (7x -13)/(3x -5) ) -7 ]Compute numerator:5*(7x -13)/(3x -5) -13 = [35x -65 -13*(3x -5)] / (3x -5)= [35x -65 -39x +65] / (3x -5)= (-4x)/ (3x -5)Denominator:3*(7x -13)/(3x -5) -7 = [21x -39 -7*(3x -5)] / (3x -5)= [21x -39 -21x +35] / (3x -5)= (-4)/ (3x -5)Therefore, f(f(f(x))) = [ -4x / (3x -5) ] / [ -4 / (3x -5) ] = (-4x)/(-4) = x.So f composed three times is the identity function. Therefore, applying f three times brings us back to the original value. Therefore, the sequence must have period 3. That is, aₙ₊₃ = aₙ for all n. Therefore, the period is 3.Therefore, the answer should be that the sequence is periodic with period 3.But wait, let me check if there's any condition where this might not hold. For example, if any denominator becomes zero. The denominators in the recursive formula are 3aₙ -7. Let's check if any term aₙ could be 7/3, which would make the denominator zero.Looking at the sequence we calculated: 2, 3, 1, 2, 3, 1,... None of these terms are 7/3. Since the sequence cycles through 2, 3, 1, none of these are 7/3, so we won't have division by zero in the recursion. Therefore, the sequence is well-defined and periodic with period 3.Therefore, the periodicity is 3.Final AnswerThe sequence is periodic with period boxed{3}.</think>Given the sequence left{a_{n}right} defined by a_{1}=2 and a_{n+1}=frac{5 a_{n}-13}{3 a_{n}-7} for n geq 1, we need to determine its periodicity.First, we calculate the initial terms of the sequence:- a_1 = 2- a_2 = frac{5a_1 - 13}{3a_1 - 7} = frac{10 - 13}{6 - 7} = frac{-3}{-1} = 3- a_3 = frac{5a_2 - 13}{3a_2 - 7} = frac{15 - 13}{9 - 7} = frac{2}{2} = 1- a_4 = frac{5a_3 - 13}{3a_3 - 7} = frac{5 - 13}{3 - 7} = frac{-8}{-4} = 2We observe that a_4 = a_1, suggesting a repeating pattern. Continuing this calculation:- a_5 = 3 (same as a_2)- a_6 = 1 (same as a_3)- a_7 = 2 (same as a_4 and a_1)Thus, the sequence repeats every three terms: 2, 3, 1, 2, 3, 1, ldotsTo confirm this periodicity, we consider the function f(x) = frac{5x - 13}{3x - 7} and its compositions:- f(f(x)) = frac{7x - 13}{3x - 5}- f(f(f(x))) = x, showing that applying f three times results in the identity function.Since f(f(f(x))) = x, the sequence must be periodic with period 3. Additionally, none of the terms in the sequence result in a denominator of zero, ensuring the sequence is well-defined.Thus, the sequence is periodic with period boxed{3}.]

🔑:Solving, from a_{1}=2, we geta_{2}=frac{5 a_{1}-13}{3 a_{1}-7}=3, a_{3}=frac{5 a_{2}-13}{3 a_{2}-7}=1,which means a_{1}, a_{2}, a_{3} are all distinct.From the given conditions, we havebegin{array}{l}a_{n+1}=frac{5 a_{n}-13}{3 a_{n}-7}, a_{n+2}=frac{5 a_{n+1}-13}{3 a_{n+1}-7}=frac{7 a_{n}-13}{3 a_{n}-5}, a_{n+3}=frac{5 a_{n+2}-13}{3 a_{n+2}-7}=a_{n},end{array}which means a_{n+3}=a_{n}(n=1,2, cdots).In summary, the sequence left{a_{n}right} is a purely periodic sequence, with the smallest positive period being 3.

🧠:[<think>Okay, so I need to find the intervals where the function y = (1/4)^x - (1/2)^x + 1 is increasing or decreasing, and also determine its range when x is between -3 and 2. Hmm, let's start by understanding the function better.First, I know that (1/4)^x can be rewritten as 4^(-x), and (1/2)^x is 2^(-x). Maybe that's helpful? Let me write it down:y = 4^(-x) - 2^(-x) + 1.Alternatively, since 4 is 2 squared, 4^(-x) is the same as (2^2)^(-x) = 2^(-2x). So maybe expressing everything in terms of 2^(-x) will help. Let's try that:y = 2^(-2x) - 2^(-x) + 1.Hmm, that might be easier to handle. Let me substitute t = 2^(-x). Then, since 2^(-x) is the same as e^(-x ln 2), which is a decreasing function because the exponent -x ln 2 decreases as x increases. So as x increases, t decreases. That might be useful later when considering the relationship between t and x.But first, let's express y in terms of t. Since t = 2^(-x), then 2^(-2x) is t squared. So:y = t^2 - t + 1.Okay, that seems simpler. So y is a quadratic function in terms of t. Let me note that down. So y(t) = t² - t + 1. Now, to analyze the function's monotonicity, I need to find its derivative with respect to x and see where it's positive or negative. But since y is expressed in terms of t, which is a function of x, I should use the chain rule.So dy/dx = dy/dt * dt/dx.First, compute dy/dt. Since y = t² - t + 1, dy/dt is 2t - 1.Next, compute dt/dx. Since t = 2^(-x), then dt/dx = d/dx [2^(-x)] = -ln 2 * 2^(-x) = -ln 2 * t.Therefore, dy/dx = (2t - 1)(-ln 2 * t) = -ln 2 * t * (2t - 1).So the derivative of y with respect to x is -ln 2 * t * (2t - 1). Since ln 2 is a positive constant (approximately 0.693), the sign of the derivative depends on -t*(2t - 1). Let's analyze that.Let me factor out the negative sign: -t*(2t - 1) = - [t*(2t - 1)]. So the sign of dy/dx is the same as the sign of - [t*(2t - 1)].To find when dy/dx is positive or negative, we need to determine when - [t*(2t - 1)] is positive or negative. Let's set up inequalities.First, let's find when - [t*(2t - 1)] > 0. That is equivalent to t*(2t - 1) < 0.Similarly, when - [t*(2t - 1)] < 0, then t*(2t - 1) > 0.So we need to solve t*(2t - 1) < 0 and t*(2t - 1) > 0.Let me solve the inequality t*(2t - 1) < 0.First, find the critical points where t*(2t - 1) = 0. That happens when t = 0 or 2t - 1 = 0.Solving 2t - 1 = 0 gives t = 1/2. So the critical points are t = 0 and t = 1/2.Since t is equal to 2^(-x), and 2^(-x) is always positive for any real x (since an exponential function is always positive), t can never be zero. So the only critical point in the domain of t is t = 1/2.So the intervals to consider for t are t > 1/2 and t < 1/2.For t*(2t - 1):- When t < 1/2: Let's pick t = 0. Then 0*(2*0 -1) = 0*(-1) = 0. Wait, but t can't be zero. So for t between 0 and 1/2 (since t is positive), let's pick t = 1/4. Then (1/4)*(2*(1/4) -1) = (1/4)*(1/2 -1) = (1/4)*(-1/2) = -1/8 < 0. So t*(2t -1) is negative when t is between 0 and 1/2.- When t > 1/2: Let's pick t = 1. Then 1*(2*1 -1) = 1*(2 -1) = 1*1 =1 >0. So positive.Therefore, t*(2t -1) is negative when t ∈ (0, 1/2) and positive when t ∈ (1/2, ∞). Since t can't be zero or negative, those are the only intervals.Therefore, going back to dy/dx, which is -ln2 * t*(2t -1):- When t ∈ (0, 1/2), t*(2t -1) is negative, so dy/dx = -ln2 * (negative) = positive. So the function is increasing.- When t ∈ (1/2, ∞), t*(2t -1) is positive, so dy/dx = -ln2 * (positive) = negative. So the function is decreasing.Therefore, the function y is increasing when t < 1/2 and decreasing when t > 1/2.But t = 2^(-x). So we need to translate this back to x values.Given that t = 2^(-x), let's solve for x when t =1/2:2^(-x) = 1/2.Taking log base 2: -x = log2(1/2) = -1. Therefore, x = 1.So t =1/2 when x =1.So when x <1, since t =2^(-x), and x is in [-3,2], then:For x <1, t =2^(-x). Let's see, when x increases, -x decreases, so t decreases. So for x <1, t >1/2, because when x approaches 1 from the left, t approaches 1/2. Wait, no:Wait, when x increases, t =2^(-x) decreases. So for x <1, t is greater than 1/2? Let's check x=0: t=1. x=1: t=1/2. x= -3: t=2^(3)=8. So yes, when x is less than 1, t is greater than 1/2, and when x is greater than1, t is less than1/2.Wait, but our original domain is x ∈ [-3,2]. So when x is in [-3,1), t is in (1/2, 8], and when x is in (1,2], t is in [1/4,1/2). At x=1, t=1/2.But earlier, we found that the function is decreasing when t >1/2, so when x ∈ [-3,1), and increasing when t <1/2, which would correspond to x ∈ (1,2]. But wait, t <1/2 is when x >1, as x increases beyond1, t decreases below1/2.Wait, but when t is less than1/2, x is greater than1. So:- For x ∈ [-3,1), t ∈ (1/2,8], so dy/dx is negative (since t >1/2). Thus, y is decreasing on [-3,1).- For x ∈ (1,2], t ∈ [1/4,1/2), so dy/dx is positive (since t <1/2). Thus, y is increasing on (1,2].But wait, at x=1, t=1/2. What's happening at x=1? Let's check the derivative at x=1:dy/dx = -ln2 * t*(2t -1). At t=1/2, it's -ln2*(1/2)*(2*(1/2)-1) = -ln2*(1/2)*(0) =0. So the derivative is zero at x=1. Therefore, x=1 is a critical point.So the function is decreasing on [-3,1) and increasing on (1,2]. At x=1, the derivative is zero, so it's a critical point. Therefore, the intervals of monotonicity are:- Decreasing on [-3,1]Wait, but when we approach x=1 from the left, the function is decreasing, and from the right, it starts increasing. But since at x=1 the derivative is zero, we need to include x=1 in one of the intervals. Typically, when the derivative is zero at a point, we can include it in both intervals, but since it's a single point, it doesn't affect the monotonicity. However, in terms of classification, the function changes from decreasing to increasing at x=1, so x=1 is a local minimum.So to write the intervals properly, since the derivative is negative on [-3,1) and positive on (1,2], the function is decreasing on [-3,1] and increasing on [1,2]? Wait, but the derivative at x=1 is zero. So, technically, the function is decreasing up to x=1 and increasing after x=1. But whether the point x=1 is included in both intervals?In calculus, when we have intervals of increase or decrease, we include the endpoints where the derivative is zero if the function doesn't change direction there. But in this case, since the derivative changes sign from negative to positive at x=1, x=1 is a local minimum. Therefore, the function is decreasing before x=1 and increasing after x=1. So the intervals of monotonicity are:- Decreasing on [-3,1]- Increasing on [1,2]But to be precise, since at x=1 the derivative is zero, but the function changes from decreasing to increasing. So in terms of intervals, whether open or closed at x=1.In some textbooks, they might write that the function is decreasing on [-3,1] and increasing on [1,2], including x=1 in both intervals. However, in reality, at x=1, the function is neither increasing nor decreasing, but since it's a single point, it doesn't affect the overall interval. So for the purpose of the answer, I think it's acceptable to write decreasing on [-3,1] and increasing on [1,2].Therefore, the intervals of monotonicity are: decreasing on [-3,1] and increasing on [1,2].Now, moving on to finding the range of the function on x ∈ [-3,2].Since the function is continuous on a closed interval, by the Extreme Value Theorem, it must attain its maximum and minimum values on this interval. The extrema can occur either at critical points or at the endpoints.We already found that there's a critical point at x=1. So we need to evaluate the function at x=-3, x=1, and x=2. Then compare the values to find the minimum and maximum, and hence the range.Let's compute y(-3), y(1), and y(2).First, y(x) = (1/4)^x - (1/2)^x + 1.Compute y(-3):(1/4)^(-3) - (1/2)^(-3) + 1 = 4^3 - 2^3 + 1 = 64 - 8 +1 = 57.Compute y(1):(1/4)^1 - (1/2)^1 +1 = 1/4 -1/2 +1 = (1/4 -2/4) +1 = (-1/4) +1 = 3/4.Compute y(2):(1/4)^2 - (1/2)^2 +1 = 1/16 -1/4 +1 = (1/16 -4/16) +1 = (-3/16) +1 = 13/16.Wait, but hold on. The function is decreasing from x=-3 to x=1, so the maximum at x=-3 (since it's decreasing) and then at x=1, it's a minimum. Then increasing from x=1 to x=2, so the value at x=2 should be higher than at x=1, but in the calculations above, y(2)=13/16≈0.8125 and y(1)=3/4=0.75. So indeed, y(2) is slightly higher than y(1), which makes sense because it's increasing from x=1 to x=2.But wait, we need to check if there are any other critical points. However, we only found one critical point at x=1. So the extrema should be at x=-3, x=1, and x=2.But let me confirm by calculating some intermediate points to ensure there are no other extrema.For example, let's compute y(0):(1/4)^0 - (1/2)^0 +1 =1 -1 +1=1.y(0)=1.Compute y(0.5):(1/4)^0.5 - (1/2)^0.5 +1 = (1/2) - (1/√2) +1 ≈0.5 -0.7071 +1≈0.7929.Similarly, y(1.5):(1/4)^1.5 - (1/2)^1.5 +1= (1/4)^(3/2) - (1/2)^(3/2) +1= (1/(8)) - (1/(2√2)) +1≈0.125 -0.3536 +1≈0.7714.Wait, but according to the earlier reasoning, the function should be increasing from x=1 to x=2. However, y(1.5)≈0.7714, which is less than y(1)=0.75? Wait, no:Wait, y(1)=0.75, y(1.5)=0.7714, which is higher. So yes, from x=1 to x=1.5, it's increasing from 0.75 to ~0.7714, and then to x=2, which is 13/16≈0.8125. So that's increasing. Similarly, from x=-3 (y=57) down to x=1 (y=0.75), then up to x=2 (y≈0.8125). So the minimum is at x=1, which is 0.75, and the maximum is at x=-3, which is 57. Then the function's range is [0.75,57].But let's check another point, say x=2:As calculated earlier, y(2)=13/16≈0.8125. So the function increases from 0.75 at x=1 to 0.8125 at x=2.So the maximum value is at x=-3, which is 57, and the minimum is at x=1, which is 0.75. Therefore, the range is [0.75,57].But wait, let me confirm with another point. For instance, x= -1:y(-1)= (1/4)^(-1) - (1/2)^(-1) +1=4 -2 +1=3.x=-2:y(-2)= (1/4)^(-2) - (1/2)^(-2) +1=16 -4 +1=13.So y(-3)=57, y(-2)=13, y(-1)=3, y(0)=1, y(1)=0.75, y(2)=0.8125.Thus, the function decreases from x=-3 to x=1, reaching the minimum at x=1, then increases slightly to x=2. So the maximum is at x=-3, and the minimum is at x=1.Therefore, the range of the function is [3/4, 57].Expressed as fractions, 3/4 is 0.75 and 57 is an integer. So the range is all real numbers y such that 3/4 ≤ y ≤57.Therefore, the range is [3/4,57].But let's verify once more if there are any other critical points. Since we found only one critical point at x=1, and the derivative doesn't cross zero elsewhere. Let's check the derivative expression again:dy/dx = -ln2 * t * (2t -1). Since t is always positive, the critical points occur when 2t -1 =0, which is t=1/2, corresponding to x=1. So there are no other critical points. Therefore, the only extrema are at the endpoints and x=1. Hence, the range is [3/4,57].So summarizing:Intervals of monotonicity: decreasing on [-3,1], increasing on [1,2].Range: [3/4,57].But to present this properly, let's write the intervals with proper notation. The function is decreasing from x=-3 to x=1 and increasing from x=1 to x=2. So in interval notation, the function is decreasing on [-3,1] and increasing on [1,2].For the range, since the function attains all values from its minimum 3/4 up to the maximum 57, the range is [3/4,57].Therefore, the final answers are:Monotonicity: decreasing on [-3,1], increasing on [1,2].Range: [3/4,57].Final AnswerThe function is decreasing on the interval boxed{[-3, 1]} and increasing on the interval boxed{[1, 2]}. The range of the function is boxed{left[ frac{3}{4}, 57 right]}.</think>To find the intervals of monotonicity and the range of the function ( y = left(frac{1}{4}right)^x - left(frac{1}{2}right)^x + 1 ) where ( x in [-3, 2] ), we start by rewriting the function in terms of ( t = 2^{-x} ):[y = t^2 - t + 1]where ( t = 2^{-x} ). The derivative of ( y ) with respect to ( x ) is calculated using the chain rule:[frac{dy}{dx} = frac{dy}{dt} cdot frac{dt}{dx}]First, compute ( frac{dy}{dt} ):[frac{dy}{dt} = 2t - 1]Next, compute ( frac{dt}{dx} ):[frac{dt}{dx} = -ln 2 cdot 2^{-x} = -ln 2 cdot t]Thus, the derivative ( frac{dy}{dx} ) is:[frac{dy}{dx} = (2t - 1)(-ln 2 cdot t) = -ln 2 cdot t cdot (2t - 1)]The sign of ( frac{dy}{dx} ) depends on ( -t(2t - 1) ). Solving ( -t(2t - 1) = 0 ) gives critical points at ( t = 0 ) (not possible since ( t > 0 )) and ( t = frac{1}{2} ). Converting ( t = frac{1}{2} ) back to ( x ):[2^{-x} = frac{1}{2} implies x = 1]Analyzing the sign of ( frac{dy}{dx} ):- For ( t > frac{1}{2} ) (i.e., ( x < 1 )), ( frac{dy}{dx} ) is negative, so the function is decreasing.- For ( t < frac{1}{2} ) (i.e., ( x > 1 )), ( frac{dy}{dx} ) is positive, so the function is increasing.Thus, the function is decreasing on ( [-3, 1] ) and increasing on ( [1, 2] ).To find the range, evaluate the function at the critical point and endpoints:- ( y(-3) = 57 )- ( y(1) = frac{3}{4} )- ( y(2) = frac{13}{16} approx 0.8125 )The maximum value is at ( x = -3 ) (57) and the minimum value is at ( x = 1 ) ((frac{3}{4})). Therefore, the range is ([ frac{3}{4}, 57 ]).Final Answer:The function is decreasing on the interval boxed{[-3, 1]} and increasing on the interval boxed{[1, 2]}. The range of the function is boxed{left[ frac{3}{4}, 57 right]}.,<think>Okay, so I need to find the intervals where the function y = (1/4)^x - (1/2)^x + 1 is increasing or decreasing, and also determine its range when x is between -3 and 2. Hmm, let me start by understanding the function better.First, I know that (1/4)^x can be rewritten as 4^(-x), and similarly (1/2)^x is 2^(-x). Maybe rewriting them with exponents to the base 2 will help. Let's see:4^(-x) is the same as (2^2)^(-x) = 2^(-2x). So the function becomes y = 2^(-2x) - 2^(-x) + 1. That might be easier to work with when taking derivatives.Now, since the problem is about monotonicity, I need to find the derivative of y with respect to x and determine where it's positive or negative. Let's compute the derivative:dy/dx = derivative of 2^(-2x) minus derivative of 2^(-x) plus derivative of 1.The derivative of 2^u with respect to x is 2^u * ln(2) * du/dx. So applying that:Derivative of 2^(-2x) is 2^(-2x) * ln(2) * (-2) = -2 ln(2) * 2^(-2x).Derivative of -2^(-x) is - [2^(-x) * ln(2) * (-1)] = 2^(-x) * ln(2).Derivative of 1 is 0.Putting it all together:dy/dx = -2 ln(2) * 2^(-2x) + ln(2) * 2^(-x).I can factor out ln(2) * 2^(-2x) from both terms. Let's see:dy/dx = ln(2) * 2^(-2x) [ -2 + 2^x ].Wait, let me check that again. If I factor out ln(2) * 2^(-2x), then the first term is -2 ln(2) * 2^(-2x) and the second term is ln(2) * 2^(-x). So:Take ln(2) * 2^(-2x) as a common factor. So:-2 ln(2) * 2^(-2x) + ln(2) * 2^(-x) = ln(2) * 2^(-2x) [ -2 + 2^(x) * 2^(2x) / 2^(2x) ) ].Wait, no. Let me re-express the second term with exponent -x in terms of exponent -2x:2^(-x) = 2^(-2x + x) = 2^(-2x) * 2^x. So, ln(2) * 2^(-x) = ln(2) * 2^x * 2^(-2x). Therefore, the derivative becomes:ln(2) * 2^(-2x) [ -2 + 2^x ].Yes, that's correct. So:dy/dx = ln(2) * 2^(-2x) [ -2 + 2^x ].Simplify inside the brackets:-2 + 2^x = 2^x - 2. So:dy/dx = ln(2) * 2^(-2x) * (2^x - 2).Now, let's analyze the sign of dy/dx. Since ln(2) is positive (because 2 > 1), and 2^(-2x) is always positive (since any exponential function is positive), the sign of dy/dx depends on the term (2^x - 2).Therefore:dy/dx > 0 when 2^x - 2 > 0 => 2^x > 2 => x > 1.dy/dx < 0 when 2^x - 2 < 0 => 2^x < 2 => x < 1.At x = 1, dy/dx = 0, so that's a critical point.Therefore, the function is decreasing on intervals where x < 1 and increasing on intervals where x > 1. However, we have to consider the domain x ∈ [-3, 2]. So within this interval, the function decreases from x = -3 up to x = 1, then increases from x = 1 to x = 2.Therefore, the monotonicity intervals are:Decreasing on [-3, 1]Increasing on [1, 2]Now, moving on to finding the range of the function on the interval [-3, 2]. To find the range, we need to determine the minimum and maximum values the function can take. Since the function is continuous on a closed interval, it will attain its maximum and minimum either at critical points or at the endpoints.We already found a critical point at x = 1. So we need to evaluate the function at x = -3, x = 1, and x = 2.Let's compute y(-3), y(1), and y(2).First, compute y(-3):y(-3) = (1/4)^(-3) - (1/2)^(-3) + 1.Negative exponents mean reciprocals, so:(1/4)^(-3) = 4^3 = 64.(1/2)^(-3) = 2^3 = 8.Therefore, y(-3) = 64 - 8 + 1 = 57.Next, compute y(1):y(1) = (1/4)^1 - (1/2)^1 + 1 = (1/4) - (1/2) + 1 = (1/4 - 2/4) + 1 = (-1/4) + 1 = 3/4.Then, compute y(2):y(2) = (1/4)^2 - (1/2)^2 + 1 = (1/16) - (1/4) + 1 = (1/16 - 4/16) + 1 = (-3/16) + 1 = 13/16 ≈ 0.8125.Wait, hold on. At x=2, (1/4)^2 is 1/16, (1/2)^2 is 1/4, so 1/16 - 1/4 +1 = 1/16 - 4/16 + 16/16 = (1 - 4 + 16)/16 = 13/16. Correct.Now, since the function decreases from x=-3 to x=1 and then increases from x=1 to x=2, the minimum value is at x=1, which is 3/4. Wait, but y(1) is 3/4, which is 0.75, and y(2) is 13/16 ≈ 0.8125, which is less than y(1). Wait, that can't be right. If the function is increasing from x=1 to x=2, then y(2) should be greater than y(1). But according to our calculations, y(2) is 13/16 which is 0.8125, and y(1) is 3/4 which is 0.75. So 0.8125 is actually greater than 0.75. Wait, 13/16 is 0.8125, which is indeed greater than 3/4 (0.75). So that's correct. So when x increases from 1 to 2, the function increases from 0.75 to 0.8125. So the minimum is at x=1, and the maximum at x=-3 (57). But wait, let's check another point. Wait, the function was decreasing from -3 to 1, so the highest value is at x=-3 (57) and decreasing to 0.75 at x=1, then increasing again but only up to 0.8125 at x=2. So the maximum value is 57 at x=-3, the minimum is 3/4 at x=1. So the range is [3/4, 57].But let me verify. Let's compute y(0), maybe there's another maximum or minimum. Wait, but since the function is decreasing until x=1 and then increasing, so the only critical point is x=1, so the extrema should be at the endpoints or at x=1. Let's check y(0):y(0) = (1/4)^0 - (1/2)^0 +1 = 1 -1 +1 = 1. So y(0)=1, which is between 3/4 and 57. So that seems okay. Similarly, y(-1):y(-1) = (1/4)^(-1) - (1/2)^(-1) +1 = 4 -2 +1=3. Which is still between 3/4 and 57. So that's correct.Therefore, the range of the function is [3/4, 57].Wait, but let me double-check y(1). Let me compute again:y(1) = (1/4)^1 - (1/2)^1 +1 = 1/4 - 1/2 +1 = (1/4 - 2/4) +1 = (-1/4) +1 = 3/4. Correct.And y(2) is 13/16 ≈0.8125. So 13/16 is indeed greater than 3/4 (which is 12/16). So yes, 13/16 is 0.8125.So the function reaches its minimum at x=1 with y=3/4, and its maximum at x=-3 with y=57. Therefore, the range is [3/4, 57].Hence, putting it all together:Monotonicity:- Decreasing on [-3, 1]- Increasing on [1, 2]Range: [3/4, 57]Wait, but let me confirm once more. Since the function is continuous on a closed interval, and we checked the endpoints and critical points, it's correct. Also, since there is only one critical point at x=1, which is a minimum, and the function was decreasing before that and increasing after, then yes, the maximum must be at the left endpoint x=-3, which is 57, and the minimum at x=1, which is 3/4, then the right endpoint x=2 gives 13/16, which is 0.8125, which is higher than the minimum but lower than the maximum. Therefore, the range is from 3/4 up to 57.So that's it.Final AnswerThe function is decreasing on the interval boxed{[-3, 1]} and increasing on the interval boxed{[1, 2]}. The range of the function is boxed{left[frac{3}{4}, 57right]}.</think>To find the intervals of monotonicity and the range of the function ( y = left(frac{1}{4}right)^x - left(frac{1}{2}right)^x + 1 ) where ( x in [-3, 2] ), we start by rewriting the function using base 2:[ y = 2^{-2x} - 2^{-x} + 1 ]Next, we compute the derivative to determine the intervals of monotonicity:[ frac{dy}{dx} = -2 ln(2) cdot 2^{-2x} + ln(2) cdot 2^{-x} ][ = ln(2) cdot 2^{-2x} (2^x - 2) ]The sign of the derivative depends on the term ( 2^x - 2 ). Since ( ln(2) > 0 ) and ( 2^{-2x} > 0 ), the derivative is positive when ( 2^x > 2 ) (i.e., ( x > 1 )) and negative when ( 2^x < 2 ) (i.e., ( x < 1 )). Therefore, the function is decreasing on ( [-3, 1] ) and increasing on ( [1, 2] ).To find the range, we evaluate the function at the critical point ( x = 1 ) and the endpoints ( x = -3 ) and ( x = 2 ):- ( y(-3) = left(frac{1}{4}right)^{-3} - left(frac{1}{2}right)^{-3} + 1 = 64 - 8 + 1 = 57 )- ( y(1) = left(frac{1}{4}right)^1 - left(frac{1}{2}right)^1 + 1 = frac{1}{4} - frac{1}{2} + 1 = frac{3}{4} )- ( y(2) = left(frac{1}{4}right)^2 - left(frac{1}{2}right)^2 + 1 = frac{1}{16} - frac{1}{4} + 1 = frac{13}{16} )The minimum value is ( frac{3}{4} ) at ( x = 1 ) and the maximum value is 57 at ( x = -3 ). Therefore, the range of the function is ( left[frac{3}{4}, 57right] ). Final AnswerThe function is decreasing on the interval boxed{[-3, 1]} and increasing on the interval boxed{[1, 2]}. The range of the function is boxed{left[frac{3}{4}, 57right]}.]

🔑:Given y=left(frac{1}{4}right)^x - left(frac{1}{2}right)^x + 1, let t=left(frac{1}{2}right)^x. Since x in [-3, 2], it follows that t in left[frac{1}{4}, 8right]. Thus, the original function can be rewritten as y=t^2 - t + 1 = left(t - frac{1}{2}right)^2 + frac{3}{4}, where t in left[frac{1}{4}, 8right]. Therefore, t = frac{1}{2} is the axis of symmetry.Since x in [-3, 1], as x increases, t=left(frac{1}{2}right)^x decreases, and t in left[frac{1}{2}, 8right], which implies y=left(t - frac{1}{2}right)^2 + frac{3}{4} decreases.Therefore, [-3, 1] is the interval of decrease for the function y=left(frac{1}{4}right)^x - left(frac{1}{2}right)^x + 1. Similarly, [1, 2] is the interval of increase for the function.Thus, the minimum value of y is frac{3}{4}, and the maximum value of y is 57.Hence, the interval of decrease for the original function is [-3, 1], the interval of increase is [1, 2], and the range is boxed{left[frac{3}{4}, 57right]}.

🧠:[<think>Okay, let's see. The problem is asking if we can divide 5 apples equally among 6 children, with the condition that at least one apple isn't cut into more than 3 pieces. Hmm. So, each child should get 5/6 of an apple. But we need to do this by cutting the apples into pieces, right? And at least one of the apples should only be cut into 1, 2, or 3 pieces. The rest can be cut into more pieces if needed, but at least one apple must have no more than 3 pieces.First, let me think about how to divide 5 apples into 6 equal parts. Each child needs 5/6 of an apple. If we cut each apple into sixths, then each child would get five 1/6 pieces. But that would require cutting each apple into 6 pieces, which is more than 3, and since the problem says at least one apple shouldn't be cut into more than 3 pieces, maybe this approach isn't the best. Because if we do that, all apples are cut into 6 pieces, which violates the condition.So, perhaps we need a different method where some apples are cut into fewer pieces. Let's see. Maybe we can use some apples cut into halves, thirds, or other fractions so that when combined, each child gets 5/6. Let me try to find a combination.Let me start by considering cutting some apples into thirds. If an apple is cut into 3 pieces, each piece is 1/3. If we have another apple cut into halves, each piece is 1/2. But 1/3 + 1/2 is 5/6. Wait, that's exactly what we need. So if each child gets one 1/2 piece and one 1/3 piece, that would give them 5/6. But how many apples would that require?Each child needs a 1/2 piece and a 1/3 piece. So, for 6 children, we need 6*(1/2) = 3 apples cut into halves and 6*(1/3) = 2 apples cut into thirds. That totals 3 + 2 = 5 apples. Perfect! So, if we cut 3 apples into halves (each apple gives 2 pieces) and 2 apples into thirds (each apple gives 3 pieces), then each child can receive one half (1/2) and one third (1/3), totaling 5/6. But wait, in this case, the apples that are cut into halves are only cut into 2 pieces, which is less than or equal to 3. The apples cut into thirds are cut into 3 pieces. So, all apples are cut into 2 or 3 pieces. Therefore, all apples are not cut into more than 3 pieces. But the problem states that "at least one apple will not be cut into more than 3 pieces." So actually, if all apples are cut into 2 or 3 pieces, then all of them satisfy the condition, which is even better. But the problem only requires that at least one apple isn't cut into more than 3. So in this case, the answer is yes, it's possible, and even all apples can be cut into 3 or fewer pieces.Wait, but let me verify again. If we have 3 apples cut into halves (so each apple is 2 pieces) and 2 apples cut into thirds (each apple is 3 pieces), that gives 3*2 + 2*3 = 6 + 6 = 12 pieces. There are 6 children, each getting 2 pieces (a half and a third). So 6 children * 2 pieces = 12 pieces. That adds up. Then each child gets 1/2 + 1/3 = 5/6. So that works. Therefore, all apples are cut into 2 or 3 pieces, so indeed, all of them satisfy the condition of not being cut into more than 3. Therefore, the answer is yes, and even more, all apples can be cut into 3 or fewer pieces.But let me check again if this is correct. Let me count the number of apples. 3 apples are cut into halves (2 pieces each) and 2 apples are cut into thirds (3 pieces each). So total apples: 3 + 2 = 5. Correct. Each child receives one half (from the first type) and one third (from the second type). So each child gets 1/2 + 1/3 = 5/6. Perfect. So this works, and all apples are cut into 2 or 3 pieces, which meets the requirement that at least one apple isn't cut into more than 3. In fact, all apples meet that. So the answer is yes.But maybe I made a mistake. Let me think again. Wait, if you cut 3 apples into halves, that gives 3*2=6 half pieces. Then 2 apples into thirds gives 2*3=6 third pieces. Then each child gets one half and one third, so 6 children get 6 halves and 6 thirds. That's exactly the number of pieces. So no leftover pieces. Perfect division.Therefore, this seems to work. Therefore, the answer is yes. Moreover, not only does at least one apple not need to be cut into more than 3 pieces, but all apples can be cut into 3 or fewer pieces.Wait, but the problem says "at least one apple will not be cut into more than 3 pieces". So even if all are cut into 3 or fewer, that's fine. So the answer is yes, it's possible. Therefore, the justification is that by cutting 3 apples into 2 pieces each and 2 apples into 3 pieces each, each child receives a half and a third, totaling 5/6, and all apples are cut into no more than 3 pieces. Hence, satisfying the condition.But let me check if there's another way. Suppose instead we have different cuts. For example, maybe some apples are cut into 3 pieces, some into 2, and maybe some into 1 (but that's not useful). Alternatively, perhaps one apple is kept whole. But 5 apples divided among 6 children can't have a whole apple because each child needs 5/6, which is less than a whole apple. So keeping an apple whole would require someone to get it, but then they would need only 5/6 -1 = -1/6, which doesn't make sense. Therefore, all apples must be cut.Alternatively, maybe one apple is cut into 3 pieces, and the rest into smaller pieces. But if we use the previous method, all apples are cut into 2 or 3. So maybe that's the optimal.Wait, another approach: Let's say we have 5 apples. Let's give each child 5/6. To do this with minimal cuts. If we take 5 apples, and divide each into 1/2 and 1/3. Wait, no. Wait, if we can have different pieces. For example, take 3 apples, cut each into 2 pieces (halves), so 6 halves. Then take the remaining 2 apples, cut each into 3 pieces (thirds), so 6 thirds. Then each child gets one half (1/2) and one third (1/3), which adds to 5/6. That's exactly the same as before.Therefore, this method works. So all apples are cut into 2 or 3 pieces. Therefore, the answer is yes. Therefore, the answer is yes, and the justification is as above.But maybe there's a mistake here. Let me check with fractions. Each half is 3/6, each third is 2/6. Adding them gives 5/6. Correct. So each child gets 3/6 + 2/6 = 5/6. The total number of halves is 6 (from 3 apples), and the total number of thirds is 6 (from 2 apples). So 6 children each get one half and one third. Perfect.Therefore, this division satisfies the condition, and all apples are cut into no more than 3 pieces. Therefore, at least one apple (in fact, all) are not cut into more than 3 pieces. So the answer is yes.Alternatively, maybe there's a different method where some apples are cut into more than 3 pieces, but at least one isn't. But the question allows that. However, the method I found actually satisfies that all apples are cut into 3 or fewer pieces, which is even better. So the answer is definitely yes.Therefore, the conclusion is that it is possible, and the method is cutting 3 apples into 2 pieces each and 2 apples into 3 pieces each, distributing one half and one third to each child. This way, all apples are cut into 3 or fewer pieces, satisfying the condition.Final Answerboxed{Yes}</think>To determine if it is possible to divide 5 apples equally among 6 children such that at least one apple is not cut into more than 3 pieces, we can use the following approach:1. Divide 3 apples into halves: Each apple is cut into 2 pieces, resulting in 3 apples × 2 pieces = 6 half pieces. Each half piece is ( frac{1}{2} ) of an apple.2. Divide 2 apples into thirds: Each apple is cut into 3 pieces, resulting in 2 apples × 3 pieces = 6 third pieces. Each third piece is ( frac{1}{3} ) of an apple.By distributing one half piece (( frac{1}{2} )) and one third piece (( frac{1}{3} )) to each child, each child receives a total of ( frac{1}{2} + frac{1}{3} = frac{3}{6} + frac{2}{6} = frac{5}{6} ) of an apple, which is the required amount. - Number of pieces: Each child receives 2 pieces, and there are 6 children, resulting in 6 × 2 = 12 pieces in total.- Apples used: 3 apples cut into halves (2 pieces each) and 2 apples cut into thirds (3 pieces each) sum up to 5 apples.In this method, all apples are cut into either 2 or 3 pieces, satisfying the condition that at least one apple is not cut into more than 3 pieces. In fact, all apples meet this condition.[boxed{Yes}],<think>Okay, so I need to figure out if I can divide 5 apples equally among 6 children, making sure that at least one apple isn't cut into more than 3 pieces. Hmm, let's start by understanding the problem better.First, each child should get an equal amount of apple. Since there are 5 apples and 6 children, each child should get 5/6 of an apple. Now, the challenge is to distribute these 5 apples into 6 portions of 5/6 each, with the condition that at least one apple remains whole or is cut into no more than 3 pieces. The other apples can be cut into more pieces if needed, but at least one has to be in 3 or fewer pieces.Let me think about how to divide the apples. If I were to cut each apple into six equal pieces, each piece would be 1/6 of an apple. Then, each child would get 5 pieces (since 5 apples divided into sixths give 30 pieces total, and 30 divided by 6 is 5). But in this case, every apple is cut into 6 pieces, which violates the condition that at least one apple isn't cut into more than 3 pieces. So that approach doesn't work.Alternatively, maybe I can find a way to cut some apples into fewer pieces and others into more, such that the total adds up. Let's consider how to get 5/6 per child without cutting all apples into too many pieces.If I have one apple that's cut into 3 equal pieces, each piece is 1/3. Then, if I have another apple cut into 2 equal pieces, each piece is 1/2. But 1/3 + 1/2 is 5/6. Wait, maybe that's a way. Let me see:Suppose we take one apple and cut it into 3 pieces (each 1/3). Then, another apple cut into 2 pieces (each 1/2). Then, each child could get one 1/2 piece and one 1/3 piece. But wait, we have 5 apples. Let's see:If we do that, each child gets 1/2 + 1/3 = 5/6, which is correct. But how many apples does this use? For each child, 1/2 and 1/3. So for 6 children, we need 6*(1/2) = 3 apples cut into halves, and 6*(1/3) = 2 apples cut into thirds. So total apples used would be 3 + 2 = 5. Perfect! That's exactly 5 apples.So, if we take 3 apples and cut each into 2 equal pieces (halves), that gives 6 half pieces. Then, take 2 apples and cut each into 3 equal pieces (thirds), giving 6 third pieces. Each child gets one half and one third, which adds up to 5/6. In this case, the 2 apples that are cut into thirds are each only cut into 3 pieces, satisfying the condition that at least one apple isn't cut into more than 3 pieces. In fact, here we have two apples that are only cut into 3 pieces each. The other three apples are cut into 2 pieces each. So yes, this method works.Wait, but the question says "at least one apple will not be cut into more than 3 pieces". So even if we have one apple cut into 3 pieces and the rest cut into more, that would satisfy the condition. But in my solution, two apples are cut into 3 pieces each, and three apples are cut into 2 pieces each. So actually, all apples are cut into 3 or fewer pieces. So even better, but the condition is still met. So this method works.But let me verify again. Each child gets 1/2 + 1/3 = 5/6. Total apples used: 3 apples cut into halves (each apple gives 2 halves, so 3 apples give 6 halves), and 2 apples cut into thirds (each apple gives 3 thirds, so 2 apples give 6 thirds). Total apples: 3 + 2 = 5. Perfect. So each child gets a half and a third. All apples are cut into either 2 or 3 pieces. Therefore, the apples cut into 3 pieces meet the condition of not being cut into more than 3. So yes, this works.Alternatively, maybe there's another way. Let's see. Suppose we leave one apple whole. Then we have 4 apples left to distribute 5/6 - 1 = -1/6 per child. Wait, that doesn't make sense. If we give one apple to one child, then the remaining 4 apples need to be divided among 6 children. But that's not equal. So that approach probably doesn't work.Alternatively, if we have one apple uncut, how would that affect the distribution? If one child gets a whole apple, then the remaining 4 apples need to be split among the remaining 5 children. Each of those 5 children needs 5/6 - 1 = -1/6, which isn't possible. So we can't have an uncut apple. So the apples need to be cut. But the problem allows cutting, just that at least one apple is not cut into more than 3 pieces.But in the first method I thought of, all apples are cut into 2 or 3 pieces. So that satisfies the condition. So the answer is yes, it's possible.But let me check another approach to ensure. Suppose we have one apple cut into 3 pieces. Each piece is 1/3. Then, each child needs 5/6. So 5/6 - 1/3 = 1/2. So each child needs an additional 1/2. So we need to distribute 1/2 to each child. So that would require 6*(1/2) = 3 apples. So if we have 3 apples cut into halves (2 pieces each), then each child gets a half. So in total, each child gets 1/3 + 1/2 = 5/6. The remaining apple (since total apples used are 1 + 3 = 4) but we have 5 apples. Wait, so there's one apple left. Hmm, so that approach only uses 4 apples. So we need to use all 5. So maybe my previous calculation was wrong.Wait, no. Wait, if we take one apple, cut into 3 pieces (each 1/3). Then, each child gets 1/3, so we need 6*(1/3) = 2 apples. Wait, no. Wait, each apple cut into 3 pieces gives 3*(1/3) = 1 apple. So to get 6*(1/3) pieces, that's 2 apples cut into 3 pieces each. Then, the remaining 3 apples are cut into halves, each giving 2*(1/2) = 1 apple. So 3 apples cut into halves gives 6*(1/2). Then, each child gets 1/2 + 1/3, which is 5/6. So total apples used: 2 + 3 = 5. So yes, that's the same as the first method. Therefore, in this method, 2 apples are cut into 3 pieces each, 3 into 2 pieces each, so indeed, at least one (actually two) apples are not cut into more than 3 pieces. Therefore, the condition is satisfied.So the answer is yes, and the method is as described.But let me check if there's a way where only one apple is not cut into more than 3, and the others are cut into more. For example, suppose one apple is cut into 3 pieces (each 1/3). Then, the remaining 4 apples need to be divided into 6 portions of 5/6 - 1/3 = 1/2. So each child needs an additional 1/2. So 6*(1/2) = 3 apples. So cutting 3 apples into halves (2 pieces each) gives 6 halves. Then, we have used 1 + 3 = 4 apples, but we have 5 apples total. So the remaining apple needs to be distributed. Wait, that's a problem. Because if we have one apple left, how do we distribute it? If we cut that apple into, say, 6 pieces, each 1/6, then each child could get an additional 1/6. But then each child's total would be 1/3 + 1/2 + 1/6 = 1/3 + 2/3 = 1. But they only need 5/6. So that's too much. Alternatively, maybe adjust the portions.Alternatively, if we have one apple cut into 3 pieces (1/3 each), and 4 apples. Let's see:Total required per child: 5/6.If one piece is 1/3, then the remaining per child is 5/6 - 1/3 = 1/2. So each child needs 1/2. For 6 children, that's 3 apples. So if we have 3 apples cut into halves, that's 6 halves. Then, the remaining apple (5 - 1 -3 =1) would need to be divided. But how? If we cut that last apple into 6 pieces of 1/6 each, then each child could get 1/6. But then each child would have 1/3 + 1/2 + 1/6 = 1, which is too much. So that doesn't work. Alternatively, maybe we can adjust the cuts.Wait, maybe instead of giving each child 1/3 + 1/2, we need to use the last apple to adjust the portions. But I think this complicates things. It seems like the initial method where 2 apples are cut into thirds and 3 into halves works perfectly without any leftover. So that's the correct approach.Another way to think about it: 5 apples divided into 6 equal parts. Each part is 5/6. To create these parts, we can combine different fractions. For example, 1/2 + 1/3 = 5/6. So each portion is made by combining a half and a third. Therefore, if we have enough halves and thirds, we can distribute them. Each half comes from an apple cut into 2, each third from an apple cut into 3. Since we need 6 halves (3 apples) and 6 thirds (2 apples), totaling 5 apples. This works and meets the condition.Therefore, the answer is yes, it's possible. The justification is that by cutting 3 apples into 2 pieces each and 2 apples into 3 pieces each, each child can receive one half and one third, totaling 5/6 of an apple. Since the 2 apples cut into 3 pieces each are not cut into more than 3 pieces, the condition is satisfied.I think that's solid. Let me just check for any miscalculations. Total pieces from 3 apples cut into halves: 3*2=6 halves. Total pieces from 2 apples cut into thirds: 2*3=6 thirds. Each child gets one half and one third: 6 children * (1/2 +1/3) = 6*(5/6)=5 apples. Perfect. So all apples are accounted for, each child gets 5/6, and two apples are only cut into three pieces. So the answer is yes.</think>Yes, it is possible to divide 5 apples equally among 6 children such that at least one apple is not cut into more than 3 pieces. Here's how:1. Cut 3 apples into halves: Each apple is divided into 2 equal pieces, yielding (3 times 2 = 6) halves (( frac{1}{2} ) each).2. Cut 2 apples into thirds: Each apple is divided into 3 equal pieces, yielding (2 times 3 = 6) thirds (( frac{1}{3} ) each).3. Distribute the pieces: Each child receives one half (( frac{1}{2} )) and one third (( frac{1}{3} )), totaling ( frac{1}{2} + frac{1}{3} = frac{5}{6} ) of an apple per child.In this method, the two apples cut into thirds satisfy the condition of not being cut into more than 3 pieces. All 5 apples are used, and every child receives an equal share. Answer: boxed{Yes}]

🔑:Solution. Besides the ambiguity between the permission in parentheses and the imposed restriction, the requirement is too weak; we can perform the division so that no apple is cut into more than three pieces, and this in a trivial way, by halving three apples and dividing the other two apples - each into thirds. frac{1}{2}+frac{1}{3}=frac{5}{6} is the mysterious equation (six times) ...

🧠:[<think>Okay, let's see. I need to find the value of f(lg lg 3) given that f(lg log₃ 10) = 5. The function f(x) is defined as a sin x + b ∛x + c ln(x + √(x² + 1)) + 4, where a, b, c are real numbers. The answer choices are A. -5, B. -3, C. 3, D. Varies with a, b, c.First, I need to understand the structure of the function f(x). It has three terms with coefficients a, b, c, and then a constant term 4. The terms are a sine function, a cube root function, and a logarithmic function. Then, there's the constant 4. The problem gives me that when x is lg log₃ 10, f(x) is 5. Then I need to find f(lg lg 3).Wait, first, let's parse these logarithmic expressions. The notation "lg" typically stands for logarithm base 10, right? So lg is log₁₀. And log₃ 10 is logarithm base 3 of 10. Then, lg log₃ 10 would be log base 10 of (log base 3 of 10). Similarly, lg lg 3 is log base 10 of (log base 10 of 3). Let me confirm that.Yes, in many contexts, especially in computer science and mathematics, "lg" is log base 2, but in some contexts, particularly in Chinese literature, "lg" is log base 10. But since the problem uses "log₃" for logarithm base 3, it's likely that "lg" here is log base 10. So I'll proceed under that assumption.So, the input to the function f is x = lg(log₃ 10) = log₁₀(log₃ 10). Then, we need to compute f at x = lg(lg 3) = log₁₀(log₁₀ 3). The question is, what's the value of f at this second x?Given that f(x) = a sin x + b ∛x + c ln(x + √(x² + 1)) + 4. The problem gives that f(lg log₃ 10) = 5. So when x is log₁₀(log₃ 10), the sum of the terms a sin x + b ∛x + c ln(...) + 4 equals 5. Therefore, the terms with a, b, c sum up to 1 because 4 + [something] = 5, so [something] = 1.But then we need to compute f at another x, which is log₁₀(log₁₀ 3). The key here is to see if there's a relationship between the two x values that can make the terms a sin x + b ∛x + c ln(...) invert their sign or something. Let's see.First, let me compute log₃ 10. Remember that log_b a = 1 / log_a b. So log₃ 10 = 1 / log₁₀ 3. Therefore, log₁₀(log₃ 10) = log₁₀(1 / log₁₀ 3) = log₁₀(1) - log₁₀(log₁₀ 3) = 0 - log₁₀(log₁₀ 3) = -log₁₀(log₁₀ 3). Therefore, the first x value is -log₁₀(log₁₀ 3). Let's denote y = log₁₀(log₁₀ 3). Then the first x is -y, and the second x is y.So f(-y) = 5. Then we need to find f(y).So let's analyze the function f(x) when x is replaced by -x. Let's see:f(-x) = a sin(-x) + b ∛(-x) + c ln(-x + √((-x)² + 1)) + 4Simplify each term:sin(-x) = -sin x, so the first term becomes -a sin x.∛(-x) = -∛x, so the second term becomes -b ∛x.For the logarithmic term, ln(-x + √(x² + 1)). Let's analyze the argument inside the ln:-x + √(x² + 1). Let's rationalize this expression. Multiply numerator and denominator by (x + √(x² + 1)):Wait, but to rationalize, let's see:If we have -x + √(x² + 1), multiply by (x + √(x² + 1))/(x + √(x² + 1)):[-x + √(x² + 1)] * [x + √(x² + 1)] / [x + √(x² + 1)]The numerator becomes (-x)(x) + (-x)(√(x² + 1)) + √(x² + 1)(x) + √(x² + 1)*√(x² + 1)Simplify:- x² - x√(x² + 1) + x√(x² + 1) + (x² + 1)Simplify term by term:- x² + x² + 1 + (-x√ + x√)(they cancel out) => 1Therefore, the numerator is 1, and denominator is x + √(x² + 1). Therefore:[-x + √(x² + 1)] = 1 / [x + √(x² + 1)]Therefore, ln(-x + √(x² + 1)) = ln(1 / [x + √(x² + 1)]) = -ln(x + √(x² + 1)).Therefore, the logarithmic term becomes -c ln(x + √(x² + 1)).Therefore, putting it all together:f(-x) = -a sin x - b ∛x - c ln(x + √(x² + 1)) + 4.But the original f(x) is:f(x) = a sin x + b ∛x + c ln(x + √(x² + 1)) + 4.Therefore, f(-x) = -[a sin x + b ∛x + c ln(x + √(x² + 1))] + 4.But from the original function, the expression in brackets is f(x) - 4. Therefore:f(-x) = -[f(x) - 4] + 4 = -f(x) + 4 + 4 = -f(x) + 8.Therefore, we have the relationship f(-x) = -f(x) + 8.Given that, let's use the given information. Let x = y = log₁₀(log₁₀ 3). Then the first x value is -y, and the second is y. So f(-y) = 5. Then using the relationship:f(-y) = -f(y) + 8But f(-y) = 5, so:5 = -f(y) + 8Therefore, solving for f(y):-f(y) = 5 - 8 = -3Multiply both sides by -1:f(y) = 3Therefore, the value of f(log₁₀(log₁₀ 3)) is 3, which is option C.Wait, but let me verify if this is correct. Let me check the steps again.1. Noticing that the first x is -y and the second is y, where y = log₁₀(log₁₀ 3). Then f(-y) = 5.2. Derive the relationship f(-x) = -f(x) + 8.3. Substitute x = y into the relationship: f(-y) = -f(y) + 8.4. Then, since f(-y) = 5, 5 = -f(y) + 8 => f(y) = 3. So the answer is 3, which is option C.Is there any assumption here? The critical step is establishing that f(-x) = -f(x) + 8, which relies on the properties of the function components. Let's confirm that.Given f(x) = a sin x + b ∛x + c ln(x + √(x² + 1)) + 4.Then f(-x) = a sin(-x) + b ∛(-x) + c ln(-x + √(x² + 1)) + 4.Which simplifies to:- a sin x - b ∛x + c ln(-x + √(x² + 1)) + 4.Then we showed that ln(-x + √(x² + 1)) = -ln(x + √(x² + 1)).Therefore, f(-x) = -a sin x - b ∛x - c ln(x + √(x² + 1)) + 4.Which can be written as -[a sin x + b ∛x + c ln(x + √(x² + 1))] + 4.But the original f(x) is [a sin x + b ∛x + c ln(x + √(x² + 1))] + 4.Therefore, the expression [a sin x + b ∛x + c ln(...)] is f(x) - 4.Therefore, substituting:f(-x) = - (f(x) - 4) + 4 = -f(x) + 4 + 4 = -f(x) + 8.Therefore, the relationship holds. Therefore, as long as x is in the domain of both f(x) and f(-x), this relationship is valid.Now, we need to check if y is in the domain. The domain of f(x) requires that x + √(x² + 1) > 0, because of the logarithm. But since √(x² + 1) ≥ |x|, so if x is positive, √(x² +1) > x, so x + √(x² +1) > 0. If x is negative, then √(x² +1) > |x| = -x, so x + √(x² +1) = √(x² +1) + x > 0 because √(x² +1) > -x. Therefore, the argument of the logarithm is always positive, so the domain is all real numbers. Therefore, both y and -y are in the domain.Therefore, the steps are valid, and hence f(y) = 3. Therefore, the answer is C.But wait, the problem gives f(lg log₃ 10) = 5. Let's check if lg log₃ 10 is indeed -y.Given y = lg lg 3 = log₁₀(log₁₀ 3). Then log₃ 10 = 1 / log₁₀ 3, so lg log₃ 10 = log₁₀(1 / log₁₀ 3) = - log₁₀(log₁₀ 3) = -y. Therefore, f(-y) = 5, so f(y) = 3. Therefore, this is correct.So the answer should be C, 3.Therefore, the answer is boxed{C}Final Answerboxed{C}</think>Given the function ( f(x) = a sin x + b sqrt[3]{x} + c ln left( x + sqrt{x^2 + 1} right) + 4 ) where ( a ), ( b ), and ( c ) are real numbers, and the information that ( f(lg log_3 10) = 5 ), we need to find the value of ( f(lg lg 3) ).First, we note that ( lg ) denotes logarithm base 10. We need to express the arguments of the function in terms of a common variable. Let ( y = lg lg 3 ). Then, we need to find ( f(y) ).We start by analyzing the given argument ( lg log_3 10 ):- ( log_3 10 = frac{1}{log_{10} 3} )- Therefore, ( lg log_3 10 = lg left( frac{1}{log_{10} 3} right) = - lg log_{10} 3 )Let ( y = lg log_{10} 3 ). Then, the given argument becomes ( -y ), so we have ( f(-y) = 5 ).Next, we analyze the function ( f(-x) ):- ( f(-x) = a sin(-x) + b sqrt[3]{-x} + c ln left( -x + sqrt{(-x)^2 + 1} right) + 4 )- Simplifying each term: - ( sin(-x) = -sin x ) - ( sqrt[3]{-x} = -sqrt[3]{x} ) - ( ln left( -x + sqrt{x^2 + 1} right) = ln left( frac{1}{x + sqrt{x^2 + 1}} right) = -ln left( x + sqrt{x^2 + 1} right) )Thus, ( f(-x) = -a sin x - b sqrt[3]{x} - c ln left( x + sqrt{x^2 + 1} right) + 4 ). This can be rewritten as:[ f(-x) = - left( a sin x + b sqrt[3]{x} + c ln left( x + sqrt{x^2 + 1} right) right) + 4 ][ f(-x) = - (f(x) - 4) + 4 ][ f(-x) = -f(x) + 8 ]Given ( f(-y) = 5 ), we substitute into the relationship:[ 5 = -f(y) + 8 ]Solving for ( f(y) ):[ -f(y) = 5 - 8 ][ -f(y) = -3 ][ f(y) = 3 ]Therefore, the value of ( f(lg lg 3) ) is (boxed{C}).,<think>Okay, let's try to figure out this problem step by step. The function given is ( f(x) = a sin x + b sqrt[3]{x} + c ln left(x + sqrt{x^2 + 1}right) + 4 ), and we know that ( f(lg log_{3} 10) = 5 ). We need to find the value of ( f(lg lg 3) ) from the options provided.First, let's parse the function and the given information. The function ( f(x) ) is composed of several terms: a sine term, a cube root term, a natural logarithm term, and a constant 4. The coefficients ( a ), ( b ), and ( c ) are real numbers. We are told that when we input ( lg log_{3} 10 ) into ( f ), the output is 5. We need to find the output when we input ( lg lg 3 ).Wait, the notation here might be a bit confusing. The problem uses ( lg ), which typically stands for base 10 logarithm. However, in some contexts, especially in different countries, ( lg ) can mean base 2 logarithm. But given the presence of ( log_{3} 10 ), I think here ( lg ) is base 10, so ( lg x = log_{10} x ).So, first, let's compute the arguments of the function ( f ). The first argument is ( lg log_{3} 10 ), and the second is ( lg lg 3 ).Let me compute ( lg log_{3} 10 ). Let's start with the inner logarithm: ( log_{3} 10 ). This is the logarithm of 10 with base 3. To compute this, we can use the change of base formula: ( log_{3} 10 = frac{ln 10}{ln 3} ) or ( frac{lg 10}{lg 3} = frac{1}{lg 3} ). Wait, actually, if ( lg ) is base 10, then ( log_{3} 10 = frac{log_{10} 10}{log_{10} 3} = frac{1}{lg 3} ). Therefore, ( lg log_{3} 10 = lg left( frac{1}{lg 3} right) = lg left( frac{1}{log_{10} 3} right) ).Similarly, the other argument is ( lg lg 3 ). Let's compute that as well. First, compute ( lg 3 ), which is ( log_{10} 3 ), and then take the logarithm base 10 of that result. So, ( lg lg 3 = log_{10} (log_{10} 3) ).Hmm, interesting. So, the two arguments ( lg log_{3} 10 ) and ( lg lg 3 ) might be related. Let me check:From earlier, we saw that ( log_{3} 10 = frac{1}{log_{10} 3} ), so ( lg log_{3} 10 = lg left( frac{1}{lg 3} right) = lg left( (lg 3)^{-1} right) = - lg (lg 3) = - log_{10} (log_{10} 3) ).Therefore, ( lg log_{3} 10 = - lg lg 3 ). That means the two arguments are negatives of each other. Let's denote ( y = lg lg 3 ), then the first argument is ( -y ).So, we have ( f(-y) = 5 ), and we need to find ( f(y) ).Looking at the function ( f(x) ), let's analyze its components:1. ( a sin x ): Sine is an odd function, so ( sin(-x) = - sin x ).2. ( b sqrt[3]{x} ): The cube root function is also odd, since ( sqrt[3]{-x} = - sqrt[3]{x} ).3. ( c ln left(x + sqrt{x^2 + 1}right) ): Let's check if this term is odd or even. Let me compute ( ln(-x + sqrt{(-x)^2 + 1}) ). Wait, ( sqrt{(-x)^2 + 1} = sqrt{x^2 +1} ), so the argument inside the logarithm becomes ( -x + sqrt{x^2 +1} ). Let's see:Note that ( x + sqrt{x^2 +1} ) is a known expression. In fact, ( ln(x + sqrt{x^2 +1}) ) is the inverse hyperbolic sine function, ( sinh^{-1} x ). The hyperbolic sine function is odd, so its inverse is also odd. Therefore, ( sinh^{-1}(-x) = - sinh^{-1} x ). Therefore, ( ln(-x + sqrt{x^2 +1}) = ln left( sqrt{x^2 +1} - x right) ). Let me check if this is equal to ( - ln(x + sqrt{x^2 +1}) ).Multiply ( sqrt{x^2 +1} - x ) by ( sqrt{x^2 +1} + x ):( (sqrt{x^2 +1} - x)(sqrt{x^2 +1} + x) = (x^2 +1) - x^2 = 1 ).Therefore, ( sqrt{x^2 +1} - x = frac{1}{sqrt{x^2 +1} + x} ). Taking natural logarithm of both sides:( ln(sqrt{x^2 +1} - x) = ln left( frac{1}{sqrt{x^2 +1} + x} right) = - ln(sqrt{x^2 +1} + x) ).Hence, ( ln(-x + sqrt{x^2 +1}) = - ln(x + sqrt{x^2 +1}) ). Therefore, the logarithmic term is also odd.So, putting all together, each of the three terms ( a sin x ), ( b sqrt[3]{x} ), and ( c ln(x + sqrt{x^2 +1}) ) is an odd function. Therefore, their combination ( a sin x + b sqrt[3]{x} + c ln(x + sqrt{x^2 +1}) ) is also an odd function. Let's denote ( g(x) = a sin x + b sqrt[3]{x} + c ln(x + sqrt{x^2 +1}) ). Then, ( g(-x) = -g(x) ).Therefore, the function ( f(x) = g(x) + 4 ), so:( f(-x) = g(-x) + 4 = -g(x) + 4 ).Given that ( f(-y) = 5 ), we have:( -g(y) + 4 = 5 )Solving for ( g(y) ):( -g(y) = 1 )( g(y) = -1 )Now, we need to find ( f(y) = g(y) + 4 = -1 + 4 = 3 ).Therefore, the value of ( f(lg lg 3) ) is 3, which is option C.Wait, let me verify once again to make sure I didn't skip any steps or make a mistake.First, confirm that ( lg log_{3} 10 = - lg lg 3 ):Yes, because ( log_{3} 10 = 1 / log_{10} 3 ), so ( lg (1 / log_{10} 3) = - lg log_{10} 3 ), which is ( - lg lg 3 ), assuming ( lg 3 = log_{10} 3 ). So that's correct.Then, since ( f(-y) = 5 ), and ( f(-y) = -g(y) + 4 ), so ( -g(y) = 1 ), hence ( g(y) = -1 ). Then ( f(y) = -1 + 4 = 3 ). That seems correct.Therefore, the answer should be C. 3.But wait, let me check if the function ( ln(x + sqrt{x^2 +1}) ) is indeed odd. We did the computation earlier, and found that ( ln(-x + sqrt{x^2 +1}) = - ln(x + sqrt{x^2 +1}) ). So that term is odd. So all three terms in ( g(x) ) are odd, so their linear combination is also odd. Therefore, ( g(-x) = -g(x) ). Hence, the reasoning holds.Therefore, the answer is 3, which is option C.Final Answerboxed{C}</think>Given the function ( f(x) = a sin x + b sqrt[3]{x} + c ln left(x + sqrt{x^2 + 1}right) + 4 ) where ( a ), ( b ), and ( c ) are real numbers, and knowing that ( f(lg log_{3} 10) = 5 ), we need to find the value of ( f(lg lg 3) ).First, we note that the arguments (lg log_{3} 10) and (lg lg 3) are related. Using the change of base formula, we find:[log_{3} 10 = frac{1}{log_{10} 3}]Thus,[lg log_{3} 10 = lg left( frac{1}{log_{10} 3} right) = -lg (log_{10} 3) = -lg lg 3]Let ( y = lg lg 3 ). Then, the given argument becomes ( -y ), and we know ( f(-y) = 5 ).Next, we analyze the function ( f(x) ). The terms ( a sin x ), ( b sqrt[3]{x} ), and ( c ln left(x + sqrt{x^2 + 1}right) ) are all odd functions. Therefore, their combination ( g(x) = a sin x + b sqrt[3]{x} + c ln left(x + sqrt{x^2 + 1}right) ) is also an odd function, meaning ( g(-x) = -g(x) ).Given ( f(-y) = 5 ), we can write:[f(-y) = g(-y) + 4 = -g(y) + 4 = 5]Solving for ( g(y) ):[-g(y) + 4 = 5 implies -g(y) = 1 implies g(y) = -1]Finally, we find ( f(y) ):[f(y) = g(y) + 4 = -1 + 4 = 3]Thus, the value of ( f(lg lg 3) ) is (boxed{C}).]

🔑:4 C. Since the function y=ln (x+sqrt{x+1}) is an odd function.Therefore, g(x)=a sin x+b sqrt[3]{x}+c ln left(x+sqrt{x^{2}+1}right) is an odd function, which means f(x)-4 is an odd function.Also,5=fleft(lg log _{3}^{10}right)=fleft(lg ^{-1} 3right)=f(-lg lg 3),which meansf(lg lg 3)-4=-(f(-lg lg 3)-4)=-1 .Thus, f(lg lg 3)=3.