💡Okay, so I need to find all number triples (x, y, z) such that when any of these numbers is added to the product of the other two, the result is 2. Hmm, let me try to understand the problem first.So, the problem is giving me three equations, right? Because it says "when any of these numbers is added to the product of the other two, the result is 2." That means for each variable, if I add it to the product of the other two, I get 2. So, translating that into equations, I think it would be:1. x + y*z = 22. y + x*z = 23. z + x*y = 2Yes, that makes sense. So, I have a system of three equations with three variables. My goal is to find all possible triples (x, y, z) that satisfy all three equations simultaneously.Alright, so let me write down the equations again:1. x + yz = 22. y + xz = 23. z + xy = 2Now, I need to solve this system. Since all three equations are similar, maybe I can find some symmetry or find a way to reduce the number of variables.Let me see. Maybe I can subtract some equations to eliminate variables. For example, subtract equation 1 from equation 2:(y + xz) - (x + yz) = 2 - 2Simplifying that:y + xz - x - yz = 0Let me factor terms:y - x + xz - yz = 0Hmm, I can factor out (y - x) from the first two terms and z from the last two:(y - x) + z(x - y) = 0Wait, notice that (x - y) is the same as -(y - x), so:(y - x) - z(y - x) = 0Factor out (y - x):(y - x)(1 - z) = 0So, this gives me two possibilities:Either y - x = 0, which means y = x, or 1 - z = 0, which means z = 1.Okay, so I have two cases to consider: Case 1 where z = 1, and Case 2 where y = x.Let me tackle Case 1 first: z = 1.If z = 1, then I can substitute z = 1 into the original equations.Substituting into equation 1: x + y*1 = 2 => x + y = 2Substituting into equation 2: y + x*1 = 2 => y + x = 2Substituting into equation 3: 1 + x*y = 2 => x*y = 1So, from equations 1 and 2, I have x + y = 2, and from equation 3, x*y = 1.So, now I have a system of two equations:x + y = 2x*y = 1This is a system of equations that I can solve. It's similar to solving for the roots of a quadratic equation. Let me think.If x + y = 2, then y = 2 - x. Substitute this into the second equation:x*(2 - x) = 1Expanding:2x - x^2 = 1Rearranging:x^2 - 2x + 1 = 0This is a quadratic equation. Let me factor it:(x - 1)^2 = 0So, x = 1 is the only solution. Therefore, y = 2 - x = 2 - 1 = 1.So, in this case, x = 1, y = 1, z = 1.So, one solution is (1, 1, 1).Okay, that's straightforward. Now, let's move on to Case 2: y = x.So, if y = x, then I can substitute y with x in the original equations.Substituting into equation 1: x + x*z = 2 => x(1 + z) = 2Substituting into equation 2: x + x*z = 2 => same as equation 1, so nothing new here.Substituting into equation 3: z + x*x = 2 => z + x^2 = 2So, from equation 1, x(1 + z) = 2, which can be rewritten as z = (2/x) - 1, provided that x ≠ 0.From equation 3, z + x^2 = 2, so substituting z from above:(2/x - 1) + x^2 = 2Simplify:2/x - 1 + x^2 = 2Combine constants:2/x + x^2 - 1 = 2Bring all terms to one side:2/x + x^2 - 3 = 0Hmm, this looks a bit complicated. Maybe I can multiply both sides by x to eliminate the denominator. Let's try that.Multiplying each term by x:2 + x^3 - 3x = 0So, we have:x^3 - 3x + 2 = 0Now, I need to solve this cubic equation: x^3 - 3x + 2 = 0Let me try to factor this. Maybe I can find rational roots using the Rational Root Theorem. The possible rational roots are ±1, ±2.Let's test x = 1:1^3 - 3*1 + 2 = 1 - 3 + 2 = 0Yes, x = 1 is a root. So, (x - 1) is a factor.Let me perform polynomial division or factor by grouping.Divide x^3 - 3x + 2 by (x - 1):Using synthetic division:1 | 1 0 -3 2 1 1 -2 1 1 -2 0So, the cubic factors as (x - 1)(x^2 + x - 2)Now, factor the quadratic: x^2 + x - 2Looking for two numbers that multiply to -2 and add to 1. Those numbers are 2 and -1.So, x^2 + x - 2 = (x + 2)(x - 1)Therefore, the cubic factors as:(x - 1)^2 (x + 2) = 0So, the roots are x = 1 (double root) and x = -2.So, x can be 1 or -2.Now, let's find the corresponding z for each x.First, x = 1:From equation 1: x(1 + z) = 2 => 1*(1 + z) = 2 => 1 + z = 2 => z = 1So, when x = 1, z = 1, and since y = x, y = 1. So, we get the solution (1, 1, 1), which we already found in Case 1.Now, x = -2:From equation 1: x(1 + z) = 2 => (-2)*(1 + z) = 2Divide both sides by -2:1 + z = -1So, z = -2Since y = x, y = -2Therefore, the solution is (-2, -2, -2)So, in Case 2, we have two solutions: (1, 1, 1) and (-2, -2, -2)Wait, but (1, 1, 1) was already found in Case 1, so the new solution is (-2, -2, -2)So, combining both cases, the solutions are (1, 1, 1) and (-2, -2, -2)Let me verify these solutions to make sure they satisfy all three original equations.First, (1, 1, 1):1. 1 + 1*1 = 1 + 1 = 2 ✔️2. 1 + 1*1 = 1 + 1 = 2 ✔️3. 1 + 1*1 = 1 + 1 = 2 ✔️Good.Now, (-2, -2, -2):1. (-2) + (-2)*(-2) = -2 + 4 = 2 ✔️2. (-2) + (-2)*(-2) = -2 + 4 = 2 ✔️3. (-2) + (-2)*(-2) = -2 + 4 = 2 ✔️Perfect, both solutions satisfy all three equations.So, are there any other solutions? Let me think.In Case 1, when z = 1, we only got (1, 1, 1). In Case 2, when y = x, we got (1, 1, 1) and (-2, -2, -2). So, these are the only solutions.Wait, but is there a possibility that both z = 1 and y = x could lead to another solution? Let me check.If z = 1 and y = x, then from equation 1: x + x*1 = 2 => 2x = 2 => x = 1, which gives y = 1, z = 1, so that's the same solution as before.So, no new solutions from that overlap.Is there any other case? Let me think.In the beginning, when I subtracted equation 1 from equation 2, I got (y - x)(1 - z) = 0, which gave me two cases: y = x or z = 1.But what if I subtract equation 1 from equation 3? Let me try that.Subtract equation 1 from equation 3:(z + xy) - (x + yz) = 2 - 2Simplify:z + xy - x - yz = 0Factor:z - x + xy - yz = 0Factor terms:(z - x) + y(x - z) = 0Factor out (z - x):(z - x)(1 - y) = 0So, this gives me another two possibilities: z = x or y = 1.Wait, so now I have another case: z = x or y = 1.Hmm, so maybe I need to consider these as well.So, previously, I considered y = x and z = 1. Now, I have another case: z = x or y = 1.So, perhaps I should structure the solution more carefully, considering all possibilities.Let me try to structure it.From subtracting equation 1 from equation 2: (y - x)(1 - z) = 0 => y = x or z = 1.From subtracting equation 1 from equation 3: (z - x)(1 - y) = 0 => z = x or y = 1.So, combining these, the possibilities are:1. y = x and z = x (i.e., x = y = z)2. y = x and y = 13. z = 1 and z = x4. z = 1 and y = 1Wait, that might complicate things. Alternatively, perhaps it's better to consider that from the first subtraction, we have either y = x or z = 1, and from the second subtraction, we have either z = x or y = 1.So, to cover all cases, we can have:Case 1: y = x and z = x (i.e., all variables equal)Case 2: y = x and y = 1Case 3: z = 1 and z = xCase 4: z = 1 and y = 1But perhaps some of these cases overlap or are redundant.Alternatively, maybe it's better to consider that from the first subtraction, either y = x or z = 1, and from the second subtraction, either z = x or y = 1.So, the possibilities are:- If y = x, then from the second subtraction, either z = x or y = 1.Similarly, if z = 1, then from the second subtraction, either z = x or y = 1.So, let's structure it as:Case 1: y = xSubcase 1a: z = xSubcase 1b: y = 1Case 2: z = 1Subcase 2a: z = xSubcase 2b: y = 1But perhaps this is getting too complicated. Maybe a better approach is to consider all possible combinations.Alternatively, perhaps I can consider that if y = x and z = x, then all variables are equal, so x = y = z.Let me check that.If x = y = z, then substituting into equation 1: x + x*x = 2 => x + x^2 = 2 => x^2 + x - 2 = 0Solving this quadratic: x = [-1 ± sqrt(1 + 8)] / 2 = [-1 ± 3]/2So, x = 1 or x = -2Therefore, the solutions are (1, 1, 1) and (-2, -2, -2), which are the same as before.So, that's one set of solutions.Now, what if y = x but z ≠ x? Then, from the first subtraction, y = x, and from the second subtraction, since z ≠ x, we must have y = 1.So, y = x = 1.Then, substituting into equation 1: 1 + 1*z = 2 => z = 1.But that brings us back to z = 1, so x = y = z = 1, which is already covered.Similarly, if z = 1, then from the second subtraction, either z = x or y = 1.If z = 1 and z = x, then x = 1, and substituting into equation 1: 1 + y*1 = 2 => y = 1, so again x = y = z = 1.If z = 1 and y = 1, then substituting into equation 1: x + 1*1 = 2 => x = 1, so again x = y = z = 1.Therefore, all cases lead back to the two solutions we found earlier: (1, 1, 1) and (-2, -2, -2).Wait, but earlier, when we considered y = x, we got x = 1 and x = -2, leading to (1,1,1) and (-2,-2,-2). So, that seems consistent.Therefore, I think we've covered all possible cases, and the only solutions are (1,1,1) and (-2,-2,-2).Let me just double-check if there are any other possibilities.Suppose that none of y = x, z = 1, z = x, or y = 1 hold. Is that possible?From the first subtraction, we have (y - x)(1 - z) = 0, so either y = x or z = 1.Similarly, from the second subtraction, (z - x)(1 - y) = 0, so either z = x or y = 1.Therefore, if neither y = x nor z = 1, then from the first subtraction, we must have z = x, and from the second subtraction, we must have y = 1.But if z = x and y = 1, then substituting into equation 1: x + y*z = x + 1*x = 2 => x + x = 2 => 2x = 2 => x = 1Therefore, x = 1, y = 1, z = x = 1, which is again the solution (1,1,1).So, even in this case, we don't get any new solutions.Therefore, I think we've exhausted all possibilities, and the only solutions are (1,1,1) and (-2,-2,-2).I think that's it. So, the final answer is these two triples.

💡Alright, so I have this problem here about an even function. Let me try to figure it out step by step. Okay, the function f(x) is even, which I remember means that f(-x) = f(x) for all x in its domain. That's the definition of an even function. So, if I plug in a negative x, it should give me the same result as when I plug in a positive x.The problem says that when x is greater than or equal to 0, f(x) is equal to x cubed plus x. So, f(x) = x³ + x for x ≥ 0. Now, they're asking me what f(x) is when x is less than 0. Since the function is even, I can use the property that f(-x) = f(x). Let me think about how to apply this. If x is negative, then -x would be positive, right? So, for example, if x is -2, then -x is 2, which is positive. Since we know f(x) for positive numbers, we can use that to find f(x) for negative numbers.So, if I take f(x) when x is negative, that's the same as f(-x) because f is even. But f(-x) is just the expression for f when the input is positive, which is (-x)³ + (-x). Wait, hold on, is that right? Let me write that out.If x is negative, then f(x) = f(-x) because it's even. But f(-x) is equal to (-x)³ + (-x) because when the input is positive, it's x³ + x. So substituting -x in place of x, we get (-x)³ + (-x). Let me compute that.(-x)³ is equal to -x³ because when you cube a negative, it stays negative. Similarly, (-x) is just -x. So, putting it together, f(-x) = -x³ - x. Therefore, since f(x) = f(-x), f(x) must be equal to -x³ - x when x is negative.Let me double-check that. If I take a specific example, say x = 1, which is positive. Then f(1) = 1³ + 1 = 2. Now, if x = -1, which is negative, then f(-1) should be equal to f(1) because it's even. So, f(-1) should also be 2. Let me plug x = -1 into the expression I got for negative x: -(-1)³ - (-1). That would be -(-1) - (-1) = 1 + 1 = 2. Okay, that checks out.Wait, let me try another example to be sure. Let's take x = 2. Then f(2) = 2³ + 2 = 8 + 2 = 10. Now, f(-2) should also be 10 because it's even. Plugging x = -2 into my expression: -(-2)³ - (-2) = -(-8) - (-2) = 8 + 2 = 10. Perfect, that works too.Hmm, so it seems like my reasoning is correct. When x is negative, f(x) is equal to -x³ - x. Looking back at the options given:A: f(x) = x³ - x B: f(x) = -x³ - x C: f(x) = -x³ + x D: f(x) = x³ + xSo, the expression I derived is -x³ - x, which is option B. That must be the correct answer.Just to make sure I didn't make any mistakes, let me think about the properties of even functions again. An even function is symmetric about the y-axis. So, if I plot f(x) for positive x, it should mirror on the negative side. The given function for positive x is x³ + x. Let me see what that looks like.For positive x, x³ is a cubic term that grows rapidly, and x is a linear term. So, the function is increasing and becomes steeper as x increases. If it's even, then for negative x, it should mirror that behavior. So, when x is negative, the function should also be increasing as x becomes more negative, but since it's mirrored, the expression should have the same magnitude but perhaps different signs.Wait, actually, when you reflect over the y-axis, the x becomes -x, so the function should behave similarly but with the signs adjusted accordingly. Let me see, if I have f(-x) = (-x)³ + (-x) = -x³ - x, which is exactly what I got before. So, that makes sense.Another way to think about it is that for even functions, all the exponents of x in the polynomial should be even. But in this case, the given function for positive x is x³ + x, which has odd exponents. That seems contradictory because even functions are supposed to have only even exponents. Wait, is that always the case?Hmm, actually, no. Wait, that's not necessarily true. Wait, if a function is even, it doesn't have to be a polynomial. But if it is a polynomial, then all the exponents must be even. But in this case, the function is defined piecewise. For x ≥ 0, it's x³ + x, which is an odd function on its own. But since it's extended to negative x by being even, it's not a polynomial overall, but just an even function.So, maybe that's why it's allowed to have odd exponents in the piecewise definition. Because when you make it even, you effectively create a function that isn't a polynomial but still satisfies the evenness condition.Wait, but if I consider the entire function, f(x) is even, so f(-x) = f(x). So, for negative x, f(x) = f(-x) = (-x)³ + (-x) = -x³ - x. So, that's consistent.But hold on, if I think about the graph, for positive x, it's x³ + x, which is an odd function if considered alone. But when you make it even, you reflect it over the y-axis, so the negative side becomes a mirror image, which would be -x³ - x. So, that makes sense.Let me also think about the derivative. If f(x) is even, then its derivative should be odd, right? Because the derivative of an even function is odd. So, let's compute the derivative for positive x and see.For x > 0, f(x) = x³ + x, so f'(x) = 3x² + 1. If the function is even, then f'(x) should be odd. So, f'(-x) = -f'(x). Let's check that.If x is negative, then f(x) = -x³ - x, so f'(x) = -3x² - 1. Now, f'(-x) would be f'(-x) = -3(-x)² - 1 = -3x² - 1. On the other hand, -f'(x) for positive x is -(3x² + 1) = -3x² - 1. So, f'(-x) = -f'(x), which confirms that the derivative is odd, as expected for an even function. That gives me more confidence that my expression for f(x) when x < 0 is correct.Another thing I can check is the value at x = 0. For x = 0, f(0) = 0³ + 0 = 0. Since the function is even, f(-0) should also be 0, which it is. So, that's consistent.Wait, what if I plug in x = -1 into the original expression for positive x? That would be f(-1) = (-1)³ + (-1) = -1 - 1 = -2. But since the function is even, f(-1) should equal f(1) = 2. So, clearly, I can't just use the same expression for negative x. I have to adjust it to make sure that f(-x) = f(x). That's why we have to use f(x) = f(-x) = (-x)³ + (-x) = -x³ - x for x < 0.So, putting it all together, the expression for x < 0 is -x³ - x, which is option B. That seems to satisfy all the conditions: it's even, it matches the given expression for positive x, and it behaves correctly under differentiation.I think I've covered all the bases here. I've checked specific examples, thought about the properties of even functions, considered the derivative, and made sure that the expression makes sense both algebraically and graphically. So, I'm pretty confident that option B is the correct answer.**Final Answer**boxed{B}

💡Okay, so I've got this geometry problem here, and I need to figure out the length of DP. Let me try to understand what's given and what I need to find.First, there are some angle equalities:- Angle APB is equal to angle BAC.- Angle APB is also equal to angle AKC.- Angle AKC is equal to angle BAC.- Angle KAC is equal to angle ABC.Hmm, okay, so a bunch of angles are equal. That probably means some triangles are similar or something like that. Also, it's mentioned that AC is a tangent to the circle. I remember that when a tangent touches a circle, it creates a right angle with the radius at the point of contact. Maybe that will come in handy later.Next, it says that triangle ABC is similar to triangle AKC. Similar triangles have their sides in proportion. So, the ratio of AB to AK is the same as AC to KC, and also the same as BC to AC. They give specific ratios: AB over 4 equals AC over 3 equals 12 over AC. Let me write that down:AB/4 = AC/3 = 12/AC.From this, I can solve for AC first. Let's set AC/3 equal to 12/AC. Cross-multiplying gives AC squared equals 36, so AC must be 6. Then, using AB/4 equals AC/3, which is 6/3=2, so AB is 8. Got it, AC is 6 and AB is 8.Now, moving on to the median CD. They mention using the cosine theorem for triangles ADC and BDC. The cosine theorem is like the law of cosines, right? It relates the sides of a triangle to the cosine of one of its angles.For triangle ADC:AC² = AD² + CD² - 2*AD*CD*cos(angle ADC).For triangle BDC:BC² = BD² + CD² + 2*BD*CD*cos(angle ADC).Since AD equals BD (because CD is a median), we can add these two equations together. That gives AC² + BC² = 2AD² + 2CD². Plugging in the known values, AC is 6 and BC is 12, so 6² + 12² = 2AD² + 2CD². That simplifies to 36 + 144 = 2AD² + 2CD², which is 180 = 2AD² + 2CD². Dividing both sides by 2, we get 90 = AD² + CD².But wait, AD is half of AB because CD is a median, right? AB is 8, so AD is 4. Therefore, AD² is 16. Plugging that back in, 16 + CD² = 90, so CD² is 74, and CD is the square root of 74. Okay, so CD is √74.Now, let's look at DP and DN. DP is a segment from D to P, and DN is from D to N, where N is another intersection point of line CD with the circle. The quadrilateral ANBP is inscribed in a circle, which means it's cyclic. For cyclic quadrilaterals, there's a property that the product of the lengths of the diagonals is equal to the sum of the products of the opposite sides. But here, they mention AD*DB = PD*DN. Since AD is 4 and DB is also 4 (because D is the midpoint), AD*DB is 16. So, PD*DN equals 16.Also, since AC is a tangent to the circle, there's a tangent-secant theorem which says that the square of the tangent is equal to the product of the whole secant and its external part. So, CN*CP equals AC², which is 36. CN is CD minus DN, and CP is CD plus DP. So, (CD - DN)*(CD + DP) = 36.We already know CD is √74, so plugging that in, we get (√74 - y)*(√74 + x) = 36, where x is DP and y is DN. We also know from earlier that x*y = 16.So, now we have two equations:1. x*y = 162. (√74 - y)*(√74 + x) = 36I need to solve these two equations to find x, which is DP. Let me expand the second equation:(√74 - y)*(√74 + x) = (√74)^2 + √74*x - √74*y - x*y = 74 + √74*x - √74*y - x*y.We know that x*y is 16, so substituting that in:74 + √74*x - √74*y - 16 = 36Simplify:74 - 16 + √74*(x - y) = 3658 + √74*(x - y) = 36√74*(x - y) = 36 - 58√74*(x - y) = -22So, x - y = -22 / √74But from the first equation, y = 16 / xSo, substituting y into the equation:x - (16 / x) = -22 / √74Multiply both sides by x to eliminate the denominator:x² - 16 = (-22 / √74)*xBring all terms to one side:x² + (22 / √74)*x - 16 = 0This is a quadratic equation in terms of x. Let me write it as:x² + (22 / √74)x - 16 = 0To solve this quadratic equation, I'll use the quadratic formula:x = [-b ± √(b² - 4ac)] / (2a)Here, a = 1, b = 22 / √74, and c = -16.First, calculate the discriminant:b² - 4ac = (22 / √74)² - 4*1*(-16)= (484 / 74) + 64= (484 / 74) + (64*74)/74= (484 + 4736) / 74= 5220 / 74= 70.5405...Wait, let me double-check that:(22 / √74)^2 = (22^2) / (√74)^2 = 484 / 74 = 6.5405...Then, -4ac = -4*1*(-16) = 64.So, discriminant is 6.5405 + 64 = 70.5405.So, √(70.5405) ≈ 8.4.But let me keep it exact for now.So, x = [ - (22 / √74) ± √(70.5405) ] / 2But let's express √(70.5405) in terms of √74.Wait, 70.5405 is approximately 70.54, which is roughly (8.4)^2.But maybe there's a better way to write it.Alternatively, let's rationalize the denominator for b:b = 22 / √74 = (22√74) / 74 = (11√74) / 37So, b² = (121*74) / (37²) = (121*74) / 1369Calculate 121*74:121*70 = 8470121*4 = 484Total: 8470 + 484 = 8954So, b² = 8954 / 1369 ≈ 6.5405Then, discriminant is 6.5405 + 64 = 70.5405So, √(70.5405) ≈ 8.4But let's see if we can write it more precisely.Wait, 70.5405 is approximately 70.54, which is close to 70.5405.But maybe we can write it as √(70.5405) = √(70 + 0.5405) ≈ 8.4.Alternatively, let's see if 70.5405 is a perfect square.But 8.4^2 = 70.56, which is very close to 70.5405. So, √70.5405 ≈ 8.4.So, x ≈ [ - (22 / √74) ± 8.4 ] / 2But let's compute it more accurately.First, compute 22 / √74:√74 ≈ 8.6023So, 22 / 8.6023 ≈ 2.557So, b ≈ 2.557Then, discriminant ≈ 70.5405, so √ discriminant ≈ 8.4Thus, x ≈ [ -2.557 ± 8.4 ] / 2We have two solutions:x ≈ ( -2.557 + 8.4 ) / 2 ≈ 5.843 / 2 ≈ 2.9215andx ≈ ( -2.557 - 8.4 ) / 2 ≈ -10.957 / 2 ≈ -5.4785Since length can't be negative, we take the positive solution: x ≈ 2.9215But let's express it exactly.Wait, let's go back to the quadratic equation:x² + (22 / √74)x - 16 = 0Let me rationalize the coefficients:Multiply both sides by √74 to eliminate the denominator:√74*x² + 22x - 16√74 = 0Now, it's a bit messy, but let's write it as:√74 x² + 22x - 16√74 = 0Using the quadratic formula:x = [ -22 ± √(22² - 4*√74*(-16√74)) ] / (2*√74)Calculate discriminant:22² = 4844*√74*(-16√74) = 4*(-16)*(√74)^2 = 4*(-16)*74 = -4736But since it's under the square root, it's:√(484 - (-4736)) = √(484 + 4736) = √5220Simplify √5220:5220 = 4*1305 = 4*5*261 = 4*5*3*87 = 4*5*3*3*29 = 2²*3²*5*29So, √5220 = 2*3*√(5*29) = 6√145So, discriminant is 6√145Thus, x = [ -22 ± 6√145 ] / (2√74)Simplify numerator and denominator:Factor out 2 in numerator:x = [ -11 ± 3√145 ] / √74Since length can't be negative, we take the positive solution:x = [ -11 + 3√145 ] / √74To rationalize the denominator, multiply numerator and denominator by √74:x = [ (-11 + 3√145 ) * √74 ] / 74But it's fine as is, so the exact value is:x = ( -11 + 3√145 ) / √74So, DP is ( -11 + 3√145 ) / √74I think that's the answer they're looking for.

💡Okay, so I have this function f(x) = x² - 4x + (2 - a)ln x, and I need to do a few things with it. Let's break it down step by step.First, part 1a: When a = 8, find the interval where f(x) is increasing. Hmm, okay. So, when a is 8, the function becomes f(x) = x² - 4x + (2 - 8)ln x, which simplifies to f(x) = x² - 4x - 6 ln x. To find where the function is increasing, I remember that I need to look at the derivative of f(x). The derivative f'(x) will tell me the slope of the function at any point x. If f'(x) is positive, the function is increasing there.So, let's compute f'(x). The derivative of x² is 2x, the derivative of -4x is -4, and the derivative of -6 ln x is -6*(1/x). So putting it all together, f'(x) = 2x - 4 - 6/x. Hmm, that's a bit messy. Maybe I can combine the terms over a common denominator to make it easier to analyze.Let's rewrite f'(x) as (2x² - 4x - 6)/x. That way, I have a single fraction. Now, to find where f'(x) is positive, I need to solve the inequality (2x² - 4x - 6)/x > 0. Since x is in the domain of f(x), which is x > 0 because of the ln x term, I don't have to worry about x being negative or zero. So, the sign of f'(x) depends on the numerator: 2x² - 4x - 6.Let me factor the numerator if possible. 2x² - 4x - 6. Let's factor out a 2 first: 2(x² - 2x - 3). Now, can I factor x² - 2x - 3? Let's see, looking for two numbers that multiply to -3 and add to -2. Hmm, -3 and +1. So, x² - 2x - 3 = (x - 3)(x + 1). Therefore, the numerator factors to 2(x - 3)(x + 1). So, f'(x) = [2(x - 3)(x + 1)] / x.Now, since x > 0, the denominator is always positive. So, the sign of f'(x) depends on the numerator: 2(x - 3)(x + 1). Since 2 is positive, we can ignore it for the sign analysis. So, the sign is determined by (x - 3)(x + 1). Let's analyze this product.The critical points are x = 3 and x = -1. But since x > 0, x = -1 is not in our domain. So, we only consider x = 3. Let's test intervals around x = 3.For x < 3 (but x > 0), let's pick x = 1: (1 - 3)(1 + 1) = (-2)(2) = -4, which is negative. So, f'(x) is negative here.For x > 3, let's pick x = 4: (4 - 3)(4 + 1) = (1)(5) = 5, which is positive. So, f'(x) is positive here.Therefore, f(x) is increasing when f'(x) > 0, which is when x > 3. So, the interval where f(x) is increasing is (3, ∞).Wait, let me double-check. When x approaches 0 from the right, ln x goes to negative infinity, so f(x) tends to negative infinity. At x = 1, f(1) = 1 - 4 + (2 - 8)ln 1 = -3 + 0 = -3. At x = 3, f(3) = 9 - 12 - 6 ln 3 ≈ -3 - 6*1.0986 ≈ -3 - 6.5916 ≈ -9.5916. So, it's decreasing until x = 3, then increasing after that. That makes sense.Okay, moving on to part 1b: Find the equation of the tangent line to the curve y = f(x) at the point (1, -3). Hmm, so I need to find the tangent line at x = 1. The tangent line equation is given by y = f(a) + f'(a)(x - a), where a is the x-coordinate of the point. Here, a = 1, f(a) = -3, so I need f'(1).From earlier, f'(x) = 2x - 4 - 6/x. So, f'(1) = 2*1 - 4 - 6/1 = 2 - 4 - 6 = -8. So, the slope of the tangent line is -8.Therefore, the equation is y = -3 + (-8)(x - 1). Simplifying, y = -3 -8x + 8, which is y = -8x + 5. Alternatively, in standard form, 8x + y = 5, or 8x + y - 5 = 0.Let me verify: At x = 1, y = -8(1) + 5 = -3, which matches the point. The slope is correct as we calculated f'(1) = -8. So, that seems right.Now, part 2: Find the minimum value of the function f(x) on the interval [e, e²]. Okay, so we need to find the minimum of f(x) = x² - 4x + (2 - a)ln x on [e, e²]. To find the minimum, I should check the critical points inside the interval and also evaluate the function at the endpoints.First, let's find the critical points by setting f'(x) = 0. From earlier, f'(x) = (2x² - 4x + 2 - a)/x. So, setting numerator equal to zero: 2x² - 4x + 2 - a = 0.This is a quadratic equation in x: 2x² - 4x + (2 - a) = 0. Let's compute the discriminant: D = (-4)² - 4*2*(2 - a) = 16 - 8*(2 - a) = 16 - 16 + 8a = 8a.So, discriminant D = 8a. Depending on the value of a, the quadratic can have two real roots, one real root, or no real roots.Case 1: If D < 0, which is when a < 0, then there are no real roots. So, f'(x) does not cross zero, meaning f'(x) is always positive or always negative. Let's check the sign.Looking back at f'(x) = (2x² - 4x + 2 - a)/x. If a < 0, then 2 - a > 2, so the constant term is positive. The quadratic 2x² - 4x + (2 - a) has a positive leading coefficient and a positive constant term. The vertex of the parabola is at x = -b/(2a) = 4/(4) = 1. The value at x = 1 is 2(1)² - 4(1) + (2 - a) = 2 - 4 + 2 - a = (0) - a. Since a < 0, this is positive. So, the quadratic is always positive, meaning f'(x) > 0 for all x > 0. Therefore, f(x) is increasing on [e, e²], so the minimum is at x = e.Case 2: If D = 0, which is when a = 0. Then, the quadratic has one real root. Let's compute it: x = [4 ± sqrt(0)]/(2*2) = 4/4 = 1. So, x = 1 is the critical point. But x = 1 is not in [e, e²] since e ≈ 2.718. So, on [e, e²], f'(x) is positive because the quadratic is positive everywhere except at x = 1. So, again, f(x) is increasing on [e, e²], minimum at x = e.Case 3: If D > 0, which is when a > 0. Then, there are two real roots. Let's compute them: x = [4 ± sqrt(8a)]/(4) = [4 ± 2√(2a)]/4 = [2 ± √(2a)]/2 = 1 ± (√(2a))/2.So, the critical points are x = 1 + (√(2a))/2 and x = 1 - (√(2a))/2. Since x must be positive, and 1 - (√(2a))/2 must be positive. Let's see when that is: 1 - (√(2a))/2 > 0 ⇒ √(2a) < 2 ⇒ 2a < 4 ⇒ a < 2. So, if a < 2, the smaller root is positive; otherwise, it's negative.But our interval is [e, e²], which is approximately [2.718, 7.389]. So, we need to see if the critical points lie within this interval.First, let's consider x = 1 + (√(2a))/2. Let's see when this is less than e or greater than e².Compute 1 + (√(2a))/2:If 1 + (√(2a))/2 < e, then (√(2a))/2 < e - 1 ⇒ √(2a) < 2(e - 1) ⇒ 2a < 4(e - 1)² ⇒ a < 2(e - 1)².Similarly, if 1 + (√(2a))/2 > e², then (√(2a))/2 > e² - 1 ⇒ √(2a) > 2(e² - 1) ⇒ 2a > 4(e² - 1)² ⇒ a > 2(e² - 1)².So, depending on the value of a, the critical point x = 1 + (√(2a))/2 can be inside or outside the interval [e, e²].So, let's break it down:Subcase 3a: If a ≥ 2(e² - 1)², then x = 1 + (√(2a))/2 ≥ e². So, the critical point is to the right of our interval. Therefore, on [e, e²], f'(x) is negative because the quadratic 2x² - 4x + 2 - a is negative between its roots. Wait, but if a is very large, the quadratic might be negative throughout the interval? Let me think.Wait, when a is large, 2x² - 4x + 2 - a is dominated by the -a term, so it's negative. Therefore, f'(x) = (negative)/x, which is negative. So, f(x) is decreasing on [e, e²], so the minimum is at x = e².Subcase 3b: If a ≤ 2(e - 1)², then x = 1 + (√(2a))/2 ≤ e. So, the critical point is to the left of our interval. Then, on [e, e²], f'(x) is positive because the quadratic is positive beyond its larger root. So, f(x) is increasing on [e, e²], so the minimum is at x = e.Subcase 3c: If 2(e - 1)² < a < 2(e² - 1)², then x = 1 + (√(2a))/2 is between e and e². So, the critical point is inside the interval. Therefore, f(x) has a minimum at x = 1 + (√(2a))/2.So, summarizing:- If a ≤ 2(e - 1)², minimum at x = e.- If a ≥ 2(e² - 1)², minimum at x = e².- If 2(e - 1)² < a < 2(e² - 1)², minimum at x = 1 + (√(2a))/2.Now, let's compute the minimum value in each case.Case 1 and 2 (a ≤ 2(e - 1)²): Minimum at x = e.f(e) = e² - 4e + (2 - a)ln e = e² - 4e + (2 - a)*1 = e² - 4e + 2 - a.Case 3a (a ≥ 2(e² - 1)²): Minimum at x = e².f(e²) = (e²)² - 4e² + (2 - a)ln(e²) = e⁴ - 4e² + (2 - a)*2 = e⁴ - 4e² + 4 - 2a.Case 3c (2(e - 1)² < a < 2(e² - 1)²): Minimum at x = 1 + (√(2a))/2.Let me denote x₀ = 1 + (√(2a))/2.Compute f(x₀):f(x₀) = x₀² - 4x₀ + (2 - a)ln x₀.But since x₀ is a critical point, f'(x₀) = 0 ⇒ 2x₀² - 4x₀ + 2 - a = 0 ⇒ a = 2x₀² - 4x₀ + 2.So, we can substitute a in terms of x₀.Thus, f(x₀) = x₀² - 4x₀ + (2 - (2x₀² - 4x₀ + 2))ln x₀.Simplify the expression inside the ln:2 - (2x₀² - 4x₀ + 2) = 2 - 2x₀² + 4x₀ - 2 = -2x₀² + 4x₀.So, f(x₀) = x₀² - 4x₀ + (-2x₀² + 4x₀)ln x₀.Factor out terms:= x₀² - 4x₀ + (-2x₀² + 4x₀)ln x₀.Alternatively, factor out 2x₀:= x₀² - 4x₀ + 2x₀(-x₀ + 2)ln x₀.But maybe it's better to leave it as is.Alternatively, since a = 2x₀² - 4x₀ + 2, we can write f(x₀) in terms of a.But perhaps it's more straightforward to leave it in terms of x₀.So, f(x₀) = x₀² - 4x₀ + (-2x₀² + 4x₀)ln x₀.Alternatively, factor out (-2x₀² + 4x₀):= x₀² - 4x₀ + (-2x₀² + 4x₀)ln x₀.Hmm, not sure if it can be simplified further. Maybe we can express it in terms of a.Since a = 2x₀² - 4x₀ + 2, let's solve for x₀ in terms of a.But x₀ = 1 + (√(2a))/2, so plugging that back in might not lead to a simpler expression.Alternatively, let's express f(x₀) in terms of a.We have:f(x₀) = x₀² - 4x₀ + (2 - a)ln x₀.But a = 2x₀² - 4x₀ + 2, so 2 - a = -2x₀² + 4x₀.Therefore, f(x₀) = x₀² - 4x₀ + (-2x₀² + 4x₀)ln x₀.Which is the same as before.Alternatively, factor out x₀² and x₀:= x₀²(1 - 2 ln x₀) + x₀(-4 + 4 ln x₀).But I don't think that helps much.Alternatively, let's compute it numerically for specific a, but since a is a variable, we need to express it symbolically.So, perhaps the expression is as simplified as it can be.Therefore, the minimum value is:- If a ≤ 2(e - 1)²: e² - 4e + 2 - a.- If a ≥ 2(e² - 1)²: e⁴ - 4e² + 4 - 2a.- If 2(e - 1)² < a < 2(e² - 1)²: x₀² - 4x₀ + (-2x₀² + 4x₀)ln x₀, where x₀ = 1 + (√(2a))/2.Alternatively, expressing in terms of a:Since x₀ = 1 + (√(2a))/2, we can write:f(x₀) = [1 + (√(2a))/2]² - 4[1 + (√(2a))/2] + (2 - a)ln[1 + (√(2a))/2].Expanding [1 + (√(2a))/2]²:= 1 + √(2a) + (2a)/4 = 1 + √(2a) + (a)/2.So, f(x₀) = 1 + √(2a) + (a)/2 - 4 - 2√(2a) + (2 - a)ln[1 + (√(2a))/2].Simplify:= (1 - 4) + (√(2a) - 2√(2a)) + (a)/2 + (2 - a)ln[1 + (√(2a))/2].= -3 - √(2a) + (a)/2 + (2 - a)ln[1 + (√(2a))/2].So, f(x₀) = (a/2) - √(2a) - 3 + (2 - a)ln[1 + (√(2a))/2].That seems as simplified as it can get.So, putting it all together:- When a ≤ 2(e - 1)², minimum is e² - 4e + 2 - a.- When a ≥ 2(e² - 1)², minimum is e⁴ - 4e² + 4 - 2a.- When 2(e - 1)² < a < 2(e² - 1)², minimum is (a/2) - √(2a) - 3 + (2 - a)ln[1 + (√(2a))/2].Let me compute 2(e - 1)² and 2(e² - 1)² numerically to have a sense:e ≈ 2.718, so e - 1 ≈ 1.718. (e - 1)² ≈ 2.952. So, 2*(2.952) ≈ 5.904.Similarly, e² ≈ 7.389, so e² - 1 ≈ 6.389. (e² - 1)² ≈ 40.81. So, 2*(40.81) ≈ 81.62.So, for a between approximately 5.904 and 81.62, the minimum is at the critical point inside the interval.Alright, I think that's the breakdown.

💡Alright, so I have this problem here: If |y - 2x| = x², where -1 < x < 0, then I need to find the range of the real number y. Hmm, okay. Let me try to figure this out step by step.First, I remember that absolute value equations can have two cases: one where the expression inside is positive, and one where it's negative. So, for |y - 2x| = x², that means either y - 2x = x² or y - 2x = -x². That makes sense because the absolute value of something equals a number means that the something is either equal to that number or its negative.So, let's write that down:Case 1: y - 2x = x² Case 2: y - 2x = -x²Now, I can solve each case for y to express y in terms of x.Starting with Case 1: y - 2x = x² Adding 2x to both sides: y = x² + 2xOkay, so in this case, y is equal to x squared plus 2x. Now, let's move on to Case 2.Case 2: y - 2x = -x² Adding 2x to both sides: y = -x² + 2xAlright, so in this case, y is equal to negative x squared plus 2x.Now, I have two expressions for y in terms of x: 1. y = x² + 2x 2. y = -x² + 2xSince we're given that -1 < x < 0, I need to find the range of y for each of these expressions and then combine them to find the overall range of y.Let me tackle each case separately.**Case 1: y = x² + 2x**I can rewrite this as y = x² + 2x. Maybe completing the square would help me find the maximum or minimum values of y in this interval.Completing the square: y = x² + 2x y = (x² + 2x + 1) - 1 y = (x + 1)² - 1So, y = (x + 1)² - 1. This is a parabola that opens upwards with vertex at x = -1, y = -1.Since the vertex is at (-1, -1), and the parabola opens upwards, the minimum value of y is -1, and as x moves away from -1, y increases.But our interval is -1 < x < 0. So, x is just to the right of -1, up to 0. Let's see what y does in this interval.At x = -1: y = (-1 + 1)² - 1 = 0 - 1 = -1 But x is greater than -1, so we don't include x = -1.As x increases from -1 towards 0, (x + 1) increases from 0 to 1, so (x + 1)² increases from 0 to 1, and thus y increases from -1 to 0.Wait, let me check that.At x approaching -1 from the right: x approaches -1, so (x + 1) approaches 0, so y approaches -1.At x = 0: y = (0 + 1)² - 1 = 1 - 1 = 0 But x is less than 0, so we don't include x = 0.So, in this case, y ranges from just above -1 to just below 0.So, for Case 1, y is in (-1, 0).**Case 2: y = -x² + 2x**Again, let me try to analyze this function. Maybe I can complete the square here as well.y = -x² + 2x Factor out the negative sign: y = -(x² - 2x) Now, complete the square inside the parentheses: y = -(x² - 2x + 1 - 1) y = -( (x - 1)² - 1 ) y = -(x - 1)² + 1So, y = -(x - 1)² + 1. This is a parabola that opens downward with vertex at x = 1, y = 1.Since the vertex is at (1, 1), and it opens downward, the maximum value is 1, and as x moves away from 1, y decreases.But our interval is -1 < x < 0. So, x is between -1 and 0, which is to the left of the vertex at x = 1.Since the parabola opens downward, the function is increasing as we move from left to right up to the vertex. So, in the interval -1 < x < 0, which is to the left of the vertex, the function is increasing.Therefore, as x increases from -1 to 0, y increases.Let's compute the values at the endpoints.At x approaching -1 from the right: y = -(-1 - 1)² + 1 = -(-2)² + 1 = -4 + 1 = -3 But x is greater than -1, so y approaches -3 from above.At x = 0: y = -(0 - 1)² + 1 = -1 + 1 = 0 But x is less than 0, so y approaches 0 from below.So, in this case, y ranges from just above -3 to just below 0.Therefore, for Case 2, y is in (-3, 0).**Combining Both Cases**Now, from Case 1, y is in (-1, 0), and from Case 2, y is in (-3, 0). So, combining these two intervals, the overall range of y is the union of (-1, 0) and (-3, 0), which is (-3, 0).Wait, let me make sure. Since (-1, 0) is entirely within (-3, 0), the union is just (-3, 0). That makes sense because in Case 2, y can go as low as -3, and in Case 1, it can only go down to -1. So, the lowest y can be is -3, and the highest is just below 0.Therefore, the range of y is all real numbers between -3 and 0, not including -3 and 0 themselves.**Double-Checking**Let me verify this with some test points.For Case 1: y = x² + 2x.Take x = -0.5: y = (-0.5)² + 2*(-0.5) = 0.25 - 1 = -0.75, which is in (-1, 0).Take x approaching -1: y approaches (-1)^2 + 2*(-1) = 1 - 2 = -1.Take x approaching 0: y approaches 0 + 0 = 0.Okay, that seems consistent.For Case 2: y = -x² + 2x.Take x = -0.5: y = -(-0.5)^2 + 2*(-0.5) = -0.25 -1 = -1.25, which is in (-3, 0).Take x approaching -1: y approaches -(-1)^2 + 2*(-1) = -1 - 2 = -3.Take x approaching 0: y approaches -(0)^2 + 2*0 = 0.So, that also seems consistent.Therefore, combining both cases, y can take any value between -3 and 0, not including the endpoints.**Conclusion**So, after analyzing both cases and checking with test points, I'm confident that the range of y is (-3, 0).**Final Answer**The range of the real number ( y ) is boxed{(-3, 0)}.

💡First, evaluate the exponent ((sqrt{2} - 1)^0). Any non-zero number raised to the power of 0 is 1, so this simplifies to 1.Next, evaluate the inverse (3^{-1}). The inverse of 3 is (frac{1}{3}).Finally, subtract (frac{1}{3}) from 1. To do this, express 1 as (frac{3}{3}) and then subtract (frac{1}{3}) to get (frac{2}{3}).