💡First, determine the work rates of Rahul and Rajesh. Rahul can complete the work in 3 days, so his work rate is ( frac{1}{3} ) of the work per day. Rajesh can complete the work in 2 days, so his work rate is ( frac{1}{2} ) of the work per day.Next, calculate the combined work rate when they work together:[frac{1}{3} + frac{1}{2} = frac{2}{6} + frac{3}{6} = frac{5}{6}]This means together they can complete ( frac{5}{6} ) of the work in one day.To find out how many days they take to complete the work together, take the reciprocal of the combined work rate:[frac{6}{5} text{ days}]Determine the ratio of their contributions to the work:[frac{1}{3} : frac{1}{2} = 2 : 3]This means Rahul contributes 2 parts and Rajesh contributes 3 parts of the work.Given Rahul's share is 900, which corresponds to 2 parts, calculate the value of one part:[frac{900}{2} = 450]Finally, calculate the total payment by summing up all parts:[5 times 450 = 2250]

💡Okay, so I have this problem about the annual incomes of 1,500 families. The incomes range from 8,200 to 98,000. But there was a mistake where the largest income was entered as 1,480,000 instead of the actual 98,000. I need to find the difference between the mean of the incorrect data and the mean of the actual data. The options are given, so I need to figure out which one is correct.First, let me understand what the problem is asking. There are 1,500 families, and their incomes are recorded. One of these incomes was incorrectly entered as a much larger number, which is 1,480,000 instead of 98,000. So, the incorrect data set has this one very high income, while the actual data set has the correct highest income of 98,000. I need to find how much the mean (average) changes because of this error.To find the difference in means, I think I can approach it by calculating the mean with the incorrect income and then the mean with the correct income, and then subtract the two to find the difference.Let me denote the sum of all the incomes except the largest one as S. So, in the actual data, the total sum would be S + 98,000, and in the incorrect data, it would be S + 1,480,000. Then, the mean of the actual data would be (S + 98,000)/1,500, and the mean of the incorrect data would be (S + 1,480,000)/1,500.So, the difference between the incorrect mean and the actual mean would be:[(S + 1,480,000)/1,500] - [(S + 98,000)/1,500]Since both terms have the same denominator, I can combine them:[(S + 1,480,000) - (S + 98,000)] / 1,500Simplifying the numerator:S + 1,480,000 - S - 98,000 = 1,480,000 - 98,000 = 1,382,000So, the difference is 1,382,000 / 1,500Now, I need to compute this division. Let me do that step by step.First, divide 1,382,000 by 1,500.I can simplify this by dividing both numerator and denominator by 100 to make it easier:1,382,000 ÷ 100 = 13,8201,500 ÷ 100 = 15So now, the problem becomes 13,820 ÷ 15Let me compute this division.15 × 900 = 13,500Subtracting that from 13,820: 13,820 - 13,500 = 320Now, 15 × 21 = 315Subtracting that from 320: 320 - 315 = 5So, we have 900 + 21 = 921, and a remainder of 5.So, 13,820 ÷ 15 = 921 with a remainder of 5, which is 921 and 5/15, which simplifies to 921 and 1/3.So, the difference is 921 and 1/3 dollars.Looking at the options, B is 921.overline{3} dollars, which is the same as 921 and 1/3.So, the answer should be B.Wait, let me double-check my calculations to make sure I didn't make a mistake.First, the difference in the total sum is 1,480,000 - 98,000 = 1,382,000. That seems correct.Then, dividing 1,382,000 by 1,500. I simplified it to 13,820 ÷ 15, which is correct.15 × 900 = 13,500. Subtracting from 13,820 gives 320. Then, 15 × 21 = 315. Subtracting gives 5. So, total is 921 and 1/3. That seems correct.Yes, I think that's right. So, the difference is 921 and 1/3 dollars, which is option B.

💡Okay, so I have this problem about the function ( f(x) = frac{x^2}{a} - 2ln x ) where ( a ) is a real number and not zero. There are two parts: first, I need to discuss the monotonicity of ( f(x) ), which means figuring out where it's increasing or decreasing. Second, I have to find the range of ( m ) such that the equation involving the minimum value of ( f(x) ) has three different real roots in terms of ( a ). Starting with part (1). Monotonicity involves looking at the derivative of the function. So I should compute ( f'(x) ). Let's do that step by step.The function is ( f(x) = frac{x^2}{a} - 2ln x ). The derivative of ( frac{x^2}{a} ) with respect to ( x ) is ( frac{2x}{a} ). The derivative of ( -2ln x ) is ( -frac{2}{x} ). So putting it together, the derivative ( f'(x) ) is:( f'(x) = frac{2x}{a} - frac{2}{x} )Simplify this expression. Let's factor out the 2:( f'(x) = 2left( frac{x}{a} - frac{1}{x} right) )To analyze the sign of ( f'(x) ), which determines where the function is increasing or decreasing, I need to look at ( frac{x}{a} - frac{1}{x} ). Let's combine these terms over a common denominator:( frac{x^2 - a}{a x} )So, ( f'(x) = 2 cdot frac{x^2 - a}{a x} ). The sign of ( f'(x) ) depends on the numerator ( x^2 - a ) and the denominator ( a x ). Since ( x > 0 ) (because of the ( ln x ) term, which is only defined for positive ( x )), the denominator ( a x ) is positive if ( a > 0 ) and negative if ( a < 0 ).Case 1: ( a > 0 )In this case, the denominator ( a x ) is positive. So the sign of ( f'(x) ) is determined by ( x^2 - a ). - When ( x^2 - a > 0 ), which is when ( x > sqrt{a} ), ( f'(x) > 0 ), so the function is increasing.- When ( x^2 - a < 0 ), which is when ( x < sqrt{a} ), ( f'(x) < 0 ), so the function is decreasing.Therefore, for ( a > 0 ), ( f(x) ) is decreasing on ( (0, sqrt{a}) ) and increasing on ( (sqrt{a}, +infty) ).Case 2: ( a < 0 )Here, the denominator ( a x ) is negative because ( a ) is negative and ( x ) is positive. So the sign of ( f'(x) ) is the opposite of the sign of ( x^2 - a ).But ( x^2 ) is always positive, and ( a ) is negative, so ( x^2 - a ) is always positive because subtracting a negative is adding. Therefore, ( x^2 - a > 0 ) for all ( x > 0 ).Since the numerator is positive and the denominator is negative, ( f'(x) ) is negative for all ( x > 0 ). So the function is decreasing on its entire domain ( (0, +infty) ).So summarizing part (1):- If ( a > 0 ), ( f(x) ) decreases on ( (0, sqrt{a}) ) and increases on ( (sqrt{a}, +infty) ).- If ( a < 0 ), ( f(x) ) is decreasing on ( (0, +infty) ).Moving on to part (2). It says that if ( f(x) ) has a minimum value, denoted as ( g(a) ), and the equation ( g(a) + a - frac{2}{9a} - 1 = m ) has three different real roots with respect to ( a ), find the range of ( m ).First, from part (1), we know that ( f(x) ) has a minimum only when ( a > 0 ) because when ( a < 0 ), the function is always decreasing and doesn't have a minimum. So ( a > 0 ) is necessary for ( g(a) ) to exist.So, let's find the minimum value ( g(a) ). From part (1), the function has a critical point at ( x = sqrt{a} ). Let's compute ( f(sqrt{a}) ):( f(sqrt{a}) = frac{(sqrt{a})^2}{a} - 2ln(sqrt{a}) )Simplify:( f(sqrt{a}) = frac{a}{a} - 2 cdot frac{1}{2} ln a = 1 - ln a )So, ( g(a) = 1 - ln a ).Now, substitute ( g(a) ) into the given equation:( g(a) + a - frac{2}{9a} - 1 = m )Plugging in ( g(a) = 1 - ln a ):( (1 - ln a) + a - frac{2}{9a} - 1 = m )Simplify:( 1 - ln a + a - frac{2}{9a} - 1 = m )The 1 and -1 cancel out:( a - ln a - frac{2}{9a} = m )So, the equation becomes:( m = a - ln a - frac{2}{9a} )We need to find the range of ( m ) such that this equation has three different real roots for ( a > 0 ). To analyze this, let's define a function ( F(a) = a - ln a - frac{2}{9a} ) for ( a > 0 ). We need to find the values of ( m ) for which the horizontal line ( y = m ) intersects the graph of ( F(a) ) three times. To do this, we should analyze the behavior of ( F(a) ). Specifically, we need to find its critical points by taking its derivative and determining where it's increasing or decreasing. Then, we can find its local maxima and minima, which will help us determine the range of ( m ) for which the equation has three real roots.Compute the derivative ( F'(a) ):( F(a) = a - ln a - frac{2}{9a} )Differentiate term by term:- The derivative of ( a ) is 1.- The derivative of ( -ln a ) is ( -frac{1}{a} ).- The derivative of ( -frac{2}{9a} ) is ( frac{2}{9a^2} ).So,( F'(a) = 1 - frac{1}{a} + frac{2}{9a^2} )Let me write that as:( F'(a) = 1 - frac{1}{a} + frac{2}{9a^2} )To find critical points, set ( F'(a) = 0 ):( 1 - frac{1}{a} + frac{2}{9a^2} = 0 )Multiply both sides by ( 9a^2 ) to eliminate denominators:( 9a^2 - 9a + 2 = 0 )Now, solve this quadratic equation for ( a ):Quadratic equation: ( 9a^2 - 9a + 2 = 0 )Using the quadratic formula:( a = frac{9 pm sqrt{81 - 72}}{18} = frac{9 pm sqrt{9}}{18} = frac{9 pm 3}{18} )So,( a = frac{9 + 3}{18} = frac{12}{18} = frac{2}{3} )and( a = frac{9 - 3}{18} = frac{6}{18} = frac{1}{3} )So, the critical points are at ( a = frac{1}{3} ) and ( a = frac{2}{3} ).Now, let's analyze the intervals where ( F(a) ) is increasing or decreasing.We can test the sign of ( F'(a) ) in the intervals determined by ( a = frac{1}{3} ) and ( a = frac{2}{3} ).Let me rewrite ( F'(a) ):( F'(a) = 1 - frac{1}{a} + frac{2}{9a^2} )Alternatively, we can factor ( F'(a) ). Let me see:Let me write ( F'(a) ) as:( F'(a) = frac{9a^2 - 9a + 2}{9a^2} )Since the numerator is ( 9a^2 - 9a + 2 ), which factors as ( (3a - 1)(3a - 2) ). Let me check:( (3a - 1)(3a - 2) = 9a^2 - 6a - 3a + 2 = 9a^2 - 9a + 2 ). Yes, correct.So,( F'(a) = frac{(3a - 1)(3a - 2)}{9a^2} )Therefore, the sign of ( F'(a) ) depends on the numerator ( (3a - 1)(3a - 2) ) because the denominator ( 9a^2 ) is always positive for ( a > 0 ).So, the critical points are at ( a = frac{1}{3} ) and ( a = frac{2}{3} ). Let's analyze the sign of ( F'(a) ) in the intervals:1. ( 0 < a < frac{1}{3} )2. ( frac{1}{3} < a < frac{2}{3} )3. ( a > frac{2}{3} )For ( 0 < a < frac{1}{3} ):- ( 3a - 1 < 0 ) (since ( 3a < 1 ))- ( 3a - 2 < 0 ) (since ( 3a < 2 ))- So, the product ( (3a - 1)(3a - 2) ) is positive (negative times negative)- Therefore, ( F'(a) > 0 ): ( F(a) ) is increasing.For ( frac{1}{3} < a < frac{2}{3} ):- ( 3a - 1 > 0 ) (since ( 3a > 1 ))- ( 3a - 2 < 0 ) (since ( 3a < 2 ))- So, the product ( (3a - 1)(3a - 2) ) is negative (positive times negative)- Therefore, ( F'(a) < 0 ): ( F(a) ) is decreasing.For ( a > frac{2}{3} ):- ( 3a - 1 > 0 )- ( 3a - 2 > 0 )- So, the product is positive- Therefore, ( F'(a) > 0 ): ( F(a) ) is increasing.So, summarizing:- ( F(a) ) is increasing on ( (0, frac{1}{3}) )- Decreasing on ( (frac{1}{3}, frac{2}{3}) )- Increasing again on ( (frac{2}{3}, +infty) )This means that ( F(a) ) has a local maximum at ( a = frac{1}{3} ) and a local minimum at ( a = frac{2}{3} ).To find the range of ( m ) such that ( F(a) = m ) has three real roots, we need to consider the behavior of ( F(a) ). Since ( F(a) ) is increasing, then decreasing, then increasing, it will have a "hill" and a "valley". For certain values of ( m ), the horizontal line ( y = m ) will intersect the graph of ( F(a) ) three times: once on ( (0, frac{1}{3}) ), once on ( (frac{1}{3}, frac{2}{3}) ), and once on ( (frac{2}{3}, +infty) ).But for this to happen, ( m ) must be between the local minimum and the local maximum. If ( m ) is equal to the local maximum or the local minimum, the equation will have fewer roots. So, the range of ( m ) is between the local minimum value and the local maximum value.Therefore, we need to compute ( F(frac{1}{3}) ) and ( F(frac{2}{3}) ).Compute ( F(frac{1}{3}) ):( Fleft( frac{1}{3} right) = frac{1}{3} - ln left( frac{1}{3} right) - frac{2}{9 cdot frac{1}{3}} )Simplify each term:- ( frac{1}{3} ) is straightforward.- ( ln left( frac{1}{3} right) = -ln 3 )- ( frac{2}{9 cdot frac{1}{3}} = frac{2}{3} )So,( Fleft( frac{1}{3} right) = frac{1}{3} - (-ln 3) - frac{2}{3} = frac{1}{3} + ln 3 - frac{2}{3} = -frac{1}{3} + ln 3 )Compute ( F(frac{2}{3}) ):( Fleft( frac{2}{3} right) = frac{2}{3} - ln left( frac{2}{3} right) - frac{2}{9 cdot frac{2}{3}} )Simplify each term:- ( frac{2}{3} ) is straightforward.- ( ln left( frac{2}{3} right) = ln 2 - ln 3 )- ( frac{2}{9 cdot frac{2}{3}} = frac{2}{6} = frac{1}{3} )So,( Fleft( frac{2}{3} right) = frac{2}{3} - (ln 2 - ln 3) - frac{1}{3} = frac{2}{3} - ln 2 + ln 3 - frac{1}{3} = frac{1}{3} - ln 2 + ln 3 )So, the local maximum is ( Fleft( frac{1}{3} right) = -frac{1}{3} + ln 3 ) and the local minimum is ( Fleft( frac{2}{3} right) = frac{1}{3} - ln 2 + ln 3 ).Therefore, for the equation ( F(a) = m ) to have three different real roots, ( m ) must lie strictly between these two values. So, the range of ( m ) is:( frac{1}{3} - ln 2 + ln 3 < m < -frac{1}{3} + ln 3 )To make sure, let me compute these values numerically to get an idea.Compute ( ln 3 approx 1.0986 ), ( ln 2 approx 0.6931 ).Compute ( Fleft( frac{1}{3} right) = -frac{1}{3} + ln 3 approx -0.3333 + 1.0986 approx 0.7653 )Compute ( Fleft( frac{2}{3} right) = frac{1}{3} - ln 2 + ln 3 approx 0.3333 - 0.6931 + 1.0986 approx 0.3333 + 0.4055 approx 0.7388 )Wait, that's interesting. So, ( Fleft( frac{2}{3} right) approx 0.7388 ) and ( Fleft( frac{1}{3} right) approx 0.7653 ). So, actually, ( Fleft( frac{2}{3} right) < Fleft( frac{1}{3} right) ). That makes sense because ( F(a) ) is increasing up to ( a = frac{1}{3} ), then decreasing to ( a = frac{2}{3} ), then increasing again. So, the local maximum is at ( a = frac{1}{3} ) and the local minimum is at ( a = frac{2}{3} ).Therefore, the range of ( m ) is between approximately 0.7388 and 0.7653. But in exact terms, it's ( frac{1}{3} - ln 2 + ln 3 < m < -frac{1}{3} + ln 3 ).So, the answer should be expressed in terms of logarithms, not numerically. Therefore, the range is:( frac{1}{3} - ln 2 + ln 3 < m < -frac{1}{3} + ln 3 )I think that's the final answer.**Final Answer**The range of ( m ) is boxed{left( frac{1}{3} - ln 2 + ln 3, -frac{1}{3} + ln 3 right)}.

💡Okay, so I have this problem with vectors in the plane. Let me try to figure it out step by step. First, the vectors given are:- Vector a = (3, 2)- Vector b = (-1, 2)- Vector c = (4, 1)There are two parts to this problem. Let me tackle them one by one.**Problem (1):** If (a + k*c) is perpendicular to (2b - a), find the value of the real number k.Alright, so I remember that two vectors are perpendicular if their dot product is zero. So, I need to compute the vectors (a + k*c) and (2b - a), then set their dot product equal to zero and solve for k.Let me compute each vector first.Starting with (a + k*c):- Vector a is (3, 2), and vector c is (4, 1). So, multiplying c by k gives (4k, k). Adding this to a: - x-component: 3 + 4k - y-component: 2 + kSo, (a + k*c) = (3 + 4k, 2 + k)Next, compute (2b - a):- Vector b is (-1, 2). Multiplying by 2 gives (-2, 4). Now subtract vector a, which is (3, 2): - x-component: -2 - 3 = -5 - y-component: 4 - 2 = 2So, (2b - a) = (-5, 2)Now, since these two vectors are perpendicular, their dot product should be zero.Dot product formula is:(a + k*c) · (2b - a) = (3 + 4k)*(-5) + (2 + k)*(2) = 0Let me compute each term:- First term: (3 + 4k)*(-5) = -15 - 20k- Second term: (2 + k)*(2) = 4 + 2kAdding both terms together:-15 - 20k + 4 + 2k = 0Combine like terms:(-15 + 4) + (-20k + 2k) = 0-11 - 18k = 0Now, solve for k:-18k = 11k = 11 / (-18)k = -11/18Hmm, that seems straightforward. Let me double-check my calculations.Wait, in the dot product, I had (3 + 4k)*(-5) which is indeed -15 -20k. Then (2 + k)*2 is 4 + 2k. Adding them: -15 -20k +4 +2k = (-15 +4) + (-20k +2k) = -11 -18k. Setting equal to zero: -11 -18k = 0, so -18k = 11, k = -11/18. Yeah, that seems right.**Problem (2):** If vector d satisfies d is parallel to c, and |d| = sqrt(34), find the coordinates of vector d.Okay, so d is parallel to c. That means d is a scalar multiple of c. So, d = t*c for some scalar t.Given that |d| = sqrt(34). Let me recall that the magnitude of a vector (x, y) is sqrt(x² + y²). So, |d| = sqrt( (4t)² + (t)² ) = sqrt(16t² + t²) = sqrt(17t²) = |t|*sqrt(17).We are given that |d| = sqrt(34). So:|t|*sqrt(17) = sqrt(34)Solving for |t|:|t| = sqrt(34) / sqrt(17) = sqrt(34/17) = sqrt(2)So, t = sqrt(2) or t = -sqrt(2)Therefore, vector d can be:If t = sqrt(2):d = (4*sqrt(2), 1*sqrt(2)) = (4√2, √2)If t = -sqrt(2):d = (4*(-sqrt(2)), 1*(-sqrt(2))) = (-4√2, -√2)So, the coordinates of vector d are either (4√2, √2) or (-4√2, -√2).Let me verify if this makes sense. The magnitude of d should be sqrt(34). Let's compute |d| for t = sqrt(2):|d| = sqrt( (4√2)^2 + (√2)^2 ) = sqrt(16*2 + 2) = sqrt(32 + 2) = sqrt(34). Perfect.Similarly, for t = -sqrt(2), the magnitude is the same because squaring removes the negative. So, that checks out.Wait, but is there another way to approach this? Maybe by setting up equations.Let me try that. Let vector d = (x, y). Since d is parallel to c, which is (4,1), the direction ratios must be proportional. So, x/4 = y/1, which implies x = 4y.Also, |d| = sqrt(34), so sqrt(x² + y²) = sqrt(34). Squaring both sides: x² + y² = 34.Substituting x = 4y into the equation:(4y)^2 + y^2 = 3416y² + y² = 3417y² = 34y² = 2y = sqrt(2) or y = -sqrt(2)Therefore, x = 4y = 4*sqrt(2) or x = -4*sqrt(2)So, vector d is (4√2, √2) or (-4√2, -√2). Yep, same result as before. So, that's consistent.I think that's solid. I don't see any mistakes in my reasoning.**Final Answer**(1) The value of ( k ) is (boxed{-dfrac{11}{18}}).(2) The coordinates of vector ( overrightarrow{d} ) are (boxed{(4sqrt{2}, sqrt{2})}) or (boxed{(-4sqrt{2}, -sqrt{2})}).

💡Alright, so I have this problem with two functions, f(x) and g(x), and I need to find the range of values for m such that there exist x1 and x2 in the interval [0, π/4] where f(x1) equals g(x2). Hmm, okay, let's break this down step by step.First, let me write down the functions again to make sure I have them right:f(x) = sin(2x) + 2√3 cos²x - √3g(x) = m cos(2x - π/6) - 2m + 3, where m > 0.Alright, so I need to find m such that there are some x1 and x2 in [0, π/4] where f(x1) = g(x2). That means the ranges of f and g must overlap somewhere in that interval. So, I think I need to find the ranges of f(x) and g(x) over [0, π/4] and then find the m that makes their ranges intersect.Let me start with f(x). It looks a bit complicated, but maybe I can simplify it. I see sin(2x) and cos²x. Hmm, I remember that cos²x can be written using the double-angle identity: cos²x = (1 + cos(2x))/2. Let me try that.So, f(x) = sin(2x) + 2√3 * (1 + cos(2x))/2 - √3Simplify that:= sin(2x) + √3*(1 + cos(2x)) - √3= sin(2x) + √3 + √3 cos(2x) - √3The √3 and -√3 cancel out, so we have:f(x) = sin(2x) + √3 cos(2x)Hmm, that's simpler. Now, this looks like a sinusoidal function. Maybe I can write it in the form A sin(2x + φ) or something like that. Let me recall the formula: a sinθ + b cosθ = R sin(θ + φ), where R = √(a² + b²) and tanφ = b/a.Here, a is 1 (coefficient of sin(2x)) and b is √3 (coefficient of cos(2x)). So, R = √(1 + 3) = √4 = 2. And tanφ = √3/1 = √3, so φ = π/3.Therefore, f(x) can be rewritten as:f(x) = 2 sin(2x + π/3)Okay, that's much cleaner. Now, let's find the range of f(x) when x is in [0, π/4]. So, let's find the range of 2 sin(2x + π/3) over x ∈ [0, π/4].First, let's find the range of the argument inside the sine function: 2x + π/3.When x = 0: 2*0 + π/3 = π/3When x = π/4: 2*(π/4) + π/3 = π/2 + π/3 = (3π + 2π)/6 = 5π/6So, the argument 2x + π/3 ranges from π/3 to 5π/6 as x goes from 0 to π/4.Now, let's recall the sine function between π/3 and 5π/6.At π/3, sin(π/3) = √3/2 ≈ 0.866At π/2, sin(π/2) = 1At 5π/6, sin(5π/6) = 1/2So, the sine function increases from π/3 to π/2 and then decreases from π/2 to 5π/6.Therefore, the maximum value of sin(2x + π/3) is 1, and the minimum value is 1/2.Thus, f(x) = 2 sin(2x + π/3) ranges from 2*(1/2) = 1 to 2*1 = 2.So, the range of f(x) over [0, π/4] is [1, 2].Alright, that's f(x) done. Now, let's move on to g(x).g(x) = m cos(2x - π/6) - 2m + 3Hmm, again, this is a cosine function scaled and shifted. Let's see if we can find its range over x ∈ [0, π/4].First, let's analyze the argument of the cosine: 2x - π/6.When x = 0: 2*0 - π/6 = -π/6When x = π/4: 2*(π/4) - π/6 = π/2 - π/6 = (3π - π)/6 = 2π/6 = π/3So, the argument 2x - π/6 ranges from -π/6 to π/3 as x goes from 0 to π/4.Now, let's recall the cosine function over this interval.At -π/6, cos(-π/6) = cos(π/6) = √3/2 ≈ 0.866At 0, cos(0) = 1At π/3, cos(π/3) = 1/2So, the cosine function starts at √3/2 when x=0, increases to 1 at x=π/12 (since 2x - π/6 = 0 when x=π/12), and then decreases to 1/2 at x=π/4.Therefore, the maximum value of cos(2x - π/6) is 1, and the minimum is 1/2.Thus, the range of cos(2x - π/6) over [0, π/4] is [1/2, 1].Now, let's plug this back into g(x):g(x) = m * [cos(2x - π/6)] - 2m + 3So, since cos(2x - π/6) ∈ [1/2, 1], then:g(x) ∈ [m*(1/2) - 2m + 3, m*1 - 2m + 3]Simplify these:Lower bound: (m/2) - 2m + 3 = (-3m/2) + 3Upper bound: m - 2m + 3 = (-m) + 3So, the range of g(x) is [3 - (3m/2), 3 - m]Alright, so f(x) has a range of [1, 2], and g(x) has a range of [3 - (3m/2), 3 - m].We need these two ranges to overlap, meaning there must be some value y such that y ∈ [1, 2] and y ∈ [3 - (3m/2), 3 - m].So, for the ranges to overlap, the lower bound of g(x) must be less than or equal to the upper bound of f(x), and the upper bound of g(x) must be greater than or equal to the lower bound of f(x).In other words:3 - (3m/2) ≤ 2 ...(1)and3 - m ≥ 1 ...(2)Let me solve these inequalities one by one.Starting with inequality (1):3 - (3m/2) ≤ 2Subtract 3 from both sides:- (3m/2) ≤ -1Multiply both sides by (-1), remembering to reverse the inequality:3m/2 ≥ 1Multiply both sides by (2/3):m ≥ 2/3Okay, so m must be at least 2/3.Now, inequality (2):3 - m ≥ 1Subtract 3 from both sides:- m ≥ -2Multiply both sides by (-1), reversing the inequality:m ≤ 2So, m must be at most 2.Putting these together, m must satisfy both m ≥ 2/3 and m ≤ 2. Therefore, m ∈ [2/3, 2].Wait, but let me double-check to make sure I didn't make a mistake.First, for inequality (1):3 - (3m/2) ≤ 2Subtract 3: -3m/2 ≤ -1Multiply by (-1): 3m/2 ≥ 1Multiply by 2/3: m ≥ 2/3That seems correct.For inequality (2):3 - m ≥ 1Subtract 3: -m ≥ -2Multiply by (-1): m ≤ 2That also seems correct.So, combining them, m must be between 2/3 and 2, inclusive.Looking back at the options:A: (0,1]B: [1,2]C: [2/3,2]D: [2/3,4/3]So, the correct answer should be C: [2/3,2].But wait, let me think again. Is there a possibility that the ranges could overlap even if one is entirely within the other? For example, if the range of g(x) is entirely within [1,2], then they would still overlap. So, perhaps I should also check if 3 - (3m/2) ≤ 2 and 3 - m ≥ 1, but also consider if the entire range of g(x) is within [1,2].Wait, but if the lower bound of g(x) is greater than 1 and the upper bound is less than 2, that would also satisfy the condition. So, perhaps I need to ensure that either the ranges overlap or one is entirely within the other.But in this case, since f(x) is [1,2], and g(x) is [3 - 3m/2, 3 - m], we need to make sure that [3 - 3m/2, 3 - m] overlaps with [1,2].So, to have overlap, the following must hold:3 - 3m/2 ≤ 2 and 3 - m ≥ 1Which is exactly what I did earlier, leading to m ∈ [2/3, 2].Alternatively, if 3 - 3m/2 ≤ 1 and 3 - m ≥ 2, but that would require:3 - 3m/2 ≤ 1 => -3m/2 ≤ -2 => m ≥ 4/3and3 - m ≥ 2 => -m ≥ -1 => m ≤ 1But m cannot be both ≥4/3 and ≤1, so that case is impossible.Therefore, the only possible overlap is when 3 - 3m/2 ≤ 2 and 3 - m ≥ 1, which gives m ∈ [2/3,2].So, I think my initial conclusion was correct.Therefore, the range of m is [2/3,2], which is option C.**Final Answer**boxed{C}

💡Okay, so I have this problem here about averages, and I need to find the value of a - c. Let me try to understand what's being asked and figure out how to approach it step by step.First, the problem says that the average of a, b, and d is 110. I know that the average of three numbers is the sum of those numbers divided by three. So, if I let S1 be the sum of a, b, and d, then S1 divided by 3 equals 110. That means S1, the sum of a, b, and d, is 110 multiplied by 3, which is 330. So, a + b + d = 330. Got that down.Next, the problem also says that the average of b, c, and d is 150. Using the same logic, if I let S2 be the sum of b, c, and d, then S2 divided by 3 equals 150. So, S2, which is b + c + d, is 150 multiplied by 3, which is 450. Therefore, b + c + d = 450. Okay, so now I have two equations:1. a + b + d = 3302. b + c + d = 450I need to find a - c. Hmm, how can I get a - c from these two equations? Maybe if I subtract one equation from the other, I can eliminate some variables.Let me try subtracting the first equation from the second one. So, (b + c + d) - (a + b + d) = 450 - 330. Let's see what that gives me.On the left side, b - b cancels out, d - d cancels out, and I'm left with c - a. On the right side, 450 - 330 is 120. So, c - a = 120.But the problem asks for a - c, not c - a. Well, if c - a is 120, then a - c would just be the negative of that, right? So, a - c = -120.Wait, let me double-check that to make sure I didn't make a mistake. If I have c - a = 120, then multiplying both sides by -1 gives me a - c = -120. Yeah, that seems correct.Just to recap, I set up the two equations based on the averages given, subtracted one from the other to eliminate b and d, and then solved for a - c by taking the negative of the result. It all seems to add up.I think that's the answer. a - c equals -120.