💡Okay, so I have this problem where I need to find ( a^3 + b^3 + c^3 ) given that ( a, b, c ) are the roots of the polynomial ( p(x) = x^3 + x^2 - 333x - 1001 ). Hmm, I remember that when dealing with polynomials and their roots, Vieta's formulas can be really helpful. Let me try to recall what Vieta's formulas say.For a cubic polynomial ( x^3 + px^2 + qx + r ), the sum of the roots ( a + b + c ) is equal to ( -p ). The sum of the products of the roots taken two at a time ( ab + bc + ca ) is equal to ( q ), and the product of the roots ( abc ) is equal to ( -r ). So, applying this to the given polynomial ( p(x) = x^3 + x^2 - 333x - 1001 ):- The sum of the roots ( a + b + c = -1 ) because the coefficient of ( x^2 ) is 1.- The sum of the products ( ab + bc + ca = -333 ) because the coefficient of ( x ) is -333.- The product of the roots ( abc = -(-1001) = 1001 ) because the constant term is -1001.Alright, so I have these three pieces of information. Now, I need to find ( a^3 + b^3 + c^3 ). I remember there's a formula for the sum of cubes of roots in terms of the sums and products of the roots. Let me try to recall that formula.I think the formula is ( a^3 + b^3 + c^3 = (a + b + c)^3 - 3(a + b + c)(ab + bc + ca) + 3abc ). Let me check if that makes sense. If I expand ( (a + b + c)^3 ), it should give me ( a^3 + b^3 + c^3 + 3(a + b)(b + c)(a + c) ) or something like that. Wait, maybe I'm mixing it up with another formula.Alternatively, I recall that ( a^3 + b^3 + c^3 ) can be expressed using the identity ( a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc ). That seems more familiar. Let me verify this identity.If I expand ( (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) ), I get:( a(a^2 + b^2 + c^2 - ab - bc - ca) + b(a^2 + b^2 + c^2 - ab - bc - ca) + c(a^2 + b^2 + c^2 - ab - bc - ca) )Which simplifies to:( a^3 + ab^2 + ac^2 - a^2b - abc - a^2c + ba^2 + b^3 + bc^2 - ab^2 - b^2c - b c a + ca^2 + cb^2 + c^3 - abc - bc^2 - c^2a )Hmm, this seems complicated. Maybe I should look for a better way to verify this identity. Alternatively, I can use the standard formula for ( a^3 + b^3 + c^3 ) in terms of the sums and products.Wait, I think another approach is to use Newton's identities, which relate power sums to elementary symmetric sums. Newton's identities might be the way to go here. Let me recall them.For a monic polynomial of degree 3, the power sums ( p_k = a^k + b^k + c^k ) can be expressed using the elementary symmetric sums ( s_1 = a + b + c ), ( s_2 = ab + bc + ca ), and ( s_3 = abc ). The formula for ( p_3 ) is:( p_3 = s_1 p_2 - s_2 p_1 + s_3 p_0 )Where ( p_0 = 3 ) because there are three roots, each raised to the power of 0, which is 1. So, ( p_0 = 1 + 1 + 1 = 3 ).But to use this, I need ( p_2 ), which is ( a^2 + b^2 + c^2 ). I don't have that yet, so maybe I should compute ( p_2 ) first.I know that ( p_2 = (a + b + c)^2 - 2(ab + bc + ca) ). Plugging in the known values:( p_2 = (-1)^2 - 2(-333) = 1 + 666 = 667 ).Okay, so ( p_2 = 667 ). Now, going back to Newton's identities:( p_3 = s_1 p_2 - s_2 p_1 + s_3 p_0 )Plugging in the known values:( p_3 = (-1)(667) - (-333)(-1) + (1001)(3) )Let me compute each term step by step:1. ( s_1 p_2 = (-1)(667) = -667 )2. ( -s_2 p_1 = -(-333)(-1) = -333 ) (because negative times negative is positive, and then positive times negative is negative)3. ( s_3 p_0 = (1001)(3) = 3003 )Now, adding these up:( p_3 = -667 - 333 + 3003 )Let me compute this:First, ( -667 - 333 = -1000 )Then, ( -1000 + 3003 = 2003 )So, ( a^3 + b^3 + c^3 = 2003 ).Wait, let me double-check my calculations to make sure I didn't make any mistakes. Sometimes when dealing with negative signs, it's easy to slip up.Starting again:( p_3 = s_1 p_2 - s_2 p_1 + s_3 p_0 )Given:- ( s_1 = -1 )- ( p_2 = 667 )- ( s_2 = -333 )- ( p_1 = -1 )- ( s_3 = 1001 )- ( p_0 = 3 )So,( p_3 = (-1)(667) - (-333)(-1) + (1001)(3) )Calculating each term:1. ( (-1)(667) = -667 )2. ( -(-333)(-1) = - (333) = -333 ) (because two negatives make a positive, but then multiplied by -1 again)3. ( (1001)(3) = 3003 )Adding them together:( -667 - 333 + 3003 = (-667 - 333) + 3003 = (-1000) + 3003 = 2003 )Yes, that seems correct. So, the value of ( a^3 + b^3 + c^3 ) is 2003.Alternatively, I can think about the fact that each root ( a, b, c ) satisfies the equation ( x^3 + x^2 - 333x - 1001 = 0 ). So, for each root, we have:( a^3 = -a^2 + 333a + 1001 )Similarly,( b^3 = -b^2 + 333b + 1001 )( c^3 = -c^2 + 333c + 1001 )Therefore, adding these up:( a^3 + b^3 + c^3 = -(a^2 + b^2 + c^2) + 333(a + b + c) + 3 times 1001 )We already know ( a + b + c = -1 ) and ( a^2 + b^2 + c^2 = 667 ). Plugging these in:( a^3 + b^3 + c^3 = -667 + 333(-1) + 3003 )Calculating each term:1. ( -667 )2. ( 333(-1) = -333 )3. ( 3 times 1001 = 3003 )Adding them together:( -667 - 333 + 3003 = (-1000) + 3003 = 2003 )So, this method also gives the same result, which is reassuring. It seems like 2003 is the correct answer.I think I've verified it sufficiently. Both methods lead to the same conclusion, so I'm confident that ( a^3 + b^3 + c^3 = 2003 ).**Final Answer**boxed{2003}

💡Okay, so I need to find the domain of the expression (frac{sqrt{3x - 6}}{sqrt{7 - 2x}}). Hmm, the domain of an expression refers to all the real numbers (x) for which the expression is defined. Since this expression involves square roots and a denominator, I need to make sure that both the numerator and the denominator are defined and that the denominator isn't zero because division by zero is undefined.Let me start by looking at the numerator, which is (sqrt{3x - 6}). For a square root to be defined, the expression inside the square root must be greater than or equal to zero. So, I can write the inequality:(3x - 6 geq 0)Solving for (x), I add 6 to both sides:(3x geq 6)Then, divide both sides by 3:(x geq 2)Okay, so (x) has to be greater than or equal to 2 for the numerator to be defined.Now, moving on to the denominator, which is (sqrt{7 - 2x}). Similarly, the expression inside the square root must be greater than or equal to zero. So, I set up the inequality:(7 - 2x geq 0)Solving for (x), I subtract 7 from both sides:(-2x geq -7)Then, divide both sides by -2. Wait, I remember that when you divide or multiply both sides of an inequality by a negative number, the inequality sign flips. So, dividing by -2:(x leq frac{7}{2})So, (x) has to be less than or equal to (frac{7}{2}) for the denominator to be defined.But wait, the denominator is a square root, and it's in the denominator of the entire expression. That means the denominator can't just be zero because division by zero is undefined. So, I need to make sure that (7 - 2x) is strictly greater than zero, not just greater than or equal to. Let me write that:(7 - 2x > 0)Solving this:(-2x > -7)Divide both sides by -2, flipping the inequality:(x < frac{7}{2})So, combining the two conditions from the numerator and the denominator, (x) must be greater than or equal to 2 and less than (frac{7}{2}). Let me write that as an interval. The lower bound is 2, which is included because (x) can be equal to 2, and the upper bound is (frac{7}{2}), which is not included because (x) must be less than (frac{7}{2}). So, in interval notation, this is:([2, frac{7}{2}))Just to double-check, let me test a value in this interval, say (x = 3). Plugging into the numerator: (sqrt{3*3 - 6} = sqrt{9 - 6} = sqrt{3}), which is defined. Plugging into the denominator: (sqrt{7 - 2*3} = sqrt{7 - 6} = sqrt{1} = 1), which is also defined and not zero. So, the expression is defined at (x = 3).What about the endpoints? At (x = 2), the numerator becomes (sqrt{3*2 - 6} = sqrt{6 - 6} = sqrt{0} = 0), which is fine, and the denominator is (sqrt{7 - 2*2} = sqrt{7 - 4} = sqrt{3}), which is defined and not zero. So, (x = 2) is included.At (x = frac{7}{2}), which is 3.5, the denominator becomes (sqrt{7 - 2*(3.5)} = sqrt{7 - 7} = sqrt{0} = 0), which would make the expression undefined. So, (x = frac{7}{2}) is excluded, which matches our interval.I think that covers all the necessary conditions. I don't think I missed anything. So, the domain should be all real numbers from 2 up to but not including (frac{7}{2}).

💡Okay, so I've got this problem with two parts. Let me try to tackle them one by one. First, the functions are given as f(x) = x² - 1 and g(x) = a|x - 1|. **Problem (1):** Find the range of values for the real number a such that the inequality f(x) ≥ g(x) holds true for all x ∈ ℝ.Alright, so I need to find all real numbers a where x² - 1 is always greater than or equal to a|x - 1| for every real x. Hmm, that sounds like I need to analyze the inequality x² - 1 ≥ a|x - 1| for all x.Let me think about how to approach this. Maybe I can rewrite the inequality as x² - 1 - a|x - 1| ≥ 0 for all x. So, I need to ensure that this expression is non-negative everywhere.I remember that when dealing with absolute values, it's often useful to split into cases based on the sign inside the absolute value. So, let's consider two cases: x ≥ 1 and x < 1.**Case 1: x ≥ 1**In this case, |x - 1| = x - 1. So, the inequality becomes:x² - 1 - a(x - 1) ≥ 0Let me factor x² - 1 as (x - 1)(x + 1). So,(x - 1)(x + 1) - a(x - 1) ≥ 0Factor out (x - 1):(x - 1)(x + 1 - a) ≥ 0Since x ≥ 1, (x - 1) is non-negative. Therefore, the inequality reduces to:x + 1 - a ≥ 0Which simplifies to:x + 1 ≥ aBut this has to hold for all x ≥ 1. So, the smallest value of x + 1 in this interval is when x = 1, which gives 2. Therefore, to satisfy x + 1 ≥ a for all x ≥ 1, we must have a ≤ 2.Wait, hold on. Is that correct? Because if a is less than or equal to 2, then x + 1 - a will be non-negative for all x ≥ 1. Hmm, seems okay.**Case 2: x < 1**Here, |x - 1| = -(x - 1) = 1 - x. So, the inequality becomes:x² - 1 - a(1 - x) ≥ 0Again, factor x² - 1 as (x - 1)(x + 1):(x - 1)(x + 1) - a(1 - x) ≥ 0Note that (1 - x) = -(x - 1), so:(x - 1)(x + 1) + a(x - 1) ≥ 0Factor out (x - 1):(x - 1)(x + 1 + a) ≥ 0Now, since x < 1, (x - 1) is negative. Therefore, the inequality becomes:(x + 1 + a) ≤ 0Because when you divide both sides by a negative number, the inequality flips. So,x + 1 + a ≤ 0Which simplifies to:x + 1 ≤ -aBut this has to hold for all x < 1. The maximum value of x + 1 in this interval is when x approaches 1 from the left, which is 2. Therefore, to satisfy x + 1 ≤ -a for all x < 1, we must have 2 ≤ -a, which implies a ≤ -2.Wait, so in Case 1, we had a ≤ 2, and in Case 2, a ≤ -2. But since both cases must hold simultaneously for the inequality to be true for all x, we need to take the intersection of these conditions. So, a must satisfy both a ≤ 2 and a ≤ -2. Therefore, the stricter condition is a ≤ -2.But hold on, let me verify this. If a is less than or equal to -2, does the inequality hold for all x?Let me test a = -2.For x = 1: f(1) = 1 - 1 = 0, g(1) = -2|0| = 0. So, 0 ≥ 0, which holds.For x > 1: Let's pick x = 2. f(2) = 4 - 1 = 3, g(2) = -2|2 - 1| = -2. So, 3 ≥ -2, which is true.For x < 1: Let's pick x = 0. f(0) = 0 - 1 = -1, g(0) = -2|0 - 1| = -2. So, -1 ≥ -2, which is true.What about x approaching infinity? For x very large, f(x) is x², which dominates over g(x) which is linear. So, f(x) will be much larger than g(x). Similarly, for x approaching negative infinity, f(x) is still x², which is positive, and g(x) is linear, so f(x) will dominate.Wait, but what if a is greater than -2? Let's say a = -1. Then, in Case 2, when x approaches 1 from the left, x + 1 approaches 2, so 2 ≤ -a would mean 2 ≤ 1, which is false. Hence, the inequality would not hold for x near 1. So, a must indeed be ≤ -2.Therefore, the range of a is all real numbers less than or equal to -2.**Problem (2):** Find the maximum value of the function h(x) = |f(x)| + g(x) in the interval [0, 2].So, h(x) = |x² - 1| + a|x - 1|. We need to find its maximum on [0, 2].First, let's analyze |x² - 1|. Since x² - 1 is negative when |x| < 1 and positive otherwise. But in the interval [0, 2], x² - 1 is negative when x ∈ [0, 1) and positive when x ∈ [1, 2]. So, |x² - 1| is equal to:- 1 - x² when x ∈ [0, 1)- x² - 1 when x ∈ [1, 2]Similarly, |x - 1| is:- 1 - x when x ∈ [0, 1)- x - 1 when x ∈ [1, 2]So, let's write h(x) as a piecewise function:For x ∈ [0, 1):h(x) = (1 - x²) + a(1 - x)For x ∈ [1, 2]:h(x) = (x² - 1) + a(x - 1)So, let's write that out:h(x) = {1 - x² + a(1 - x), 0 ≤ x < 1x² - 1 + a(x - 1), 1 ≤ x ≤ 2}Now, to find the maximum of h(x) on [0, 2], we can analyze each piece separately and then compare the maximums.Let me handle each interval.**Interval [0, 1):**h(x) = 1 - x² + a(1 - x)Let me expand this:h(x) = 1 - x² + a - a xCombine like terms:h(x) = (-x²) + (-a x) + (1 + a)This is a quadratic function in x, opening downward (since the coefficient of x² is negative). Therefore, it has a maximum at its vertex.The vertex of a quadratic ax² + bx + c is at x = -b/(2a). Here, a = -1, b = -a.So, x = -(-a)/(2*(-1)) = a / (-2) = -a/2But wait, our interval is [0, 1). So, we need to check if the vertex is within this interval.So, if -a/2 is in [0, 1), then the maximum is at x = -a/2. Otherwise, the maximum is at the endpoints.So, let's find when -a/2 ∈ [0, 1):- 0 ≤ -a/2 < 1Multiply all parts by -2 (remembering to reverse inequalities when multiplying by negative):0 ≥ a > -2So, if a > -2, then the vertex is in [0, 1). If a ≤ -2, the vertex is at x ≥ 1 or x < 0, so outside the interval.Therefore, for a > -2, the maximum in [0,1) is at x = -a/2.For a ≤ -2, the maximum in [0,1) is at x = 0 or x approaching 1.Let me compute h(x) at x = -a/2:h(-a/2) = 1 - (-a/2)^2 + a(1 - (-a/2)) = 1 - (a²)/4 + a(1 + a/2) = 1 - (a²)/4 + a + (a²)/2 = 1 + a + (a²)/4Similarly, compute h(0):h(0) = 1 - 0 + a(1 - 0) = 1 + aCompute h(1) from the left:h(1-) = 1 - 1 + a(1 - 1) = 0 + 0 = 0But since x approaches 1 from the left, h(x) approaches 0.So, for a > -2, the maximum in [0,1) is 1 + a + (a²)/4.For a ≤ -2, the maximum in [0,1) is max{h(0), h(1-)} = max{1 + a, 0}. Since a ≤ -2, 1 + a ≤ -1, so the maximum is 0.**Interval [1, 2]:**h(x) = x² - 1 + a(x - 1)Expand:h(x) = x² - 1 + a x - a = x² + a x - (1 + a)This is a quadratic function in x, opening upward (since the coefficient of x² is positive). Therefore, it has a minimum at its vertex, and the maximum will be at one of the endpoints.The vertex is at x = -b/(2a) where the quadratic is ax² + bx + c. Here, a = 1, b = a.So, x = -a/(2*1) = -a/2But our interval is [1, 2]. So, we need to check if the vertex is within [1, 2].So, if -a/2 ∈ [1, 2), then the function has a minimum there, but the maximum will still be at the endpoints.Wait, since it's opening upward, the maximum will be at the endpoints regardless of where the vertex is. So, we just need to evaluate h(x) at x = 1 and x = 2.Compute h(1):h(1) = 1 - 1 + a(1 - 1) = 0 + 0 = 0Compute h(2):h(2) = 4 - 1 + a(2 - 1) = 3 + aSo, on [1, 2], the maximum is 3 + a.**Putting it all together:**Now, we have to compare the maximums from both intervals.From [0,1):- If a > -2: max is 1 + a + (a²)/4- If a ≤ -2: max is 0From [1,2]: max is 3 + aSo, overall maximum on [0,2] is the larger of the two.Let's analyze:Case 1: a > -2Compare 1 + a + (a²)/4 and 3 + a.Compute 1 + a + (a²)/4 vs 3 + a.Subtract (3 + a) from both:(1 + a + (a²)/4) - (3 + a) = (1 - 3) + (a - a) + (a²)/4 = -2 + (a²)/4So, if -2 + (a²)/4 ≥ 0, then 1 + a + (a²)/4 ≥ 3 + a.Solve -2 + (a²)/4 ≥ 0:(a²)/4 ≥ 2 => a² ≥ 8 => |a| ≥ 2√2 ≈ 2.828But in this case, a > -2. So, a must be ≥ 2√2 or ≤ -2√2. But since a > -2, the only possibility is a ≥ 2√2.Therefore:- If a ≥ 2√2, then the maximum is 1 + a + (a²)/4- If -2 < a < 2√2, then the maximum is 3 + aCase 2: a ≤ -2From [0,1), the maximum is 0, and from [1,2], the maximum is 3 + a.But since a ≤ -2, 3 + a ≤ 1. So, compare 0 and 3 + a.Since 3 + a could be positive or negative.Wait, 3 + a ≥ 0 when a ≥ -3.So:- If -3 ≤ a ≤ -2, then 3 + a ≥ 0, so the maximum is 3 + a- If a < -3, then 3 + a < 0, so the maximum is 0Therefore, putting it all together:- If a ≥ 2√2: max is 1 + a + (a²)/4- If -3 ≤ a < 2√2: max is 3 + a- If a < -3: max is 0But wait, let me check if 1 + a + (a²)/4 is indeed larger than 3 + a when a ≥ 2√2.Compute 1 + a + (a²)/4 - (3 + a) = (a²)/4 - 2Set this ≥ 0:(a²)/4 - 2 ≥ 0 => a² ≥ 8 => |a| ≥ 2√2So yes, when a ≥ 2√2, 1 + a + (a²)/4 is larger.But what about when a is negative? For a < -2, we have:- If -3 ≤ a ≤ -2: max is 3 + a- If a < -3: max is 0So, overall, the maximum value of h(x) on [0,2] is:- 1 + a + (a²)/4 when a ≥ 2√2- 3 + a when -3 ≤ a < 2√2- 0 when a < -3But wait, let's test a = -2:From [0,1), the maximum is 0, and from [1,2], h(2) = 3 + (-2) = 1. So, the maximum is 1. But according to our earlier analysis, for a = -2, which is in -3 ≤ a < 2√2, the maximum is 3 + a = 1, which matches.Similarly, for a = -4, which is < -3, the maximum is 0.For a = 3, which is ≥ 2√2 (since 2√2 ≈ 2.828), the maximum is 1 + 3 + (9)/4 = 4 + 2.25 = 6.25, which is indeed larger than 3 + 3 = 6.Wait, but 1 + a + (a²)/4 when a = 3 is 1 + 3 + 9/4 = 4 + 2.25 = 6.25, which is correct.But let me check a = 2√2:Compute 1 + 2√2 + (8)/4 = 1 + 2√2 + 2 = 3 + 2√2 ≈ 3 + 2.828 ≈ 5.828And 3 + 2√2 ≈ 5.828, so both expressions give the same value at a = 2√2, which is consistent.Therefore, the maximum value is:- 0 when a < -3- 3 + a when -3 ≤ a < 2√2- 1 + a + (a²)/4 when a ≥ 2√2But the problem asks for the maximum value, so perhaps we can express it as a piecewise function.However, the problem doesn't specify a particular a, so maybe we need to express it in terms of a.Alternatively, perhaps I made a mistake in the earlier analysis. Let me double-check.Wait, in the interval [0,1), for a > -2, the maximum is at x = -a/2, which is 1 + a + (a²)/4. For a ≤ -2, the maximum is 0.In [1,2], the maximum is always 3 + a.So, the overall maximum is the maximum between 1 + a + (a²)/4 (if a > -2) and 3 + a.But when a > -2, we have to compare 1 + a + (a²)/4 and 3 + a.As we saw earlier, 1 + a + (a²)/4 is greater than 3 + a when a²/4 ≥ 2, i.e., a ≥ 2√2 or a ≤ -2√2. But since a > -2, only a ≥ 2√2 applies.Therefore, the maximum is:- 1 + a + (a²)/4 when a ≥ 2√2- 3 + a when -2 < a < 2√2- 0 when a ≤ -2Wait, but earlier I considered a < -3 separately. Hmm, perhaps I need to adjust.Wait, when a ≤ -2, in [0,1), the maximum is 0, and in [1,2], the maximum is 3 + a. So, if 3 + a > 0, then the maximum is 3 + a; otherwise, it's 0.So, 3 + a > 0 when a > -3. Therefore:- If -3 < a ≤ -2: maximum is 3 + a- If a ≤ -3: maximum is 0Therefore, combining all cases:- If a ≥ 2√2: max is 1 + a + (a²)/4- If -3 ≤ a < 2√2: max is 3 + a- If a < -3: max is 0Yes, that seems correct.But let me check a = -3:h(x) = |x² -1| + (-3)|x -1|In [0,1): h(x) = 1 - x² + (-3)(1 - x) = 1 - x² -3 + 3x = -x² + 3x - 2This is a downward opening parabola. Its maximum is at x = -b/(2a) = -3/(2*(-1)) = 1.5, which is outside [0,1). So, evaluate at endpoints:h(0) = -0 + 0 -2 = -2h(1-) = -1 + 3 -2 = 0In [1,2]: h(x) = x² -1 + (-3)(x -1) = x² -1 -3x +3 = x² -3x +2This is upward opening. Vertex at x = 3/2 = 1.5. Compute h(1.5):(2.25) - 4.5 + 2 = -0.25So, maximum at endpoints:h(1) = 1 -3 + 2 = 0h(2) = 4 -6 +2 = 0So, maximum is 0. Hence, for a = -3, the maximum is 0.Similarly, for a < -3, the maximum is 0.Therefore, the final answer is:The maximum value of h(x) on [0,2] is:- 1 + a + (a²)/4 when a ≥ 2√2- 3 + a when -3 ≤ a < 2√2- 0 when a < -3But the problem asks for the maximum value, so perhaps we can write it as a piecewise function.Alternatively, since the problem doesn't specify a particular a, maybe we need to express it in terms of a.But looking back, the problem is part (2), which is separate from part (1). So, perhaps in part (2), a is given as in part (1), but no, part (2) is a separate question.Wait, no, part (2) is just to find the maximum of h(x) on [0,2], given f(x) and g(x). So, a is a parameter, and we need to find the maximum in terms of a.Therefore, the answer is as above.But let me see if I can simplify it further.Note that 1 + a + (a²)/4 can be written as (a² + 4a + 4)/4 = (a + 2)² / 4.So, 1 + a + (a²)/4 = (a + 2)² / 4.Therefore, the maximum is:- (a + 2)² / 4 when a ≥ 2√2- 3 + a when -3 ≤ a < 2√2- 0 when a < -3But 2√2 is approximately 2.828, so it's a specific threshold.Alternatively, perhaps we can write it as:max{ (a + 2)² / 4, 3 + a, 0 }But with the conditions on a.Alternatively, since (a + 2)² / 4 is greater than 3 + a when a ≥ 2√2, and less otherwise.But perhaps the answer expects a specific value, but since a is a parameter, it's likely expressed in terms of a.Wait, but the problem didn't specify a particular a, so maybe I need to find the maximum over all a? No, that doesn't make sense.Wait, no, the function h(x) is defined as |f(x)| + g(x), with a being a parameter. So, for each a, h(x) is a function on [0,2], and we need to find its maximum.Therefore, the answer is as above, expressed in terms of a.But perhaps the problem expects a numerical answer, but since a is a variable, it's piecewise.Alternatively, maybe I made a mistake in the analysis. Let me think again.Wait, in the interval [0,1), for a > -2, the maximum is at x = -a/2, which is 1 + a + (a²)/4. For a ≤ -2, the maximum is 0.In [1,2], the maximum is 3 + a.So, the overall maximum is the maximum of these two.Therefore, the maximum is:- If 1 + a + (a²)/4 ≥ 3 + a, then 1 + a + (a²)/4- Else, 3 + aBut 1 + a + (a²)/4 ≥ 3 + a simplifies to (a²)/4 ≥ 2, so a² ≥ 8, which is a ≥ 2√2 or a ≤ -2√2.But in the context of the problem, a can be any real number.But in the interval [0,1), for a ≤ -2, the maximum is 0, and in [1,2], the maximum is 3 + a.So, when a ≤ -2, we have:- If 3 + a ≥ 0, i.e., a ≥ -3, then the maximum is 3 + a- If 3 + a < 0, i.e., a < -3, then the maximum is 0Therefore, combining all:- If a ≥ 2√2, max is (a + 2)² / 4- If -3 ≤ a < 2√2, max is 3 + a- If a < -3, max is 0Yes, that seems correct.But let me check for a = 0:h(x) = |x² -1| + 0 = |x² -1|On [0,2], the maximum is at x=2: |4 -1| = 3, which is 3 + 0 = 3, which matches.For a = 4 (which is ≥ 2√2):Compute (4 + 2)² /4 = 36/4 = 9Compute h(2) = 3 + 4 = 7But wait, h(x) at x = -a/2 = -4/2 = -2, which is outside [0,1). So, in [0,1), the maximum is at x=0: 1 + 0 + 0 =1, and at x=1, it's 0. But in [1,2], the maximum is 7. However, according to our earlier analysis, the maximum should be 9. Wait, that's a contradiction.Wait, no, when a =4, which is ≥ 2√2, the maximum should be (a + 2)² /4 = 36/4=9. But h(x) at x = -a/2 = -2, which is outside [0,1). So, in [0,1), the maximum is at x=0: h(0)=1 +4=5, and in [1,2], the maximum is 3 +4=7. So, the overall maximum is 7, not 9.Wait, that contradicts our earlier conclusion. So, perhaps my earlier analysis was wrong.Wait, I think I made a mistake in assuming that for a > -2, the maximum in [0,1) is at x = -a/2, but when a is large positive, x = -a/2 is negative, which is outside [0,1). Therefore, in that case, the maximum in [0,1) is at x=0.Wait, that's a crucial point I missed.So, let's correct that.In [0,1), for a > -2, the vertex is at x = -a/2. But if -a/2 < 0, i.e., a > 0, then the maximum in [0,1) is at x=0.Similarly, if -a/2 ∈ [0,1), i.e., a ∈ (-2, 0], then the maximum is at x = -a/2.If a >0, then x = -a/2 <0, so the maximum in [0,1) is at x=0.Therefore, we need to split the analysis further.So, let's correct the earlier analysis.**Revised Analysis:**For [0,1):h(x) =1 -x² +a(1 -x)This is a downward opening parabola.Vertex at x = -a/2.Case 1: a >0Then, x = -a/2 <0, so outside [0,1). Therefore, maximum in [0,1) is at x=0: h(0)=1 +aCase 2: -2 <a ≤0Then, x = -a/2 ∈ [0,1). Therefore, maximum at x = -a/2: h(-a/2)=1 +a + (a²)/4Case 3: a ≤-2Maximum in [0,1) is 0 (as before)For [1,2], maximum is 3 +aTherefore, overall maximum is:- If a >0: compare h(0)=1 +a and h(2)=3 +a. Since 3 +a >1 +a, maximum is 3 +a- If -2 <a ≤0: compare h(-a/2)=1 +a + (a²)/4 and h(2)=3 +a- If a ≤-2: compare 0 and 3 +aSo, let's analyze:**Case 1: a >0**Maximum is 3 +a**Case 2: -2 <a ≤0**Compare 1 +a + (a²)/4 and 3 +aAs before, 1 +a + (a²)/4 - (3 +a)= (a²)/4 -2Set this ≥0: a² ≥8, which is |a| ≥2√2≈2.828But in this case, a ≤0, so a ≤-2√2. But our current case is -2 <a ≤0, so a² <4 <8. Therefore, (a²)/4 -2 <0, so 1 +a + (a²)/4 <3 +aTherefore, the maximum is 3 +a**Case 3: a ≤-2**Compare 0 and 3 +aIf 3 +a ≥0, i.e., a ≥-3, maximum is 3 +aIf 3 +a <0, i.e., a < -3, maximum is 0Therefore, putting it all together:- If a >0: maximum is 3 +a- If -3 ≤a ≤0: maximum is 3 +a- If a < -3: maximum is 0Wait, but what about a between 0 and 2√2? Earlier, I thought for a ≥2√2, the maximum is (a +2)² /4, but now I see that for a >0, the maximum is 3 +a, regardless of a.Wait, but when a is large positive, say a=10, then h(2)=3 +10=13, and h(0)=1 +10=11, so maximum is 13.But according to the earlier analysis, when a ≥2√2, the maximum should be (a +2)² /4, but in reality, it's 3 +a.Wait, so perhaps my initial assumption that for a > -2, the maximum in [0,1) is at x=-a/2 is only valid when -a/2 ∈ [0,1), i.e., when a ∈ (-2,0]. For a >0, the maximum in [0,1) is at x=0.Therefore, the maximum in [0,1) is:- For a >0: 1 +a- For -2 <a ≤0: 1 +a + (a²)/4- For a ≤-2: 0And in [1,2], the maximum is always 3 +a.Therefore, the overall maximum is:- For a >0: max{1 +a, 3 +a} = 3 +a- For -2 <a ≤0: max{1 +a + (a²)/4, 3 +a} = 3 +a (since 1 +a + (a²)/4 <3 +a)- For a ≤-2: max{0, 3 +a} = 3 +a if a ≥-3, else 0Therefore, the maximum value of h(x) on [0,2] is:- 3 +a when a ≥-3- 0 when a < -3Wait, that simplifies things. So, regardless of a, as long as a ≥-3, the maximum is 3 +a, and when a < -3, the maximum is 0.But let me verify with a=4:h(x)=|x² -1| +4|x -1|In [0,1): h(x)=1 -x² +4(1 -x)=1 -x² +4 -4x=5 -x² -4xThis is a downward opening parabola. Vertex at x = -b/(2a)=4/(2*(-1))=-2, outside [0,1). So, maximum at x=0: h(0)=5In [1,2]: h(x)=x² -1 +4(x -1)=x² -1 +4x -4=x² +4x -5This is upward opening. Vertex at x=-2, outside [1,2). So, maximum at x=2: h(2)=4 +8 -5=7Therefore, overall maximum is 7, which is 3 +4=7. Correct.For a= -4:h(x)=|x² -1| -4|x -1|In [0,1): h(x)=1 -x² -4(1 -x)=1 -x² -4 +4x= -x² +4x -3This is downward opening. Vertex at x=2, outside [0,1). So, evaluate at endpoints:h(0)= -0 +0 -3=-3h(1-)= -1 +4 -3=0In [1,2]: h(x)=x² -1 -4(x -1)=x² -1 -4x +4=x² -4x +3This is upward opening. Vertex at x=2, which is in [1,2]. Compute h(2)=4 -8 +3=-1So, maximum at endpoints:h(1)=1 -4 +3=0h(2)=-1Therefore, maximum is 0. Correct.For a= -2:h(x)=|x² -1| -2|x -1|In [0,1): h(x)=1 -x² -2(1 -x)=1 -x² -2 +2x= -x² +2x -1This is downward opening. Vertex at x=1, which is the endpoint. So, h(1-)= -1 +2 -1=0In [1,2]: h(x)=x² -1 -2(x -1)=x² -1 -2x +2=x² -2x +1This is upward opening. Vertex at x=1, h(1)=0At x=2: h(2)=4 -4 +1=1So, maximum is 1, which is 3 + (-2)=1. Correct.Therefore, the maximum value is:- 3 +a when a ≥-3- 0 when a < -3So, the answer is:The maximum value of h(x) on [0,2] is 3 +a when a ≥-3, and 0 when a < -3.But let me check a= -3:h(x)=|x² -1| -3|x -1|In [0,1): h(x)=1 -x² -3(1 -x)=1 -x² -3 +3x= -x² +3x -2This is downward opening. Vertex at x=3/2=1.5, outside [0,1). So, evaluate at endpoints:h(0)= -0 +0 -2=-2h(1-)= -1 +3 -2=0In [1,2]: h(x)=x² -1 -3(x -1)=x² -1 -3x +3=x² -3x +2This is upward opening. Vertex at x=1.5. Compute h(1.5)=2.25 -4.5 +2= -0.25So, maximum at endpoints:h(1)=1 -3 +2=0h(2)=4 -6 +2=0Therefore, maximum is 0, which is 3 + (-3)=0. Correct.Therefore, the final answer is:The maximum value of h(x) on [0,2] is:- 3 +a when a ≥-3- 0 when a < -3So, in boxed form:If a ≥ -3, the maximum is boxed{a + 3}If a < -3, the maximum is boxed{0}But since the problem asks for the maximum value, and a is a parameter, perhaps we can write it as a piecewise function. However, since the problem is in two parts, and part (1) is about a ≤-2, but part (2) is separate, so the answer is as above.But looking back, in part (1), we found that a ≤-2. So, in part (2), perhaps a is constrained by part (1). Wait, no, part (2) is a separate question, so a can be any real number.Therefore, the answer is:The maximum value of h(x) on [0,2] is:- boxed{a + 3} when a ≥ -3- boxed{0} when a < -3But since the problem asks for the maximum value, perhaps it's better to write it as a piecewise function. However, the system might expect a single boxed answer, but since it's conditional, perhaps we need to write both cases.But in the initial problem statement, part (2) is separate, so perhaps the answer is simply the maximum value, which depends on a. Therefore, the answer is:The maximum value is boxed{a + 3} when a ≥ -3, and boxed{0} when a < -3.But since the problem didn't specify to write it in terms of a, but just to find the maximum value, perhaps it's expecting a numerical answer, but since a is a parameter, it's expressed in terms of a.Alternatively, perhaps the problem expects the maximum over all a, which would be unbounded as a increases. But no, the problem is to find the maximum of h(x) on [0,2], which depends on a.Therefore, the answer is as above.

💡Okay, so I have this problem where I'm given a function f(x) = 2 + log base 3 of x, and x is in the interval [1, 9]. I need to find the maximum value of another function y, which is defined as [f(x)] squared plus f(x squared). Hmm, that sounds a bit complicated, but let me try to break it down step by step.First, I should probably write down what y is in terms of x. So, y = [f(x)]² + f(x²). Since f(x) is given as 2 + log₃x, let me substitute that into the equation. That would make y = [2 + log₃x]² + [2 + log₃(x²)]. Okay, that looks better.Now, I need to simplify this expression to make it easier to work with. Let's start by expanding [2 + log₃x]². Using algebra, I know that (a + b)² = a² + 2ab + b², so applying that here, I get:[2 + log₃x]² = 2² + 2*2*log₃x + (log₃x)² = 4 + 4 log₃x + (log₃x)².So now, y becomes:y = 4 + 4 log₃x + (log₃x)² + 2 + log₃(x²).Wait, I can simplify log₃(x²). I remember that log_b(a^n) = n log_b a, so log₃(x²) is 2 log₃x. So substituting that in, we have:y = 4 + 4 log₃x + (log₃x)² + 2 + 2 log₃x.Now, let's combine like terms. The constants are 4 and 2, which add up to 6. The terms with log₃x are 4 log₃x and 2 log₃x, which add up to 6 log₃x. So now, y simplifies to:y = (log₃x)² + 6 log₃x + 6.Hmm, that looks like a quadratic equation in terms of log₃x. Maybe I can make a substitution to make it easier. Let me let t = log₃x. Then, y becomes:y = t² + 6t + 6.Now, I need to find the maximum value of this quadratic function. But before I do that, I should figure out the domain of t. Since x is in [1, 9], and t = log₃x, I can find the range of t.When x = 1, log₃1 = 0, because 3^0 = 1. When x = 9, log₃9 = 2, because 3² = 9. So, t is in [0, 2].Wait a minute, but hold on. The original function y is defined as [f(x)]² + f(x²). So, x² must also be within the domain of f, which is [1, 9]. So, x² must be ≤ 9, which means x ≤ 3, because if x is 3, x² is 9. So, actually, the domain of x isn't [1, 9], but rather [1, 3], because x² has to be within [1, 9]. So, x is in [1, 3], which means t = log₃x is in [0, 1], since log₃3 = 1.Okay, so t is in [0, 1]. So, now I have y = t² + 6t + 6, where t is between 0 and 1. I need to find the maximum value of this quadratic function on the interval [0, 1].Quadratic functions have their extrema either at the vertex or at the endpoints of the interval. The vertex of a quadratic function y = at² + bt + c is at t = -b/(2a). In this case, a = 1, b = 6, so the vertex is at t = -6/(2*1) = -3. But wait, t is in [0, 1], so the vertex at t = -3 is outside of our interval. That means the maximum must occur at one of the endpoints, either t = 0 or t = 1.Let me evaluate y at t = 0 and t = 1.At t = 0:y = 0² + 6*0 + 6 = 0 + 0 + 6 = 6.At t = 1:y = 1² + 6*1 + 6 = 1 + 6 + 6 = 13.So, y is 6 at t = 0 and 13 at t = 1. Therefore, the maximum value of y on the interval [0, 1] is 13.Wait, let me double-check my steps to make sure I didn't make a mistake. I started by substituting f(x) into y, expanded it, simplified using logarithm properties, made a substitution to turn it into a quadratic, found the domain of t, and then evaluated the quadratic at the endpoints since the vertex was outside the interval. It seems correct.But just to be thorough, let me verify the substitution step. I had y = [2 + log₃x]² + [2 + log₃(x²)]. Expanding [2 + log₃x]² gives 4 + 4 log₃x + (log₃x)². Then, log₃(x²) is 2 log₃x, so adding 2 + 2 log₃x. Combining all terms, 4 + 2 is 6, 4 log₃x + 2 log₃x is 6 log₃x, and then (log₃x)². So, y = (log₃x)² + 6 log₃x + 6. Yes, that seems right.Then, substituting t = log₃x, so y = t² + 6t + 6. The domain of x is [1, 3], so log₃x is [0, 1]. So t is in [0, 1]. Quadratic function with a positive coefficient on t² opens upwards, so the minimum is at the vertex, but since the vertex is at t = -3, which is outside our interval, the maximum occurs at the right endpoint, t = 1.Calculating y at t = 1: 1 + 6 + 6 = 13. Yep, that seems correct.Just to make sure, let me plug x = 3 back into the original y expression to verify.f(3) = 2 + log₃3 = 2 + 1 = 3.So, [f(3)]² = 3² = 9.f(3²) = f(9) = 2 + log₃9 = 2 + 2 = 4.So, y = 9 + 4 = 13. Perfect, that matches.And if I plug x = 1 into the original y:f(1) = 2 + log₃1 = 2 + 0 = 2.[f(1)]² = 4.f(1²) = f(1) = 2.So, y = 4 + 2 = 6, which is the minimum value.Therefore, my conclusion is that the maximum value of y is 13, achieved when x = 3.**Final Answer**The maximum value of the function is boxed{13}.

💡Alright, so I have this problem here where I need to find the ratio p over q, given that when I multiply (3 - 4i) by (p + qi), the result is a pure imaginary number. Hmm, okay, let me break this down step by step.First, I remember that a pure imaginary number has no real part. That means if I expand the product (3 - 4i)(p + qi), the real parts should cancel out, leaving only the imaginary part. So, my goal is to make sure that the real part of the product is zero.Let me write out the multiplication. When I multiply two complex numbers, I use the distributive property, also known as the FOIL method for binomials. So, let's do that:(3 - 4i)(p + qi) = 3*p + 3*qi - 4i*p - 4i*qi.Simplifying each term:- The first term is 3p, which is real.- The second term is 3qi, which is imaginary.- The third term is -4pi, which is also imaginary.- The fourth term is -4qi^2. Wait, I remember that i squared is equal to -1, so this term becomes -4q*(-1) = 4q, which is real.So, putting it all together, the product is:3p + 4q + (3q - 4p)i.Now, since the result is supposed to be a pure imaginary number, the real part must be zero. That means 3p + 4q = 0. The imaginary part can be anything, but the real part has to vanish.So, I have the equation:3p + 4q = 0.I need to find the ratio p/q. Let me solve for p in terms of q:3p = -4qDivide both sides by 3:p = (-4/3)qSo, p over q is (-4/3). Therefore, the ratio p/q is -4/3.Wait, let me double-check my steps to make sure I didn't make a mistake. I expanded the product correctly, right? 3p, 3qi, -4pi, and -4qi^2. Then, since i^2 is -1, that term becomes +4q. So, combining the real parts: 3p + 4q, and the imaginary parts: (3q - 4p)i. Yeah, that seems right.Setting the real part equal to zero: 3p + 4q = 0. Solving for p/q, I get p/q = -4/3. Hmm, that seems straightforward. I don't think I missed anything here.Just to be thorough, let me plug this ratio back into the original expression to see if it indeed results in a pure imaginary number. Suppose p = (-4/3)q. Then, let's substitute:(3 - 4i)( (-4/3)q + qi )First, factor out q:q*(3 - 4i)*(-4/3 + i)Let me compute (3 - 4i)*(-4/3 + i):Multiply 3 by (-4/3): 3*(-4/3) = -4.Multiply 3 by i: 3i.Multiply -4i by (-4/3): (-4i)*(-4/3) = (16/3)i.Multiply -4i by i: -4i^2 = -4*(-1) = 4.So, adding all these together:-4 + 3i + (16/3)i + 4.Combine like terms:Real parts: -4 + 4 = 0.Imaginary parts: 3i + (16/3)i = (9/3 + 16/3)i = (25/3)i.So, the product is 0 + (25/3)i, which is indeed a pure imaginary number. That checks out.Therefore, my solution seems correct. The ratio p/q is -4/3.**Final Answer**The ratio (frac{p}{q}) is boxed{-dfrac{4}{3}}.

💡First, I need to calculate the cost of each item Sophie buys.For the cupcakes, she buys 5 cupcakes at 2 each. So, 5 multiplied by 2 equals 10.Next, for the doughnuts, she buys 6 doughnuts at 1 each. Multiplying 6 by 1 gives 6.Then, for the apple pie slices, she buys 4 slices at 2 each. This results in 4 times 2, which is 8.Finally, for the cookies, she buys 15 cookies at 0.60 each. Multiplying 15 by 0.60 equals 9.After calculating the cost of each item, I add them all together: 10 plus 6 plus 8 plus 9 equals 33.Therefore, Sophie spends a total of 33.