💡Okay, so I have this problem here with four different operations labeled A, B, C, and D. I need to figure out which one is correct. Let me go through each option one by one and see if I can spot any mistakes or confirm if they're right.Starting with option A: ( a^{2} cdot a^{3} = a^{6} ). Hmm, I remember that when you multiply two exponents with the same base, you add the exponents. So, ( a^{2} cdot a^{3} ) should be ( a^{2+3} = a^{5} ). But here it's saying ( a^{6} ), which doesn't match. So, A seems incorrect.Moving on to option B: ( 3a - 2a = 1 ). This looks like combining like terms. If I have 3a minus 2a, that should be ( (3 - 2)a = a ). But the answer here is 1, which doesn't make sense because it should still have the variable 'a'. So, B is also incorrect.Now, option C: ( (-2a^{2})^{3} = -8a^{6} ). Okay, raising a product to a power means raising each factor to that power. So, ( (-2)^{3} = -8 ) and ( (a^{2})^{3} = a^{6} ). Multiplying those together gives ( -8a^{6} ), which matches what's given. So, C looks correct.Finally, option D: ( a^{6} div a^{2} = a^{3} ). When dividing exponents with the same base, you subtract the exponents. So, ( a^{6 - 2} = a^{4} ). But here it's saying ( a^{3} ), which is wrong. So, D is incorrect.After checking all four options, only option C is correct.

💡Alright, let's tackle this problem step by step. So, we have the universal set U as the set of all real numbers, ℝ. The set M is defined as all real numbers x such that x squared minus 2x is greater than zero. We need to find the complement of M in ℝ, which is denoted as ∁ℝM.First, let's understand what the set M is. M is the set of all x where x² - 2x > 0. To find this set, I think I need to solve the inequality x² - 2x > 0. Let me write that down: x² - 2x > 0.Hmm, this looks like a quadratic inequality. I remember that to solve quadratic inequalities, it's helpful to factor them if possible. Let's try factoring x² - 2x.Factoring out an x from both terms, we get: x(x - 2) > 0.Okay, so the inequality becomes x(x - 2) > 0. Now, to solve this inequality, I need to find the values of x for which the product of x and (x - 2) is positive.I recall that a product of two numbers is positive if both numbers are positive or both are negative. So, let's consider the critical points where each factor equals zero. Here, x = 0 and x - 2 = 0, which gives x = 2.These critical points divide the real number line into three intervals:1. x < 02. 0 < x < 23. x > 2Now, I need to test each interval to see where the product x(x - 2) is positive.Starting with the first interval, x < 0. Let's pick a test point, say x = -1.Plugging into x(x - 2): (-1)(-1 - 2) = (-1)(-3) = 3, which is positive. So, the inequality holds for x < 0.Next, the interval 0 < x < 2. Let's choose x = 1.Plugging into x(x - 2): 1(1 - 2) = 1(-1) = -1, which is negative. So, the inequality does not hold here.Lastly, the interval x > 2. Let's take x = 3.Plugging into x(x - 2): 3(3 - 2) = 3(1) = 3, which is positive. So, the inequality holds for x > 2.Therefore, the solution to the inequality x² - 2x > 0 is x < 0 or x > 2. In interval notation, this is (-∞, 0) ∪ (2, ∞).So, the set M is (-∞, 0) ∪ (2, ∞).Now, we need to find the complement of M in ℝ, which is everything in ℝ that's not in M. Since M is (-∞, 0) ∪ (2, ∞), the complement would be the interval between 0 and 2, including the endpoints because M doesn't include 0 or 2.Wait, let me think about that. If M is x < 0 or x > 2, then the complement should be x ≥ 0 and x ≤ 2. So, that's the interval [0, 2].But hold on, let me double-check. If M is (-∞, 0) ∪ (2, ∞), then the complement would be the closure of that set, which is [0, 2]. Because in the complement, we include all points not in M, which are exactly the points from 0 to 2, including 0 and 2 since they were excluded from M.Yes, that makes sense. So, the complement of M is [0, 2].Looking back at the options:A: [0,2]B: (0,2)C: (-∞,0)∪(2,+∞)D: (-∞,0]∪[2,+∞)So, the complement is [0,2], which is option A.Wait a second, let me make sure I didn't make a mistake. The original set M was x² - 2x > 0, which we solved as x < 0 or x > 2. So, M is (-∞,0) ∪ (2, ∞). Therefore, the complement should be the closure of that, which is [0,2]. So, yes, option A is correct.But just to be thorough, let me consider the endpoints. At x = 0, x² - 2x = 0 - 0 = 0, which is not greater than zero, so 0 is not in M. Similarly, at x = 2, x² - 2x = 4 - 4 = 0, which is also not greater than zero, so 2 is not in M. Therefore, in the complement, we include 0 and 2 because they are not in M.Therefore, the complement is [0,2], which is option A.I think that's solid. I don't see any mistakes in my reasoning. So, the answer should be A.**Final Answer**boxed{A}

💡Alright, so I have this expression: (x + 3)/(x - 1) + (x - 2)^0. I need to figure out for which values of x this expression is meaningful. Hmm, okay, let's break it down step by step.First, I know that for a fraction like (x + 3)/(x - 1), the denominator can't be zero because division by zero is undefined. So, I need to make sure that x - 1 is not zero. That means x can't be 1. Got it, so x ≠ 1 is one condition.Next, there's this term (x - 2)^0. I remember that any number raised to the power of zero is 1, right? So, (x - 2)^0 should just be 1 as long as x - 2 is not zero. Wait, is that true? If x - 2 is zero, then we're looking at 0^0, which is undefined. So, actually, x - 2 can't be zero either. That means x can't be 2. So, now I have two conditions: x ≠ 1 and x ≠ 2.But hold on, I thought any number to the power of zero is 1, but does that include zero? I think 0^0 is undefined, so yeah, x can't be 2 because that would make the base zero. So, x ≠ 2 is another restriction.Putting it all together, the expression is meaningful when x is any real number except 1 and 2. So, the range of x is all real numbers except 1 and 2.Wait, but in the initial problem, the user mentioned that the standard solution had an error regarding x ≠ -2. Did I miss something? Let me double-check. The expression is (x + 3)/(x - 1) + (x - 2)^0. There's no x + 2 in the denominator or anywhere else, so why would x ≠ -2 be a condition? Maybe that was a mistake in the standard solution. I think I'm confident with x ≠ 1 and x ≠ 2.So, to summarize, the expression is meaningful for all real numbers x except x = 1 and x = 2.

💡Alright, so I have this problem about the function ( f(x) = x^2 + ax + b ) where ( a ) and ( b ) are real numbers. It's divided into two main parts, I and II, each with subparts. Let me try to tackle each part step by step.Starting with part (Ⅰ):(i) If ( a = b = 1 ), find the range of the function ( f(x) ) for ( x in [0, 1] ).Okay, so substituting ( a = 1 ) and ( b = 1 ) into the function, we get:( f(x) = x^2 + x + 1 )Hmm, this is a quadratic function. Since the coefficient of ( x^2 ) is positive, it opens upwards. The vertex of this parabola will give me the minimum value. The vertex occurs at ( x = -frac{a}{2} ), which in this case is ( x = -frac{1}{2} ). But wait, our domain is ( x in [0, 1] ), so the vertex at ( x = -frac{1}{2} ) is outside this interval. That means the function is increasing on the interval [0, 1] because the vertex is to the left of this interval.So, to find the range, I just need to evaluate the function at the endpoints of the interval.At ( x = 0 ):( f(0) = 0^2 + 0 + 1 = 1 )At ( x = 1 ):( f(1) = 1^2 + 1 + 1 = 3 )Therefore, the function increases from 1 to 3 as ( x ) goes from 0 to 1. So, the range of ( f(x) ) is [1, 3].Wait, let me double-check. Since the function is increasing on [0,1], the minimum is at x=0 and the maximum at x=1. Yep, that seems right.So, part (Ⅰ)(i) is done.(ii) If the range of the function ( f(x) ) is [0, 1] for ( x in [0, 1] ), find the values of ( a ) and ( b ).Alright, so now the function ( f(x) = x^2 + ax + b ) has a range [0,1] on the interval [0,1]. That means the minimum value is 0 and the maximum is 1.Since it's a quadratic function, its graph is a parabola. Depending on the position of the vertex relative to the interval [0,1], the function could be increasing, decreasing, or have a minimum or maximum within the interval.First, let's recall that the vertex of ( f(x) = x^2 + ax + b ) is at ( x = -frac{a}{2} ). So, depending on where this vertex is, the behavior of the function on [0,1] changes.Case 1: The vertex is to the left of the interval [0,1], i.e., ( -frac{a}{2} leq 0 ) which implies ( a geq 0 ). In this case, the function is increasing on [0,1]. Therefore, the minimum is at x=0 and the maximum at x=1.So, we have:( f(0) = b = 0 )( f(1) = 1 + a + b = 1 )But since ( b = 0 ), substituting into the second equation:( 1 + a + 0 = 1 ) => ( a = 0 )So, one solution is ( a = 0 ), ( b = 0 ).Case 2: The vertex is to the right of the interval [0,1], i.e., ( -frac{a}{2} geq 1 ) which implies ( a leq -2 ). In this case, the function is decreasing on [0,1]. Therefore, the minimum is at x=1 and the maximum at x=0.So, we have:( f(1) = 1 + a + b = 0 )( f(0) = b = 1 )From ( f(0) = 1 ), we get ( b = 1 ). Substituting into the first equation:( 1 + a + 1 = 0 ) => ( a = -2 )So, another solution is ( a = -2 ), ( b = 1 ).Case 3: The vertex is inside the interval [0,1], i.e., ( 0 < -frac{a}{2} < 1 ) which implies ( -2 < a < 0 ). In this case, the function has its minimum at the vertex. So, the minimum value is 0, and the maximum could be at either x=0 or x=1.So, let's set the minimum at the vertex to 0.The vertex is at ( x = -frac{a}{2} ), and the value there is:( fleft(-frac{a}{2}right) = left(-frac{a}{2}right)^2 + aleft(-frac{a}{2}right) + b = frac{a^2}{4} - frac{a^2}{2} + b = -frac{a^2}{4} + b )Set this equal to 0:( -frac{a^2}{4} + b = 0 ) => ( b = frac{a^2}{4} )Now, we need to find the maximum value of f(x) on [0,1]. Since the function is a parabola opening upwards, the maximum will be at one of the endpoints.Compute ( f(0) = b = frac{a^2}{4} )Compute ( f(1) = 1 + a + b = 1 + a + frac{a^2}{4} )We are told the range is [0,1], so the maximum must be 1. Therefore, either ( f(0) = 1 ) or ( f(1) = 1 ).Subcase 3a: ( f(0) = 1 )( frac{a^2}{4} = 1 ) => ( a^2 = 4 ) => ( a = pm 2 )But in this case, ( -2 < a < 0 ), so ( a = -2 ). But wait, ( a = -2 ) is the boundary case, which we already considered in Case 2. So, this might not be a valid solution here.Subcase 3b: ( f(1) = 1 )( 1 + a + frac{a^2}{4} = 1 ) => ( a + frac{a^2}{4} = 0 ) => ( aleft(1 + frac{a}{4}right) = 0 )So, ( a = 0 ) or ( a = -4 )But ( a = 0 ) is the boundary case of Case 1, and ( a = -4 ) is outside the interval ( -2 < a < 0 ). So, this also doesn't give a valid solution in this case.Hmm, so maybe there's no solution in this case? Or perhaps I made a mistake.Wait, let's check the calculations again.We had ( b = frac{a^2}{4} ), and then ( f(1) = 1 + a + b = 1 + a + frac{a^2}{4} ). Setting this equal to 1:( 1 + a + frac{a^2}{4} = 1 ) => ( a + frac{a^2}{4} = 0 ) => ( a(1 + frac{a}{4}) = 0 )So, ( a = 0 ) or ( a = -4 ). But both are outside the interval ( -2 < a < 0 ). So, indeed, there are no solutions in this case.Therefore, the only valid solutions are from Case 1 and Case 2, which are ( a = 0, b = 0 ) and ( a = -2, b = 1 ).Wait, but let me think again. The function is quadratic, so maybe there's another scenario where the maximum is achieved at the vertex? But since the parabola opens upwards, the vertex is the minimum, not the maximum. So, the maximum must be at one of the endpoints. So, I think my reasoning is correct.Therefore, the possible pairs are ( (a, b) = (0, 0) ) and ( (-2, 1) ).Moving on to part (Ⅱ):When ( |x| geq 2 ), it always holds that ( f(x) geq 0 ), and the maximum value of ( f(x) ) in the interval ( (2, 3] ) is 1. Find the maximum and minimum values of ( a^2 + b^2 ).Alright, so this seems more complex. Let me try to break it down.First, the function ( f(x) = x^2 + ax + b ) is a quadratic, opening upwards since the coefficient of ( x^2 ) is positive. So, it has a minimum at its vertex.Given that ( f(x) geq 0 ) for ( |x| geq 2 ). That means for ( x leq -2 ) and ( x geq 2 ), the function is non-negative.Additionally, in the interval ( (2, 3] ), the maximum value of ( f(x) ) is 1.We need to find the maximum and minimum of ( a^2 + b^2 ).Let me think about the conditions.First, ( f(x) geq 0 ) for ( |x| geq 2 ). Since the parabola opens upwards, this implies that the function does not cross the x-axis beyond ( |x| geq 2 ). So, either the function is always non-negative (i.e., no real roots), or if it has real roots, they must lie within ( |x| < 2 ).But wait, if the function is non-negative for ( |x| geq 2 ), it could have roots within ( |x| < 2 ) or not. So, two possibilities:1. The quadratic has no real roots, so it's always positive. Then, the discriminant ( a^2 - 4b < 0 ).2. The quadratic has real roots, both lying within ( (-2, 2) ). So, the roots are inside this interval, and outside of ( |x| geq 2 ), the function is positive.Additionally, in the interval ( (2, 3] ), the maximum of ( f(x) ) is 1.Since the function is a parabola opening upwards, its maximum on an interval can occur either at an endpoint or at the vertex if the vertex is within the interval.But in ( (2, 3] ), the function is increasing or decreasing? Let's see.The vertex is at ( x = -frac{a}{2} ). So, if ( -frac{a}{2} ) is less than 2, then on ( (2, 3] ), the function is increasing. If ( -frac{a}{2} ) is between 2 and 3, then the function would have a minimum at the vertex, and the maximum would be at one of the endpoints. But since the maximum is given as 1, which is at the endpoint.Wait, actually, since the function is opening upwards, on the interval ( (2, 3] ), if the vertex is to the left of 2, then the function is increasing on ( (2, 3] ), so the maximum is at x=3. If the vertex is within ( (2, 3] ), then the function would have a minimum there, and the maximum would still be at x=3 or x=2, but since it's a maximum, it would be at the endpoints.But the problem says the maximum in ( (2, 3] ) is 1. So, likely, the maximum is at x=3, so ( f(3) = 1 ).Wait, but let me verify.Suppose the vertex is at x = c. If c < 2, then on (2,3], the function is increasing, so maximum at x=3.If c is in (2,3), then the function has a minimum at c, and the maximum would still be at the endpoints, either x=2 or x=3.But since the maximum is 1, and f(3) is given as 1, perhaps f(3)=1 and f(2) <=1.But let's formalize this.Given that the maximum on (2,3] is 1, so f(3)=1, and f(x) <=1 for x in (2,3].But also, f(x) >=0 for |x| >=2.So, let's write down the conditions.First, f(3)=1:( f(3) = 9 + 3a + b = 1 ) => ( 3a + b = -8 ) => ( b = -3a -8 )Second, f(x) >=0 for |x| >=2.So, for x=2 and x=-2, f(x) >=0.Compute f(2):( f(2) = 4 + 2a + b )But since b = -3a -8, substitute:( f(2) = 4 + 2a + (-3a -8) = 4 - a -8 = -a -4 )Similarly, f(-2):( f(-2) = 4 - 2a + b )Again, substitute b:( f(-2) = 4 - 2a + (-3a -8) = 4 -5a -8 = -5a -4 )So, we have:f(2) = -a -4 >=0 => -a -4 >=0 => -a >=4 => a <= -4f(-2) = -5a -4 >=0 => -5a >=4 => a <= -4/5But from f(2) >=0, we have a <= -4, which is more restrictive than a <= -4/5.So, combining both, a <= -4.But also, we need to ensure that f(x) >=0 for all |x| >=2, not just at x=2 and x=-2. Since the function is a parabola opening upwards, if it has real roots, they must lie within (-2, 2). Otherwise, if the roots are outside, the function would be negative beyond them, which contradicts f(x) >=0 for |x| >=2.So, either the quadratic has no real roots (discriminant <0) or it has real roots within (-2,2).Case 1: Quadratic has no real roots.Then, discriminant D = a^2 -4b <0.But b = -3a -8, so substitute:D = a^2 -4*(-3a -8) = a^2 +12a +32 <0Solve the inequality:a^2 +12a +32 <0Find roots of a^2 +12a +32=0:a = [-12 ± sqrt(144 -128)] /2 = [-12 ± sqrt(16)] /2 = [-12 ±4]/2So, a = (-12 +4)/2 = -8/2 = -4, and a = (-12 -4)/2 = -16/2 = -8So, the quadratic a^2 +12a +32 is positive outside (-8, -4) and negative inside (-8, -4).But we have a <= -4 from earlier.So, for discriminant <0, a must be in (-8, -4). But since a <= -4, the overlap is a in (-8, -4].But wait, a <= -4, so a in (-8, -4] satisfies discriminant <0.Case 2: Quadratic has real roots within (-2,2).So, discriminant D >=0, which implies a <= -8 or a >= -4, but since a <= -4, we have a <= -8.But wait, from earlier, a <= -4, so a <= -8 is a subset.But also, the roots must lie within (-2,2). So, let's denote the roots as r1 and r2, with r1 < r2.We need r1 > -2 and r2 < 2.Given the quadratic f(x) = x^2 +ax +b, with roots r1 and r2.So, sum of roots: r1 + r2 = -aProduct of roots: r1*r2 = bGiven that both roots are in (-2,2), so:-2 < r1 < r2 < 2So, sum of roots: -4 < r1 + r2 <4 => -4 < -a <4 => -4 < -a <4 => -4 < -a => a <4 and -a <4 => a > -4But from earlier, a <= -4, so the only overlap is a = -4.But wait, if a = -4, then sum of roots = -a =4, but since both roots are less than 2, their sum would be less than 4. So, actually, a cannot be -4 because that would require the sum of roots to be 4, but each root is less than 2, so the maximum sum is less than 4. So, a cannot be -4.Wait, this is getting a bit tangled. Let me think.If the roots are in (-2,2), then:r1 > -2r2 < 2So, sum of roots: r1 + r2 = -aSince r1 > -2 and r2 < 2, the sum r1 + r2 > -2 + (-2) = -4 and r1 + r2 < 2 + 2 =4But since r1 < r2, and both are within (-2,2), the sum is between -4 and 4.But r1 + r2 = -a, so:-4 < -a <4 => -4 < -a => a <4 and -a <4 => a > -4So, -4 < a <4But from earlier, a <= -4, so the only overlap is a = -4.But when a = -4, the sum of roots is 4, but since each root is less than 2, their sum is less than 4. So, a cannot be -4.Therefore, there is no solution in this case where the quadratic has real roots within (-2,2). So, only Case 1 is valid, where the quadratic has no real roots, i.e., discriminant <0, and a in (-8, -4].But wait, earlier we had a <= -4, and discriminant <0 when a is in (-8, -4). So, combining these, a must be in (-8, -4].But let's verify.If a is in (-8, -4), then discriminant D = a^2 +12a +32 <0, so no real roots, hence f(x) >=0 for all x, which satisfies the condition f(x) >=0 for |x| >=2.But also, when a = -4, discriminant D = (-4)^2 +12*(-4) +32 = 16 -48 +32 = 0, so repeated root. So, f(x) = (x +2)^2, which is non-negative everywhere, so f(x) >=0 for |x| >=2 is satisfied.But when a = -4, f(x) = x^2 -4x + b. Wait, b = -3a -8, so b = -3*(-4) -8 =12 -8=4.So, f(x) =x^2 -4x +4 = (x-2)^2, which is non-negative everywhere, so indeed, f(x) >=0 for |x| >=2.But wait, when a = -4, f(x) = (x-2)^2, which is zero at x=2. So, at x=2, f(x)=0, which is allowed since f(x) >=0.But in the interval (2,3], the maximum is 1. Let's check f(3):f(3) = (3-2)^2 =1, which is correct.Similarly, for a in (-8, -4), f(x) is always positive, so f(x) >=0 for |x| >=2, and f(3)=1.So, overall, a is in [-8, -4], because when a = -8, let's check:b = -3*(-8) -8 =24 -8=16f(x)=x^2 -8x +16=(x-4)^2, which is non-negative everywhere, and f(3)= (3-4)^2=1, which is correct.So, a ranges from -8 to -4.Now, we need to find the maximum and minimum of ( a^2 + b^2 ).Given that b = -3a -8, so ( a^2 + b^2 = a^2 + (-3a -8)^2 )Let me compute this:( a^2 + (-3a -8)^2 = a^2 + 9a^2 + 48a +64 =10a^2 +48a +64 )So, we have ( a^2 + b^2 =10a^2 +48a +64 )This is a quadratic in terms of a, opening upwards (since coefficient of a^2 is positive). So, its minimum occurs at the vertex, and the maximum occurs at the endpoints of the interval.The interval for a is [-8, -4].First, find the vertex of this quadratic.The vertex occurs at ( a = -frac{B}{2A} = -frac{48}{2*10} = -frac{48}{20} = -12/5 = -2.4 )But -2.4 is not in our interval [-8, -4]. So, the minimum must occur at one of the endpoints.Wait, but since the parabola opens upwards, the minimum is at the vertex if the vertex is within the interval. If the vertex is outside, then the minimum is at the closest endpoint.But our interval is [-8, -4], and the vertex is at a = -2.4, which is to the right of the interval. So, the function is decreasing on (-infty, -2.4) and increasing on (-2.4, +infty). But our interval is [-8, -4], which is entirely to the left of the vertex. So, on [-8, -4], the function is decreasing because it's to the left of the vertex.Wait, actually, no. Wait, the function ( 10a^2 +48a +64 ) is a parabola opening upwards, so it decreases until a = -2.4 and then increases. But on the interval [-8, -4], which is entirely to the left of -2.4, the function is decreasing as a increases.Wait, no, as a increases from -8 to -4, which is moving to the right, the function is decreasing until a = -2.4, but since -4 is still to the left of -2.4, the function is decreasing throughout the interval.Wait, let me compute the derivative to confirm.The derivative of ( 10a^2 +48a +64 ) with respect to a is 20a +48.Set derivative to zero: 20a +48=0 => a= -48/20= -12/5= -2.4So, the function has a minimum at a= -2.4, which is outside our interval.Therefore, on the interval [-8, -4], the function is decreasing because the derivative is negative throughout.Compute derivative at a=-8:20*(-8) +48= -160 +48= -112 <0At a=-4:20*(-4)+48= -80 +48= -32 <0So, the function is decreasing on [-8, -4], meaning the maximum occurs at a=-8 and the minimum at a=-4.Therefore, compute ( a^2 + b^2 ) at a=-8 and a=-4.At a=-8:b= -3*(-8) -8=24 -8=16So, ( a^2 + b^2 = (-8)^2 +16^2=64 +256=320 )Wait, but earlier I thought the maximum was 74 and minimum 32. Hmm, seems conflicting.Wait, no, wait, I think I made a mistake in substitution.Wait, ( a^2 + b^2 =10a^2 +48a +64 )At a=-8:10*(-8)^2 +48*(-8) +64=10*64 + (-384) +64=640 -384 +64=640 -384=256 +64=320At a=-4:10*(-4)^2 +48*(-4) +64=10*16 + (-192) +64=160 -192 +64= (160 +64) -192=224 -192=32So, the minimum is 32 at a=-4, and the maximum is 320 at a=-8.But wait, in the initial problem statement, the maximum value of f(x) in (2,3] is 1, and we found that a is in [-8, -4]. So, the maximum of ( a^2 + b^2 ) is 320, and the minimum is 32.But in the initial solution, it was stated as 74 and 32. Hmm, perhaps I made a mistake in the earlier steps.Wait, let me double-check the expression for ( a^2 + b^2 ).Given b= -3a -8, so:( a^2 + b^2 = a^2 + (-3a -8)^2 = a^2 +9a^2 +48a +64=10a^2 +48a +64 ). That seems correct.Then, evaluating at a=-4:10*(-4)^2 +48*(-4) +64=10*16 + (-192) +64=160 -192 +64=32At a=-8:10*(-8)^2 +48*(-8) +64=10*64 + (-384) +64=640 -384 +64=320So, yes, 32 and 320.But in the initial solution, it was 32 and 74. So, perhaps the initial solution had a miscalculation.Wait, let me check the initial solution again.In the initial solution, it was stated:"From the problem, the graph of the function is an upward-opening parabola, and the maximum value of f(x) in the interval (2, 3] can only be obtained at the closed endpoint, thus we have f(2) ≤ f(3) = 1, thereby a ≥ -5 and b = -3a -8."Wait, that seems different from my approach.Wait, in the initial solution, they concluded that a >= -5, but in my reasoning, I concluded a <= -4.Hmm, perhaps there's a discrepancy.Wait, let's go back.In the initial solution, they said:"From the problem, the graph of the function is an upward-opening parabola, and the maximum value of f(x) in the interval (2, 3] can only be obtained at the closed endpoint, thus we have f(2) ≤ f(3) = 1, thereby a ≥ -5 and b = -3a -8."Wait, so they set f(2) <= f(3)=1, which gives:f(2)=4 +2a +b <=1But since f(3)=1, and b= -3a -8, substitute:f(2)=4 +2a + (-3a -8)=4 -a -8= -a -4 <=1So, -a -4 <=1 => -a <=5 => a >= -5So, they concluded a >= -5, but in my earlier reasoning, I concluded a <= -4.Wait, that seems conflicting.Wait, perhaps I made a mistake in the direction of the inequality.Let me re-examine.Given that f(3)=1, and f(x) <=1 in (2,3].But f(x) is increasing on (2,3] if the vertex is to the left of 2, which would mean that f(3) is the maximum.But if the vertex is within (2,3], then f(x) would have a minimum at the vertex, and the maximum would still be at x=3 or x=2.But since f(3)=1 is the maximum, we have f(2) <=1.So, f(2)= -a -4 <=1 => -a <=5 => a >= -5So, a >= -5.But earlier, from f(2) >=0, we had a <= -4.So, combining these, a must be in [-5, -4].Wait, that makes more sense.So, a is between -5 and -4.Therefore, a in [-5, -4].Then, b= -3a -8.So, when a=-5, b= -3*(-5) -8=15 -8=7When a=-4, b= -3*(-4) -8=12 -8=4So, a ranges from -5 to -4, and b from 7 to4.Now, compute ( a^2 + b^2 ).Expressed as:( a^2 + b^2 = a^2 + (-3a -8)^2 =10a^2 +48a +64 )Now, with a in [-5, -4].Find the minimum and maximum of this quadratic in a.The quadratic is 10a^2 +48a +64.The vertex is at a= -48/(2*10)= -2.4, which is outside the interval [-5, -4].Since the parabola opens upwards, and the interval is to the left of the vertex, the function is decreasing on [-5, -4].Therefore, the maximum occurs at a=-5, and the minimum at a=-4.Compute at a=-5:10*(-5)^2 +48*(-5) +64=10*25 + (-240) +64=250 -240 +64=10 +64=74At a=-4:10*(-4)^2 +48*(-4) +64=10*16 + (-192) +64=160 -192 +64=32So, the minimum of ( a^2 + b^2 ) is 32, and the maximum is 74.Therefore, the initial solution was correct, and my earlier mistake was not considering the correct interval for a, which should be [-5, -4] instead of [-8, -4].So, to summarize:For part (Ⅱ), after correctly considering the conditions, a is in [-5, -4], leading to ( a^2 + b^2 ) ranging from 32 to 74.**Final Answer**Ⅰ(i) The range of ( f(x) ) is boxed{[1, 3]}.Ⅰ(ii) The values of ( a ) and ( b ) are ( a = 0 ) and ( b = 0 ), or ( a = -2 ) and ( b = 1 ).Ⅱ The minimum value of ( a^2 + b^2 ) is boxed{32}, and the maximum value is boxed{74}.

💡Okay, so Lucy has 212 fish right now, and she needs to buy 68 more to reach a certain number. I need to find out what that certain number is. Hmm, let me think about how to approach this.First, I know that if Lucy has 212 fish and she wants to have more, she needs to add the number of fish she's going to buy to her current amount. So, it's like she's starting with 212 and then adding 68 to get to her goal.Let me write that down: 212 plus 68 equals what? Okay, so I can do this addition step by step to make sure I don't make a mistake. Maybe I'll break it down into smaller parts.Starting with 212, if I add 60 first, that should be easier. So, 212 plus 60 is 272. Now, I have to add the remaining 8. So, 272 plus 8 equals 280. That seems right.Wait, let me double-check to make sure I didn't skip any steps or make an error in my calculation. If I add 212 and 68 together, does it really equal 280? Let me try adding them another way to confirm.I can add the units place first. 2 plus 8 is 10, so I write down 0 and carry over 1. Then, in the tens place, 1 plus 6 is 7, plus the carried-over 1 makes 8. Finally, in the hundreds place, 2 plus 0 is 2. So, putting it all together, I get 280. Yep, that matches what I got before.Just to be extra sure, maybe I can use a number line or visualize it. Starting at 212, if I move up 68 units, where do I land? Let's see, from 212, adding 60 gets me to 272, and then adding 8 more gets me to 280. Yep, that checks out.I think I'm confident now that Lucy wants to have a total of 280 fish. It makes sense because she needs to add 68 to her current 212 to reach that number. No mistakes in my calculations, so I'm good.

💡Let's denote the total number of stickers Oliver had initially as S.First, Oliver used 1/3 of his stickers, so he was left with 2/3 of his stickers.Then, he gave 2/5 of the remaining stickers to his friend. This means he kept 3/5 of the remaining stickers after using 1/3.So, we can write the equation as follows:(3/5) * (2/3) * S = 54To find S, we need to solve for it:(3/5) * (2/3) * S = 54(6/15) * S = 54(2/5) * S = 54Now, we multiply both sides of the equation by the reciprocal of (2/5) to isolate S:S = 54 * (5/2)S = 54 * 2.5S = 135Therefore, Oliver initially had boxed{135} stickers.