🧠:[<think>Okay, let me tackle these two problems one by one. Starting with the first one: Calculate (π - 1)^0 - √8 + |-2√2|. Hmm, let's break this down step by step. First, I remember that any number raised to the power of 0 is 1, right? So (π - 1)^0 should be 1, regardless of what π - 1 is. That part seems straightforward. Next up is √8. I need to simplify that. Let me think, 8 can be broken down into 4 * 2, and since 4 is a perfect square, the square root of 4 is 2. So √8 simplifies to √(4*2) = √4 * √2 = 2√2. Got that down. Then there's the absolute value part: |-2√2|. Absolute value makes sure the result is non-negative, so |-2√2| would just be 2√2. The negative sign inside the absolute value disappears. So putting it all together: 1 (from the first term) minus 2√2 (from √8) plus 2√2 (from the absolute value). Wait, the -2√2 and +2√2 would cancel each other out, right? So that leaves us with just 1. Let me check that again:First term: 1Second term: -2√2Third term: +2√2Adding those together: 1 - 2√2 + 2√2 = 1 + 0 = 1. Yeah, that seems correct. The square roots cancel each other, leaving only 1. That was easier than I thought once broken down. Now moving on to the second problem: Solve the inequality 3x - 2 > x + 4. Okay, inequalities can sometimes trip me up, but let's take it step by step. First, I want to get all the x terms on one side and the constants on the other. Let me subtract x from both sides to start. That would give:3x - 2 - x > x + 4 - xSimplifying both sides:2x - 2 > 4Now, I need to isolate the term with x. So add 2 to both sides:2x - 2 + 2 > 4 + 2Which simplifies to:2x > 6Then, divide both sides by 2 to solve for x:(2x)/2 > 6/2So, x > 3. Let me verify that. If x is greater than 3, let's pick a number like 4. Plugging into the original inequality:Left side: 3*4 - 2 = 12 - 2 = 10Right side: 4 + 4 = 810 > 8, which is true. What if x is 3? Then left side: 9 - 2 = 7, right side: 7. 7 > 7 is false, so the inequality doesn't hold at x = 3, which matches x > 3. How about a number less than 3, like 2? Left side: 6 - 2 = 4, right side: 6. 4 > 6 is false. So, it seems correct that x must be greater than 3. I think that's solid. Let me just recap the steps again to be sure. Subtract x from both sides, add 2 to both sides, divide by 2. Yep, straightforward linear inequality. So, first problem simplifies to 1, second problem solution is x > 3. Final Answer(1) boxed{1}(2) boxed{x > 3}</think>(1) Calculate: ((pi - 1)^0 - sqrt{8} + |-2sqrt{2}|)First, any number raised to the power of 0 is 1, so ((pi - 1)^0 = 1).Next, simplify (sqrt{8}). Since (8 = 4 times 2), we have (sqrt{8} = sqrt{4 times 2} = sqrt{4} times sqrt{2} = 2sqrt{2}).Then, evaluate the absolute value (|-2sqrt{2}|). The absolute value of a negative number is its positive counterpart, so (|-2sqrt{2}| = 2sqrt{2}).Putting it all together:[1 - 2sqrt{2} + 2sqrt{2}]The terms (-2sqrt{2}) and (2sqrt{2}) cancel each other out, leaving:[1](2) Solve the inequality: (3x - 2 > x + 4)Subtract (x) from both sides:[3x - 2 - x > x + 4 - x]Simplifying both sides:[2x - 2 > 4]Add 2 to both sides:[2x - 2 + 2 > 4 + 2]Simplifying both sides:[2x > 6]Divide both sides by 2:[x > 3] Final Answer(1) (boxed{1})(2) (boxed{x > 3}),<think>Okay, let's tackle these two problems one by one. Starting with part (1): Calculate (π - 1)^0 - √8 + |-2√2|. Hmm, let me break this down step by step. First, I remember that any number raised to the power of 0 is 1, right? So (π - 1)^0 should be 1, regardless of what π - 1 is. That part seems straightforward. Next, we have -√8. Let me think about √8. Since 8 is 4 times 2, and the square root of 4 is 2, so √8 is √(4*2) which is √4 * √2 = 2√2. So √8 is 2√2. Therefore, -√8 would be -2√2. Then, the last term is |-2√2|. The absolute value of a number is its distance from zero on the number line, so it's always non-negative. Since -2√2 is negative, its absolute value would be 2√2. Putting it all together, the expression becomes: 1 - 2√2 + 2√2. Wait, the -2√2 and +2√2 are like terms. If I combine them, they cancel each other out. So that leaves me with just 1. Let me check that again to be sure. (π - 1)^0 is definitely 1. Then -√8 is -2√2, and |-2√2| is 2√2. So 1 - 2√2 + 2√2. The -2√2 and +2√2 add up to zero. So yes, the result is 1. That seems right. Maybe I was overcomplicating it at first, but breaking it down step by step makes sense.Okay, moving on to part (2): Solve the inequality 3x - 2 > x + 4. Alright, solving inequalities is similar to solving equations, but I need to remember to reverse the inequality sign if I multiply or divide by a negative number. Let me start by getting all the x terms on one side and the constants on the other.First, subtract x from both sides: 3x - 2 - x > 4. That simplifies to 2x - 2 > 4. Then, add 2 to both sides to isolate the term with x: 2x > 6. Finally, divide both sides by 2: x > 3. Wait, let me verify that. Starting with 3x - 2 > x + 4. Subtract x: 2x - 2 > 4. Add 2: 2x > 6. Divide by 2: x > 3. Yep, that seems correct. Since we didn't multiply or divide by a negative number, the inequality direction stays the same. So the solution is all real numbers greater than 3. Let me just check with a number greater than 3, say x = 4. Plugging into the original inequality: 3(4) - 2 = 12 - 2 = 10. On the right side: 4 + 4 = 8. 10 > 8, which is true. What about x = 3? Left side: 9 - 2 = 7. Right side: 3 + 4 = 7. 7 > 7 is false, so x=3 is not included, which matches x > 3. If I pick x = 0, which is less than 3: left side is -2, right side is 4. -2 is not greater than 4. So the solution x > 3 seems correct. I think that's all for these problems. The first one simplified nicely once I recognized the exponents and absolute value, and the inequality was straightforward algebra.</think>(1) To calculate ((π - 1)^0 - sqrt{8} + |-2sqrt{2}|):- Any non-zero number to the power of 0 is 1: ((π - 1)^0 = 1).- Simplify (sqrt{8}): (sqrt{8} = 2sqrt{2}), so (- sqrt{8} = -2sqrt{2}).- Absolute value of (-2sqrt{2}) is (2sqrt{2}): (|-2sqrt{2}| = 2sqrt{2}).Combining these:[1 - 2sqrt{2} + 2sqrt{2} = 1 + (-2sqrt{2} + 2sqrt{2}) = 1 + 0 = 1](2) To solve the inequality (3x - 2 > x + 4):- Subtract (x) from both sides: (2x - 2 > 4).- Add 2 to both sides: (2x > 6).- Divide by 2: (x > 3).Answers:(1) (boxed{1})(2) (boxed{x > 3})]

🔑: Problem 1: Calculate (π-1)^{0}-sqrt{8}+|-2sqrt{2}|Step 1: Evaluate the exponent (π-1)^{0} = 1 Step 2: Simplify the square root sqrt{8} = sqrt{4 cdot 2} = 2sqrt{2} Step 3: Evaluate the absolute value |-2sqrt{2}| = 2sqrt{2} Step 4: Combine the results 1 - 2sqrt{2} + 2sqrt{2} = 1 Therefore, the final answer is boxed{1}. Problem 2: Solve the inequality 3x-2 gt x+4Step 1: Rearrange the inequality to isolate terms involving x on one side 3x - x gt 4 + 2 Step 2: Simplify both sides 2x gt 6 Step 3: Divide both sides by 2 to solve for x x gt 3 Therefore, the solution set of the original inequality is boxed{x gt 3}.

🧠:[<think>Okay, let's see. The problem says that 'a' is a rational number, and we need to determine which of the options A, B, C, or D is the correct conclusion. Let's go through each option one by one.Starting with option A: "-a is a negative rational number." Hmm, wait. If 'a' is a rational number, then '-a' would just be the additive inverse of 'a'. But whether it's negative depends on the original value of 'a'. For example, if 'a' is positive, then '-a' is negative. But if 'a' is negative, then '-a' becomes positive. So this statement isn't always true. Like if 'a' is -3, then '-a' is 3, which is positive. So A can't be the right answer because it's not necessarily true for all rational numbers.Moving on to option B: "|a| is a positive number." The absolute value of 'a' is always non-negative, right? Because absolute value measures the distance from zero on the number line, which can't be negative. But does that mean it's always positive? Wait, if 'a' is zero, then |a| is zero. And zero is neither positive nor negative. So in that case, |a| would be zero, which is not positive. So B says it's a positive number, but if 'a' is zero, that's not true. So B is incorrect because it might be zero.Option C: "|a| is a non-negative number." Let's think. The absolute value of any real number, including rational numbers, is always non-negative. That's the definition. Even if 'a' is zero, |a| is zero, which is non-negative. So whether 'a' is positive, negative, or zero, |a| is always non-negative. So C seems correct. But let me check the other options to be sure.Option D: "|a| = a." Hmm, this is only true if 'a' is non-negative. If 'a' is positive or zero, then |a| equals 'a'. But if 'a' is negative, then |a| is the opposite of 'a', so |a| = -a, which isn't equal to 'a' unless 'a' is zero. So for example, if 'a' is -5, then |a| is 5, which is not equal to -5. Therefore, D is only true when 'a' is non-negative, but since the problem only states that 'a' is rational, which could be negative, positive, or zero, D isn't necessarily always true.So summarizing:- A is false because if 'a' is negative, -a is positive.- B is false because |a| could be zero.- C is true because |a| is always non-negative regardless of 'a'.- D is false when 'a' is negative.Therefore, the correct answer should be option C.Final Answerboxed{C}</think>Given that ( a ) is a rational number, we need to determine the correct conclusion among the options provided.- Option A: (-a) is a negative rational number. This is not necessarily true because if ( a ) is negative, (-a) would be positive. For example, if ( a = -3 ), then (-a = 3 ), which is positive. Hence, A is incorrect.- Option B: (|a|) is a positive number. This is false because if ( a = 0 ), then (|a| = 0 ), which is not positive. Hence, B is incorrect.- Option C: (|a|) is a non-negative number. This is true because the absolute value of any rational number is always non-negative. For example, if ( a = 0 ), (|a| = 0 ); if ( a = 5 ), (|a| = 5 ); and if ( a = -5 ), (|a| = 5 ). All these values are non-negative. Hence, C is correct.- Option D: (|a| = a). This is only true if ( a ) is non-negative. If ( a ) is negative, (|a| = -a neq a ). For example, if ( a = -5 ), (|a| = 5 neq -5 ). Hence, D is incorrect.Thus, the correct conclusion is (boxed{C}).,<think>Okay, let's see here. The problem says that "a is a rational number, the correct conclusion is ()" with options A through D. Hmm, I need to figure out which of these statements must always be true if a is a rational number. Let me go through each option one by one.Starting with option A: "-a is a negative rational number." Hmm, so if a is rational, then -a is also rational because the negative of a rational number is still rational. But does that make -a necessarily negative? Well, if a is positive, then -a is negative. But what if a is negative? Then -a would be positive, right? For example, if a is -3, then -a would be 3, which is positive. So, -a is only negative if a is positive. But since a could be either positive or negative (since the problem just states a is rational, not specifying its sign), we can't say for sure that -a is negative. Therefore, option A is not always true. So A is incorrect.Moving on to option B: "|a| is a positive number." The absolute value of a number is always non-negative, right? So |a| is either positive or zero. But if a is zero, then |a| is zero. So, is zero considered a positive number? No, zero is neither positive nor negative. Therefore, |a| is non-negative, but not necessarily positive. So if a is zero, |a| is zero, which is not positive. Therefore, B is not always true. So B is incorrect.Option C: "|a| is a non-negative number." As I thought in B, the absolute value of any real number (and hence any rational number) is always non-negative. That means |a| is either zero or positive. Since non-negative includes zero and positive numbers, this statement holds true regardless of the value of a. Even if a is zero, |a| is zero, which is non-negative. If a is positive or negative, |a| is positive. So C is always true. Therefore, C is a correct conclusion.Option D: "|a| = a." Hmm, this would mean that the absolute value of a is equal to a itself. But that's only true if a is non-negative. If a is positive or zero, then |a| = a. However, if a is negative, |a| = -a, which is not equal to a. For example, if a is -5, |a| is 5, which is not equal to -5. So this statement is only true when a is non-negative. But since a could be negative (as the problem doesn't restrict a to being non-negative), this isn't always true. Therefore, D is incorrect.So, going through all options:- A is not always true because -a could be positive if a is negative.- B is not always true because |a| could be zero.- C is always true because |a| is non-negative regardless of a's value.- D is not always true because if a is negative, |a| ≠ a.Therefore, the correct answer must be option C.Final Answerboxed{C}</think>Given that (a) is a rational number, we need to determine the correct conclusion among the options provided.- Option A: (-a) is a negative rational number. This is not necessarily true because if (a) is negative, (-a) becomes positive. Therefore, (-a) is not always negative. Hence, A is incorrect.- Option B: (|a|) is a positive number. The absolute value (|a|) is non-negative, which includes zero. However, zero is not a positive number. Therefore, B is incorrect.- Option C: (|a|) is a non-negative number. The absolute value of any rational number is always non-negative (either zero or positive). This is always true regardless of the value of (a). Hence, C is correct.- Option D: (|a| = a). This is true only when (a) is non-negative. If (a) is negative, (|a| = -a neq a). Therefore, D is incorrect.Thus, the correct answer is (boxed{C}).]

🔑:To analyze each option step-by-step:A: -a is a negative rational number- Consider a=-1, which is a rational number.- Then, -a = -(-1) = 1, which is a positive rational number.- Therefore, statement A is incorrect because -a can be positive.B: |a| is a positive number- Consider a=0, which is a rational number.- Then, |a| = |0| = 0, which is not positive.- Therefore, statement B is incorrect because |a| can be 0.C: |a| is a non-negative number- If a=0, then |a| = |0| = 0, which is non-negative.- If a is a positive number, then |a| = a, which is non-negative.- If a is a negative number, then |a| = -a, which is non-negative because -a is positive in this case.- Therefore, statement C is correct as |a| is always non-negative regardless of a being positive, negative, or zero.D: |a|=a- If a is a negative number, then |a| = -a neq a.- Therefore, statement D is incorrect because |a| equals a only if a is non-negative.Based on the analysis, the correct choice is boxed{C}.

🧠:[<think>Okay, so I need to find the second derivative of the function y(x) = x² ln(1 + sin x). Hmm, let me start by recalling how to take derivatives of logarithmic functions and products. Since this function is a product of x squared and the natural log of (1 + sin x), I think I'll have to use the product rule for differentiation. First, let me note that the first derivative y'(x) will involve the product rule. The product rule states that if you have a function h(x) = f(x)g(x), then h'(x) = f'(x)g(x) + f(x)g'(x). Here, f(x) is x², and g(x) is ln(1 + sin x). So, let's compute f'(x) first. The derivative of x² is straightforward: f'(x) = 2x. Now, g(x) = ln(1 + sin x). To find g'(x), I need the chain rule. The derivative of ln(u) is (1/u) * u', where u = 1 + sin x. The derivative of u = 1 + sin x is cos x. Therefore, g'(x) = [1/(1 + sin x)] * cos x = cos x / (1 + sin x). Putting this into the product rule formula, y'(x) = f'(x)g(x) + f(x)g'(x) = 2x * ln(1 + sin x) + x² * [cos x / (1 + sin x)]. Okay, so that's the first derivative. Now, I need to find the second derivative y''(x), which is the derivative of y'(x). This means I have to differentiate the sum of two terms: 2x ln(1 + sin x) and x² [cos x / (1 + sin x)]. Let me break this down into two parts. Let's call the first term A = 2x ln(1 + sin x) and the second term B = x² [cos x / (1 + sin x)]. Then, y''(x) will be A' + B'. Starting with term A: A = 2x ln(1 + sin x). To differentiate this, I again need the product rule. Let f1(x) = 2x and g1(x) = ln(1 + sin x). Then, A' = f1'(x)g1(x) + f1(x)g1'(x). The derivative of f1(x) = 2x is f1'(x) = 2. The derivative of g1(x) = ln(1 + sin x) is the same as before, which is cos x / (1 + sin x). So, g1'(x) = cos x / (1 + sin x). Therefore, A' = 2 * ln(1 + sin x) + 2x * [cos x / (1 + sin x)]. Now, moving on to term B: B = x² [cos x / (1 + sin x)]. This is another product of two functions: f2(x) = x² and g2(x) = cos x / (1 + sin x). Again, apply the product rule: B' = f2'(x)g2(x) + f2(x)g2'(x). First, f2'(x) = 2x. Now, g2(x) is cos x over (1 + sin x). To differentiate this, I might need the quotient rule. The quotient rule states that if we have a function h(x) = numerator / denominator, then h'(x) = [num' * den - num * den'] / den². Let me set numerator = cos x and denominator = 1 + sin x. Then, numerator’ = -sin x, and denominator’ = cos x. Therefore, g2'(x) = [ -sin x (1 + sin x) - cos x * cos x ] / (1 + sin x)^2. Let me compute the numerator:- sin x (1 + sin x) - cos² x = -sin x - sin² x - cos² x. I notice that - sin² x - cos² x is equal to - (sin² x + cos² x) = -1. Therefore, the numerator simplifies to -sin x - 1. So, g2'(x) = (-sin x - 1) / (1 + sin x)^2. Alternatively, factoring out a negative sign in the numerator: -(sin x + 1) / (1 + sin x)^2 = -1 / (1 + sin x). Wait, hold on, let's check that again. The numerator was -sin x - sin² x - cos² x. But sin² x + cos² x = 1, so substituting:- sin x - (sin² x + cos² x) = - sin x - 1. Therefore, the numerator is - (sin x + 1). Therefore, g2'(x) = - (sin x + 1) / (1 + sin x)^2. Notice that sin x + 1 is the same as 1 + sin x, so we can write this as:g2'(x) = - (1 + sin x) / (1 + sin x)^2 = -1 / (1 + sin x). Ah, that simplifies nicely! So, g2'(x) = -1 / (1 + sin x). Therefore, going back to B':B' = f2'(x)g2(x) + f2(x)g2'(x) = 2x * [cos x / (1 + sin x)] + x² * [-1 / (1 + sin x)]. So, simplifying:B' = [2x cos x / (1 + sin x)] - [x² / (1 + sin x)]. Factor out 1/(1 + sin x):B' = [2x cos x - x²] / (1 + sin x). Now, putting it all together, y''(x) = A' + B' = [2 ln(1 + sin x) + 2x cos x / (1 + sin x)] + [ (2x cos x - x²) / (1 + sin x) ]. Let me combine the terms with the same denominator. The terms with denominator (1 + sin x) are:2x cos x / (1 + sin x) + (2x cos x - x²) / (1 + sin x) = [2x cos x + 2x cos x - x²] / (1 + sin x) = [4x cos x - x²] / (1 + sin x). Therefore, y''(x) = 2 ln(1 + sin x) + [4x cos x - x²] / (1 + sin x). Wait, let me check that step again. So A' has two terms: 2 ln(1 + sin x) and 2x cos x / (1 + sin x). Then B' is [2x cos x - x²] / (1 + sin x). So when we add A' and B', the 2x cos x / (1 + sin x) from A' and the [2x cos x - x²] / (1 + sin x) from B' can be combined. So, adding those fractions:[2x cos x + 2x cos x - x²] / (1 + sin x) = [4x cos x - x²] / (1 + sin x). Yes, that seems right. Then, the remaining term from A' is 2 ln(1 + sin x). Therefore, the second derivative is:y''(x) = 2 ln(1 + sin x) + (4x cos x - x²) / (1 + sin x). Hmm, I need to check if this can be simplified further. Let's see. Looking at the numerator of the fractional term: 4x cos x - x². Perhaps factor out an x? That would give x(4 cos x - x). Not sure if that helps. Alternatively, maybe check if there's a common factor with the denominator or some trigonometric identity. The denominator is 1 + sin x. Recall that 1 + sin x can be written as 2 sin²(x/2 + π/4) or something like that, but I don't know if that would be helpful here. Alternatively, multiplying numerator and denominator by 1 - sin x to rationalize, but that might complicate things. Alternatively, perhaps the expression is already simplified enough. Let me see. Alternatively, maybe we can write the entire expression over a common denominator? But since there is a logarithmic term, that might not be feasible. So perhaps the answer is best left as:y''(x) = 2 ln(1 + sin x) + (4x cos x - x²)/(1 + sin x). But let me verify the computations once more to be sure. Starting with the first derivative: y'(x) = 2x ln(1 + sin x) + x² [cos x / (1 + sin x)]. Correct. Then, differentiating term A: 2x ln(1 + sin x). Its derivative is 2 ln(1 + sin x) + 2x [cos x / (1 + sin x)]. Correct. Differentiating term B: x² [cos x / (1 + sin x)]. The derivative is 2x [cos x / (1 + sin x)] + x² [ -1 / (1 + sin x) ]. Correct. Then combining the terms: The 2x [cos x / (1 + sin x)] from A' and 2x [cos x / (1 + sin x)] from B' sum to 4x cos x / (1 + sin x). Then, subtracting x² / (1 + sin x). So total fractional term is (4x cos x - x²)/(1 + sin x). Yes, that seems correct. Therefore, the second derivative is indeed 2 ln(1 + sin x) + (4x cos x - x²)/(1 + sin x). I think that's the final answer. Let me just check if there's any possibility of further simplification. Alternatively, note that (4x cos x - x²) can be written as x(4 cos x - x), but I don't see a way to combine that with the logarithmic term. Therefore, I think the expression is simplified as much as possible. So, the second derivative of y(x) = x² ln(1 + sin x) is y''(x) = 2 ln(1 + sin x) + (4x cos x - x²)/(1 + sin x). I believe that's the answer.Final AnswerThe second derivative is boxed{2 ln(1 + sin x) + frac{4x cos x - x^2}{1 + sin x}}.</think>To find the second derivative of the function ( y(x) = x^2 ln(1 + sin x) ), we start by computing the first derivative using the product rule.The first derivative ( y'(x) ) is calculated as follows:- Let ( f(x) = x^2 ) and ( g(x) = ln(1 + sin x) ).- Then, ( f'(x) = 2x ) and ( g'(x) = frac{cos x}{1 + sin x} ).Applying the product rule:[y'(x) = f'(x)g(x) + f(x)g'(x) = 2x ln(1 + sin x) + x^2 left( frac{cos x}{1 + sin x} right)]Next, we find the second derivative ( y''(x) ) by differentiating ( y'(x) ). We break ( y'(x) ) into two terms:- Term ( A = 2x ln(1 + sin x) )- Term ( B = x^2 left( frac{cos x}{1 + sin x} right) )Differentiating term ( A ):- Let ( f_1(x) = 2x ) and ( g_1(x) = ln(1 + sin x) ).- Then, ( f_1'(x) = 2 ) and ( g_1'(x) = frac{cos x}{1 + sin x} ).Applying the product rule:[A' = 2 ln(1 + sin x) + 2x left( frac{cos x}{1 + sin x} right)]Differentiating term ( B ):- Let ( f_2(x) = x^2 ) and ( g_2(x) = frac{cos x}{1 + sin x} ).- Then, ( f_2'(x) = 2x ) and using the quotient rule, ( g_2'(x) = -frac{1}{1 + sin x} ).Applying the product rule:[B' = 2x left( frac{cos x}{1 + sin x} right) + x^2 left( -frac{1}{1 + sin x} right)]Combining the results for ( A' ) and ( B' ):[y''(x) = 2 ln(1 + sin x) + frac{2x cos x}{1 + sin x} + frac{2x cos x - x^2}{1 + sin x}]Simplifying the fractional terms:[frac{2x cos x + 2x cos x - x^2}{1 + sin x} = frac{4x cos x - x^2}{1 + sin x}]Thus, the second derivative is:[boxed{2 ln(1 + sin x) + frac{4x cos x - x^2}{1 + sin x}}],<think>Okay, so I need to find the second derivative of the function y(x) = x² ln(1 + sin x). Hmm, let's start by recalling how to take derivatives. Since this is a second derivative, I guess I need to differentiate the function twice. The function is a product of x squared and the natural logarithm of (1 + sin x). So, first, I should probably use the product rule for differentiation when taking the first derivative.Alright, the product rule says that if you have two functions multiplied together, say u(x) and v(x), then the derivative is u'(x)v(x) + u(x)v'(x). In this case, u(x) is x² and v(x) is ln(1 + sin x). So, let's find the first derivative y'(x).First, compute u'(x). The derivative of x² is 2x. Got that. Now, v(x) is ln(1 + sin x). To find v'(x), I need the chain rule. The derivative of ln(something) is 1 over that something times the derivative of the something. So, derivative of ln(1 + sin x) is [1/(1 + sin x)] times the derivative of (1 + sin x). The derivative of 1 is 0, and the derivative of sin x is cos x. Therefore, v'(x) = cos x / (1 + sin x).Putting that into the product rule: y'(x) = u'(x)v(x) + u(x)v'(x) = 2x ln(1 + sin x) + x² [cos x / (1 + sin x)]. Okay, so that's the first derivative. Now, need to take the derivative of that to get the second derivative y''(x). Hmm, this might get a bit complicated, but let's take it step by step.So, y'(x) has two terms: the first term is 2x ln(1 + sin x), and the second term is x² times cos x divided by (1 + sin x). So, to find y''(x), we need to differentiate each term separately and then add the results.Starting with the first term: d/dx [2x ln(1 + sin x)]. Again, this is a product of 2x and ln(1 + sin x), so we need the product rule here too. Let's set u1(x) = 2x and v1(x) = ln(1 + sin x). Then, the derivative is u1'(x)v1(x) + u1(x)v1'(x). The derivative of 2x is 2, so u1'(x) = 2. The derivative of v1(x) = ln(1 + sin x) is the same as before, which is cos x / (1 + sin x). So, putting this together, the derivative of the first term is 2 ln(1 + sin x) + 2x [cos x / (1 + sin x)].Now, moving on to the second term of y'(x): d/dx [x² cos x / (1 + sin x)]. This looks more complicated because it's a quotient involving x² and cos x in the numerator and (1 + sin x) in the denominator. Maybe we can use the quotient rule here, but sometimes it's easier to write it as a product and use the product rule along with the chain rule. Let me think.Alternatively, we can split this into two parts: first, x² times [cos x / (1 + sin x)]. Let's denote this as u2(x) = x² and v2(x) = cos x / (1 + sin x). Then, the derivative is u2'(x)v2(x) + u2(x)v2'(x). So, first, compute u2'(x) = 2x. Then, we need to compute v2'(x).So, v2(x) = cos x / (1 + sin x). To differentiate this, we can use the quotient rule. The quotient rule says that if we have a function f(x)/g(x), its derivative is [f'(x)g(x) - f(x)g'(x)] / [g(x)]². Applying that here:f(x) = cos x, so f'(x) = -sin x.g(x) = 1 + sin x, so g'(x) = cos x.Therefore, v2'(x) = [(-sin x)(1 + sin x) - cos x * cos x] / (1 + sin x)^2.Let's compute the numerator:First term: -sin x (1 + sin x) = -sin x - sin² xSecond term: -cos x * cos x = -cos² xSo, numerator is -sin x - sin² x - cos² x.Hmm, can we simplify this? Well, note that sin² x + cos² x = 1, so -sin² x - cos² x = - (sin² x + cos² x) = -1. Therefore, the numerator becomes -sin x - 1.Therefore, v2'(x) = (-sin x - 1) / (1 + sin x)^2.Alternatively, factor out a negative sign: - (sin x + 1) / (1 + sin x)^2 = -1 / (1 + sin x). Because (sin x + 1) is the same as (1 + sin x), so (1 + sin x) cancels one from the denominator. So, v2'(x) simplifies to -1 / (1 + sin x). That's a nice simplification. So, v2'(x) = -1 / (1 + sin x).So, putting this back into the derivative of the second term:u2'(x)v2(x) + u2(x)v2'(x) = 2x * [cos x / (1 + sin x)] + x² * [-1 / (1 + sin x)].Therefore, the derivative of the second term is [2x cos x / (1 + sin x)] - [x² / (1 + sin x)].So, combining both derivatives from the first and second terms of y'(x), the second derivative y''(x) is:[2 ln(1 + sin x) + 2x cos x / (1 + sin x)] + [2x cos x / (1 + sin x) - x² / (1 + sin x)].Now, let's simplify this expression.First, let's list all the terms:1. 2 ln(1 + sin x)2. 2x cos x / (1 + sin x)3. 2x cos x / (1 + sin x)4. -x² / (1 + sin x)So, terms 2 and 3 are the same, so they can be combined. Terms 1 and 4 are unique.Combining terms 2 and 3: 2x cos x / (1 + sin x) + 2x cos x / (1 + sin x) = 4x cos x / (1 + sin x)So, putting it all together:y''(x) = 2 ln(1 + sin x) + 4x cos x / (1 + sin x) - x² / (1 + sin x)We can factor out 1/(1 + sin x) from the last two terms:= 2 ln(1 + sin x) + [4x cos x - x²] / (1 + sin x)Alternatively, we can write this as:y''(x) = 2 ln(1 + sin x) + (4x cos x - x²) / (1 + sin x)But maybe we can simplify this further? Let's check if there's any trigonometric identity that can help here. For instance, 1 + sin x is in the denominator. Recall that 1 + sin x can be written as 2 sin²(x/2 + π/4) or something like that, but I don't think that would necessarily help here. Alternatively, multiplying numerator and denominator by 1 - sin x to rationalize? Let's see:But perhaps there's no need. Alternatively, check if the numerator can be expressed in terms of derivatives of the denominator or something. Let me see:But given that 4x cos x - x² is just a polynomial multiplied by cos x and x squared, I don't see an immediate simplification. So maybe this is the simplest form. However, let's check if we made any mistakes in the calculation.Wait a minute, when I simplified the numerator for v2'(x), let's double-check that step. The original numerator was -sin x - sin² x - cos² x. Then I noted that sin² x + cos² x = 1, so -sin² x - cos² x = -1, leading to -sin x -1. Then, we had (-sin x -1) / (1 + sin x)^2. Then, factoring out a negative sign: -(sin x +1) / (1 + sin x)^2 = -1 / (1 + sin x). That step seems correct because (sin x +1) cancels one (1 + sin x) in the denominator, leaving -1 / (1 + sin x). So that's correct.So, v2'(x) is indeed -1/(1 + sin x). Therefore, the derivative of the second term is [2x cos x / (1 + sin x)] - [x² / (1 + sin x)].Then, when combining with the derivative of the first term, which is 2 ln(1 + sin x) + 2x cos x / (1 + sin x), the total becomes 2 ln(...) plus 2x cos x/(...) plus another 2x cos x/(...) minus x²/(...). So, 2 + 2 = 4x cos x/(...). So that seems correct.Therefore, the final expression for y''(x) is 2 ln(1 + sin x) + (4x cos x - x²)/(1 + sin x). So, unless there's a way to combine those terms further, which I don't see, that should be the answer.Wait, but let me check if the problem expects any particular form or if there's a trigonometric identity that can be applied. For example, 1 + sin x can be written as [sin(x/2) + cos(x/2)]². Let me verify that:[sin(x/2) + cos(x/2)]² = sin²(x/2) + 2 sin(x/2) cos(x/2) + cos²(x/2) = (sin² + cos²)(x/2) + 2 sin(x/2) cos(x/2) = 1 + sin x. Yes, that's correct. So, 1 + sin x = [sin(x/2) + cos(x/2)]². But I don't know if that helps here. Maybe if we have to simplify the denominator, but unless the problem requires it, it's probably not necessary.Alternatively, maybe writing cos x as sqrt(1 - sin²x), but that introduces square roots which complicate things. So, perhaps it's best left as is.Therefore, the second derivative is:y''(x) = 2 ln(1 + sin x) + (4x cos x - x²)/(1 + sin x)I should check if I can factor anything else here. For example, the numerator of the fraction is 4x cos x - x². Maybe factor out an x:x(4 cos x - x)But unless that's helpful, which I don't see immediately, perhaps it's better left as is.Alternatively, maybe combine the logarithmic term with the fraction. But since one term is a logarithm and the other is a rational function, they can't be combined algebraically.Therefore, the answer is as simplified as possible. Let me just recap the steps to make sure I didn't skip anything or make any errors.First derivative:y'(x) = 2x ln(1 + sin x) + x² [cos x / (1 + sin x)]Second derivative:Differentiate the first term: 2 ln(1 + sin x) + 2x [cos x / (1 + sin x)]Differentiate the second term: 2x [cos x / (1 + sin x)] + x² [-1 / (1 + sin x)]Combine all terms:2 ln(1 + sin x) + 2x [cos x / (1 + sin x)] + 2x [cos x / (1 + sin x)] - x² / (1 + sin x)Combine like terms:2 ln(1 + sin x) + 4x [cos x / (1 + sin x)] - x² / (1 + sin x)Which is the same as:2 ln(1 + sin x) + (4x cos x - x²) / (1 + sin x)Yes, that looks correct. I don't see any mistakes in the differentiation steps or simplifications. So, I think that's the final answer.Final AnswerThe second derivative is boxed{2 ln(1 + sin x) + frac{4x cos x - x^{2}}{1 + sin x}}.</think>To find the second derivative of the function ( y(x) = x^2 ln(1 + sin x) ), we start by computing the first derivative using the product rule and chain rule.1. First Derivative: - Let ( u(x) = x^2 ) and ( v(x) = ln(1 + sin x) ). - Compute ( u'(x) = 2x ) and ( v'(x) = frac{cos x}{1 + sin x} ). - Applying the product rule: [ y'(x) = 2x ln(1 + sin x) + x^2 left( frac{cos x}{1 + sin x} right) ]2. Second Derivative: - Differentiate each term of ( y'(x) ). - First term: ( 2x ln(1 + sin x) ) - Let ( u_1(x) = 2x ) and ( v_1(x) = ln(1 + sin x) ). - Compute ( u_1'(x) = 2 ) and ( v_1'(x) = frac{cos x}{1 + sin x} ). - Applying the product rule: [ frac{d}{dx} left[ 2x ln(1 + sin x) right] = 2 ln(1 + sin x) + 2x left( frac{cos x}{1 + sin x} right) ] - Second term: ( x^2 frac{cos x}{1 + sin x} ) - Let ( u_2(x) = x^2 ) and ( v_2(x) = frac{cos x}{1 + sin x} ). - Compute ( u_2'(x) = 2x ) and ( v_2'(x) ) using the quotient rule: [ v_2'(x) = frac{-sin x(1 + sin x) - cos^2 x}{(1 + sin x)^2} = frac{-sin x - 1}{(1 + sin x)^2} = frac{-1}{1 + sin x} ] - Applying the product rule: [ frac{d}{dx} left[ x^2 frac{cos x}{1 + sin x} right] = 2x left( frac{cos x}{1 + sin x} right) + x^2 left( frac{-1}{1 + sin x} right) ]3. Combine the results: - Summing the derivatives of both terms: [ y''(x) = 2 ln(1 + sin x) + 2x left( frac{cos x}{1 + sin x} right) + 2x left( frac{cos x}{1 + sin x} right) - x^2 left( frac{1}{1 + sin x} right) ] - Simplify by combining like terms: [ y''(x) = 2 ln(1 + sin x) + frac{4x cos x - x^2}{1 + sin x} ]The second derivative of the function is:[boxed{2 ln(1 + sin x) + frac{4x cos x - x^{2}}{1 + sin x}}]]

🔑:Solution.begin{gathered}y^{prime}(x)=2 x ln (1+sin x)+x^{2} cdot frac{cos x}{1+sin x} y^{prime prime}(x)=2 ln (1+sin x)+frac{2 x cdot cos x}{1+sin x}+frac{2 x cos x}{1+sin x}+ +x^{2} cdot frac{-sin x(1+sin x)-cos ^{2} x}{(1+sin x)^{2}}=2 ln (1+sin x)+ +frac{4 x cdot cos x}{1+sin x}-x^{2} frac{(1+sin x)}{(1+sin x)^{2}}=2 ln (1+sin x)+frac{4 x cdot cos x-x^{2}}{1+sin x}end{gathered}

🧠:[<think>Okay, so I need to find the equation of the line that is the height from vertex A in triangle ABC. The points given are A(6, 1), B(-5, -4), and C(-2, 5). Hmm, let's think. The height from a vertex in a triangle is the perpendicular line from that vertex to the opposite side. So, the height from A should be perpendicular to the side BC, right? Because in a triangle, each height is perpendicular to the opposite side. Wait, is BC the opposite side of vertex A? Let me confirm. Yes, because triangle ABC has vertices A, B, C. So the sides are AB, BC, and CA. The side opposite vertex A is BC. Therefore, the height from A is the line that passes through A and is perpendicular to BC. Okay, so first, I need to find the equation of line BC. Then, find the slope of BC, take the negative reciprocal to get the slope of the height (since perpendicular lines have slopes that multiply to -1). Then, use point A to write the equation of the height.Let me start by finding the coordinates of B and C. B is (-5, -4) and C is (-2, 5). To find the slope of BC, I can use the formula (y2 - y1)/(x2 - x1). Let me compute that.Slope of BC: (5 - (-4))/(-2 - (-5)) = (5 + 4)/(-2 +5) = 9/3 = 3. So the slope of BC is 3. Therefore, the slope of the height from A, which is perpendicular to BC, should be -1/3. Because the negative reciprocal of 3 is -1/3. So now, the height from A has slope -1/3 and passes through point A(6, 1). To write the equation of this line, I can use the point-slope form: y - y1 = m(x - x1). Plugging in the values: y - 1 = (-1/3)(x - 6). Let me check that.Alternatively, I can use the slope-intercept form. Let's see. If I use point-slope, then expanding that equation: y - 1 = (-1/3)x + 2 (since -1/3 times -6 is +2). Then adding 1 to both sides: y = (-1/3)x + 3. So the equation is y = -1/3 x + 3. Wait, but let me verify if this is correct. Let's double-check the slope. The slope of BC was calculated as 3, so the perpendicular slope is -1/3. That seems right. Then using point A(6,1). Plugging into point-slope: yes, y - 1 = -1/3 (x -6). Distribute the slope: y -1 = (-1/3)x + 2, so y = (-1/3)x +3. That seems correct.But maybe I should check if this line is indeed perpendicular to BC. Let's confirm. If BC has slope 3, and the height has slope -1/3, then 3 * (-1/3) = -1, which confirms they are perpendicular. That's good.Wait, but maybe I should also verify that this line passes through point A. If I plug x=6 into the equation y = -1/3 x +3, then y = -2 +3 = 1, which matches point A(6,1). Correct.Alternatively, another way to approach this is by using vectors. The direction vector of BC is ( -2 - (-5), 5 - (-4) ) = (3, 9). So the direction vector is (3,9), which can be simplified to (1,3). So the slope is 3, as before. The perpendicular direction vector would be (-3,1) or (3,-1), but to get the slope, it would be -1/3. So that's consistent with the previous result.Alternatively, maybe I should use the formula for the equation of a line given a point and a slope. Which is what I did. But just to make sure, perhaps I can approach this problem using another method, like finding the equation of BC first and then writing the equation of the perpendicular line.The equation of BC: We have two points, B(-5, -4) and C(-2,5). The slope is 3, so using point B(-5, -4), the equation is y - (-4) = 3(x - (-5)) => y +4 = 3(x +5) => y = 3x +15 -4 => y = 3x +11. So equation of BC is y = 3x +11.Then, the height from A is perpendicular to BC, so its slope is -1/3, and passes through A(6,1). So equation is y = -1/3 x + b. To find b, substitute (6,1): 1 = -1/3*6 + b => 1 = -2 + b => b = 3. So equation is y = -1/3 x +3. Same result as before. So that's consistent. Therefore, the equation of the height from A is y = -1/3 x +3. If needed, we can write it in standard form. Multiply both sides by 3 to eliminate the fraction: 3y = -x +9 => x + 3y = 9. So the standard form is x + 3y -9 =0. But the problem just says "form the equation of the line", so either form is acceptable unless specified. The question doesn't specify, so perhaps either is okay. But maybe we can check the problem statement again.The problem says "Form the equation of the line on which the height of triangle ABC, drawn from vertex A, lies." So they just want the equation. In the plane, equations can be written in various forms. Since the user probably expects a simplified form, either slope-intercept or standard is fine. But maybe the answer should be in a specific form. Let me see. In Russian education (since the problem mentions the coordinate method on the plane, and the original points have semicolons in coordinates, which is a European format), often linear equations are written in general form Ax + By + C =0. So maybe convert it to that. Starting from y = -1/3 x +3. Multiply both sides by 3: 3y = -x +9. Then bring all terms to the left: x + 3y -9 =0. So that's the general form. Alternatively, x + 3y =9. Both are correct. So maybe present the answer in the form x +3y -9=0 or x +3y=9. But the problem says "form the equation", so either is acceptable. Let me check if there are any traditional preferences here. In many textbooks, the standard form is Ax + By = C. So x +3y =9 would be standard. Alternatively, x +3y -9 =0. Both are correct. Let me check for possible simplifications. The coefficients are integers, and there's no common factor other than 1, so it's simplified. So, yes, x +3y =9 is a valid answer, as is y = -1/3 x +3. But perhaps the problem expects the answer in a specific form. Since the original points are given with coordinates separated by semicolons, maybe the answer is expected in a similar format. But I think either form is acceptable. The problem doesn't specify, so both forms are okay. However, since the user is likely expecting a boxed answer, and in many cases, the standard form is preferred, I think writing it in Ax + By + C =0 form or Ax + By = C. Let me check with the initial thought process. The assistant might present the slope-intercept form, but the problem might expect the general form. But since in the step-by-step, both were derived, perhaps the answer is to be given in one of these forms.Alternatively, maybe parametrize it? But probably not necessary. The height is a straight line, so a linear equation suffices. To make sure, let's verify with another point. Suppose we take a point on the height. Let's say, when x=0, y=3. So (0,3). Is this point on the height? Yes, because plugging into y = -1/3 x +3, when x=0, y=3. So that point is (0,3). Let me check if this point is indeed on the height. The height is from A(6,1) perpendicular to BC. Let's see, the vector from A(6,1) to (0,3) is (-6,2). The direction vector of BC is (3,9) as before. The dot product between (-6,2) and (3,9) should be zero if they are perpendicular. So (-6)(3) + 2(9) = -18 +18=0. Yes, so the vector from A to (0,3) is perpendicular to BC. Therefore, (0,3) is indeed on the height. Therefore, the equation is correct.Therefore, the equation of the height from A is y = -1/3 x +3, or equivalently x +3y =9. Either form is correct. But since the problem says "form the equation of the line", perhaps the general form is more appropriate. So x +3y -9=0. But let me check with an example. If the user expects one answer, maybe the slope-intercept is okay, but in many academic settings, the general form is preferred. Let me recall, in problems like this, the answer is often given in Ax + By + C =0 form. Therefore, converting to that, x +3y -9=0. So I think that's the answer. Alternatively, x +3y =9. Either is acceptable. But since the problem didn't specify, maybe we can present both. But probably, the standard form is better. Therefore, the final answer is x +3y =9, which can be written as x +3y -9=0. So boxed, either boxed{x + 3y = 9} or boxed{x + 3y - 9 = 0}. Both are correct. Let me check once again.Using point A(6,1): Plugging into x +3y =9: 6 +3(1)=6 +3=9. Correct. Plugging into x +3y -9=0: 6 +3*1 -9=0. 6+3-9=0. 0=0. Correct. Therefore, both equations are valid. But in the problem statement, it's common to write the equation in the form Ax + By + C =0. So maybe the latter. But in some contexts, the former is preferred. Since the problem is translated into English and the original might have different conventions, but given that, both are correct. However, looking at similar problems, often the answer is given in the form Ax + By = C. So I think boxed{x + 3y = 9} is appropriate.Final AnswerThe equation of the height from vertex ( A ) is boxed{x + 3y = 9}.</think>To find the equation of the line on which the height of triangle (ABC) drawn from vertex (A) lies, we start by determining the slope of side (BC). The coordinates of points (B(-5, -4)) and (C(-2, 5)) are used to calculate the slope:[text{slope of } BC = frac{5 - (-4)}{-2 - (-5)} = frac{9}{3} = 3]The slope of the height from (A) is the negative reciprocal of 3, which is (-frac{1}{3}). Using point (A(6, 1)) and the slope (-frac{1}{3}), we apply the point-slope form of the equation of a line:[y - 1 = -frac{1}{3}(x - 6)]Simplifying this, we get:[y - 1 = -frac{1}{3}x + 2 implies y = -frac{1}{3}x + 3]Converting this to standard form, we multiply through by 3:[3y = -x + 9 implies x + 3y = 9]Thus, the equation of the height from vertex (A) is (boxed{x + 3y = 9}).,<think>Okay, so I need to find the equation of the line that is the height from vertex A in triangle ABC. The points given are A(6, 1), B(-5, -4), and C(-2, 5). Hmm, let's see. The height from a vertex in a triangle is the perpendicular line from that vertex to the opposite side. So in this case, the height from A should be the line that starts at A and is perpendicular to side BC. Wait, is the opposite side BC? Let me confirm. Yes, since the triangle is ABC, the sides are AB, BC, and AC. So the height from A would be perpendicular to BC. So I need to find the equation of the line that passes through point A and is perpendicular to BC.First, maybe I should find the slope of BC. Then, the slope of the height would be the negative reciprocal of that. Let me do that step by step. So, coordinates of B are (-5, -4) and C are (-2, 5). The slope of BC is (y2 - y1)/(x2 - x1). Let's calculate that. So, (5 - (-4))/(-2 - (-5)) which is (5 + 4)/(-2 + 5) = 9/3 = 3. So the slope of BC is 3. Therefore, the slope of the height from A, which is perpendicular to BC, should be -1/3. Because perpendicular lines have slopes that are negative reciprocals.Now, I have a point that the height passes through, which is A(6, 1), and the slope of the height is -1/3. So I can use the point-slope form to write the equation. Point-slope is y - y1 = m(x - x1). Plugging in the values: y - 1 = (-1/3)(x - 6). Let me simplify this equation. Multiply both sides by 3 to eliminate the fraction: 3(y - 1) = -1(x - 6). Distribute both sides: 3y - 3 = -x + 6. Then, bring all terms to one side: x + 3y - 3 - 6 = 0 → x + 3y - 9 = 0. So the equation of the line is x + 3y = 9. Alternatively, writing it in slope-intercept form, it would be y = (-1/3)x + 3. But since the problem just asks for the equation of the line, either form should be acceptable, but maybe the standard form is preferred here.Wait, but let me double-check. Did I compute the slope correctly? BC: from B(-5, -4) to C(-2, 5). The difference in y is 5 - (-4) = 9. Difference in x is -2 - (-5) = 3. So 9/3 = 3. That's correct. Then the negative reciprocal is -1/3. Correct. Then using point A(6,1). So substituting into point-slope: y - 1 = -1/3 (x - 6). Yes. Expanding that: y = -1/3 x + 2 + 1 → y = -1/3 x + 3. So the equation is y = -1/3 x + 3. If I convert that to standard form: multiply by 3: 3y = -x + 9 → x + 3y = 9. Yes, that's the same as before. So that seems right.Alternatively, I can check if this line is perpendicular to BC. The slope of BC is 3, slope of height is -1/3. Multiplying them: 3 * (-1/3) = -1, which confirms they are perpendicular. So that checks out.Also, check if point A is on the line. Plugging x=6, y=1 into x + 3y: 6 + 3*1 = 9. 9=9, yes. So A is on the line. Good. So that seems correct.Therefore, the equation of the height from vertex A is x + 3y = 9. Alternatively, written as x + 3y - 9 = 0. But the problem says "form the equation of the line," so any correct form should be okay. But perhaps they want it in a standard form like Ax + By + C = 0. So x + 3y - 9 = 0. Alternatively, maybe they accept both. But in the box, maybe write it as x + 3y = 9. Let me check if the problem has any specific form requirements. The original problem says "form the equation of the line," so standard form is likely acceptable.Alternatively, perhaps parametric equations or something else, but in most cases, the standard linear equation is expected. So I think x + 3y = 9 is the answer here. Let me just confirm once again. If I compute the altitude from A, it's perpendicular to BC. So using slope -1/3, passing through (6,1). Yes, that gives the correct line. So I think that's correct.Final AnswerThe equation of the height from vertex ( A ) is boxed{x + 3y - 9 = 0}.</think>To find the equation of the line on which the height of triangle (ABC) drawn from vertex (A) lies, we need to determine the slope of the line segment (BC) and then find the negative reciprocal of that slope, as the height from (A) must be perpendicular to (BC).1. Calculate the slope of (BC): - Coordinates of (B(-5, -4)) and (C(-2, 5)). - Slope of (BC) is (frac{5 - (-4)}{-2 - (-5)} = frac{9}{3} = 3).2. Determine the slope of the height from (A): - The slope of the height, being perpendicular to (BC), is the negative reciprocal of 3, which is (-frac{1}{3}).3. Use the point-slope form to find the equation of the line: - Point (A(6, 1)) and slope (-frac{1}{3}). - Using point-slope form: (y - 1 = -frac{1}{3}(x - 6)). - Simplifying this equation: [ y - 1 = -frac{1}{3}x + 2 ] [ y = -frac{1}{3}x + 3 ] - Converting to standard form: [ 3y = -x + 9 implies x + 3y - 9 = 0 ]4. Verification: - Check that the line passes through (A(6, 1)): (6 + 3(1) = 9), which is true. - Confirm the slopes are perpendicular: (3 times -frac{1}{3} = -1), which confirms perpendicularity.Thus, the equation of the height from vertex (A) is (boxed{x + 3y - 9 = 0}).]

🔑:Apply the condition of perpendicularity of two lines ( k_{1} cdot k_{2}=-1 ).## Answerx+3 y-9=0Let's find the equation of line BC through two points:frac{y-(-4)}{5-(-4)}=frac{x-(-5)}{-2-(-5)}, text { or } y=3 x+11Then its slope k_{1}=3. If k_{2} is the slope of the line perpendicular to the given one, then k_{1} cdot k_{2}=-1. Therefore,k_{2}=-frac{1}{k_{1}}=-frac{1}{3}The equation of the desired line, containing the height A H of triangle A B C, will be found using point A(6 ; 1) and the slope k_{2}=-frac{1}{3} :y-1=-frac{1}{3}(x-6)Write the equation of this line in general form:x+3 y-9=0

🧠:[<think>Okay, so I need to find the number of permutations of the numbers 1 through 5 such that there's no increasing subsequence of length 3. That means, in any permutation, you can't find three numbers where each one is bigger than the previous and they appear in order. Hmm, this seems related to forbidden patterns in permutations. Maybe I need to use the concept of permutation patterns here. Specifically, avoiding the 123 pattern? Wait, no, not exactly. It's avoiding any increasing triplet, regardless of their positions. So even if they're not consecutive, as long as they appear in order, it's bad.Let me think. So for example, the permutation 3,2,1,5,4 would be valid because if you check all possible triplets: 3,2,1; 3,2,5; 3,2,4; 3,1,5; 3,1,4; 3,5,4; 2,1,5; 2,1,4; 2,5,4; 1,5,4. None of these are increasing. But if you have a permutation like 1,3,2,5,4, then the triplet 1,3,5 is increasing, so that's invalid.So how do I count all permutations of 1-5 without any such increasing triplet? I remember something called the Erdős–Szekeres theorem, which says that any permutation of length n will have an increasing subsequence of length at least √n or something? But here n is 5, so √5 is roughly 2.23, so maybe that's not directly helpful. Wait, maybe not. The theorem is about sequences in general, not permutations specifically. Let me recall: Erdős–Szekeres theorem states that any sequence of more than (r-1)(s-1) distinct real numbers contains an increasing subsequence of length r or a decreasing subsequence of length s. So if we set r=3, s=3, then any sequence longer than 4 must have either an increasing subsequence of length 3 or a decreasing subsequence of length 3. But here we have permutations of length 5, so according to the theorem, every permutation of length 5 must have either an increasing subsequence of length 3 or a decreasing subsequence of length 3. Wait, but the problem is only forbidding increasing subsequences of length 3. So does that mean that such permutations must have a decreasing subsequence of length 3? But the question is just asking about permutations without increasing subsequences of length 3, so regardless of decreasing ones.But Erdős–Szekeres might not directly help here, but maybe the number can be found using inclusion-exclusion or recursion. Let me think. Alternatively, these permutations are called 123-avoiding permutations. Wait, but 123-avoiding usually means that there is no subsequence that is order-isomorphic to 123. So yes, exactly what we need here. So maybe the number of 123-avoiding permutations of length 5 is known. But how to compute it?I remember that the number of 123-avoiding permutations of length n is the same as the Catalan numbers. Wait, is that true? Wait, no, the Catalan numbers count the number of 132-avoiding permutations, or is it different? Wait, actually, for any permutation pattern of length 3, the number of permutations avoiding it is the same, given by the Catalan numbers. So whether it's 123, 132, etc., the number is Catalan(n). But wait, no, that's not correct. Wait, actually, the number of 132-avoiding permutations is Catalan(n), and similarly for other single 3-patterns, they are all counted by Catalan numbers. So if that's the case, then the number of 123-avoiding permutations of length 5 should be the 5th Catalan number. Let me check.Catalan numbers are given by C_n = (1/(n+1)) * binomial(2n, n). For n=5, C_5 = (1/6)*252 = 42. So if that's the case, the answer would be 42. But I need to confirm this. Wait, but I thought that Catalan numbers count Dyck paths, correct? And they also count 132-avoiding permutations. But does that extend to all single 3-patterns?Wait, actually, according to some references, the number of permutations avoiding any given 3-letter pattern is the same, and equal to the Catalan numbers. For example, avoiding 123, 132, 213, etc., all give the same count. So then, if that's true, the number should be 42. But wait, let me test with smaller n. For n=3, the number of 123-avoiding permutations should be Catalan(3)=5. But the total permutations of 3 elements are 6, and only one of them is 123, so the number of 123-avoiding permutations is 5. Which is indeed Catalan(3)=5. For n=4, Catalan(4)=14. The total permutations are 24. Let's count how many 123-avoiding permutations there are.Alternatively, if n=4, the number is 14. Let's see. For example, all permutations that are 123-avoiding. Let me list them. Hmm, this might take time, but perhaps I can recall that the number is indeed 14. So if that's the case, then for n=5, it should be 42. But I need to make sure.Alternatively, maybe the number is different? Wait, maybe not. Let me check another way. If the number is given by the Catalan numbers, then the answer is 42. However, I have a doubt because sometimes the 123-avoiding permutations might not correspond directly to Dyck paths. Wait, but according to the Kitaev's book "Patterns in Permutations and Words," it is stated that all single 3-patterns are counted by Catalan numbers. So perhaps that's the case.Alternatively, maybe the answer is 16? Wait, wait. Let me think. For n=3, as we saw, it's 5. For n=4, 14. For n=5, 42. But if I try to calculate it another way. Maybe using recursion. Let me see.The number of 123-avoiding permutations can be calculated using the following recursion. Let S(n) be the number of 123-avoiding permutations of length n. Then, S(n) = sum_{k=1 to n} S(k-1) * S(n - k). This is similar to the Catalan recursion. Wait, yes, that's exactly the Catalan recursion. So S(n) = C_n, the nth Catalan number. Hence, the answer should be 42.But to be thorough, let me check another way. Let's try to count manually for n=3. All permutations of 1,2,3 except 1,2,3. So 5 permutations, which is Catalan(3)=5. For n=4, the Catalan number is 14, so there are 14 permutations of 1,2,3,4 that avoid 123. Let me try to count them.Alternatively, I can use the formula for 123-avoiding permutations. If they are counted by the Catalan numbers, then for n=5, it's 42. So maybe the answer is 42. But wait, the problem says "if there do not exist 1 ≤ i < j < k ≤ 5 such that a_i < a_j < a_k", so permutations avoiding the 123 pattern. So the answer is the 5th Catalan number, 42. However, I recall that when I studied permutation patterns, the number of 123-avoiding permutations of length n is indeed the Catalan number. So I think that is correct.Alternatively, maybe it's different? Wait, let me check the OEIS. The number of 123-avoiding permutations for n=1 is 1, n=2 is 2, n=3 is 5, n=4 is 14, n=5 is 42. Yes, OEIS A000245: 1, 1, 2, 5, 14, 42, 132, ... Wait, but A000245 is the number of permutations avoiding 123 with n elements. Wait, actually, A000245 is the number of 123-avoiding permutations of n elements. Let me check the OEIS entry. Hmm, A000245 starts as 1, 1, 2, 5, 14, 42, 132, ... So yes, for n=5, it's 42. Hence, the answer is 42. Therefore, the number of such permutations is boxed{42}.Wait, but let me check again with another source. If n=5, the Catalan number is 42. So if the number of 123-avoiding permutations is equal to the Catalan number, then yes, it's 42. Therefore, the answer is 42.Final Answerboxed{42}</think>To solve the problem of finding the number of permutations of the numbers 1 through 5 that do not contain an increasing subsequence of length 3, we need to consider the concept of 123-avoiding permutations. A permutation is 123-avoiding if there does not exist a subsequence (a_i < a_j < a_k) for indices (1 leqslant i < j < k leqslant 5). The number of such permutations is given by the Catalan numbers. The Catalan numbers are a sequence of natural numbers that have many applications in combinatorial mathematics, including counting the number of 123-avoiding permutations. The (n)-th Catalan number is given by the formula:[C_n = frac{1}{n+1} binom{2n}{n}]For (n = 5), the 5th Catalan number is calculated as follows:[C_5 = frac{1}{5+1} binom{10}{5} = frac{1}{6} times 252 = 42]Thus, the number of permutations of the numbers 1 through 5 that do not contain an increasing subsequence of length 3 is (boxed{42}).,<think>Okay, so I need to figure out how many permutations of the numbers 1 through 5 don't have an increasing subsequence of length 3. Let me break this down. First, a permutation is just any rearrangement of the numbers 1 to 5. The problem says that in such a permutation, there shouldn't be three numbers where each one is bigger than the previous, and they appear in order. For example, if the permutation is 3,1,4,2,5, then there's a subsequence 1,2,5 which is increasing, so that permutation would be invalid. We need to count all permutations where such a subsequence doesn't exist.Hmm, this seems related to something called the Erdős–Szekeres theorem. I remember that theorem talks about sequences and monotonic subsequences. Let me recall... Oh right, the theorem states that any sequence of more than (m-1)(n-1) numbers contains either an increasing subsequence of length m or a decreasing subsequence of length n. But here, our case is specific: we have permutations of length 5, and we want to avoid an increasing subsequence of length 3. Wait, so maybe we can use this theorem to get a bound or something?But wait, maybe there's a more direct approach. If we need permutations with no increasing subsequence of length 3, these are called 321-avoiding permutations? Or is it 123-avoiding? Wait, no. 321-avoiding refers to avoiding a decreasing subsequence of length 3. So maybe 123-avoiding permutations are the ones that don't have an increasing subsequence of length 3. Let me check that.Yes, actually, permutations that avoid an increasing subsequence of length 3 are called 123-avoiding permutations. So perhaps there's a known count for these. I think the number of 123-avoiding permutations of length n is given by the Catalan numbers. Wait, is that true?Wait, Catalan numbers count the number of Dyck paths, correct parentheses, etc. Also, the number of 132-avoiding permutations is the Catalan number. Wait, hold on. The number of permutations avoiding a specific 3-letter pattern is given by the Catalan numbers. So whether it's 123-avoiding or 132-avoiding, does it matter? Or are they all counted by Catalan numbers?Wait, no, actually, it's known that for any permutation of length 3, the number of n-permutations avoiding it is the same, and that's the Catalan number. But is that true?Wait, let me check. Suppose we have a permutation avoiding 123. Is the number of such permutations for n=5 the 5th Catalan number? Catalan numbers go like 1, 1, 2, 5, 14, 42... So the 5th Catalan number is 42. But wait, the total number of permutations of 5 elements is 120. 42 is less than 120, so maybe that's possible.Alternatively, maybe it's not the Catalan number. Wait, I might be confused. Let me verify. For example, the number of 132-avoiding permutations of length n is indeed the nth Catalan number. But does the same apply for 123-avoiding permutations? Hmm, maybe not. Because different patterns might have different counts. Wait, actually, no. I think that all permutations avoiding a specific 3-letter pattern are counted by the Catalan numbers. Wait, no, that's not correct.Wait, let me recall. The number of 123-avoiding permutations is the same as the number of 132-avoiding permutations due to the Erdős–Szekeres theorem and some bijections. But is that true? Wait, actually, no. The number of permutations avoiding different 3-patterns can be different. Wait, no, actually, according to some results, for any two patterns of length 3, the number of n-permutations avoiding them is the same. So 123, 132, 213, etc., all have the same count, which is the Catalan number. So maybe it is the Catalan number.But wait, let me test for small n. For n=3, how many 123-avoiding permutations are there? All permutations of 3 elements except the identity permutation. So there are 5 permutations. The 3rd Catalan number is 5. That's correct. For n=3, the number is 5, which is the Catalan number. Similarly, for n=4, the 4th Catalan number is 14. Let's check. The number of 123-avoiding permutations of 4 elements: is it 14? Total permutations are 24. If we subtract those that have a 1234 subsequence... Wait, but avoiding a 123 subsequence is different.Wait, actually, avoiding a 123 subsequence (not necessarily consecutive) in a permutation. For n=3, as mentioned, there are 5 permutations avoiding 123, which are all except 123 itself. For n=4, the number is 14. Let's see: the 4th Catalan number is 14. So if that's the case, then perhaps yes. So for n=5, the 5th Catalan number is 42, so the answer would be 42? But the problem is asking for permutations of 5 elements, so n=5, then the answer is 42. But wait, let me check with another approach.Alternatively, maybe we can use the formula for the number of 123-avoiding permutations. I think there's a recursion here. Let me recall. The number of 123-avoiding permutations of length n satisfies the recursion C_n = C_{n-1} * 2(2n - 1)/(n + 1)}. Wait, no, that's the recursion for Catalan numbers. Alternatively, the recursion for Catalan numbers is C_n = sum_{i=0}^{n-1} C_i C_{n-1-i}. But maybe the number of 123-avoiding permutations follows a similar recursion.Alternatively, another way to think: any 123-avoiding permutation can be constructed by inserting n into a 123-avoiding permutation of n-1 in such a way that it doesn't create an increasing subsequence of length 3. But I might need to think more carefully.Alternatively, perhaps using dynamic programming. Let's think recursively. Suppose we have a permutation of length n that avoids 123. Then, the position of the number n must be such that there are not two increasing elements before it. Wait, because if there were two increasing elements before n, then those two plus n would form a 123 subsequence. Similarly, after n, there can't be two increasing elements, but wait, actually, since n is the largest element, any increasing subsequence after it can only be of length 1. Wait, maybe not. Let me think.Suppose we have a permutation π of 1 to n that avoids 123. If we remove the element n, which is at some position k, then the permutation is split into two parts: left part (length k-1) and right part (length n - k). Since n is the largest element, any increasing subsequence cannot include n and two elements from either the left or the right. Wait, but if there's an increasing subsequence in the left part, that's okay as long as it's not length 2. Wait, no, actually, since the entire permutation must not have an increasing subsequence of length 3.So, if we have n in position k, then the left part (elements 1 to k-1) must not have an increasing subsequence of length 2, and the right part (elements k+1 to n) must also not have an increasing subsequence of length 2. Wait, but if the left part has an increasing subsequence of length 2 and the right part also has one, then together with n, would that make a 3-length subsequence? No, because the left subsequence would be before n, and the right subsequence would be after n. So inserting n in between doesn't connect them.Wait, actually, the problem is that if either the left part or the right part has an increasing subsequence of length 2, then combining with n could create a longer subsequence. Wait, no. If the left part has an increasing subsequence of length 2, then adding n would make a subsequence of length 3 if those two elements are less than n and in increasing order. But since n is the maximum element, the two elements in the left are less than n, so if they are increasing, then combined with n, they form an increasing subsequence of length 3. Therefore, to prevent that, the left part must not have an increasing subsequence of length 2. Similarly, the right part, if it has an increasing subsequence of length 2, then those two elements are after n, but since they are less than n, they can't form a 123 subsequence with n. Wait, because 123 would require three elements increasing, but n is the maximum. So if there's an increasing subsequence in the right part, which is two elements, then those are both less than n, so combined with n, they would form an increasing subsequence of length 3. For example, if the right part has elements a and b where a < b, then n followed by a and b isn't increasing, but if a and b are after n, then n, a, b is decreasing. Wait, no. If the permutation has n in position k, and then in positions k+1 to n, we have elements a and b with a < b, then the subsequence n, a, b would be decreasing, since n is the largest. So that's not an increasing subsequence. However, if the elements a and b are in positions k+1 and k+2 with a < b, then the subsequence a, b is increasing, but since they come after n, they can't form a 123 subsequence with n. Wait, actually, 123 requires three elements increasing in order. So if you have two elements after n that are increasing, that's a subsequence of length 2. If you have two elements before n that are increasing, then combined with n, they form an increasing subsequence of length 3. Therefore, the left part (before n) must not have an increasing subsequence of length 2, but the right part (after n) can have increasing subsequences of length 2, as long as they don't combine with something else.Wait, maybe I need to formalize this. Let's denote that in order to avoid an increasing subsequence of length 3, the permutation must be such that when you place n in position k, the elements to the left of n must not have an increasing subsequence of length 2, and the elements to the right can be any 123-avoiding permutation. Wait, no. Because if the right part has an increasing subsequence of length 3, then that would already be invalid. But the right part is of length n - k, which is less than n. So actually, the recursion would be that the number of 123-avoiding permutations of length n is the sum over k=1 to n of the number of 123-avoiding permutations of length k-1 with no increasing subsequence of length 2 multiplied by the number of 123-avoiding permutations of length n - k.But this seems complicated. Maybe there's a better way. Wait, if I consider that the permutation can be decomposed in such a way that it first decreases and then increases, but avoiding the 123. Wait, maybe such permutations are called layered permutations or something similar.Alternatively, perhaps the number of 123-avoiding permutations is equal to the Catalan numbers, so for n=5, it's 42. But I need to verify this.Wait, let me check n=3. For n=3, the Catalan number is 5. The total permutations are 6. The only permutation that has an increasing subsequence of length 3 is 123 itself. So there are 5 permutations that avoid it. That matches the Catalan number. For n=4, the Catalan number is 14. Let's see: total permutations is 24. How many have an increasing subsequence of length 3? Let's calculate that. The number of permutations of 4 elements with an increasing subsequence of length 3. Hmm, inclusion-exclusion maybe.But this might get complicated. Alternatively, if the number of 123-avoiding permutations is indeed the Catalan number, then for n=4 it would be 14, and for n=5, 42. Since the problem is asking for n=5, the answer would be 42.Alternatively, maybe I can think in terms of generating the permutations. For 123-avoiding permutations, they are in bijection with Dyck paths. A Dyck path is a staircase walk from (0,0) to (n,n) that never goes above the diagonal. Each such path can be associated with a 123-avoiding permutation via some bijection. Since the number of Dyck paths is the Catalan number, this would mean the number of 123-avoiding permutations is the Catalan number. However, I thought the bijection was between Dyck paths and 132-avoiding permutations. Maybe different Catalan objects correspond to different pattern-avoiding permutations.Wait, according to some references I recall, the number of 123-avoiding permutations of n elements is indeed the nth Catalan number. For example, the book "Catalan Numbers" by Richard Stanley mentions that 123-avoiding permutations are counted by Catalan numbers. So if that's the case, then the answer is the 5th Catalan number, which is 42. Therefore, the number of such permutations is 42.But just to be thorough, let me try to count manually for n=3 and see. For n=3, as we saw, it's 5 permutations. The Catalan number C3 is 5, which matches. For n=4, C4=14. Let's check a few permutations. The 123-avoiding permutations of 4 elements would be those that don't have three increasing elements. For example, 4321 is definitely avoiding, as it's decreasing. Similarly, 3421, but wait, does 3421 have an increasing subsequence of length 3? Let's see: 3,4,2,1. The subsequence 3,4 is length 2. There's 3,2,1; no. 4,2,1; no. 3,4,2; no. 3,4,1; no. 3,2,1; 4,2,1. So no, 3421 is okay. Similarly, 3214: 3,2,1,4. The subsequence 1,4 is length 2. So that's okay. So yes, there are 14 such permutations for n=4. Since n=5 would then be 42, according to the Catalan numbers.Alternatively, perhaps I can derive the Catalan recurrence here. The number of 123-avoiding permutations of length n, let's denote it as C_n, should satisfy C_n = sum_{k=1}^n C_{k-1} C_{n - k}. Because if you place the maximum element n in position k, then the elements before position k must form a 123-avoiding permutation of length k - 1, and the elements after must form a 123-avoiding permutation of length n - k. However, wait, no. Because if we just split the permutation into two parts around n, both parts need to be 123-avoiding, but the positioning relative to n can affect whether a 123 subsequence is formed.Wait, actually, the standard Catalan recurrence is C_n = sum_{k=0}^{n-1} C_k C_{n - 1 - k}. Which is similar to the structure here. If we consider that the number of 123-avoiding permutations can be counted by choosing where to place the element n, and then multiplying the number of valid permutations on each side. Since placing n in position k would mean the left part (1 to k-1) and the right part (k+1 to n) must both be 123-avoiding. But actually, since n is the largest element, any increasing subsequence that includes n can only have n at the end. Therefore, to avoid a 123 subsequence, both the left and right parts must not have an increasing subsequence of length 2. Wait, but if the left part has an increasing subsequence of length 2, then combining with n would give an increasing subsequence of length 3. Similarly, if the right part has an increasing subsequence of length 2, but since they are after n, they can't form a 123 with n. Wait, but 123 would need three increasing elements in order. If the right part has two increasing elements, say a and b, then since they are after n, the sequence n, a, b is decreasing, so that's not a problem. However, if the left part has two increasing elements, say c and d, then c, d, n is an increasing subsequence of length 3. Therefore, the left part must be 123-avoiding and also must not have an increasing subsequence of length 2. Wait, but if the left part is 123-avoiding, then it automatically doesn't have an increasing subsequence of length 3, but it could still have an increasing subsequence of length 2. So actually, we need the left part to be a permutation that avoids increasing subsequences of length 2, but that's impossible unless the permutation is strictly decreasing.Wait, this seems confusing. Let me think again. If we place n in position k, then the left part (1 to k-1) must be a permutation of 1 to k-1 that doesn't have an increasing subsequence of length 2. Because if it does, then those two elements plus n would form an increasing subsequence of length 3. But a permutation that doesn't have an increasing subsequence of length 2 is a strictly decreasing permutation. Because the only way to not have two increasing elements is to have all elements decreasing. Similarly, the right part (k+1 to n) must be a permutation of k+1 to n (wait, no, actually, the right part is the remaining numbers, which are 1 to n-1 excluding the ones in the left part). Wait, no. If we have a permutation of 1 to 5, and we place 5 in position k, then the left part is a permutation of some k-1 numbers, and the right part is a permutation of the remaining 5 - k numbers. But these numbers aren't necessarily in order. So the left part must be a permutation of some (k-1) numbers, and the right part a permutation of the remaining (5 - k) numbers, such that neither part contains an increasing subsequence of length 2 (for the left part) or any increasing subsequence of length 3 (for both parts). Wait, this is getting too convoluted.Maybe it's better to accept that the count is given by the Catalan numbers and therefore for n=5, the answer is 42. But since the problem is from a competition, maybe the answer isn't 42. Wait, let me check known values. Wait, OEIS sequence A000108 is the Catalan numbers: 1, 1, 2, 5, 14, 42, 132..., so yes, for n=5, it's 42. However, OEIS also has other sequences related to pattern-avoiding permutations. Let me check for 123-avoiding permutations. According to OEIS A000108, yes, the number of 123-avoiding permutations of n elements is the nth Catalan number. So that's solid.Therefore, the answer should be 42. However, just to be absolutely sure, let me think of another way. For example, the number of 132-avoiding permutations is the Catalan number, and there's a bijection between 132-avoiding and 123-avoiding permutations. But actually, different pattern avoidances can have different counts. Wait, but according to the references, all single 3-pattern avoidances are counted by Catalan numbers. For example, avoiding 123, 132, 213, etc., all result in Catalan numbers. So if that's the case, then 42 is the answer.But let's think of another approach. Suppose we consider that a permutation avoiding 123 is a union of two decreasing sequences. Because if a permutation can be partitioned into two decreasing subsequences, then it cannot contain an increasing subsequence of length 3. Because any three elements would have two in one of the decreasing subsequences, which would be decreasing, so you can't have three increasing. Wait, is that true?Wait, actually, the Erdős–Szekeres theorem says that any sequence of length n contains an increasing or decreasing subsequence of length sqrt(n). But here, we want permutations avoiding increasing subsequences of length 3. According to the theorem, if we avoid increasing subsequences of length 3, then we must have a decreasing subsequence of length at least... Hmm, not exactly. The theorem gives a bound, but perhaps not directly applicable here.Alternatively, the theorem by Erdős and Szekeres says that any permutation of length (r-1)(s-1) + 1 contains an increasing subsequence of length r or a decreasing subsequence of length s. So for our case, if we have a permutation of length 5, and we want to avoid increasing subsequences of length 3, then according to the theorem, (r-1)(s-1) + 1 = 5. If r=3, then (3-1)(s-1) +1 >=5 → 2(s-1) +1 >=5 → 2(s-1) >=4 → s-1 >=2 → s>=3. So this implies that any permutation of length 5 has either an increasing subsequence of length 3 or a decreasing subsequence of length 3. Therefore, the permutations we are counting (those without increasing subsequences of length 3) must have a decreasing subsequence of length 3. But the problem doesn't restrict decreasing subsequences, so they can have them. So this doesn't directly help, but it's a nice observation.Alternatively, perhaps all such permutations can be counted by considering their structure. For example, a permutation that avoids 123 must have a certain structure, maybe being a union of two decreasing sequences. If that's the case, then the number of such permutations is the Catalan number. Because the Catalan numbers count the number of such expressions.Alternatively, let's consider generating all 123-avoiding permutations for n=5. But that would take time, but maybe we can see a pattern.For n=1, there's 1 permutation.For n=2, there are 2 permutations: 21 and 12. But wait, 12 has an increasing subsequence of length 2, but we are avoiding increasing subsequences of length 3. Since n=2, it's allowed, so both permutations are valid. Wait, but 12 is increasing, but length 2, so it's allowed. So for n=2, the count is 2, which is the Catalan number C2=2.For n=3, the 5 permutations are all except 123. So 132, 213, 231, 312, 321. Wait, but let's check each:- 132: has 1,3 which is increasing, but no three.- 213: 2,1,3: subsequence 2,3? But 1,3 is also there. Wait, 213: the increasing subsequences are 2,3 and 1,3. Both length 2. So okay.- 231: 2,3,1: increasing subsequence 2,3, but that's length 2.Wait, actually, 231 has a subsequence 2,3 which is increasing. But we need to check if there's a subsequence of length 3. 231: 2,3,1. The longest increasing subsequence is 2,3 (length 2). So okay.Wait, but 132: 1,3,2. The increasing subsequence is 1,3. Length 2. So okay.Similarly, 312: 3,1,2. Increasing subsequence 1,2. Length 2.213: 2,1,3. Increasing subsequence 1,3. Length 2.321: 3,2,1. No increasing subsequences. So okay.So all 5 permutations except 123. So that works.For n=4, it's 14 permutations, which is the 4th Catalan number. So it's reasonable that for n=5, it's 42. So the answer is 42.Therefore, I think the answer is boxed{42}.Final Answerboxed{42}</think>To determine the number of permutations of the numbers (1, 2, 3, 4, 5) that do not contain an increasing subsequence of length 3, we can use the concept of 123-avoiding permutations. These permutations are counted by the Catalan numbers.The Catalan numbers form a sequence where the (n)-th Catalan number is given by the formula:[C_n = frac{1}{n+1} binom{2n}{n}]For (n = 5), the 5th Catalan number is calculated as follows:[C_5 = frac{1}{5+1} binom{10}{5} = frac{1}{6} times 252 = 42]Thus, the number of permutations of (1, 2, 3, 4, 5) that avoid an increasing subsequence of length 3 is the 5th Catalan number, which is 42.The final answer is (boxed{42}).]

🔑:According to the problem, the permutation a_{1} a_{2} a_{3} a_{4} a_{5} does not contain any 3 numbers in ascending order.For the permutations of 1,2,3,4, there are 24 cases in total:begin{tabular}{llllllllllll}1234 & 1243 & 1324 & 1342 & 1423 & 1432 & 2134 & 2143 & 2344 & 2341 & 2413 & 2431 3124 & 3142 & 3214 & 3241 & 3412 & 3421 & 4123 & 4132 & 4213 & 4231 & 4312 & 4321end{tabular}There are 14 permutations that meet the requirement: 1432,2143,2413,2431,3142,3214,3241,3412,3421,4132,4213,4231,4312,4321.When the number 5 is added to the above permutations, such as 51432, 15432, 14532, 14352, 14325 (the 3 crossed-out permutations contain the ascending sequence 145, which does not meet the requirement).Considering all 14 permutations when adding the number 5, the number of permutations that meet the requirement is 42.

🧠:[<think>Okay, so I need to find the derivative of the function f at x=0 using the definition of the derivative. The function f(x) is defined piecewise: it's (ln(cos x))/x when x isn't zero, and 0 when x is zero. Alright, let me recall that the derivative at a point, especially at x=0 here, is the limit as h approaches 0 of [f(0+h) - f(0)] / h. So in this case, that would be the limit as x approaches 0 of [f(x) - f(0)] / (x - 0), which simplifies to the limit as x approaches 0 of [ (ln(cos x)/x - 0) / x ]? Wait, hold on, maybe I messed up here. Let me double-check.The definition of the derivative at a point a is f'(a) = lim_{h->0} [f(a+h) - f(a)] / h. Here, a is 0, so it's lim_{h->0} [f(h) - f(0)] / h. Since f(h) when h ≠ 0 is (ln(cos h))/h, and f(0) is 0. Therefore, the derivative f'(0) is the limit as h approaches 0 of [ (ln(cos h)/h - 0) / h ] = lim_{h->0} ln(cos h) / h². Wait, that seems right. So f'(0) is the limit of ln(cos h) divided by h squared as h approaches 0.Now, I need to compute this limit. Hmm, ln(cos h) over h squared. Let me think. Maybe I can use Taylor series expansions here, since as h approaches 0, cos h can be expanded into its Taylor series, and then take the logarithm of that. Let's try that approach.First, recall that the Taylor series expansion of cos h around 0 is 1 - h²/2 + h^4/24 - ... So, cos h ≈ 1 - h²/2 + higher order terms. Then, ln(cos h) can be approximated by ln(1 - h²/2 + ...). Now, the Taylor series for ln(1 + x) around x=0 is x - x²/2 + x³/3 - ..., but here we have ln(1 + (-h²/2 + ...)). So substituting x with (-h²/2 + ...), we get:ln(cos h) ≈ (-h²/2 + h^4/24 - ...) - [(-h²/2)^2]/2 + ... Let's compute up to the h² term first. The first term is -h²/2. The next term in the expansion of ln(1 + x) is -x²/2, so here x is -h²/2 + ..., so x² would be (h^4)/4, and multiplying by -1/2 gives -h^4/8. But since we're looking at the limit as h approaches 0, maybe we need to consider terms up to h^4 to see if there's a contribution when divided by h². Wait, the original limit is ln(cos h)/h². If we expand ln(cos h) as -h²/2 - h^4/12 + ..., then dividing by h² gives -1/2 - h²/12 + ... So as h approaches 0, the limit would be -1/2. Hmm, but wait, let me verify this more carefully.Let me write down step by step:1. Expand cos h around 0:cos h = 1 - h²/2 + h^4/24 - h^6/720 + ... 2. Take the natural logarithm:ln(cos h) = ln(1 - h²/2 + h^4/24 - ... )3. Let’s denote u = -h²/2 + h^4/24 - ..., so ln(1 + u) ≈ u - u²/2 + u³/3 - u^4/4 + ... Since u is small as h approaches 0, we can expand it:ln(1 + u) ≈ u - u²/2 + higher order terms.Substituting u:ln(cos h) ≈ [ -h²/2 + h^4/24 ] - [ (-h²/2)^2 ] / 2 + ...Calculating each term:First term: -h²/2 + h^4/24Second term: - [ (h^4)/4 ] / 2 = - h^4/8So combining these:ln(cos h) ≈ (-h²/2 + h^4/24) - h^4/8 + ...Simplify the h^4 terms:h^4/24 - h^4/8 = h^4 (1/24 - 3/24) = h^4 (-2/24) = -h^4/12Therefore, up to h^4 terms:ln(cos h) ≈ -h²/2 - h^4/12 + ... Therefore, ln(cos h)/h² ≈ (-h²/2 - h^4/12)/h² = -1/2 - h²/12 + ... So as h approaches 0, the term -h²/12 goes to 0, so the limit is -1/2. Therefore, f’(0) is -1/2.Wait, but let me check if I considered all the terms. Alternatively, maybe I can use L’Hospital’s Rule since the limit is of the form 0/0. Let me see.Original limit: lim_{h->0} ln(cos h)/h².Let’s set L = lim_{h->0} ln(cos h)/h². As h approaches 0, numerator ln(cos h) approaches ln(1) = 0, denominator h² approaches 0. So it's 0/0, so L’Hospital’s Rule can be applied. Taking derivative of numerator and denominator:First derivative:Numerator derivative: d/dh [ln(cos h)] = (1/cos h)(-sin h) = -tan hDenominator derivative: d/dh [h²] = 2hSo L becomes lim_{h->0} (-tan h)/(2h). Let's compute this.Again, this is (-tan h)/(2h). As h approaches 0, tan h ~ h + h^3/3 + ..., so tan h ≈ h. Therefore, (-tan h)/(2h) ≈ -h/(2h) = -1/2. So the limit is -1/2, same as before. So that confirms the answer.Alternatively, if we apply L’Hospital again, but let's check:Current limit: lim_{h->0} (-tan h)/(2h). This is again 0/0, because tan h approaches 0 and 2h approaches 0. So apply L’Hospital again.Derivative of numerator: d/dh (-tan h) = -sec² hDerivative of denominator: d/dh (2h) = 2So the limit becomes lim_{h->0} (-sec² h)/2. As h approaches 0, sec² h approaches 1, so the limit is -1/2. Same result.Therefore, regardless of method, the limit is -1/2. Therefore, f’(0) is -1/2.But just to be thorough, let me check using another approach. Maybe expanding ln(cos h) using series expansion step by step.We know that cos h = 1 - h²/2 + h^4/24 - h^6/720 + ...Then, ln(cos h) = ln(1 - h²/2 + h^4/24 - ...)We can use the expansion ln(1 + x) = x - x²/2 + x³/3 - x⁴/4 + ..., but here x is -h²/2 + h^4/24 - ..., so we need to substitute that into the expansion.Let’s denote u = -h²/2 + h^4/24 - ...Then, ln(1 + u) = u - u²/2 + u³/3 - u⁴/4 + ...So compute up to h^4 terms.First, compute u:u = -h²/2 + h^4/24Then, u² = (-h²/2 + h^4/24)^2 = h^4/4 - (h^6)/24 + h^8/576, but since we're working up to h^4 terms, u² ≈ h^4/4Similarly, u³ = (-h²/2)^3 + ... which is -h^6/8, which is higher order, so negligible for our purposes. Similarly, higher powers will be higher order.Therefore, ln(cos h) ≈ u - u²/2Substituting:u = -h²/2 + h^4/24u²/2 ≈ (h^4/4)/2 = h^4/8Therefore,ln(cos h) ≈ (-h²/2 + h^4/24) - h^4/8 = -h²/2 + h^4/24 - h^4/8Combine the h^4 terms:h^4/24 - h^4/8 = h^4 (1/24 - 3/24) = -2h^4/24 = -h^4/12Thus,ln(cos h) ≈ -h²/2 - h^4/12Therefore, ln(cos h)/h² ≈ (-h²/2 - h^4/12)/h² = -1/2 - h²/12As h approaches 0, the term -h²/12 goes to 0, so the limit is -1/2. Therefore, confirming once again the same result.Alternatively, maybe using substitution. Let me try substituting t = h, so as t approaches 0, we can write:Limit as t approaches 0 of ln(cos t)/t². Let me set t approaching 0 from the right and left, but since cos t is even function, ln(cos t) is even, and t² is even, so the limit from both sides would be the same. So no problem there.Alternatively, using known limits: we know that (cos t - 1)/t² approaches -1/2 as t approaches 0. Maybe relate ln(cos t) to cos t - 1.Note that ln(1 + (cos t - 1)) = ln(cos t). So, as t approaches 0, cos t - 1 ≈ -t²/2. Then, ln(1 + (cos t - 1)) ≈ (cos t - 1) - (cos t -1)^2 / 2 + ... So, again, expanding:ln(cos t) ≈ (cos t - 1) - (cos t - 1)^2 / 2We already know that cos t - 1 ≈ -t²/2, so substitute that:≈ (-t²/2) - (-t²/2)^2 / 2 = -t²/2 - t^4/8Then, divide by t²:ln(cos t)/t² ≈ (-t²/2 - t^4/8)/t² = -1/2 - t²/8Which also tends to -1/2 as t approaches 0. So same conclusion.So, all methods lead to the conclusion that f’(0) = -1/2.Therefore, the answer should be boxed{-dfrac{1}{2}}.Final Answerboxed{-dfrac{1}{2}}</think>To find ( f'(0) ) using the definition of the derivative, we start with the given piecewise function:[f(x) = begin{cases} frac{ln(cos x)}{x} & text{if } x neq 0, 0 & text{if } x = 0.end{cases}]The derivative at ( x = 0 ) is defined as:[f'(0) = lim_{h to 0} frac{f(h) - f(0)}{h}.]Substituting ( f(h) ) and ( f(0) ):[f'(0) = lim_{h to 0} frac{frac{ln(cos h)}{h} - 0}{h} = lim_{h to 0} frac{ln(cos h)}{h^2}.]We need to evaluate the limit ( lim_{h to 0} frac{ln(cos h)}{h^2} ).Using the Taylor series expansion for ( cos h ) around 0:[cos h = 1 - frac{h^2}{2} + frac{h^4}{24} - cdots]Taking the natural logarithm of ( cos h ):[ln(cos h) = lnleft(1 - frac{h^2}{2} + frac{h^4}{24} - cdotsright).]Using the Taylor series expansion for ( ln(1 + u) ) where ( u = -frac{h^2}{2} + frac{h^4}{24} - cdots ):[ln(1 + u) approx u - frac{u^2}{2} + cdots]Substituting ( u ):[ln(cos h) approx left(-frac{h^2}{2} + frac{h^4}{24}right) - frac{left(-frac{h^2}{2}right)^2}{2} + cdots]Simplifying the terms:[ln(cos h) approx -frac{h^2}{2} + frac{h^4}{24} - frac{h^4}{8} + cdots = -frac{h^2}{2} - frac{h^4}{12} + cdots]Dividing by ( h^2 ):[frac{ln(cos h)}{h^2} approx -frac{1}{2} - frac{h^2}{12} + cdots]As ( h to 0 ), the higher-order terms vanish, leading to:[lim_{h to 0} frac{ln(cos h)}{h^2} = -frac{1}{2}.]Thus, the derivative ( f'(0) ) is:[boxed{-dfrac{1}{2}}],<think>Okay, so I need to find f'(0) using the definition of the derivative. The function f(x) is defined piecewise: it's (ln(cos x))/x when x isn't zero, and 0 when x is zero. Alright, so to find the derivative at x=0, I should use the limit definition of the derivative, right? Because the derivative at a point is the limit of [f(a+h) - f(a)] / h as h approaches 0. Here, a is 0, so it's the limit as h approaches 0 of [f(h) - f(0)] / h. Since f(0) is 0, the expression simplifies to [f(h)/h]. Let me write that down.So, f'(0) = lim(h→0) [f(h) - f(0)] / h = lim(h→0) [ (ln(cos h)/h - 0) / h ] = lim(h→0) [ln(cos h) / h²]. Wait, is that right? Wait, hold on. Let me check. If f(h) is (ln(cos h))/h, then [f(h) - f(0)] / h would be [ (ln(cos h)/h - 0 ) ] / h, which is ln(cos h)/h². Yeah, that seems correct. So the derivative is the limit as h approaches 0 of ln(cos h) divided by h squared. Hmm. Now, I need to compute that limit.I remember that ln(cos h) can be expanded using Taylor series around h=0. Maybe that would help. Let me recall the Taylor series for cos h and ln(1 + x). The Taylor series for cos h is 1 - h²/2 + h^4/24 - ... So, cos h ≈ 1 - h²/2 when h is near 0. Then, ln(cos h) would be ln(1 - h²/2 + ...). The Taylor series for ln(1 + x) is x - x²/2 + x³/3 - ... for |x| < 1. But here, the argument inside the ln is 1 - h²/2 + ..., so x would be -h²/2 + higher order terms. So, ln(cos h) ≈ (-h²/2) - ( (-h²/2)^2 ) / 2 + ... Let's compute that.First, let me write cos h as 1 - h²/2 + h^4/24 - ... So, ln(cos h) = ln(1 - h²/2 + h^4/24 - ...). Let me denote the inside as 1 + u, where u = -h²/2 + h^4/24 - ... So, ln(1 + u) ≈ u - u²/2 + u^3/3 - u^4/4 + ... Since u is small as h approaches 0, we can expand it. So, substituting u:ln(1 + u) ≈ (-h²/2 + h^4/24) - ( (-h²/2 + h^4/24)^2 ) / 2 + ... Let's compute term by term.First term: -h²/2 + h^4/24.Second term: - [ ( (-h²/2)^2 + 2*(-h²/2)*(h^4/24) + (h^4/24)^2 ) ] / 2. Let's compute that:First, square of (-h²/2 + h^4/24):= ( (-h²/2)^2 ) + 2*(-h²/2)*(h^4/24) + (h^4/24)^2= h^4/4 - (2*h^6)/48 + h^8/576So, the second term is - [ h^4/4 - h^6/24 + h^8/576 ] / 2 = -h^4/8 + h^6/48 - h^8/1152So, combining the first two terms of the expansion:ln(cos h) ≈ (-h²/2 + h^4/24) - h^4/8 + h^6/48 - h^8/1152 + ... Let's simplify the terms up to h^4:- h²/2 + (h^4/24 - h^4/8) + higher order terms.h^4/24 - h^4/8 = h^4 (1/24 - 3/24) = h^4 (-2/24) = -h^4/12Therefore, ln(cos h) ≈ -h²/2 - h^4/12 + ... So, up to the h^4 term.Therefore, ln(cos h) ≈ -h²/2 - h^4/12 + o(h^4).Now, let's substitute this into the limit expression:lim(h→0) [ ln(cos h) / h² ] = lim(h→0) [ (-h²/2 - h^4/12 + ...) / h² ] = lim(h→0) [ -1/2 - h²/12 + ... ] = -1/2.Wait, but hold on. Wait, the original limit for f'(0) was lim(h→0) ln(cos h)/h². So according to this expansion, that limit is -1/2. So does that mean f'(0) is -1/2? Let me check if I did everything correctly.Wait, but let me verify once again. Let's do the expansion step-by-step:First, cos h = 1 - h²/2 + h^4/24 - h^6/720 + ...Then, ln(cos h) = ln(1 - h²/2 + h^4/24 - ... )Let me use the expansion ln(1 + x) ≈ x - x²/2 + x³/3 - x⁴/4 + ... for small x.But here, x = -h²/2 + h^4/24 - ..., so let's compute ln(cos h):ln(1 + (-h²/2 + h^4/24 - ...)) ≈ (-h²/2 + h^4/24) - [(-h²/2 + h^4/24)^2]/2 + [(-h²/2 + h^4/24)^3]/3 - ...First term: -h²/2 + h^4/24.Second term: - [ ( (-h²/2)^2 + 2*(-h²/2)(h^4/24) + (h^4/24)^2 ) ] / 2 = - [ h^4/4 - h^6/24 + h^8/576 ] / 2 = -h^4/8 + h^6/48 - h^8/1152.Third term: [ (-h²/2)^3 + 3*(-h²/2)^2*(h^4/24) + ... ] / 3. Since these terms will be of higher order than h^4, maybe h^6, so for our purposes up to h^4, we can ignore them.Therefore, combining up to h^4 terms:ln(cos h) ≈ (-h²/2 + h^4/24) - h^4/8 = -h²/2 + h^4/24 - h^4/8 = -h²/2 - h^4/12.Therefore, ln(cos h) ≈ -h²/2 - h^4/12.Therefore, ln(cos h)/h² ≈ (-h²/2 - h^4/12)/h² = -1/2 - h²/12.So as h approaches 0, this expression approaches -1/2. Therefore, the limit is -1/2. Hence, f'(0) = -1/2.Wait, but let me check this with another method to be sure. Maybe using L’Hospital’s Rule?The original limit is lim(x→0) ln(cos x)/x². Let's see if we can apply L’Hospital’s Rule here. Because as x approaches 0, ln(cos x) approaches ln(1) = 0, and x² approaches 0. So it's 0/0 indeterminate form, so L’Hospital’s Rule applies.Taking the derivative of the numerator and denominator:Numerator derivative: d/dx [ln(cos x)] = (1/cos x)*(-sin x) = -tan x.Denominator derivative: d/dx [x²] = 2x.Therefore, the limit becomes lim(x→0) (-tan x)/(2x) = lim(x→0) (-sin x)/(2x cos x). Since cos x approaches 1 as x→0, this simplifies to lim(x→0) (-sin x)/(2x) * 1/cos x = lim(x→0) (-sin x)/(2x) * 1. We know lim(x→0) sin x /x = 1, so this becomes (-1/2)*1 = -1/2. Therefore, same result. So this confirms that the limit is indeed -1/2.Therefore, f'(0) = -1/2.Alternatively, another approach: using the definition of derivative.Wait, but isn't that exactly what we did? The derivative at 0 is the limit as h approaches 0 of [f(h) - f(0)] / h. Since f(0) is 0, and f(h) is ln(cos h)/h, so [f(h) - 0]/h is ln(cos h)/h², which is the same as the limit we computed. So both methods confirm that the answer is -1/2.Alternatively, maybe using known limits? For example, we know that (cos x - 1)/x² approaches -1/2 as x→0. Let me recall:cos x ≈ 1 - x²/2 + x^4/24 - ...So, cos x -1 ≈ -x²/2 + x^4/24 - ..., so (cos x -1)/x² ≈ -1/2 + x²/24 - ..., so limit is -1/2. Therefore, (cos x -1)/x² → -1/2. Then, ln(cos x) can be related to this.Note that ln(cos x) = ln(1 + (cos x -1)). Let’s set t = cos x -1 ≈ -x²/2. Then, ln(1 + t) ≈ t - t²/2 + t³/3 - ... So, ln(cos x) ≈ t - t²/2 + ... where t ≈ -x²/2.Therefore, ln(cos x) ≈ (-x²/2) - (-x²/2)^2 / 2 + ... = -x²/2 - x^4/8 + ..., which matches the earlier expansion. Then, ln(cos x)/x² ≈ (-x²/2 - x^4/8)/x² = -1/2 - x²/8 + ..., so as x→0, the limit is -1/2. Hence, same result.Therefore, all methods confirm that the limit is -1/2, so f’(0) = -1/2. Therefore, the answer is boxed{-dfrac{1}{2}}.Final AnswerThe derivative ( f^{prime}(0) ) is boxed{-dfrac{1}{2}}.</think>To find ( f'(0) ) using the definition of the derivative, we start with the function:[f(x) = begin{cases} frac{ln(cos x)}{x} & text{if } x neq 0, 0 & text{if } x = 0.end{cases}]The derivative at ( x = 0 ) is given by the limit:[f'(0) = lim_{h to 0} frac{f(h) - f(0)}{h} = lim_{h to 0} frac{frac{ln(cos h)}{h} - 0}{h} = lim_{h to 0} frac{ln(cos h)}{h^2}.]To evaluate this limit, we use the Taylor series expansion of ( cos h ) around ( h = 0 ):[cos h approx 1 - frac{h^2}{2} + frac{h^4}{24} - cdots]Substituting this into ( ln(cos h) ):[ln(cos h) = lnleft(1 - frac{h^2}{2} + frac{h^4}{24} - cdotsright).]Using the Taylor series expansion for ( ln(1 + u) ) where ( u = -frac{h^2}{2} + frac{h^4}{24} - cdots ):[ln(1 + u) approx u - frac{u^2}{2} + cdots]Substituting ( u ):[ln(cos h) approx left(-frac{h^2}{2} + frac{h^4}{24}right) - frac{1}{2}left(-frac{h^2}{2} + frac{h^4}{24}right)^2 + cdots]Expanding and simplifying up to the ( h^4 ) term:[ln(cos h) approx -frac{h^2}{2} - frac{h^4}{12} + cdots]Dividing by ( h^2 ):[frac{ln(cos h)}{h^2} approx -frac{1}{2} - frac{h^2}{12} + cdots]Taking the limit as ( h to 0 ):[lim_{h to 0} frac{ln(cos h)}{h^2} = -frac{1}{2}.]This result can also be confirmed using L'Hospital's Rule. Applying L'Hospital's Rule twice to the limit ( lim_{h to 0} frac{ln(cos h)}{h^2} ) confirms the result.Thus, the derivative ( f'(0) ) is:[boxed{-dfrac{1}{2}}]]

🔑:## SolutionBy definition, the derivative at the point x=0:f^{prime}(0)=lim _{Delta x rightarrow 0} frac{f(0+Delta x)-f(0)}{Delta x}Based on the definition, we find:begin{aligned}& f^{prime}(0)=lim _{Delta x rightarrow 0} frac{f(0+Delta x)-f(0)}{Delta x}=lim _{Delta x rightarrow 0} frac{frac{ln (cos Delta x)}{Delta x}-0}{Delta x}= & =lim _{Delta x rightarrow 0} frac{ln left(1-2 sin ^{2} frac{Delta x}{2}right)}{Delta x^{2}}=end{aligned}Using the substitution of equivalent infinitesimals:ln left(1-2 sin ^{2} frac{Delta x}{2}right) sim-2 sin ^{2} frac{Delta x}{2}, as Delta x rightarrow 0left(-2 sin ^{2} frac{Delta x}{2} rightarrow 0right)We get:=lim _{Delta x rightarrow 0} frac{-2 sin ^{2} frac{Delta x}{2}}{Delta x^{2}}=Using the substitution of equivalent infinitesimals:We get:=lim _{Delta x rightarrow 0} frac{-2 cdotleft(frac{Delta x}{2}right)^{2}}{Delta x^{2}}=lim _{Delta x rightarrow 0}-frac{1}{2}=-frac{1}{2}Thus, f^{prime}(0)=-frac{1}{2}## Problem Kuznetsov Differentiation 2-17