🧠:[<think>Okay, so I need to figure out which of the given monomials is of the same type as -2mn². The options are A: 5mn, B: 2n², C: 3m²n, and D: mn². Hmm, let me recall what it means for monomials to be of the same type. I think it's about them being like terms, right? Like terms have the exact same variables raised to the same exponents. So, even if the coefficients are different, as long as the variables and their exponents match, they're considered the same type.Alright, let's break down -2mn². The variables here are m and n. The exponent of m is 1 (since it's just m) and the exponent of n is 2. So, the term is m¹n². Now, let's check each option one by one.Starting with option A: 5mn. The variables are m and n, both to the first power. So that's m¹n¹. Comparing to our original term m¹n², the exponents for n are different. So, the exponents don't match. Therefore, A is not the same type.Option B: 2n². Here, the variable is only n, raised to the second power. So, n². But the original term has both m and n. Since B doesn't have the variable m at all, it's missing a variable. So, definitely not the same type. B is out.Option C: 3m²n. Variables here are m² and n. So, m²n¹. Comparing to original m¹n², the exponents for m and n are different. The exponent for m is higher here (2 vs. 1) and the exponent for n is lower (1 vs. 2). So, different exponents on both variables. Not the same type. So C is not the answer.Option D: mn². Let's check. Variables are m and n. Exponent of m is 1, exponent of n is 2. So that's m¹n². Exactly the same variables and exponents as the original term -2mn². The coefficient here is 1 (since it's just mn²), but coefficients don't matter for being the same type. Only the variables and their exponents. Therefore, D has the same variable part. So D is the same type as -2mn².Wait, let me just double-check to make sure I didn't mix up any exponents. Original term: m is to the first power, n squared. Option D: m is also first power, n squared. Yep. So variables and exponents match. Coefficients can be different, so even though -2 and 1 are different coefficients, they are still like terms. So, D is the correct answer.Final Answerboxed{D}</think>To determine which monomial is of the same type as (-2mn^2), we need to check if they have the same variables raised to the same exponents. The given monomial (-2mn^2) has variables (m) and (n) with exponents 1 and 2 respectively.- Option A: (5mn) has variables (m) and (n) with exponents 1 and 1. The exponent of (n) is different.- Option B: (2n^2) has only the variable (n) with exponent 2. It is missing the variable (m).- Option C: (3m^2n) has variables (m) and (n) with exponents 2 and 1. The exponents of both variables are different.- Option D: (mn^2) has variables (m) and (n) with exponents 1 and 2. This matches the variables and exponents of (-2mn^2).Thus, the monomial (mn^2) is of the same type as (-2mn^2).(boxed{D}),<think>Okay, so I need to figure out which of the options A through D is the same type as -2mn². Hmm, first, what does "same type" mean here? I think it refers to like terms, right? Like terms are terms that have the same variables raised to the same powers. The coefficients can be different, but the variable parts must match exactly. Let me confirm that.So, the original monomial is -2mn². Let's break it down. The coefficient is -2, and the variables are m and n. The exponent of m is 1 (since it's just m) and the exponent of n is 2. So the variable part is m¹n².Now, I need to check each option to see if their variable parts are m¹n² as well.Starting with option A: 5mn. Let's see. The coefficient is 5, variables are m and n. Exponents: m is 1, n is 1. So variable part is m¹n¹. That's different from m¹n² because the exponent on n is 1 here instead of 2. So A is not the same type.Option B: 2n². Coefficient is 2, variable is n². Wait, there's no m here. So the variable part is just n². Comparing to the original, which has both m and n. So this is missing the m. Therefore, B is not the same type.Option C: 3m²n. Coefficient is 3. Variables are m² and n. Exponents: m is 2, n is 1. Variable part is m²n¹. Original was m¹n². So here, the exponents are swapped: m has a higher power and n has a lower one. So variable parts don't match. So C is out.Option D: mn². Coefficient here is 1 (since it's just written as mn²). Variables are m and n. Exponents: m is 1, n is 2. So variable part is m¹n², which is exactly the same as the original -2mn². The coefficients are different (-2 vs. 1), but the question is about the type, which depends on the variables and their exponents. So D should be the correct answer.Wait, let me double-check. The original is -2mn², so variables m and n with exponents 1 and 2. Option D is mn², same variables, same exponents. Yes. So even though the coefficients are different, they are considered the same type. So the answer is D.Final Answerboxed{D}</think>To determine which monomial is of the same type as (-2mn^2), we need to identify a term with the same variables raised to the same powers, regardless of the coefficient. The given monomial (-2mn^2) has the variables (m) and (n) with exponents 1 and 2, respectively. - Option A: (5mn) has variables (m) and (n) with exponents 1 and 1. This is not the same.- Option B: (2n^2) has only the variable (n) with exponent 2. This is missing the variable (m), so it is not the same.- Option C: (3m^2n) has variables (m) and (n) with exponents 2 and 1. The exponents are swapped, so this is not the same.- Option D: (mn^2) has variables (m) and (n) with exponents 1 and 2. This matches the variable part of (-2mn^2).Thus, the correct answer is (boxed{D}).]

🔑:To determine which monomial is of the same type as -2mn^{2}, we need to compare the variables and their corresponding exponents in each option with those in -2mn^{2}.- Option A: 5mn - Variables: m, n - Exponents: m^1, n^1 - Comparison: The variables match, but the exponents do not match those in -2mn^{2} (m^1n^2). Therefore, 5mn is not of the same type.- Option B: 2n^{2} - Variables: n - Exponents: n^2 - Comparison: Missing the variable m present in -2mn^{2}. Therefore, 2n^{2} is not of the same type.- Option C: 3m^{2}n - Variables: m, n - Exponents: m^2, n^1 - Comparison: The variables match, but the exponents do not match those in -2mn^{2} (m^1n^2). Therefore, 3m^{2}n is not of the same type.- Option D: mn^{2} - Variables: m, n - Exponents: m^1, n^2 - Comparison: Both the variables and their exponents match exactly with those in -2mn^{2}. Therefore, mn^{2} is of the same type.Given the analysis above, the monomial that is of the same type as -2mn^{2} is:boxed{D}

🧠:[<think>Alright, let's tackle this problem step by step. So, we have a geometric sequence {a_n} with a common ratio q > 1. The given conditions are a₁ + a₆ = 8 and a₃ * a₄ = 12. We need to find the value of a_{2018} divided by a_{2013}. Hmm, okay.First, let me recall some properties of geometric sequences. In a geometric sequence, each term after the first is found by multiplying the previous term by a constant called the common ratio, which in this case is q. So, the nth term of a geometric sequence can be written as a_n = a₁ * q^{n-1}. That's the formula I need to use here.Given that, let's write down expressions for the terms mentioned in the problem. First condition: a₁ + a₆ = 8. Using the formula, a₆ is a₁ * q^{5} (since n=6, so exponent is 6-1=5). Therefore, the first equation becomes:a₁ + a₁*q^5 = 8.We can factor out a₁:a₁(1 + q^5) = 8. Let's keep this as equation (1).Second condition: a₃ * a₄ = 12. Let's write expressions for a₃ and a₄. a₃ = a₁ * q^{2} (since 3-1=2)a₄ = a₁ * q^{3} (since 4-1=3)So, multiplying these together:(a₁ * q^2) * (a₁ * q^3) = a₁² * q^{5} = 12. Let's call this equation (2).So now, we have two equations:1) a₁(1 + q^5) = 82) a₁² * q^5 = 12Hmm. Let me see. Perhaps we can solve these two equations for a₁ and q. Let's try to express a₁ from equation (1) and substitute into equation (2).From equation (1):a₁ = 8 / (1 + q^5)Substitute this into equation (2):(8 / (1 + q^5))² * q^5 = 12Let's compute this:64 / (1 + q^5)^2 * q^5 = 12Multiply both sides by (1 + q^5)^2:64 * q^5 = 12 * (1 + q^5)^2Divide both sides by 4 to simplify:16 * q^5 = 3 * (1 + q^5)^2Hmm. Let's let’s denote x = q^5. Then, the equation becomes:16x = 3(1 + x)^2Expanding the right side:16x = 3(x² + 2x + 1)16x = 3x² + 6x + 3Bring all terms to one side:3x² + 6x + 3 - 16x = 0Simplify:3x² - 10x + 3 = 0So, quadratic equation in x: 3x² -10x +3=0. Let's solve for x.Using quadratic formula: x = [10 ± sqrt(100 - 36)] / 6Compute discriminant sqrt(64) = 8Thus, x = [10 ± 8]/6So, two solutions:x = (10 + 8)/6 = 18/6 = 3x = (10 - 8)/6 = 2/6 = 1/3So, x = 3 or x = 1/3. But x = q^5, and since q > 1, then x must be greater than 1^5 = 1. Therefore, x = 3 is the valid solution. So, x = 3 => q^5 = 3 => q = 3^{1/5}Therefore, the common ratio q is the fifth root of 3.Now, let's find a₁. From equation (1):a₁ = 8 / (1 + q^5) = 8 / (1 + 3) = 8 / 4 = 2So, a₁ = 2.Now, the problem asks for a_{2018} / a_{2013}.Let's compute these terms. Using the formula a_n = a₁ * q^{n-1}a_{2018} = 2 * q^{2017}a_{2013} = 2 * q^{2012}Therefore, the ratio is:a_{2018} / a_{2013} = (2 * q^{2017}) / (2 * q^{2012}) ) = q^{2017 - 2012} = q^5But we already found that q^5 = 3. So, the ratio is 3.Therefore, the answer is 3.Wait, that seems straightforward, but let me verify.Given that q^5 = 3, so indeed, the ratio between terms five apart is q^5. Since 2018 - 2013 = 5, so the ratio is q^5 = 3. So yes, the answer is 3. That makes sense.Let me check if all steps are correct.We set up the equations correctly. The substitution steps are correct, solved for x = q^5, found x=3. Then, computed a₁=2. Then, computed the ratio as q^5. All steps check out.Therefore, the answer should be 3.Final Answerboxed{3}</think>Given a geometric sequence ({a_n}) with a common ratio (q > 1), and the conditions (a_1 + a_6 = 8) and (a_3 a_4 = 12), we need to find the value of (dfrac{a_{2018}}{a_{2013}}).First, we express the terms using the formula for the (n)-th term of a geometric sequence, (a_n = a_1 cdot q^{n-1}):1. The first condition (a_1 + a_6 = 8) translates to: [ a_1 + a_1 q^5 = 8 implies a_1 (1 + q^5) = 8 ]2. The second condition (a_3 a_4 = 12) translates to: [ (a_1 q^2)(a_1 q^3) = 12 implies a_1^2 q^5 = 12 ]From the first equation, we solve for (a_1): [ a_1 = frac{8}{1 + q^5} ]Substituting this into the second equation: [ left(frac{8}{1 + q^5}right)^2 q^5 = 12 ] Simplifying, we get: [ frac{64 q^5}{(1 + q^5)^2} = 12 implies 16 q^5 = 3 (1 + q^5)^2 ]Let (x = q^5), then the equation becomes: [ 16 x = 3 (1 + x)^2 implies 3x^2 - 10x + 3 = 0 ]Solving this quadratic equation using the quadratic formula: [ x = frac{10 pm sqrt{64}}{6} = frac{10 pm 8}{6} ] This gives solutions (x = 3) or (x = frac{1}{3}). Since (q > 1), we have (x = 3), thus (q^5 = 3).Substituting (q^5 = 3) back into the expression for (a_1): [ a_1 = frac{8}{1 + 3} = 2 ]To find (dfrac{a_{2018}}{a_{2013}}): [ frac{a_{2018}}{a_{2013}} = frac{a_1 cdot q^{2017}}{a_1 cdot q^{2012}} = q^5 = 3 ]Thus, the answer is (boxed{3}).,<think>Okay, so I have this problem about a geometric sequence. Let me see. The problem states that in a geometric sequence {a_n} with a common ratio q > 1, it's given that a₁ + a₆ = 8 and a₃a₄ = 12. Then, I need to find the value of a₂₀₁₈ divided by a₂₀₁₃. Hmm, okay. Let me try to break this down step by step.First, recalling that in a geometric sequence, each term is the previous term multiplied by the common ratio q. So, the nth term can be written as a_n = a₁ * q^(n-1). That seems right. So, for example, a₆ would be a₁ * q^(5), since (6-1)=5. Similarly, a₃ = a₁ * q² and a₄ = a₁ * q³. Okay.Given that, the first equation is a₁ + a₆ = 8. Substituting the formula for a₆, that would be a₁ + a₁*q⁵ = 8. Factoring out a₁, this becomes a₁(1 + q⁵) = 8. So, equation (1): a₁(1 + q⁵) = 8.The second equation is a₃a₄ = 12. Let's substitute those terms as well. a₃ is a₁*q² and a₄ is a₁*q³. So, multiplying them together: (a₁*q²)(a₁*q³) = a₁²*q^(2+3) = a₁²*q⁵ = 12. So, equation (2): a₁²*q⁵ = 12.Now, we have two equations:1) a₁(1 + q⁵) = 82) a₁²*q⁵ = 12We need to solve for a₁ and q, probably, and then use those to compute a₂₀₁₈ / a₂₀₁₃. Let's see.First, maybe we can express a₁ from equation (1) and substitute into equation (2). Let's try that.From equation (1): a₁ = 8 / (1 + q⁵). Then substitute into equation (2):(8 / (1 + q⁵))² * q⁵ = 12.Let me compute that. So, (64 / (1 + q⁵)²) * q⁵ = 12.Multiply both sides by (1 + q⁵)^2:64 q⁵ = 12 (1 + q⁵)^2.Divide both sides by 4 to simplify:16 q⁵ = 3 (1 + q⁵)^2.Hmm, that's 16q⁵ = 3(1 + q⁵)^2. Let's let’s rearrange this equation.First, expand the right-hand side: 3(1 + 2q⁵ + q^{10}) = 3 + 6q⁵ + 3q^{10}.So, the equation becomes:16q⁵ = 3 + 6q⁵ + 3q^{10}Bring all terms to one side:3q^{10} + 6q⁵ + 3 - 16q⁵ = 0Simplify:3q^{10} - 10q⁵ + 3 = 0So, we have 3q^{10} - 10q⁵ + 3 = 0.Wait, this is a quadratic in terms of q⁵. Let me substitute y = q⁵. Then the equation becomes:3y² - 10y + 3 = 0Yes, that seems manageable. Let's solve for y.Using the quadratic formula: y = [10 ± sqrt(100 - 36)] / 6 = [10 ± sqrt(64)] / 6 = [10 ± 8] / 6.So, two solutions:y = (10 + 8)/6 = 18/6 = 3y = (10 - 8)/6 = 2/6 = 1/3So, y = 3 or y = 1/3. But y = q⁵, and since q > 1 (given), then q⁵ must be greater than 1^5 = 1, so y = 3 is acceptable, y = 1/3 is less than 1, so we discard that. Therefore, y = 3, so q⁵ = 3. Therefore, q = 3^(1/5). Hmm, 3 to the power 1/5. Not a nice integer, but maybe we can keep it as is for now.So, q = 3^(1/5). Then, from equation (1), a₁(1 + q⁵) = 8. But q⁵ = 3, so 1 + q⁵ = 4. Therefore, a₁*4 = 8, so a₁ = 8/4 = 2. So, a₁ = 2.Therefore, the first term is 2, the common ratio is 3^(1/5).Now, the question is to find a₂₀₁₈ / a₂₀₁₃.Let's compute that. Since a_n = a₁ * q^(n-1), then:a₂₀₁₈ = a₁ * q^(2017)a₂₀₁₃ = a₁ * q^(2012)Therefore, their ratio is:a₂₀₁₈ / a₂₀₁₃ = [a₁ * q^2017] / [a₁ * q^2012] = q^(2017 - 2012) = q^5.But q^5 = 3, as established earlier (since q⁵ = 3). Therefore, the ratio is 3.Wait, that's straightforward. The ratio of terms five apart in a geometric sequence is q^5, which we already found to be 3. So, the answer is 3. Therefore, the value of a₂₀₁₈ divided by a₂₀₁₃ is 3. Let me just verify everything to make sure.First, found that q⁵ = 3, so q = 3^(1/5). Then a₁ = 2. So, let's check the initial equations.Check a₁ + a₆: a₁ = 2, a₆ = 2 * q^5 = 2 * 3 = 6. So, 2 + 6 = 8, which matches the first condition.Check a₃a₄: a₃ = 2 * q², a₄ = 2 * q³. Multiply them: 4 * q^(2+3) = 4 * q⁵ = 4 * 3 = 12, which matches the second condition. So, all checks out.Therefore, the ratio a₂₀₁₈ / a₂₀₁₃ is q^5 = 3. So, the answer is 3. That seems correct.Final Answerboxed{3}</think>Given a geometric sequence ({a_n}) with a common ratio (q > 1), and the conditions (a_1 + a_6 = 8) and (a_3 a_4 = 12), we need to find the value of (dfrac{a_{2018}}{a_{2013}}).First, recall the general term of a geometric sequence: (a_n = a_1 cdot q^{n-1}).From the given conditions:1. (a_1 + a_6 = 8) [ a_1 + a_1 q^5 = 8 implies a_1 (1 + q^5) = 8 ]2. (a_3 a_4 = 12) [ (a_1 q^2)(a_1 q^3) = 12 implies a_1^2 q^5 = 12 ]We solve these equations by expressing (a_1) from the first equation and substituting into the second:[a_1 = frac{8}{1 + q^5}]Substituting into the second equation:[left(frac{8}{1 + q^5}right)^2 q^5 = 12 implies frac{64 q^5}{(1 + q^5)^2} = 12 implies 16 q^5 = 3 (1 + q^5)^2]Let (y = q^5), then the equation becomes:[16y = 3(1 + y)^2 implies 3y^2 - 10y + 3 = 0]Solving this quadratic equation:[y = frac{10 pm sqrt{100 - 36}}{6} = frac{10 pm 8}{6} implies y = 3 text{ or } y = frac{1}{3}]Since (q > 1), we have (q^5 = 3). Thus, (q = 3^{1/5}).Substituting (q^5 = 3) back into the first equation:[a_1 (1 + 3) = 8 implies a_1 = 2]To find (dfrac{a_{2018}}{a_{2013}}):[dfrac{a_{2018}}{a_{2013}} = dfrac{a_1 cdot q^{2017}}{a_1 cdot q^{2012}} = q^5 = 3]Thus, the final answer is (boxed{3}).]

🔑:AnalysisThis problem mainly examines the properties of geometric sequences. The key to solving the problem is to use a_3a_4=a_1a_6, and it also tests the ability to transform, which is a basic question.According to the properties of the geometric sequence, the question can be simplified to dfrac{a_{2018}}{a_{2013}}=q^{5}. Combined with the conditions in the question, we can get a_1=2, a_6=6, and then obtain the answer.SolutionGiven: The sequence {a_n} is a geometric sequence, so dfrac{a_{2018}}{a_{2013}}=q^{5}. Since {a_n} is a geometric sequence and a_3a_4=12, we have a_3a_4=a_1a_6=12. Because a_1+a_6=8 and the common ratio q > 1, solving gives a_1=2, a_6=6, thus q^{5}=dfrac{a_{6}}{a_{1}}=3. Therefore, the answer is boxed{3}.

🧠:[<think>Alright, let's see. I need to solve problem 2a and 2b based on the given information. Let me start with part a. The problem states that the sum of the odd numbers from 1 to 2013 is equal to (2014 multiplied by 1007) divided by 2, which simplifies to 1007 squared. Similarly, the sum from 1 to 105 is (106 multiplied by 53) divided by 2, which is 53 squared. Then, the equation given is that 1007² divided by x equals 53². Solving for x, we get x equals (1007 over 53) squared, which is 19 squared, so x is 361. Hmm, okay. Let me verify each step to make sure I understand it properly. First, for the sum of consecutive odd numbers. I remember that the sum of the first n odd numbers is n squared. So, for example, 1 + 3 + 5 + ... + (2n-1) = n². Is that right? Let me check with small n. If n=1: 1=1², okay. n=2: 1+3=4=2². n=3: 1+3+5=9=3². Yeah, that seems correct. So, if we have a sequence of odd numbers starting at 1 and ending at some odd number, the sum is the square of the number of terms. But in part a, the first sum is from 1 to 2013. Let me check how many terms that is. Since the sequence is 1, 3, 5, ..., 2013. The nth term of an arithmetic sequence is given by a_n = a_1 + (n-1)d. Here, a_1=1, d=2, and a_n=2013. So, 2013 = 1 + (n-1)*2. Solving for n: 2013 -1 = (n-1)*2 => 2012 = 2(n-1) => n-1=1006 => n=1007. So, the number of terms is 1007, which means the sum should be 1007². The given solution says (2014*1007)/2. Let me check that. 2014*1007/2. Well, 2014/2 is 1007, so 1007*1007, which is 1007². Yep, that's consistent. So that part checks out.Similarly, the sum from 1 to 105. Let's find the number of terms. Again, using a_n=1 + (n-1)*2=105. So, 105-1=104=2(n-1) => n-1=52 => n=53. So, the sum is 53², and the given formula is (106*53)/2. 106/2=53, so 53*53=53². Correct. Therefore, the equation given is 1007² / x = 53². Solving for x, we cross-multiply: x = (1007²) / (53²) = (1007/53)². Then, they compute 1007 divided by 53. Let me do that division. 53*19 is 1007, right? 53*10=530, 53*20=1060, which is too much. So, 53*19: 53*10=530, 53*9=477, so 530+477=1007. Yes, 53*19=1007. So, 1007/53=19. Therefore, x=19²=361. That makes sense. Okay, part a seems solid. Now, part b. They say x=361, so (361-1):10 = y². Let's parse that. The colon is probably representing division. So, (361 -1)/10 = y². 361-1=360. 360 divided by 10 is 36. Therefore, y²=36, so y=±6, hence the solution set S={-6,6}. Let me verify each step again. Starting with x=361. Then, (x -1)/10 = y². So, (361 -1)=360, 360/10=36. Yep, 36 is a perfect square, so y=6 or y=-6. So, the solutions are -6 and 6. Wait, but in part a, the equation was 1007² / x = 53², leading to x=361. Then in part b, substituting x=361 into another equation. So, problem 2 is structured as part a leading to part b? That is, part a is solving for x, and part b uses that x to solve for y. Therefore, the steps are correct. Let me just recap to ensure no mistakes. Starting with part a: sum of odds from 1 to 2013 is 1007², sum from 1 to 105 is 53². Therefore, 1007² / x = 53² implies x=(1007/53)²=19²=361. Then, in part b, using x=361, compute (361 -1)/10 = 360/10=36=y², so y=±6. Everything checks out. The key steps here are recognizing the sum of odd numbers formula, performing the division to find how many terms there are, then solving for x by equating the ratio of the two sums squared. Then, substituting x into the next equation to solve for y. I don't see any mistakes here. The arithmetic all seems correct. 53*19 is indeed 1007, and 19 squared is 361. Then 361-1 is 360, divided by 10 is 36, which is 6 squared. So, the solutions for y are 6 and -6. Therefore, the answers are x=361 and S={-6,6}. I think that's all.Final Answera) boxed{361}b) boxed{{-6, 6}}</think>To solve part a, we start by recognizing that the sum of the first ( n ) odd numbers is ( n^2 ). - For the sum ( 1 + 3 + 5 + ldots + 2013 ): - The number of terms ( n ) is calculated as follows: [ 2013 = 1 + (n-1) cdot 2 implies 2012 = 2(n-1) implies n = 1007 ] - The sum is ( 1007^2 ).- For the sum ( 1 + 3 + 5 + ldots + 105 ): - The number of terms ( n ) is calculated as follows: [ 105 = 1 + (n-1) cdot 2 implies 104 = 2(n-1) implies n = 53 ] - The sum is ( 53^2 ).The equation given is:[frac{1007^2}{x} = 53^2]Solving for ( x ):[x = left( frac{1007}{53} right)^2 = 19^2 = 361]For part b, using ( x = 361 ):[(361 - 1) div 10 = y^2 implies frac{360}{10} = 36 = y^2 implies y = pm 6]Thus, the solution set is ( {-6, 6} ). Final Answera) boxed{361}b) boxed{{-6, 6}},<think>Okay, let me try to solve part a) first. The problem shows an equation involving the sum of odd numbers from 1 to 2013 and from 1 to 105. Then it says the equation is equivalent to 1007² divided by x equals 53², leading to x being (1007/53)², which is 19², so x is 361. Hmm, I need to understand how they got that.First, the sum of odd numbers. I remember that the sum of the first n odd numbers is n². So, for example, 1+3+5+...+(2n-1) = n². Let me verify that with the given examples.In the first sum, 1+3+5+...+2013. Let's figure out how many terms there are. The nth odd number is 2n-1. So if the last term is 2013, then 2n-1 = 2013. Solving for n: 2n = 2014, so n = 1007. So the sum of the first 1007 odd numbers is 1007². That matches the given equation where they wrote it as (2014 * 1007)/2. Wait, 2014*1007/2. Let me check that. 2014 divided by 2 is 1007, so 1007*1007, which is 1007². So that makes sense.Similarly, the sum 1+3+5+...+105. Let's find how many terms. The nth term is 2n-1 = 105. Solving for n: 2n = 106, so n=53. Therefore, the sum is 53². The given equation is (106*53)/2. 106 divided by 2 is 53, so 53*53 = 53². Correct.So the original equation must be (sum from 1 to 2013 of odds) divided by x equals (sum from 1 to 105 of odds). That is, 1007² / x = 53². Solving for x, cross-multiplying, x = (1007²) / (53²) = (1007/53)². Now, compute 1007 divided by 53. Let me do that division. 53*19 is 1007 (since 53*20=1060, subtract 53: 1060-53=1007). So 1007/53=19. Therefore, (19)²=361. So x=361. That seems straightforward.Now part b) says x=361, so (361 -1)/10 = y². Let's compute 361-1=360. Then 360 divided by 10 is 36. So y²=36. Therefore, y can be 6 or -6. So the solution set S is {-6,6}. That also seems straightforward.But wait, let me double-check each step to be sure.Starting with part a). The sum of odd numbers up to 2013: check that 2013 is indeed the 1007th odd number. Let me compute 2013+1=2014, divided by 2 is 1007. Yes, that's correct. So the sum is 1007².Similarly, the sum up to 105: 105+1=106, divided by 2 is 53. So sum is 53². Then the equation is 1007² / x = 53². Solving for x gives x = (1007/53)². Calculating 1007 divided by 53: 53*19=1007, so yes, 19. Squaring that gives 361. Perfect.Part b) uses x=361, so (361-1)/10 = (360)/10 = 36. So y²=36, so y=±6. The solutions are -6 and 6. That all checks out.Is there any step where I could have made a mistake? Let me see:- When calculating the number of terms in the odd sequences: yes, using the formula (last term +1)/2 gives the number of terms. Since the nth odd term is 2n-1, so solving 2n-1 = last term gives n=(last term +1)/2. That seems right.- Checking 53*19: 50*19=950, 3*19=57, so 950+57=1007. Correct.- Squaring 19: 19*19. Let's compute 20*20=400, subtract 20+19, which is 39, so 400-39=361. Correct.- Part b): 361-1=360, 360/10=36. Square root of 36 is 6 and -6. Correct.So all steps seem correct. The key was recognizing that the sum of the first n odd numbers is n², then applying that to find the respective sums, set up the equation, and solve for x. Then use x to find y. Seems good.Final Answera) boxed{361}b) boxed{{-6, 6}}</think>The problem involves solving two parts using the sum of odd numbers and algebraic manipulation. Part a)The sum of the first ( n ) odd numbers is ( n^2 ). 1. For the sum ( 1 + 3 + 5 + ldots + 2013 ): - The number of terms ( n ) is given by solving ( 2n - 1 = 2013 ), which gives ( n = 1007 ). - The sum is ( 1007^2 ).2. For the sum ( 1 + 3 + 5 + ldots + 105 ): - The number of terms ( n ) is given by solving ( 2n - 1 = 105 ), which gives ( n = 53 ). - The sum is ( 53^2 ).The equation given is ( frac{1007^2}{x} = 53^2 ). Solving for ( x ):[x = left( frac{1007}{53} right)^2]Since ( 1007 div 53 = 19 ), we have:[x = 19^2 = 361] Part b)Given ( x = 361 ), we need to solve ( frac{361 - 1}{10} = y^2 ):[frac{360}{10} = 36 implies y^2 = 36]The solutions are ( y = pm 6 ), so the solution set is ( {-6, 6} ). Final Answersa) (boxed{361})b) (boxed{{-6, 6}})]

🔑:2. (4p) a) Find the non-zero rational number x that satisfies the equation:frac{1}{x}+frac{3}{x}+frac{5}{x}+ldots+frac{2013}{x}=1+3+5+ldots+105(3p) b) For x determined above, solve the equation in the set of rational numbers(x-1): 10=y^{2}Doina Negrilă

🧠:[<think>Okay, so I need to figure out the range of values for the real number ( a ) such that the function ( f(x) = frac{1}{3}x^3 - x^2 + ax - 5 ) is not monotonic in the interval ([-1, 2]). Hmm, let me start by recalling what it means for a function to be monotonic. If a function is monotonic on an interval, it means it is either entirely non-increasing or non-decreasing throughout that interval. So, if the function is not monotonic, it must change its increasing or decreasing behavior at least once within that interval. In calculus terms, this relates to the derivative of the function. If the derivative doesn't change sign over the interval, the function is monotonic. Conversely, if the derivative does change sign, the function is not monotonic. So, my task is to ensure that the derivative of ( f(x) ) changes sign within ([-1, 2]). Let me compute the derivative first. The derivative of ( f(x) ) with respect to ( x ) is:( f'(x) = frac{d}{dx} left( frac{1}{3}x^3 - x^2 + ax - 5 right) )Calculating term by term:- The derivative of ( frac{1}{3}x^3 ) is ( x^2 ).- The derivative of ( -x^2 ) is ( -2x ).- The derivative of ( ax ) is ( a ).- The derivative of the constant ( -5 ) is ( 0 ).So, putting it all together:( f'(x) = x^2 - 2x + a )Alright, so ( f'(x) = x^2 - 2x + a ). Now, for the function ( f(x) ) not to be monotonic on ([-1, 2]), this derivative must take both positive and negative values within that interval. In other words, the quadratic ( x^2 - 2x + a ) must have at least one real root in the interval ([-1, 2]). Because if the quadratic crosses the x-axis within the interval, then the derivative changes sign there, making the function non-monotonic.Therefore, I need to find the values of ( a ) for which the equation ( x^2 - 2x + a = 0 ) has at least one real root in ([-1, 2]). Let me approach this step by step.First, let's consider the quadratic equation ( x^2 - 2x + a = 0 ). To find the roots, we can use the quadratic formula:( x = frac{2 pm sqrt{(2)^2 - 4 cdot 1 cdot a}}{2 cdot 1} = frac{2 pm sqrt{4 - 4a}}{2} = 1 pm sqrt{1 - a} )So the roots are ( x = 1 + sqrt{1 - a} ) and ( x = 1 - sqrt{1 - a} ). For the roots to be real numbers, the discriminant must be non-negative:( 4 - 4a geq 0 implies 1 - a geq 0 implies a leq 1 )So, if ( a > 1 ), there are no real roots, meaning the derivative doesn't cross zero, so the function is either always increasing or always decreasing. If ( a leq 1 ), there are real roots, so the derivative may cross zero, leading to a change in monotonicity.But we need these roots to lie within the interval ([-1, 2]). So, even if ( a leq 1 ), we need to ensure that at least one of the roots is within ([-1, 2]). If both roots are outside this interval, then the derivative does not change sign within ([-1, 2]), and the function remains monotonic there. Therefore, the critical point is to determine the values of ( a ) such that at least one root of ( x^2 - 2x + a = 0 ) is inside ([-1, 2]). Alternatively, another approach is to analyze the behavior of ( f'(x) ) over the interval ([-1, 2]). Since ( f'(x) ) is a quadratic function (a parabola opening upwards because the coefficient of ( x^2 ) is positive), its minimum occurs at the vertex. The vertex of the parabola ( x^2 - 2x + a ) is at ( x = frac{2}{2} = 1 ). So, the vertex is at ( x = 1 ), and the minimum value of ( f'(x) ) is ( f'(1) = (1)^2 - 2(1) + a = 1 - 2 + a = a - 1 ).Since the parabola opens upwards, the derivative has a minimum at ( x = 1 ), which is ( a - 1 ), and it tends to infinity as ( x ) moves away from 1. For the derivative to change sign in the interval ([-1, 2]), the minimum value ( a - 1 ) must be less than zero (so that the parabola dips below the x-axis), and there must be points in the interval where the derivative is positive and where it is negative. But wait, if the minimum at ( x = 1 ) is ( a - 1 ), then if ( a - 1 < 0 ), i.e., ( a < 1 ), the derivative will be negative at ( x = 1 ). However, since the parabola opens upwards, the derivative will be positive for ( x ) values sufficiently far from 1. Therefore, if the derivative is negative at ( x = 1 ) (which is inside the interval [-1, 2]) and positive at the endpoints ( x = -1 ) and ( x = 2 ), then the derivative must cross the x-axis twice within some interval. But we need to check if those crossings are within [-1, 2].Alternatively, perhaps a better approach is to check the values of the derivative at the endpoints of the interval and see where it changes sign. Let me compute ( f'(-1) ) and ( f'(2) ):First, ( f'(-1) = (-1)^2 - 2(-1) + a = 1 + 2 + a = 3 + a )Second, ( f'(2) = (2)^2 - 2(2) + a = 4 - 4 + a = 0 + a = a )So, at the left endpoint ( x = -1 ), the derivative is ( 3 + a ), and at the right endpoint ( x = 2 ), the derivative is ( a ). Since the parabola opens upwards, and the vertex at ( x = 1 ) has a value ( a - 1 ), which is the minimum. Case 1: If ( a - 1 > 0 ), which implies ( a > 1 ). Then, the entire parabola is above the x-axis (since the minimum is above zero), so the derivative is always positive on [-1, 2], so the function is increasing throughout, hence monotonic. But the problem states that the function is not monotonic, so such ( a ) values are excluded.Case 2: If ( a - 1 = 0 ), i.e., ( a = 1 ). Then, the minimum value of the derivative is zero at ( x = 1 ). The derivative is ( x^2 - 2x + 1 = (x - 1)^2 ). So, the derivative is non-negative everywhere (since it's a square), and it's zero only at ( x = 1 ). So, in this case, the derivative does not change sign; it's zero at ( x = 1 ) but positive elsewhere. Therefore, the function is non-decreasing on the interval (since the derivative is non-negative), so it's still considered monotonic. Thus, ( a = 1 ) is excluded.Case 3: If ( a - 1 < 0 ), i.e., ( a < 1 ). Then, the derivative has a minimum at ( x = 1 ) below zero. Since the parabola opens upwards, this means the derivative will cross the x-axis at two points: one to the left of ( x = 1 ) and one to the right. Now, we need to check if these crossing points (the roots of ( f'(x) = 0 )) lie within the interval [-1, 2].So, let's find the roots of ( f'(x) = x^2 - 2x + a = 0 ). As calculated earlier, the roots are ( x = 1 pm sqrt{1 - a} ).Given that ( a < 1 ), ( sqrt{1 - a} ) is a real number. So, the two roots are:( x_1 = 1 - sqrt{1 - a} )( x_2 = 1 + sqrt{1 - a} )We need at least one of these roots to lie within [-1, 2]. Let's find the conditions on ( a ) for which either ( x_1 in [-1, 2] ) or ( x_2 in [-1, 2] ).First, let's consider ( x_1 = 1 - sqrt{1 - a} ). Since ( sqrt{1 - a} ) is non-negative, ( x_1 leq 1 ). Similarly, ( x_2 = 1 + sqrt{1 - a} geq 1 ). So, ( x_1 ) is the left root and ( x_2 ) is the right root.To check if ( x_1 ) is within [-1, 2], we can write:( -1 leq 1 - sqrt{1 - a} leq 2 )Similarly, for ( x_2 ):( -1 leq 1 + sqrt{1 - a} leq 2 )But since ( x_2 geq 1 ), the lower bound ( -1 leq x_2 ) is automatically satisfied. So, we need to check if ( x_2 leq 2 ). Let's compute:For ( x_2 leq 2 ):( 1 + sqrt{1 - a} leq 2 implies sqrt{1 - a} leq 1 implies 1 - a leq 1 implies -a leq 0 implies a geq 0 )So, if ( a geq 0 ), then ( x_2 leq 2 ). But remember, ( a < 1 ) in this case.Similarly, check for ( x_1 geq -1 ):( 1 - sqrt{1 - a} geq -1 implies -sqrt{1 - a} geq -2 implies sqrt{1 - a} leq 2 implies 1 - a leq 4 implies -a leq 3 implies a geq -3 )But since ( a ) is a real number, and we already have ( a < 1 ), combining with ( a geq -3 ), we have ( -3 leq a < 1 ). Wait, but this seems conflicting with the earlier result. Let me double-check.Wait, perhaps the way I derived is a bit confusing. Let me approach it differently.First, we need at least one root in [-1, 2]. So, the quadratic equation ( x^2 - 2x + a = 0 ) must have a solution in [-1, 2]. For that, we can use the Intermediate Value Theorem, but perhaps more straightforward is to consider the location of the roots relative to the interval.Alternatively, we can consider the function ( f'(x) = x^2 - 2x + a ) over the interval [-1, 2] and find for which values of ( a ) there is a sign change. Since ( f'(x) ) is continuous, if ( f'(x) ) takes both positive and negative values over the interval, then it must cross zero somewhere in between.So, let's check the sign of ( f'(x) ) at the endpoints:At ( x = -1 ): ( f'(-1) = (-1)^2 - 2(-1) + a = 1 + 2 + a = 3 + a )At ( x = 2 ): ( f'(2) = 4 - 4 + a = 0 + a = a )So, if ( f'(-1) ) and ( f'(2) ) have opposite signs, then by the Intermediate Value Theorem, there is at least one root in [-1, 2]. Additionally, even if they have the same sign, there could still be roots within the interval if the function dips below or above the x-axis between them. But given that the parabola opens upwards and the vertex is at ( x = 1 ), which is inside the interval [-1, 2], the minimum value is at ( x = 1 ), which is ( a - 1 ). So, if ( a - 1 < 0 ), which is ( a < 1 ), the derivative at the vertex is negative, and since the parabola opens upwards, the derivative must cross the x-axis twice. Therefore, if the minimum is below zero, then the function ( f'(x) ) will cross the x-axis twice, provided the roots exist (which they do when ( a < 1 )). However, whether these roots are within the interval [-1, 2] depends on the value of ( a ).So, to ensure that at least one root is in [-1, 2], we need either the left root ( x_1 = 1 - sqrt{1 - a} ) to be greater than or equal to -1, or the right root ( x_2 = 1 + sqrt{1 - a} ) to be less than or equal to 2. Let's compute these conditions.First, for the left root ( x_1 geq -1 ):( 1 - sqrt{1 - a} geq -1 )Subtract 1 from both sides:( -sqrt{1 - a} geq -2 )Multiply both sides by -1 (reversing the inequality):( sqrt{1 - a} leq 2 )Square both sides:( 1 - a leq 4 )Hence:( -a leq 3 implies a geq -3 )So, if ( a geq -3 ), then the left root ( x_1 geq -1 ). Since ( a < 1 ), combining these gives ( -3 leq a < 1 ).Next, for the right root ( x_2 leq 2 ):( 1 + sqrt{1 - a} leq 2 )Subtract 1:( sqrt{1 - a} leq 1 )Square both sides:( 1 - a leq 1 implies -a leq 0 implies a geq 0 )Therefore, if ( a geq 0 ), then the right root ( x_2 leq 2 ). Again, since ( a < 1 ), combining gives ( 0 leq a < 1 ).Putting these together, the roots ( x_1 ) and ( x_2 ) are within the interval [-1, 2] if both ( a geq -3 ) and ( a geq 0 ). Wait, no. Wait, actually, the left root ( x_1 ) is within [-1, 2] when ( a geq -3 ), and the right root ( x_2 ) is within [-1, 2] when ( a geq 0 ). Therefore, the roots are in [-1, 2] when both conditions are satisfied. But actually, if ( a geq -3 ), then ( x_1 geq -1 ), and if ( a geq 0 ), then ( x_2 leq 2 ). But since ( x_1 ) is always less than or equal to 1 (because ( sqrt{1 - a} geq 0 )), and the interval goes up to 2, even if ( x_1 ) is in [-1, 1], as long as ( x_2 ) is also in [1, 2], then both roots are within [-1, 2]. However, for different ranges of ( a ), the roots may be inside or outside the interval.Wait, perhaps it's better to consider the different cases:Case A: Both roots are within [-1, 2].Case B: Only the left root is within [-1, 2].Case C: Only the right root is within [-1, 2].But actually, for the quadratic ( x^2 - 2x + a ), since the vertex is at ( x = 1 ), and the parabola opens upwards, if there are two real roots (i.e., ( a < 1 )), then the left root is ( x_1 = 1 - sqrt{1 - a} ) and the right root is ( x_2 = 1 + sqrt{1 - a} ). So, both roots are equidistant from the vertex at ( x = 1 ). Therefore, both roots will be within [-1, 2] if:Left root ( x_1 geq -1 ) AND Right root ( x_2 leq 2 )Which gives:( 1 - sqrt{1 - a} geq -1 ) (as before, leading to ( a geq -3 ))AND( 1 + sqrt{1 - a} leq 2 ) (leading to ( a geq 0 ))Thus, both roots are in [-1, 2] when ( a geq 0 ) and ( a < 1 ). However, if ( -3 leq a < 0 ), then only the left root ( x_1 ) is in [-1, 2], because ( x_1 = 1 - sqrt{1 - a} geq -1 ) (since ( a geq -3 )), but ( x_2 = 1 + sqrt{1 - a} ). Let's check when ( a < 0 ), say ( a = -1 ):Then, ( x_2 = 1 + sqrt{1 - (-1)} = 1 + sqrt{2} approx 2.414 ), which is greater than 2. So, in this case, ( x_2 ) is outside the interval. Therefore, for ( -3 leq a < 0 ), only the left root ( x_1 ) is within [-1, 2], and the right root ( x_2 ) is outside. For ( 0 leq a < 1 ), both roots are within [-1, 2]. But in either case, if there's at least one root in the interval, the derivative changes sign, making the function non-monotonic. Therefore, the range of ( a ) for which there is at least one root in [-1, 2] is ( -3 leq a < 1 ). Wait, but let's verify this.Wait, when ( a geq -3 ), the left root ( x_1 geq -1 ). Since ( x_1 = 1 - sqrt{1 - a} ), let's compute when ( a = -3 ):( x_1 = 1 - sqrt{1 - (-3)} = 1 - sqrt{4} = 1 - 2 = -1 ). So, at ( a = -3 ), the left root is exactly at ( x = -1 ). Similarly, when ( a ) increases from ( -3 ) to ( 1 ), the left root moves from ( -1 ) to ( 1 ), and the right root moves from ( 3 ) (when ( a = -3 ), ( x_2 = 1 + sqrt{1 - (-3)} = 1 + 2 = 3 )) down to ( 1 ) as ( a ) approaches 1.Therefore, when ( a = -3 ), the left root is at ( x = -1 ), which is the boundary of the interval. For ( a > -3 ), the left root is inside the interval [-1, 2], up until ( a = 0 ), where the right root reaches ( x = 2 ). Wait, let's check when ( a = 0 ):( x_2 = 1 + sqrt{1 - 0} = 1 + 1 = 2 ). So, at ( a = 0 ), the right root is exactly at ( x = 2 ). For ( a ) between 0 and 1, both roots ( x_1 ) and ( x_2 ) are within (-1, 2). For example, take ( a = 0.5 ):( x_1 = 1 - sqrt{1 - 0.5} = 1 - sqrt{0.5} approx 1 - 0.707 = 0.293 )( x_2 = 1 + sqrt{0.5} approx 1.707 ), which is within [-1, 2].Therefore, summarizing:- For ( a < -3 ): The left root ( x_1 < -1 ), and the right root ( x_2 > 3 ). So, both roots are outside the interval [-1, 2]. Therefore, the derivative ( f'(x) ) is positive throughout the interval (since the parabola opens upwards and the minimum at ( x = 1 ) is ( a - 1 ), which is less than -4 when ( a < -3 ), but wait, actually if both roots are outside the interval, then the sign of the derivative on the interval depends on the parabola's behavior.Wait, actually, if the parabola opens upwards and both roots are outside the interval, then the sign of the derivative on the interval will be the same as the sign at any point in the interval. Let's check for ( a < -3 ). For example, take ( a = -4 ):Then, ( f'(x) = x^2 - 2x - 4 ). Let's compute ( f'(-1) = 1 + 2 - 4 = -1 ), ( f'(2) = 4 - 4 - 4 = -4 ). So, both endpoints have negative derivatives. Since the parabola opens upwards, and the vertex is at ( x = 1 ), with ( f'(1) = 1 - 2 - 4 = -5 ). Therefore, the entire interval [-1, 2] has a negative derivative. Therefore, the function is decreasing throughout, hence monotonic. So, for ( a < -3 ), even though there are roots, they lie outside the interval, so the derivative does not change sign on the interval. Thus, the function is monotonic. Therefore, these values of ( a ) should be excluded.Wait, but earlier, we concluded that when ( a geq -3 ), the left root is at least -1. So, for ( a geq -3 ), the left root is within the interval or at the boundary. Let me check for ( a = -3 ):( f'(x) = x^2 - 2x - 3 ). Then, the roots are ( x = 1 pm sqrt{1 - (-3)} = 1 pm 2 ), so ( x = -1 ) and ( x = 3 ). Therefore, at ( a = -3 ), the derivative is zero at ( x = -1 ) and ( x = 3 ). So, on the interval [-1, 2], at ( x = -1 ), the derivative is zero. Let's check the sign just to the right of ( x = -1 ):Take ( x = -0.9 ):( f'(-0.9) = (-0.9)^2 - 2*(-0.9) - 3 = 0.81 + 1.8 - 3 = -0.39 )At ( x = 2 ):( f'(2) = 4 - 4 - 3 = -3 )But wait, if at ( x = -1 ), the derivative is zero, and just to the right it's negative, and at ( x = 1 ), the derivative is ( f'(1) = 1 - 2 - 3 = -4 ), which is still negative. Then, moving to ( x = 2 ), it's also negative. Therefore, the derivative does not cross zero in (-1, 2]; it only touches zero at ( x = -1 ), which is the boundary. Therefore, the derivative is non-positive on the entire interval [-1, 2], but since it's zero at ( x = -1 ), the function is non-increasing on [-1, 2], hence monotonic. Therefore, even at ( a = -3 ), the function is monotonic. Therefore, the previous conclusion that the range is ( -3 leq a < 1 ) is incorrect because at ( a = -3 ), the function is still monotonic.Therefore, there must be a mistake in the reasoning. Let me reconsider.The key point is that for the function to be non-monotonic, the derivative must change sign within the open interval (-1, 2). If the derivative is zero at the endpoints but does not change sign in between, the function is still monotonic. For example, if the derivative is zero at ( x = -1 ) but negative in (-1, 2], then the function is non-increasing on [-1, 2], hence monotonic.Therefore, in order for the function to not be monotonic, the derivative must cross zero within the open interval (-1, 2). So, even if one root is at ( x = -1 ), but the other root is outside, then the derivative doesn't change sign within the interval. Similarly, if a root is at ( x = 2 ), same thing. Therefore, we need the quadratic ( x^2 - 2x + a = 0 ) to have at least one root in the open interval (-1, 2).So, let's rephrase the problem: Find all real numbers ( a ) such that the equation ( x^2 - 2x + a = 0 ) has at least one root in (-1, 2). To solve this, we can use the concept that a quadratic equation ( ax^2 + bx + c = 0 ) has a root in an interval (m, n) if either:1. The quadratic changes sign over (m, n), i.e., ( f(m) ) and ( f(n) ) have opposite signs, or2. The quadratic has a root at a point where the function is zero and the quadratic is tangent to the x-axis (i.e., a repeated root), but in this case, the repeated root would need to lie within (m, n).However, for our quadratic ( x^2 - 2x + a ), since the leading coefficient is positive, the parabola opens upwards. Therefore, if there's a repeated root (discriminant zero), that root is at the vertex. So, if the vertex is within (m, n) and the discriminant is zero, then that's a case where the quadratic touches the x-axis within the interval.But in our problem, the quadratic must have at least one root in (-1, 2). Therefore, we can use the following conditions:Either:1. There is exactly one root in (-1, 2), which would happen if one root is inside and the other outside, or2. There are two roots in (-1, 2).But since the quadratic opens upwards, if there are two roots in the interval, then the quadratic must be negative between them. So, the conditions can be translated into the following:The quadratic ( x^2 - 2x + a ) has at least one root in (-1, 2) if either:- ( f'(-1) ) and ( f'(2) ) have opposite signs (so by the Intermediate Value Theorem, there is a root in (-1, 2)), or- The quadratic has a root at the vertex (but the vertex is at x=1, which is inside the interval), so if the discriminant is zero (i.e., ( a = 1 )), but as we saw earlier, at ( a = 1 ), the quadratic is non-negative everywhere, so it doesn't change sign. Therefore, even though the vertex is at x=1, which is in the interval, the function does not change sign there, so the function remains monotonic. So this case is excluded.Alternatively, we can use the following approach: The quadratic has a root in (-1, 2) if and only if either:1. ( f'(-1) cdot f'(2) < 0 ), which would imply a root in (-1, 2) by the Intermediate Value Theorem, or2. The quadratic has a root at a point where ( f'(x) = 0 ) and x is in (-1, 2), which can be checked by ensuring that at least one root is in (-1, 2).But this is somewhat circular. Maybe a better approach is to use the method of intervals and analyze the roots.Alternatively, use the theorem that a quadratic equation ( ax^2 + bx + c = 0 ) has a root in the interval (m, n) if and only if at least one of the following holds:- ( f(m) cdot f(n) < 0 )- ( af(m) < 0 ) and the discriminant is non-negative, and the vertex is within (m, n)But perhaps it's easier to use the following conditions.Given the quadratic equation ( x^2 - 2x + a = 0 ), we can write the conditions for it to have at least one root in (-1, 2):1. The discriminant ( D = 4 - 4a geq 0 implies a leq 1 ).2. And either: a. One root is in (-1, 2) and the other is outside, or b. Both roots are in (-1, 2).To check these, we can use the following criteria:For a quadratic equation with roots ( x_1 leq x_2 ):- There is exactly one root in (m, n) if ( f(m) cdot f(n) < 0 ).- There are two roots in (m, n) if: - The discriminant ( D geq 0 ), - The axis of symmetry is within (m, n), - ( f(m) > 0 ), - ( f(n) > 0 ).But since the parabola opens upwards, the quadratic will be negative between the two roots. Therefore, for two roots in (m, n), the quadratic must be negative somewhere in (m, n), which requires that the minimum (vertex) is below zero and the roots are within (m, n).Alternatively, given the quadratic ( x^2 - 2x + a ), we can use the following steps:1. The quadratic has real roots iff ( a leq 1 ).2. The number of roots in (-1, 2) depends on the position of the roots relative to the interval.To find the values of ( a ) such that at least one root is in (-1, 2):We can consider two scenarios:- The quadratic has exactly one root in (-1, 2). This happens when one root is inside the interval and the other is outside. For this, either: - ( f'(-1) < 0 ) and ( f'(2) > 0 ), or - ( f'(-1) > 0 ) and ( f'(2) < 0 ).But since the parabola opens upwards, if ( f'(-1) < 0 ) and ( f'(2) > 0 ), there is a root in (-1, 2). Similarly, if ( f'(-1) > 0 ) and ( f'(2) < 0 ), there is a root in (-1, 2). However, since the parabola opens upwards, the left side tends to infinity as ( x ) approaches infinity, but over our interval, the values at the endpoints are ( f'(-1) = 3 + a ) and ( f'(2) = a ).Therefore, ( f'(-1) = 3 + a ) and ( f'(2) = a ). For the product ( f'(-1) cdot f'(2) < 0 ), which would indicate a sign change, we need:( (3 + a) cdot a < 0 )This inequality holds when ( -3 < a < 0 ). So, if ( -3 < a < 0 ), then ( f'(-1) ) and ( f'(2) ) have opposite signs, implying there's a root in (-1, 2), making the derivative change sign, hence the function is not monotonic.Additionally, we must consider the case where the quadratic has both roots in (-1, 2). For this, we need:1. The roots ( x_1 ) and ( x_2 ) satisfy ( -1 < x_1 leq x_2 < 2 ).But given that the roots are ( 1 pm sqrt{1 - a} ), we can write:Left root: ( x_1 = 1 - sqrt{1 - a} > -1 )Right root: ( x_2 = 1 + sqrt{1 - a} < 2 )Which, as before, leads to:For ( x_1 > -1 ):( 1 - sqrt{1 - a} > -1 implies sqrt{1 - a} < 2 implies 1 - a < 4 implies a > -3 )For ( x_2 < 2 ):( 1 + sqrt{1 - a} < 2 implies sqrt{1 - a} < 1 implies 1 - a < 1 implies a > 0 )Therefore, both roots are in (-1, 2) when ( a > 0 ) and ( a leq 1 ). But since ( a leq 1 ) is already required for real roots, combining these gives ( 0 < a leq 1 ).However, when ( a = 1 ), the quadratic becomes ( x^2 - 2x + 1 = (x - 1)^2 ), which has a double root at ( x = 1 ). At ( x = 1 ), the derivative is zero, but since it's a double root, the derivative does not change sign—it touches the x-axis and remains non-negative. Therefore, the function ( f(x) ) has a critical point at ( x = 1 ) but doesn't change direction, so it's still considered non-decreasing (monotonic). Hence, ( a = 1 ) should be excluded. Therefore, both roots are in (-1, 2) when ( 0 < a < 1 ).But in this case, when ( 0 < a < 1 ), both roots are in (-1, 2), so the derivative changes sign twice within the interval, making the function non-monotonic.Additionally, when ( a = 0 ), the quadratic is ( x^2 - 2x = x(x - 2) ). The roots are ( x = 0 ) and ( x = 2 ). Here, ( x = 0 ) is in (-1, 2), and ( x = 2 ) is at the boundary. Therefore, at ( a = 0 ), the derivative is zero at ( x = 0 ) and ( x = 2 ). Checking the sign around ( x = 0 ):Just left of 0 (say ( x = -0.1 )): ( f'(-0.1) = (-0.1)^2 - 2(-0.1) + 0 = 0.01 + 0.2 = 0.21 > 0 )Just right of 0 (say ( x = 0.1 )): ( f'(0.1) = (0.1)^2 - 2(0.1) = 0.01 - 0.2 = -0.19 < 0 )So, the derivative changes sign from positive to negative at ( x = 0 ), hence the function changes from increasing to decreasing, making it non-monotonic. At ( x = 2 ), the derivative is zero, but since it's the endpoint, we don't consider the behavior beyond that. So, at ( a = 0 ), the function is non-monotonic.Similarly, when ( a ) approaches 1 from below, say ( a = 0.99 ):The roots are ( x = 1 pm sqrt{1 - 0.99} = 1 pm sqrt{0.01} = 1 pm 0.1 ), so ( x = 0.9 ) and ( x = 1.1 ), both within (-1, 2). Therefore, the derivative changes sign twice, leading to non-monotonicity.Therefore, combining the two cases:1. When ( -3 < a < 0 ), there's exactly one root in (-1, 2), causing a sign change.2. When ( 0 leq a < 1 ), there are two roots in (-1, 2) (or one at the boundary when ( a = 0 )), causing sign changes.However, at ( a = -3 ), the left root is at ( x = -1 ), which is the boundary. As checked earlier, the derivative at ( x = -1 ) is zero, but the function remains non-increasing on [-1, 2], so it's still monotonic. Similarly, at ( a = 0 ), the right root is at ( x = 2 ), but the derivative changes sign at ( x = 0 ), which is inside the interval, so even though one root is at the boundary, the other is inside, so the function is non-monotonic.Therefore, the values of ( a ) for which the function is not monotonic in [-1, 2] are those where the quadratic has at least one root in (-1, 2). This happens when:- Either ( -3 < a < 0 ), with one root in (-1, 2)- Or ( 0 leq a < 1 ), with both roots in [-1, 2] (but at ( a = 0 ), one root is at 0 and 2, but the sign change still occurs within the interval)However, we need to verify the edge case at ( a = -3 ):At ( a = -3 ), the quadratic is ( x^2 - 2x - 3 ), roots at ( x = -1 ) and ( x = 3 ). The derivative at ( x = -1 ) is zero, but since it's the left endpoint, and the derivative is negative just to the right of ( x = -1 ), the entire interval from ( x = -1 ) to ( x = 2 ) has a non-positive derivative, making the function non-increasing, hence monotonic. Therefore, ( a = -3 ) should be excluded.Similarly, at ( a = 1 ), the quadratic has a double root at ( x = 1 ), but the derivative does not change sign, so the function is non-decreasing, hence monotonic. Therefore, ( a = 1 ) is excluded.Putting it all together, the values of ( a ) for which the function is not monotonic in [-1, 2] are those where ( -3 < a < 1 ). However, we need to confirm if this is correct.Wait, earlier analysis suggests that:- For ( -3 < a < 0 ), there's one root in (-1, 2) and one outside, leading to a sign change.- For ( 0 leq a < 1 ), there are two roots in [-1, 2], leading to two sign changes.Therefore, combining both intervals gives ( -3 < a < 1 ).But wait, at ( a = 0 ), as discussed, the roots are at 0 and 2. Even though one root is at the endpoint, the derivative changes sign at ( x = 0 ), hence the function is non-monotonic. Similarly, when ( a ) is between 0 and 1, both roots are inside the interval. Therefore, the entire interval for ( a ) is ( -3 < a < 1 ).But earlier, when considering the product ( f'(-1) cdot f'(2) < 0 ), we found that this product is negative when ( -3 < a < 0 ), indicating a root in (-1, 2). For ( 0 leq a < 1 ), even though the product ( f'(-1) cdot f'(2) = (3 + a) cdot a ) is non-negative (since ( a geq 0 )), the quadratic still has two roots in [-1, 2] because both roots are within the interval. Therefore, to capture all such cases, the correct interval for ( a ) is ( -3 < a < 1 ).However, we need to confirm this by testing specific values:Test case 1: ( a = -2 ) (which is in ( -3 < a < 0 )):Quadratic is ( x^2 - 2x - 2 ). The roots are ( x = 1 pm sqrt{1 - (-2)} = 1 pm sqrt{3} approx 1 pm 1.732 ). So, roots at approximately ( -0.732 ) and ( 2.732 ). Therefore, the left root ( -0.732 ) is in (-1, 2), and the right root is outside. Therefore, the derivative changes sign once in (-1, 2), making the function non-monotonic.Test case 2: ( a = 0.5 ) (which is in ( 0 leq a < 1 )):Quadratic is ( x^2 - 2x + 0.5 ). Roots at ( x = 1 pm sqrt{1 - 0.5} = 1 pm sqrt{0.5} approx 1 pm 0.707 ). So, roots at approximately ( 0.293 ) and ( 1.707 ), both in (-1, 2). Thus, the derivative changes sign twice, so the function is non-monotonic.Test case 3: ( a = -3 ):As discussed, the roots are at ( x = -1 ) and ( x = 3 ). The derivative at ( x = -1 ) is zero, but the function remains non-increasing on [-1, 2], so it's monotonic.Test case 4: ( a = 1 ):Quadratic is ( (x - 1)^2 ), derivative is non-negative everywhere, so the function is non-decreasing, hence monotonic.Therefore, the correct range is ( -3 < a < 1 ).However, the original problem statement says the function is not monotonic in the interval [-1, 2]. Therefore, the answer should be all real numbers ( a ) such that ( -3 < a < 1 ).But let's verify this with another test case at ( a = -3 ):As before, the derivative at ( a = -3 ) is ( x^2 - 2x - 3 ), which factors to ( (x + 1)(x - 3) ). Therefore, the derivative is zero at ( x = -1 ) and ( x = 3 ). On the interval [-1, 2], the derivative at ( x = -1 ) is zero, and for all ( x > -1 ) in the interval, the derivative is negative. Therefore, the function is non-increasing on the entire interval, hence monotonic. So, ( a = -3 ) is excluded.Similarly, at ( a = 1 ), the derivative is non-negative, so the function is non-decreasing, hence monotonic. So, ( a = 1 ) is excluded.Thus, the range of ( a ) is ( -3 < a < 1 ).Therefore, the final answer is that ( a ) must lie in the open interval (-3, 1). In boxed form, this is ( boxed{(-3, 1)} ).Final AnswerThe range of values for ( a ) is boxed{(-3, 1)}.</think>To determine the range of values for the real number ( a ) such that the function ( f(x) = frac{1}{3}x^3 - x^2 + ax - 5 ) is not monotonic in the interval ([-1, 2]), we need to analyze the derivative of the function.The derivative is:[f'(x) = x^2 - 2x + a]For the function ( f(x) ) to not be monotonic, the derivative ( f'(x) ) must change sign within the interval ([-1, 2]). This requires the quadratic equation ( x^2 - 2x + a = 0 ) to have at least one root in the interval ((-1, 2)).The roots of the quadratic equation ( x^2 - 2x + a = 0 ) are:[x = 1 pm sqrt{1 - a}]For the roots to be real, the discriminant must be non-negative:[1 - a geq 0 implies a leq 1]Next, we need to ensure that at least one of the roots lies within the interval ((-1, 2)). The roots are ( x_1 = 1 - sqrt{1 - a} ) and ( x_2 = 1 + sqrt{1 - a} ).1. For the left root ( x_1 = 1 - sqrt{1 - a} ) to be within ((-1, 2)): [ -1 < 1 - sqrt{1 - a} < 2 ] Solving the left inequality: [ 1 - sqrt{1 - a} > -1 implies sqrt{1 - a} < 2 implies 1 - a < 4 implies a > -3 ]2. For the right root ( x_2 = 1 + sqrt{1 - a} ) to be within ((-1, 2)): [ -1 < 1 + sqrt{1 - a} < 2 ] Solving the right inequality: [ 1 + sqrt{1 - a} < 2 implies sqrt{1 - a} < 1 implies 1 - a < 1 implies a > 0 ]Combining these conditions, we find that:- For ( -3 < a < 0 ), the left root ( x_1 ) is within ((-1, 2)) and the right root ( x_2 ) is outside.- For ( 0 leq a < 1 ), both roots ( x_1 ) and ( x_2 ) are within ((-1, 2)).Thus, the function ( f(x) ) is not monotonic in the interval ([-1, 2]) if and only if ( a ) lies in the open interval ((-3, 1)).[boxed{(-3, 1)}],<think>Okay, so I need to find the range of values for the real number ( a ) such that the function ( f(x) = frac{1}{3}x^3 - x^2 + ax - 5 ) is not monotonic in the interval ([-1, 2]). Hmm, let's start by recalling what it means for a function to be monotonic. If a function is monotonic on an interval, it means it's either entirely non-increasing or non-decreasing throughout that interval. So, if it's not monotonic, then it must change its increasing or decreasing behavior at some point within that interval. To determine when a function is not monotonic, we can look at its derivative. If the derivative changes sign in the interval, the function has a local maximum or minimum there, which means it's not monotonic. So, my first step should be to find the derivative of ( f(x) ).Let's compute ( f'(x) ):[f'(x) = frac{d}{dx}left( frac{1}{3}x^3 - x^2 + ax - 5 right) = x^2 - 2x + a]So, the derivative is a quadratic function: ( f'(x) = x^2 - 2x + a ). Now, for the function ( f(x) ) to not be monotonic on ([-1, 2]), the derivative ( f'(x) ) must change its sign within that interval. That is, ( f'(x) ) must have at least one real root in ([-1, 2]). Therefore, the quadratic equation ( x^2 - 2x + a = 0 ) must have at least one solution in the interval ([-1, 2]).So, the problem reduces to finding the values of ( a ) for which the quadratic equation ( x^2 - 2x + a = 0 ) has at least one real root in ([-1, 2]). First, let's recall that for a quadratic equation ( x^2 + bx + c = 0 ), the roots can be found using the quadratic formula: ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ). In our case, the quadratic is ( x^2 - 2x + a = 0 ), so ( b = -2 ), ( c = a ). Wait, actually, in standard form ( ax^2 + bx + c = 0 ), here the coefficients are: leading coefficient is 1, then the linear coefficient is -2, and the constant term is ( a ). So, the quadratic equation is ( x^2 - 2x + a = 0 ).So, the discriminant ( D ) of this quadratic is ( D = (-2)^2 - 4(1)(a) = 4 - 4a ). For the equation to have real roots, the discriminant must be non-negative. Thus, ( 4 - 4a geq 0 Rightarrow a leq 1 ). So, if ( a > 1 ), the quadratic equation has no real roots, meaning the derivative ( f'(x) ) does not cross zero. Therefore, if ( a > 1 ), then ( f'(x) = x^2 - 2x + a ) is always positive (since the quadratic opens upwards and the minimum value is above the x-axis), so ( f(x) ) would be increasing on the entire interval ([-1, 2]), making it monotonic. Hence, to have the function not monotonic, ( a ) must be less than or equal to 1. But wait, even if ( a leq 1 ), the roots might lie outside the interval ([-1, 2]). So, even though there are real roots, they need to be within ([-1, 2]) for the derivative to change sign in that interval.Therefore, the next step is to find the values of ( a ) such that the quadratic ( x^2 - 2x + a = 0 ) has at least one root in ([-1, 2]). Let me denote the roots as ( x_1 ) and ( x_2 ), where ( x_1 leq x_2 ).First, let's consider the conditions required for at least one root in the interval ([-1, 2].) A quadratic equation will have at least one root in an interval if either:1. The quadratic has exactly one root in the interval (i.e., the root is on the boundary or the quadratic touches the x-axis at that point), but since quadratics are continuous and smooth, this would require the quadratic to touch the x-axis at the boundary. However, since the interval is closed, maybe this is possible.But more generally, there's a theorem in calculus called the Intermediate Value Theorem, which says that if a continuous function changes sign over an interval, there's at least one root in that interval. Since the quadratic is continuous, if ( f'(x) ) takes both positive and negative values over ([-1, 2]), then there must be a root in between. Alternatively, if one of the roots of the quadratic is within the interval, then the derivative changes sign there.Alternatively, we can use the following approach: For the quadratic ( x^2 - 2x + a ), we can analyze its behavior over the interval. Let's first find the critical points (vertex) of the quadratic. The vertex of a quadratic ( ax^2 + bx + c ) is at ( x = -frac{b}{2a} ). In our case, ( a = 1 ), ( b = -2 ), so the vertex is at ( x = -(-2)/(2*1) = 1 ). So, the vertex is at ( x = 1 ), which is within the interval ([-1, 2]].The value of the quadratic at the vertex is ( f'(1) = (1)^2 - 2*(1) + a = 1 - 2 + a = a - 1 ). Since the quadratic opens upwards (leading coefficient positive), the vertex is the minimum point. Therefore, the minimum value of ( f'(x) ) on the real line is ( a - 1 ). For the quadratic to have real roots, we need ( a - 1 leq 0 ), which gives ( a leq 1 ), as we found before.But even if ( a leq 1 ), the roots might still be outside the interval ([-1, 2]. So, we need to ensure that at least one root is within this interval. To do this, we can check if the quadratic has a root in [-1, 2] by using the following conditions:Either:1. The quadratic evaluates to zero at one of the endpoints, i.e., ( f'(-1) = 0 ) or ( f'(2) = 0 ), which would mean that one of the endpoints is a root.Or:2. The quadratic changes sign over the interval, which would imply by the Intermediate Value Theorem that there is a root in between.Alternatively, we can use the concept that if the minimum value of the quadratic in the interval is less than or equal to zero, and the quadratic has a root in the interval. But since the quadratic is convex (opens upwards), if the minimum at x=1 is less than or equal to zero, and the quadratic is positive at both endpoints, there could still be two roots within the interval. Wait, actually, if the vertex is the minimum, and the minimum is negative, then the quadratic will cross the x-axis twice, but depending on the endpoints, these crossings (roots) might be inside or outside the interval.Alternatively, perhaps a better approach is to solve for the roots and see when they lie within [-1, 2].Let me find the roots of the quadratic equation ( x^2 - 2x + a = 0 ).Using the quadratic formula:[x = frac{2 pm sqrt{4 - 4a}}{2} = frac{2 pm 2sqrt{1 - a}}{2} = 1 pm sqrt{1 - a}]So, the roots are ( x = 1 + sqrt{1 - a} ) and ( x = 1 - sqrt{1 - a} ).Since ( sqrt{1 - a} ) is real only if ( 1 - a geq 0 Rightarrow a leq 1 ), which we already established.So, the two roots are ( 1 + sqrt{1 - a} ) and ( 1 - sqrt{1 - a} ). Now, we need at least one of these roots to lie in the interval ([-1, 2]).So, let's analyze the positions of these roots relative to the interval.First, since the quadratic opens upwards, and the vertex is at x=1, which is the minimum point, then the smaller root is ( 1 - sqrt{1 - a} ) and the larger root is ( 1 + sqrt{1 - a} ).We need either:1. ( 1 - sqrt{1 - a} geq -1 ) (so the smaller root is greater than or equal to -1)OR2. ( 1 + sqrt{1 - a} leq 2 ) (so the larger root is less than or equal to 2)But actually, for at least one root in the interval, either the smaller root is within [-1, 2] or the larger root is within [-1, 2]. However, given the quadratic opens upwards, if the vertex is at x=1 and the minimum is below or equal to zero, then the roots are symmetric around x=1. So, depending on the value of ( sqrt{1 - a} ), the roots can be spread out.Alternatively, perhaps we can set up inequalities for the roots to lie within the interval.So, for the smaller root ( x_1 = 1 - sqrt{1 - a} geq -1 ):[1 - sqrt{1 - a} geq -1 Rightarrow -sqrt{1 - a} geq -2 Rightarrow sqrt{1 - a} leq 2 Rightarrow 1 - a leq 4 Rightarrow -a leq 3 Rightarrow a geq -3]Similarly, for the larger root ( x_2 = 1 + sqrt{1 - a} leq 2 ):[1 + sqrt{1 - a} leq 2 Rightarrow sqrt{1 - a} leq 1 Rightarrow 1 - a leq 1 Rightarrow -a leq 0 Rightarrow a geq 0]Wait, this is interesting. So, if we require the smaller root to be within the interval (i.e., ( x_1 geq -1 )), we get ( a geq -3 ). If we require the larger root to be within the interval (i.e., ( x_2 leq 2 )), we get ( a geq 0 ). However, we need at least one root in the interval, so either ( x_1 geq -1 ) or ( x_2 leq 2 ). But since the roots are symmetric around x=1, maybe both conditions need to be considered.Wait, let's think again. If the smaller root ( x_1 = 1 - sqrt{1 - a} ) is greater than or equal to -1, that places a lower bound on ( a ), namely ( a geq -3 ). If the larger root ( x_2 = 1 + sqrt{1 - a} ) is less than or equal to 2, that also places a lower bound on ( a ), but this time ( a geq 0 ).However, this seems contradictory. Wait, perhaps I need to approach this differently.Let me consider the positions of the roots. Since the quadratic is ( x^2 - 2x + a ), which has vertex at (1, a - 1). For the quadratic to cross the x-axis within the interval [-1, 2], we need that either:1. The vertex is below or on the x-axis (i.e., a - 1 ≤ 0 ⇒ a ≤ 1), and at least one of the roots is in [-1, 2].So, given that a ≤ 1, and the roots are ( 1 pm sqrt{1 - a} ), let's find for which values of a, one of these roots is in [-1, 2].First, consider the left root ( x_1 = 1 - sqrt{1 - a} ). We want ( x_1 in [-1, 2] ). Similarly, the right root ( x_2 = 1 + sqrt{1 - a} ). We want ( x_2 in [-1, 2] ).But since the quadratic is symmetric about x = 1, and the roots are spaced ( sqrt{1 - a} ) away from 1. Let's see:For the left root ( x_1 = 1 - sqrt{1 - a} geq -1 ). As we derived earlier:( 1 - sqrt{1 - a} geq -1 Rightarrow sqrt{1 - a} leq 2 Rightarrow 1 - a leq 4 Rightarrow a geq -3 ).Similarly, for the right root ( x_2 = 1 + sqrt{1 - a} leq 2 ):( 1 + sqrt{1 - a} leq 2 Rightarrow sqrt{1 - a} leq 1 Rightarrow 1 - a leq 1 Rightarrow a geq 0 ).So, if a ≥ -3, the left root is ≥ -1, and if a ≥ 0, the right root is ≤ 2. But since we need at least one root in [-1, 2], the possible cases are:1. a is between -3 and 1, which makes the left root between -1 and 1 (since when a = -3, the left root is 1 - sqrt(1 - (-3)) = 1 - sqrt(4) = 1 - 2 = -1, and as a increases to 1, sqrt(1 - a) decreases from 2 to 0, so the left root moves from -1 to 1).2. a is between 0 and 1, which also affects the right root. When a = 0, the right root is 1 + sqrt(1 - 0) = 1 + 1 = 2, and as a increases to 1, sqrt(1 - a) decreases to 0, so the right root moves from 2 to 1.Therefore, combining these two intervals:- For a in [-3, 1], the left root is between -1 and 1.- For a in [0, 1], the right root is between 1 and 2.Wait, but actually, when a is between -3 and 0, the left root is between -1 and 1 - sqrt(1 - a). Let's check when a = -3:a = -3, left root is -1, right root is 1 + sqrt(1 - (-3)) = 1 + 2 = 3. So, right root is at 3, which is outside the interval. So, only the left root at -1 is in the interval. As a increases from -3 to 0, the left root moves from -1 to 1 - sqrt(1 - a). For example, at a = 0, left root is 1 - 1 = 0. So, left root moves from -1 to 0 as a goes from -3 to 0. Then, when a is 0, the right root is 2, and as a increases from 0 to 1, the right root moves from 2 to 1. So, the right root is in the interval [1, 2] when a is in [0,1].Therefore, for a in [-3, 0), the left root is in [-1, 0), and for a in [0,1], both roots are in the interval (the left root in [0,1] and the right root in [1,2]). Wait, but when a is 0, left root is 0, right root is 2. So, when a is exactly 0, the roots are 0 and 2, which are both in the interval. As a increases from 0 to 1, left root moves from 0 to 1, and right root moves from 2 to 1. So, when a is 1, both roots coincide at x = 1.Therefore, combining these, for a in [-3, 1], the quadratic has at least one root in [-1, 2]. Therefore, the range of a is [-3, 1]. But wait, let's check when a is exactly -3. Then, the left root is -1, which is in the interval, so f'(x) has a root at -1, which is an endpoint. Similarly, when a is 1, the root is at x = 1, which is in the interval. So, endpoints are included.But let's verify this with some test values. Let's take a = -4. Then, the discriminant would be 4 - 4*(-4) = 4 + 16 = 20, which is positive, so roots are 1 ± sqrt(1 - (-4)) = 1 ± sqrt(5). sqrt(5) ≈ 2.236, so roots are approximately 1 + 2.236 ≈ 3.236 and 1 - 2.236 ≈ -1.236. So, the roots are at approximately 3.236 and -1.236. Both of these are outside the interval [-1, 2], so even though a = -4 is less than 1, the roots are outside the interval. Therefore, the function f'(x) would be positive throughout [-1, 2], since the quadratic opens upwards and both roots are outside. Therefore, f(x) would be increasing on [-1, 2], making it monotonic. Therefore, a = -4 is not acceptable, which contradicts the previous conclusion that a >= -3. So, there must be an error in my reasoning.Wait, when a = -3, the roots are 1 ± sqrt(1 - (-3)) = 1 ± sqrt(4) = 1 ± 2, so roots at -1 and 3. So, when a = -3, there is a root at x = -1, which is in the interval. But if a < -3, then sqrt(1 - a) becomes larger than 2, so 1 - sqrt(1 - a) becomes less than -1, and 1 + sqrt(1 - a) becomes larger than 3. Therefore, both roots are outside the interval [-1, 2] when a < -3. Therefore, in this case, f'(x) = x² - 2x + a would be positive for all x in [-1, 2] because the quadratic opens upwards and both roots are outside the interval. Therefore, f(x) would be increasing on [-1, 2], hence monotonic. Therefore, for a < -3, the function is monotonic. So, the lower bound for a is -3, since for a >= -3, there is at least one root in the interval. Wait, but when a = -3, the root is at x = -1, which is an endpoint. So, does the derivative change sign at the endpoint? If the root is at the endpoint, then in the interval [-1, 2], the derivative at x = -1 is zero. But to the right of x = -1, what is the derivative's sign? Let's compute f'(x) at x = -1: 0. Then, just to the right of x = -1, say x = -0.9, compute f'(-0.9) = (-0.9)^2 - 2*(-0.9) + a. If a = -3, f'(-0.9) = 0.81 + 1.8 - 3 = (0.81 + 1.8) - 3 = 2.61 - 3 = -0.39. Wait, that's negative. But at x = -1, it's zero. Then, moving right from x = -1 into the interval, the derivative is negative. But since the quadratic has its minimum at x = 1, and the other root is at x = 3, outside the interval. Therefore, in the interval [-1, 2], the derivative starts at zero at x = -1, becomes negative just to the right, reaches the minimum at x = 1 (which is a - 1 = -4), and then increases back up to f'(2) = 4 - 4 + a = a = -3. So, f'(2) = -3. Therefore, the derivative is negative throughout the interval except at x = -1 where it's zero. Therefore, the function f(x) is decreasing on [-1, 2], except at the very left endpoint where the derivative is zero. But since it's decreasing on (-1, 2], is it considered monotonic on [-1, 2]? Wait, the definition of monotonic allows for constant intervals, but in this case, the derivative is zero at x = -1, then negative on (-1, 2]. So, technically, the function is non-increasing on the entire interval [-1, 2], because it never increases. A function is non-increasing if for any x1 < x2, f(x1) ≥ f(x2). Here, from x = -1 to x = 2, the function is decreasing. So, even though the derivative is zero at x = -1, the function is still considered non-increasing (and hence monotonic) on the interval. Therefore, even at a = -3, the function is monotonic (non-increasing). Therefore, my previous conclusion was incorrect.This suggests that my approach has a flaw. Let's recap. If the derivative has a root at the endpoint, does that make the function non-monotonic? For example, suppose the derivative is zero at x = -1, but negative on (-1, 2]. Then, the function is decreasing on [-1, 2], hence monotonic. Similarly, if the derivative is zero at x = 2, but positive on [-1, 2), then the function is increasing on [-1, 2], hence monotonic. Therefore, in order for the function to not be monotonic, the derivative must have a root inside the interval (-1, 2), not at the endpoints.Therefore, we need the quadratic ( x^2 - 2x + a = 0 ) to have at least one root in the open interval (-1, 2). So, roots exactly at the endpoints don't count because the function remains monotonic. Therefore, we need to adjust our conditions to require that the quadratic has at least one root in (-1, 2), not including the endpoints.Therefore, let's re-express the problem: Find all real numbers ( a ) such that the equation ( x^2 - 2x + a = 0 ) has at least one solution in (-1, 2).To solve this, we can use the following method:1. Find the values of ( a ) such that there exists an ( x in (-1, 2) ) with ( x^2 - 2x + a = 0 ).Alternatively, we can express ( a = -x^2 + 2x ), so ( a = -x^2 + 2x ). Therefore, for ( x in (-1, 2) ), the corresponding ( a ) values are ( a = -x^2 + 2x ). Therefore, the range of ( a ) is the image of the function ( g(x) = -x^2 + 2x ) over the interval ( x in (-1, 2) ).So, let's analyze the function ( g(x) = -x^2 + 2x ). This is a quadratic function that opens downward (since the coefficient of ( x^2 ) is negative). Its maximum is at the vertex. The vertex of ( g(x) ) is at ( x = -b/(2a) ), where here ( a = -1 ), ( b = 2 ). So, vertex at ( x = -2/(2*(-1)) = -2 / (-2) = 1 ). So, the vertex is at x = 1, which is inside the interval (-1, 2). The value of ( g(x) ) at the vertex is ( g(1) = -1^2 + 2*1 = -1 + 2 = 1 ).Now, since the quadratic opens downward, the maximum value is 1 at x = 1, and the function decreases towards both ends. Let's evaluate ( g(x) ) at the endpoints of the interval:At x = -1: ( g(-1) = -(-1)^2 + 2*(-1) = -1 - 2 = -3 ).At x = 2: ( g(2) = -(2)^2 + 2*2 = -4 + 4 = 0 ).But since we're considering the open interval (-1, 2), we don't include the endpoints. However, the function approaches -3 as x approaches -1 from the right and approaches 0 as x approaches 2 from the left. Therefore, the range of ( g(x) ) over (-1, 2) is (-3, 1]. Wait, let's confirm:At x approaching -1 from the right, g(x) approaches -3. At x approaching 2 from the left, g(x) approaches 0. But since the maximum is 1 at x = 1. However, since x is in the open interval (-1, 2), the function ( g(x) ) can take all values from just above -3 up to 1. Wait, actually, as x approaches -1 from the right, g(x) approaches -3, and as x approaches 1, it reaches 1, then decreases to 0 as x approaches 2. So, the range should be (-3, 1]. Let me verify:Take x approaching -1 from the right: let x = -1 + ε, where ε is a small positive number. Then,g(-1 + ε) = -(-1 + ε)^2 + 2*(-1 + ε) = - (1 - 2ε + ε²) - 2 + 2ε = -1 + 2ε - ε² - 2 + 2ε = (-1 - 2) + (2ε + 2ε) - ε² = -3 + 4ε - ε². As ε approaches 0, this approaches -3.Similarly, as x approaches 2 from the left: x = 2 - ε,g(2 - ε) = -(2 - ε)^2 + 2*(2 - ε) = - (4 - 4ε + ε²) + 4 - 2ε = -4 + 4ε - ε² + 4 - 2ε = ( -4 + 4 ) + (4ε - 2ε) - ε² = 0 + 2ε - ε². As ε approaches 0, this approaches 0.Therefore, the function ( g(x) ) maps the open interval (-1, 2) to (-3, 1]. Therefore, the corresponding values of ( a ) are all real numbers ( a ) such that ( a in (-3, 1] ).But wait, when a = 1, the quadratic equation becomes ( x^2 - 2x + 1 = 0 ), which factors as ( (x - 1)^2 = 0 ), so there's a repeated root at x = 1. Since x = 1 is within the interval (-1, 2), then even though it's a repeated root, it's still a root inside the interval. Therefore, a = 1 is acceptable. However, when a = -3, as before, the root is at x = -1, which is not in the open interval (-1, 2), so a = -3 is not acceptable. Therefore, the range of a is (-3, 1].But let's check with specific a values:1. a = 0: Then, the equation ( x^2 - 2x = 0 ), roots at x = 0 and x = 2. But x = 2 is not in the open interval (-1, 2), so only x = 0 is in (-1, 2). Therefore, the derivative changes sign at x = 0, so the function is not monotonic. Therefore, a = 0 is valid.2. a = 1: The equation ( x^2 - 2x + 1 = 0 ), root at x = 1. Since x = 1 is in (-1, 2), the derivative touches zero here but doesn't cross (since it's a double root). Wait, but if the derivative has a double root at x = 1, does the sign change? Let's check the derivative on either side of x = 1. For a = 1, f'(x) = x² - 2x + 1 = (x - 1)^2, which is always non-negative. So, the derivative is zero at x = 1 but positive on either side. Therefore, the derivative does not change sign; it remains non-negative throughout. Therefore, the function f(x) is non-decreasing on [-1, 2], hence monotonic. Therefore, even though there is a root in the interval, since it's a double root, the derivative does not change sign. Therefore, the function is monotonic. Therefore, a = 1 should not be included in the valid range.Wait, this complicates things. So, even if there is a root in the interval, if it's a double root (i.e., the discriminant is zero), then the derivative does not change sign, so the function remains monotonic. Therefore, the requirement is not just that there is a root in the interval, but that the derivative actually changes sign there, i.e., the root is a simple root (multiplicity 1). Therefore, to ensure that the function is not monotonic, the derivative must have at least one simple root in the interval, i.e., the quadratic must cross the x-axis (not just touch it) within the interval.Therefore, in addition to having a root in the interval, we must exclude the case where the root is a double root (which occurs when discriminant is zero, i.e., when a = 1). Therefore, in our previous analysis, when a = 1, the quadratic has a double root at x = 1, but since the derivative does not change sign there (it's a minimum point where the derivative touches zero and remains positive on both sides), the function remains monotonic. Therefore, a = 1 should not be included in the range.Similarly, when a approaches 1 from below, the two roots approach each other and merge at a = 1. So, for a < 1, the quadratic has two distinct real roots, and for a = 1, a repeated root. Therefore, to have the derivative change sign, we need a < 1, so that the quadratic crosses the x-axis, leading to a sign change. Therefore, the range of a is (-3, 1).Wait, but when a approaches 1 from below, say a = 0.99, the roots are 1 ± sqrt(1 - 0.99) = 1 ± sqrt(0.01) = 1 ± 0.1, so 0.9 and 1.1. Both roots are in (-1, 2), so the derivative changes sign twice. However, even if one root is inside the interval, the derivative changes sign. Therefore, for a < 1, there are two distinct roots, but depending on a, either both are inside the interval or one is inside and one is outside.Wait, but earlier we saw that for a between -3 and 0, the left root is in (-1, 2) and the right root is outside. For a between 0 and 1, both roots are in (-1, 2). But in any case, as long as a is less than 1, there is at least one simple root in the interval, leading to a sign change. However, when a = 1, the root is a double root, so no sign change. Therefore, the range of a should be (-3, 1).But let's confirm with a = 1:f'(x) = (x - 1)^2, which is always non-negative. Therefore, f(x) is non-decreasing on the entire interval, hence monotonic. Therefore, a = 1 must be excluded.Similarly, when a = -3, f'(x) has a root at x = -1, which is not in the open interval (-1, 2). Therefore, the derivative does not change sign in (-1, 2), so the function is monotonic (non-increasing) on [-1, 2]. Therefore, a = -3 must also be excluded.Therefore, the correct range for a is (-3, 1). But let's check with an example where a is just above -3, say a = -2.9. Then, the roots are 1 ± sqrt(1 - (-2.9)) = 1 ± sqrt(3.9) ≈ 1 ± 1.9748. So, the roots are approximately 1 - 1.9748 ≈ -0.9748 and 1 + 1.9748 ≈ 2.9748. The left root, -0.9748, is within (-1, 2), so the derivative changes sign there, making the function non-monotonic. Similarly, if a = 0.5, the roots are 1 ± sqrt(1 - 0.5) = 1 ± sqrt(0.5) ≈ 1 ± 0.707, so roots at approximately 0.293 and 1.707, both within (-1, 2), so derivative changes sign twice, leading to a non-monotonic function. If a = -4, as before, both roots are outside the interval, so the function is monotonic.Therefore, the correct range is a ∈ (-3, 1). However, let me verify using another method. The function ( g(x) = -x^2 + 2x ) maps x ∈ (-1, 2) to a ∈ (-3, 1). Therefore, the values of a for which the equation ( x^2 - 2x + a = 0 ) has a solution in (-1, 2) are exactly the values of a in (-3, 1). Therefore, the answer should be ( -3 < a < 1 ).But let me also check the endpoints:- When a approaches -3 from above, say a = -2.999, the left root approaches -1 from the right, so it's inside the interval, causing the derivative to change sign.- When a approaches 1 from below, the roots approach 1 from both sides, leading to a sign change.Therefore, the range of a is all real numbers between -3 and 1, not including -3 and 1 themselves. Therefore, the range is ( (-3, 1) ).But to express this in the problem's requested format, which asks for the range of values for the real number a, we can write this as an interval:( -3 < a < 1 )Or, using interval notation: ( (-3, 1) ).But let's double-check with another approach. Use the Intermediate Value Theorem for the derivative. For the derivative ( f'(x) = x^2 - 2x + a ) to change sign in [-1, 2], we need f'(x) to take both positive and negative values in the interval. Therefore, there must exist points c and d in [-1, 2] such that f'(c) < 0 and f'(d) > 0.The minimum value of f'(x) on [-1, 2] is at the vertex x = 1, which is f'(1) = 1 - 2 + a = a - 1. The maximum value on the interval will be at one of the endpoints. Let's compute f'(-1) and f'(2):f'(-1) = (-1)^2 - 2*(-1) + a = 1 + 2 + a = 3 + af'(2) = (2)^2 - 2*(2) + a = 4 - 4 + a = aTherefore, the maximum value of f'(x) on [-1, 2] is the maximum of f'(-1) and f'(2). Since f'(-1) = 3 + a and f'(2) = a. So, if 3 + a > a, which is always true (since 3 > 0), the maximum is f'(-1) = 3 + a.For the derivative to take both positive and negative values, we need the minimum value (a - 1) to be less than 0 and the maximum value (3 + a) to be greater than 0. Therefore:a - 1 < 0 ⇒ a < 1and3 + a > 0 ⇒ a > -3Therefore, combining these, we have -3 < a < 1.This matches our previous conclusion. Therefore, the correct range is ( (-3, 1) ).Final AnswerThe range of values for ( a ) is boxed{(-3, 1)}.</think>To determine the range of values for the real number ( a ) such that the function ( f(x) = frac{1}{3}x^3 - x^2 + ax - 5 ) is not monotonic in the interval ([-1, 2]), we need to analyze the derivative ( f'(x) ).The derivative of ( f(x) ) is:[f'(x) = x^2 - 2x + a]For ( f(x) ) to not be monotonic in ([-1, 2]), the derivative ( f'(x) ) must change sign within the interval. This requires the quadratic equation ( x^2 - 2x + a = 0 ) to have at least one root in the open interval ((-1, 2)).The roots of the quadratic equation ( x^2 - 2x + a = 0 ) are:[x = 1 pm sqrt{1 - a}]For the roots to be real, the discriminant must be non-negative:[4 - 4a geq 0 implies a leq 1]Next, we need to ensure that at least one root lies within the interval ((-1, 2)). The roots are ( 1 + sqrt{1 - a} ) and ( 1 - sqrt{1 - a} ).1. For the root ( 1 - sqrt{1 - a} ) to be greater than (-1):[1 - sqrt{1 - a} > -1 implies sqrt{1 - a} < 2 implies 1 - a < 4 implies a > -3]2. For the root ( 1 + sqrt{1 - a} ) to be less than (2):[1 + sqrt{1 - a} < 2 implies sqrt{1 - a} < 1 implies 1 - a < 1 implies a > 0]However, we need to consider the interval ((-1, 2)) and ensure that the quadratic changes sign within this interval. Using the Intermediate Value Theorem, we check if the derivative ( f'(x) ) takes both positive and negative values in ([-1, 2]). The minimum value of ( f'(x) ) on ([-1, 2]) is at ( x = 1 ):[f'(1) = a - 1]The maximum values at the endpoints are:[f'(-1) = 3 + a quad text{and} quad f'(2) = a]For ( f'(x) ) to change sign, we need:[a - 1 < 0 implies a < 1]and[3 + a > 0 implies a > -3]Combining these conditions, we get:[-3 < a < 1]Thus, the range of values for ( a ) is (boxed{(-3, 1)}).]

🔑:AnalysisThis problem involves the application of derivative knowledge, function monotonicity, and solving inequalities. Understanding the problem correctly is crucial.SolutionThe derivative of the function is given by f'(x) = x^2 - 2x + a = (x-1)^2 + a - 1.If the function f(x) = frac{1}{3} x^3 - x^2 + ax - 5 is monotonic in the interval [-1, 2], then a - 1 geq 0 or f'(-1) = 3 + a leq 0 and f'(2) = a leq 0.Therefore, a geq 1 or a leq -3.Thus, the range of real numbers a that satisfies the condition is (-3, 1).So, the answer is boxed{(-3,1)}.

🧠:[<think>Okay, so I need to solve two parts of this problem involving the function f(x) = e^x - x + 1. Let's start with part (1): finding the equation of the tangent line at the point (2, f(2)). First, I remember that the equation of a tangent line at a point (a, f(a)) is given by y = f(a) + f’(a)(x - a). So I need to find f(2) and f’(2). Let me compute f(2) first. Plugging x = 2 into the function: f(2) = e^2 - 2 + 1. That simplifies to e^2 - 1. Let me check: e squared minus 2 plus 1 is indeed e squared minus 1. Next, I need the derivative f’(x) to find the slope at x = 2. The derivative of e^x is e^x, and the derivative of -x is -1, and the derivative of 1 is 0. So f’(x) = e^x - 1. Then f’(2) is e^2 - 1. So the slope of the tangent line at x = 2 is e^2 - 1, and the point is (2, e^2 - 1). Plugging these into the tangent line equation: y = (e^2 - 1) + (e^2 - 1)(x - 2). Let me simplify that. Distribute the slope: y = (e^2 - 1) + (e^2 - 1)x - 2(e^2 - 1). Combining the constants: (e^2 - 1) - 2(e^2 - 1) = - (e^2 - 1). So the equation becomes y = (e^2 - 1)x - (e^2 - 1) - (e^2 - 1) wait, no, hold on. Wait, let me recheck that step.Wait, no. Let's do the algebra again. The equation is y = (e² - 1) + (e² - 1)(x - 2). Let me expand the second term: (e² -1)x - 2(e² -1). Then combine with the first term: (e² -1)x - 2(e² -1) + (e² -1). That simplifies to (e² -1)x - (2(e² -1) - (e² -1)) which is (e² -1)x - (e² -1). So the equation is y = (e² -1)(x - 1). Hmm, is that correct? Let me verify:If I factor out (e² -1), then y = (e² -1)[(x - 2) + 1] = (e² -1)(x -1). Yes, that's another way to see it. So the tangent line equation is y = (e² -1)(x -1). That seems right. Let me check with a point. When x = 2, y should be e² -1. Plugging in x = 2: y = (e² -1)(2 -1) = e² -1. Correct. So that's the tangent line.So part (1) is done. Now moving to part (2): finding the maximum and minimum values of f(x) on the interval [-2, 1]. To find extrema on a closed interval, we need to check critical points inside the interval and the endpoints. Critical points occur where the derivative is zero or undefined. Since f(x) is differentiable everywhere (since it's composed of exponential and polynomial terms), critical points are where f’(x) = 0. We already found f’(x) = e^x -1. Setting that equal to zero: e^x -1 = 0 → e^x = 1 → x = 0. So the only critical point is at x = 0, which is within the interval [-2, 1]. So we need to evaluate f(x) at x = -2, x = 0, and x = 1.Compute f(-2): e^{-2} - (-2) +1 = e^{-2} + 2 +1 = e^{-2} +3. Compute f(0): e^{0} -0 +1 = 1 +1 = 2.Compute f(1): e^{1} -1 +1 = e^1 = e ≈ 2.718.So comparing these three values: f(-2) ≈ 1/e² +3 ≈ 0.135 + 3 ≈ 3.135, f(0) = 2, f(1) ≈ 2.718.Therefore, the maximum value is at x = -2 with f(-2) = 3 + e^{-2}, and the minimum value is at x = 0 with f(0) = 2. Wait, but let me confirm the values again. Wait, hold on. Let me compute f(-2) again. e^{-2} is approximately 0.135, so 0.135 +3 is approximately 3.135. Then f(1) is e ≈ 2.718. So f(-2) is the largest, and f(0) is the smallest. Wait, but wait, f(-2) is approximately 3.135, which is more than f(1) ≈ 2.718. So maximum at x = -2 and minimum at x =0. But let me check if there are any other critical points. We found only x=0. So that's correct.Therefore, the maximum value on [-2,1] is 3 + e^{-2} (since f(-2)=e^{-2} +3) and the minimum value is 2 at x=0.Wait, but let me verify f(-2) again. Original function: e^x -x +1. At x=-2, that's e^{-2} - (-2) +1 = e^{-2} +2 +1 = e^{-2} +3. So yes, 3 + e^{-2}. Correct.Similarly, at x=1, f(1)=e^1 -1 +1= e, as e -1 +1= e. Correct. So yes, the maximum is at x=-2, and the minimum is at x=0.Therefore, answers are:(1) The tangent line is y = (e² -1)(x -1).(2) The maximum value is 3 + e^{-2} at x=-2 and the minimum is 2 at x=0.I need to check if there's any mistake here. Let me check the derivative again. For f(x)=e^x -x +1, derivative is e^x -1. Correct. Setting to zero gives x=0. Correct. Then evaluating at endpoints and critical points. Yes. So seems correct. So that's the conclusion.Final Answer(1) The equation of the tangent line is boxed{y = (e^2 - 1)(x - 1)}.(2) The maximum value is boxed{3 + frac{1}{e^2}} and the minimum value is boxed{2}.</think>To find the equation of the tangent line to the function ( y = f(x) ) at the point ( (2, f(2)) ):1. Compute ( f(2) ): [ f(2) = e^2 - 2 + 1 = e^2 - 1 ]2. Find the derivative ( f'(x) ): [ f'(x) = e^x - 1 ]3. Evaluate the derivative at ( x = 2 ): [ f'(2) = e^2 - 1 ]4. Use the point-slope form of the tangent line equation: [ y = f(2) + f'(2)(x - 2) = (e^2 - 1) + (e^2 - 1)(x - 2) ] Simplifying this, we get: [ y = (e^2 - 1)(x - 1) ]For the maximum and minimum values of ( y = f(x) ) on the interval ([-2, 1]):1. Find the critical points by setting the derivative ( f'(x) = e^x - 1 ) to zero: [ e^x - 1 = 0 implies e^x = 1 implies x = 0 ]2. Evaluate ( f(x) ) at the critical point and the endpoints ( x = -2 ) and ( x = 1 ): - At ( x = -2 ): [ f(-2) = e^{-2} - (-2) + 1 = e^{-2} + 3 ] - At ( x = 0 ): [ f(0) = e^0 - 0 + 1 = 2 ] - At ( x = 1 ): [ f(1) = e^1 - 1 + 1 = e ]3. Compare the values: - The maximum value is ( 3 + e^{-2} ) at ( x = -2 ) - The minimum value is ( 2 ) at ( x = 0 ) Final Answer(1) The equation of the tangent line is boxed{y = (e^2 - 1)(x - 1)}.(2) The maximum value is boxed{3 + frac{1}{e^2}} and the minimum value is boxed{2}.,<think>Okay, so I have this problem here with the function f(x) = e^x - x + 1. There are two parts: first, finding the equation of the tangent line at the point (2, f(2)), and second, finding the maximum and minimum values of the function on the interval [-2, 1]. Let me tackle each part step by step.Starting with part (1): Find the tangent line equation at (2, f(2)). I remember that the equation of a tangent line at a point (a, f(a)) is given by y = f(a) + f’(a)(x - a). So, I need to compute f(2) and f’(2).First, let me compute f(2). The function is e^x - x + 1, so substituting x=2:f(2) = e^2 - 2 + 1 = e^2 - 1. Hmm, e^2 is approximately 7.389, so 7.389 -1 is about 6.389. But maybe I should keep it exact instead of using decimal approximations. So, f(2) = e² - 1.Next, find the derivative f’(x) to get the slope at x=2. The derivative of e^x is e^x, the derivative of -x is -1, and the derivative of 1 is 0. So, f’(x) = e^x - 1. Therefore, f’(2) = e² - 1, same as f(2). That's interesting. So the slope of the tangent line at x=2 is e² -1.Therefore, plugging into the tangent line equation:y = f(2) + f’(2)(x - 2) = (e² -1) + (e² -1)(x -2). Let me simplify this equation.First, factor out (e² -1):y = (e² -1)[1 + (x -2)] = (e² -1)(x -1). Wait, is that right? Let me check:Starting again: y = (e² -1) + (e² -1)(x -2)Expand the second term: (e² -1)x - 2(e² -1)Combine with the first term: (e² -1) + (e² -1)x - 2(e² -1) = (e² -1)x - (e² -1)Factor out (e² -1): (e² -1)(x -1). Yeah, that seems correct. So the equation simplifies to y = (e² -1)(x -1). Hmm, that seems a bit surprising, but algebraically it checks out.Alternatively, maybe it's better not to factor and leave it as y = (e² -1)(x -2) + (e² -1). Either way is correct, but perhaps the factored form is more elegant. Let me verify with a different approach.Alternatively, if I use point-slope form: slope m = e² -1, point (2, e² -1). So,y - (e² -1) = (e² -1)(x -2)Then, adding (e² -1) to both sides:y = (e² -1)(x -2) + (e² -1) = (e² -1)(x -2 +1) = (e² -1)(x -1). Yep, same result. So the tangent line is y = (e² -1)(x -1). That seems correct. So part (1) is done.Moving on to part (2): Find the maximum and minimum values of f(x) on [-2, 1]. For continuous functions on a closed interval, by Extreme Value Theorem, they should attain their max and min on the interval, either at critical points or at endpoints. So, I need to find critical points by setting derivative f’(x) =0, then evaluate f(x) at critical points and endpoints.We already know f’(x) = e^x -1. So set e^x -1 =0 => e^x =1 => x = ln(1) =0. So the only critical point in the interval [-2,1] is at x=0. Therefore, the critical points are x=0, and endpoints x=-2 and x=1.Compute f(x) at x=-2, x=0, x=1.First, x=-2: f(-2) = e^{-2} - (-2) +1 = e^{-2} +2 +1 = e^{-2} +3. e^{-2} is approximately 0.135, so f(-2) ≈ 0.135 +3 ≈3.135.But again, let me keep it exact: f(-2) = e^{-2} +3.Next, x=0: f(0) = e^{0} -0 +1 =1 +0 +1=2.x=1: f(1)= e^1 -1 +1= e -1 +1= e. So f(1)=e ≈2.718.So comparing the three values:f(-2)= e^{-2} +3 ≈3.135,f(0)=2,f(1)=e≈2.718.So the maximum value is at x=-2, which is e^{-2} +3, and the minimum value is at x=0, which is 2.Wait, but hold on. Let me double-check. At x=0, the function is 2. At x=1, it's e (~2.718), which is larger than 2. At x=-2, it's e^{-2} +3 ≈3.135, which is the largest. So maximum is at x=-2, minimum at x=0. That seems right.But let me check the critical point at x=0. Since f’(x) = e^x -1, so to the left of x=0 (negative x), e^x <1, so f’(x) <0, meaning function is decreasing. To the right of x=0 (positive x), e^x >1, so f’(x) >0, function is increasing. Therefore, x=0 is a local minimum. So on the interval [-2,1], the function decreases from x=-2 to x=0, reaching a minimum at x=0, then increases from x=0 to x=1. Therefore, the maximum occurs at the left endpoint x=-2, and the minimum at x=0.Yes, that's consistent with the computed values. Therefore, the maximum is e^{-2} +3 and the minimum is 2.So, part (2) answer: maximum at x=-2, value e^{-2} +3; minimum at x=0, value 2.Wait, but the problem says "maximum and minimum values", so maybe just the values, not the points. Let me check.The question says: "Find the maximum and minimum values of the function y=f(x) on the interval [-2,1]". So yes, just the values, not the x where they occur. So maximum value is e^{-2} +3, minimum value is 2.But let me confirm once again the calculations.At x=-2:f(-2) = e^{-2} - (-2) +1 = e^{-2} +2 +1 = e^{-2} +3. Correct.At x=0:f(0)= e^0 -0 +1=1+0+1=2. Correct.At x=1:f(1)= e -1 +1= e. Correct.Yes, so comparing those three, maximum is indeed e^{-2} +3, which is approximately 3.135, and minimum is 2. So that's part (2) done.I think that's all. Let me recap:1) Tangent line at (2, f(2)): y = (e² -1)(x -1).2) Maximum value on [-2,1] is e^{-2} +3, minimum value is 2.Final Answer(1) The equation of the tangent line is boxed{y = (e^2 - 1)(x - 1)}.(2) The maximum value is boxed{3 + frac{1}{e^2}} and the minimum value is boxed{2}.</think>(1) To find the equation of the tangent line to the function ( y = f(x) ) at the point ( (2, f(2)) ):First, compute ( f(2) ):[f(2) = e^2 - 2 + 1 = e^2 - 1]Next, find the derivative ( f'(x) ):[f'(x) = e^x - 1]Evaluate the derivative at ( x = 2 ):[f'(2) = e^2 - 1]Using the point-slope form of the tangent line equation:[y - f(2) = f'(2)(x - 2)]Substitute the values:[y - (e^2 - 1) = (e^2 - 1)(x - 2)]Simplify to get the equation of the tangent line:[y = (e^2 - 1)(x - 1)](2) To find the maximum and minimum values of ( y = f(x) ) on the interval ([-2, 1]):First, find the critical points by setting the derivative ( f'(x) = 0 ):[e^x - 1 = 0 implies e^x = 1 implies x = 0]Evaluate ( f(x) ) at the critical point ( x = 0 ) and at the endpoints ( x = -2 ) and ( x = 1 ):- At ( x = -2 ):[f(-2) = e^{-2} - (-2) + 1 = e^{-2} + 3]- At ( x = 0 ):[f(0) = e^0 - 0 + 1 = 2]- At ( x = 1 ):[f(1) = e^1 - 1 + 1 = e]Comparing the values:- The maximum value is ( e^{-2} + 3 ) at ( x = -2 )- The minimum value is ( 2 ) at ( x = 0 ) Final Answer(1) The equation of the tangent line is boxed{y = (e^2 - 1)(x - 1)}.(2) The maximum value is boxed{3 + frac{1}{e^2}} and the minimum value is boxed{2}.]

🔑:(1) First, we find the derivative of the function f(x). The derivative f'(x) = e^{x} - 1.Next, we compute f'(2) = e^{2} - 1 and f(2) = e^{2} - 2 + 1 = e^{2} - 1.The equation of the tangent line at point (2,f(2)) is given by y - (e^{2}-1) = (e^{2}-1)(x-2).This simplifies to boxed{(e^{2}-1)x - y - e^{2} + 1 = 0}.(2) To find the maximum and minimum values, we first find the critical points by setting f'(x) = e^{x} - 1 = 0. Solving for x, we get x=0.Now we determine the intervals where the function is increasing or decreasing:- For the interval (-infty, 0), the function is decreasing.- For the interval [0, +infty), the function is increasing.On the interval [-2,1], the function is decreasing on [-2,0] and increasing on [0,1]. Therefore, the function takes its minimum value at x=0 and its maximum value at one of the endpoints.Now, let's compute the function values at the critical point and the endpoints:- f(0) = e^{0} - 0 + 1 = 2- f(-2) = e^{-2} - (-2) + 1 = e^{-2} + 3- f(1) = e^{1} - 1 + 1 = eSince e^{-2} + 3 > e, the function takes its maximum value at x=-2. Therefore, the maximum value is boxed{e^{-2} + 3} and the minimum value is boxed{2}.

🧠:[<think>Okay, let me try to figure out how to find the complement of set A with respect to the universal set U. So, the problem says that the universal set U is {1, 3, 5} and set A is {1, 5}. I need to find the complement of A, which is written as ∁_U A. Hmm, I remember from my basic set theory that the complement of a set consists of all the elements in the universal set that are not in the original set. Let me make sure I remember that right.Alright, so the complement of A, denoted ∁_U A or sometimes A', should be all the elements that are in U but not in A. Let's break it down. The universal set U has the elements 1, 3, and 5. Set A has 1 and 5. So, if we remove the elements that are in A from U, what's left should be the complement. That means we take U and subtract A from it.Wait, subtraction in sets is like set difference, right? So, ∁_U A is the same as U A. So, U A would be all elements in U that are not in A. Let me write that out. U is {1, 3, 5}, A is {1, 5}. So, taking out 1 and 5 from U leaves us with just {3}. That seems straightforward. So, the complement of A with respect to U is {3}. Let me verify that. If the universal set is all the elements we're considering, and A is a subset of U, then the complement should indeed be everything in U that's not in A. Since A has 1 and 5, the only element left in U that's not in A is 3. Therefore, the complement should be {3}. Wait a second, did I miss anything here? Let's check each element of U. First element: 1. Is 1 in A? Yes, so it's excluded. Second element: 3. Is 3 in A? No, so it's included in the complement. Third element: 5. Is 5 in A? Yes, so it's excluded. So, only 3 is left. Yep, that's correct. I don't think there are any other elements to consider here. Since U only has three elements and two of them are in A, the complement must be the remaining one. I guess that's it. The complement of A with respect to U is {3}. I don't see any mistakes in my reasoning. The key was remembering that the complement is relative to the universal set, and in this case, U is clearly defined, so there's no ambiguity. If U had more elements, the complement might be larger, but here it's straightforward.Final AnswerThe complement of ( A ) with respect to ( U ) is boxed{{3}}.</think>To find the complement of set ( A ) with respect to the universal set ( U ), we need to identify all elements in ( U ) that are not in ( A ).Given:- The universal set ( U = {1, 3, 5} )- The set ( A = {1, 5} )The complement of ( A ) with respect to ( U ), denoted by ( complement_U A ), is the set of elements in ( U ) that are not in ( A ). Checking each element of ( U ):- The element ( 1 ) is in ( A ), so it is excluded.- The element ( 3 ) is not in ( A ), so it is included.- The element ( 5 ) is in ( A ), so it is excluded.Thus, the complement of ( A ) with respect to ( U ) is the set containing the remaining element ( 3 ).[complement_U A = U setminus A = {1, 3, 5} setminus {1, 5} = {3}]The final answer is (boxed{{3}}).,<think>Okay, let's see. I need to find the complement of set A with respect to the universal set U. The universal set U is given as {1, 3, 5}, and set A is {1, 5}. So, the complement of A, denoted by ∁_U A, should be all the elements in U that are not in A. Let me break this down step by step.First, I'll recall what a complement of a set means. If I have a universal set U, then the complement of a subset A, written as ∁_U A or sometimes A', is the set of all elements in U that are not in A. So, basically, it's U minus A. That makes sense. So, the complement should include everything that's in the universal set but not in the original set A.Given that U is {1, 3, 5} and A is {1, 5}, let's list out the elements. The elements of U are 1, 3, and 5. The elements of A are 1 and 5. So, if I subtract A from U, I remove the elements that are in A from U. So removing 1 and 5 from U would leave just 3. Therefore, the complement of A with respect to U is {3}.Wait, let me check if I did that right. Let me write it out in another way. The complement of A, ∁_U A, is defined as U A, which is the set difference. So, U A would be all elements x such that x is in U and x is not in A. So, in symbols, ∁_U A = {x ∈ U | x ∉ A}.Looking at each element in U:1 is in U. Is 1 in A? Yes, so we don't include it in the complement.3 is in U. Is 3 in A? Let me check set A: {1, 5}. No, 3 isn't there. So, we include 3 in the complement.5 is in U. Is 5 in A? Yes, so we exclude it from the complement.Therefore, the only element left is 3, so the complement is {3}. That seems straightforward.I want to make sure there isn't any step I missed. Sometimes when dealing with sets, especially complements, you have to confirm that the universal set is correctly defined because the complement depends entirely on the universal set. In this case, the problem clearly states that the universal set U is {1, 3, 5}, so there's no ambiguity there. If the universal set were different, say, including more elements, the complement could be different. But here, since U is limited to three elements, and two of them are in A, only the third remains.Another way to verify is using the formula for the complement: the cardinality of the complement should be the cardinality of U minus the cardinality of A. Here, |U| is 3 and |A| is 2, so |∁_U A| should be 1. And indeed, {3} has one element. So that checks out.Let me also consider if there could be any other elements. Suppose I thought that maybe numbers not mentioned in U are part of the complement? But no, the universal set is explicitly given as {1, 3, 5}, which means that all elements under consideration are only those three. So even if other numbers exist outside this universal set, they aren't part of this particular problem's universe. Therefore, the complement is strictly elements within U not in A.So, no other elements to consider. The complement is just {3}. I think that's solid. Let me recap:- Universal set U: {1, 3, 5}- Set A: {1, 5}- Complement of A: U minus A = {3}Yes, that's correct. I don't see any mistakes here. The steps all follow logically, and each element is checked against both sets. The result is a singleton set containing the only element in U that isn't in A. Therefore, the answer should be {3}, which in boxed form is boxed{{3}}.Final AnswerThe complement of ( A ) with respect to ( U ) is boxed{{3}}.</think>To find the complement of set ( A ) with respect to the universal set ( U ), we need to identify all elements in ( U ) that are not in ( A ).Given:- The universal set ( U = {1, 3, 5} )- The set ( A = {1, 5} )The complement of ( A ) with respect to ( U ), denoted by ( complement_U A ), is the set of elements in ( U ) that are not in ( A ). Checking each element of ( U ):- The element ( 1 ) is in ( U ) and also in ( A ), so it is excluded.- The element ( 3 ) is in ( U ) but not in ( A ), so it is included.- The element ( 5 ) is in ( U ) and also in ( A ), so it is excluded.Thus, the complement of ( A ) with respect to ( U ) is ( {3} ).[complement_U A = U setminus A = {1, 3, 5} setminus {1, 5} = {3}]The final answer is boxed{{3}}.]

🔑:Given the universal set U = {1, 3, 5} and the set A = {1, 5}, the complement of A with respect to U includes all elements that are in U but not in A. To find complement_U A, we look at the elements of U one by one and check if they are not in A:- 1 is in A, so it is not in complement_U A.- 3 is not in A, so it is in complement_U A.- 5 is in A, so it is not in complement_U A.Therefore, the set complement_U A = {3}. We have used the definition of complement, which consists of the elements that are in the universal set U but not in the subset A. This is a fundamental concept pertaining to set theory dealing with definitions and application problems.[boxed{complement_U A = {3}}]