answer：Alright, so I've got this problem about a curve C defined by the equation ( x^2 + y^2 - 2x - 4y + m = 0 ). There are two parts to the problem. The first part asks for the value of ( m ) such that the curve C represents a circle. The second part is a bit more involved: it says that if curve C intersects the line ( x + 2y - 3 = 0 ) at points M and N, and the vectors ( overrightarrow{OM} ) and ( overrightarrow{ON} ) are perpendicular, then we need to find the value of ( m ).Starting with the first part: determining when the given equation represents a circle. I remember that the general equation of a circle is ( x^2 + y^2 + Dx + Ey + F = 0 ), and for it to be a circle, the coefficients of ( x^2 ) and ( y^2 ) must be equal, which they are here (both are 1). Also, the radius must be a real number, which means that the expression ( D^2 + E^2 - 4F ) must be positive. In this case, ( D = -2 ), ( E = -4 ), and ( F = m ). So, plugging these into the formula for the radius squared, we get ( (-2)^2 + (-4)^2 - 4m = 4 + 16 - 4m = 20 - 4m ). For the radius to be real, ( 20 - 4m > 0 ), which simplifies to ( m < 5 ). So, the curve C is a circle when ( m ) is less than 5.Moving on to the second part: finding ( m ) such that the curve intersects the line ( x + 2y - 3 = 0 ) at points M and N, and the vectors from the origin to these points are perpendicular. This means that the dot product of vectors ( overrightarrow{OM} ) and ( overrightarrow{ON} ) should be zero. If M is ( (x_1, y_1) ) and N is ( (x_2, y_2) ), then ( x_1x_2 + y_1y_2 = 0 ).To find the points of intersection between the circle and the line, I'll substitute the expression for ( x ) from the line equation into the circle's equation. From ( x + 2y - 3 = 0 ), we get ( x = 3 - 2y ). Substituting this into the circle equation:( (3 - 2y)^2 + y^2 - 2(3 - 2y) - 4y + m = 0 )Expanding ( (3 - 2y)^2 ):( 9 - 12y + 4y^2 + y^2 - 6 + 4y - 4y + m = 0 )Combining like terms:( 9 - 6 + (-12y + 4y - 4y) + (4y^2 + y^2) + m = 0 )Simplifying:( 3 - 12y + 5y^2 + m = 0 )So, the quadratic equation in terms of ( y ) is:( 5y^2 - 12y + (3 + m) = 0 )Let’s denote this as ( 5y^2 - 12y + (3 + m) = 0 ). Let ( y_1 ) and ( y_2 ) be the roots of this quadratic equation. Then, by Vieta's formulas, we know that:( y_1 + y_2 = frac{12}{5} )and( y_1y_2 = frac{3 + m}{5} )Since ( x = 3 - 2y ), the corresponding ( x )-coordinates for points M and N are:( x_1 = 3 - 2y_1 )( x_2 = 3 - 2y_2 )Now, we need to compute ( x_1x_2 + y_1y_2 = 0 ). Let's compute ( x_1x_2 ):( x_1x_2 = (3 - 2y_1)(3 - 2y_2) )Expanding this:( 9 - 6y_1 - 6y_2 + 4y_1y_2 )So, ( x_1x_2 + y_1y_2 = 9 - 6y_1 - 6y_2 + 4y_1y_2 + y_1y_2 = 9 - 6(y_1 + y_2) + 5y_1y_2 )We know from Vieta's formulas that ( y_1 + y_2 = frac{12}{5} ) and ( y_1y_2 = frac{3 + m}{5} ). Plugging these into the equation:( 9 - 6left(frac{12}{5}right) + 5left(frac{3 + m}{5}right) = 0 )Simplify each term:First term: 9Second term: ( -6 times frac{12}{5} = -frac{72}{5} )Third term: ( 5 times frac{3 + m}{5} = 3 + m )So, putting it all together:( 9 - frac{72}{5} + 3 + m = 0 )Combine the constants:( 9 + 3 = 12 )So,( 12 - frac{72}{5} + m = 0 )Convert 12 to fifths to combine with ( -frac{72}{5} ):( frac{60}{5} - frac{72}{5} = -frac{12}{5} )Thus,( -frac{12}{5} + m = 0 )Solving for ( m ):( m = frac{12}{5} )So, the value of ( m ) that satisfies the condition is ( frac{12}{5} ).Before finalizing, let me double-check if this value of ( m ) satisfies the condition for the curve to be a circle. From the first part, we found that ( m < 5 ). Since ( frac{12}{5} = 2.4 ), which is less than 5, it does satisfy the condition. Also, checking the discriminant of the quadratic equation in ( y ):Discriminant ( D = (-12)^2 - 4 times 5 times (3 + m) = 144 - 20(3 + m) )Plugging ( m = frac{12}{5} ):( D = 144 - 20left(3 + frac{12}{5}right) = 144 - 20 times frac{27}{5} = 144 - 108 = 36 )Since the discriminant is positive, there are two real intersection points, which is consistent with the problem statement.Therefore, the value of ( m ) is ( frac{12}{5} ).**Final Answer**The value of ( m ) is boxed{dfrac{12}{5}}.

answer：Okay, so I have this problem with two functions, f(x) and g(x), and I need to find the value of a + b where all the zeros of F(x) = f(x+4)·g(x-4) lie in the intervals [a-1, a] ∪ [b-1, b], with a < b and b being an integer. Hmm, let me try to break this down step by step.First, let me write down the functions again to make sure I have them right:f(x) = 1 + x - (x²)/2 + (x³)/3 - (x⁴)/4 + … + (x²⁰¹⁷)/2017g(x) = 1 - x + (x²)/2 - (x³)/3 + (x⁴)/4 - … - (x²⁰¹⁷)/2017So, both f(x) and g(x) are polynomials of degree 2017, with alternating signs. Interesting. And F(x) is the product of f(x+4) and g(x-4). So, F(x) = f(x+4)·g(x-4). I need to find the zeros of F(x), which means the zeros of f(x+4) and the zeros of g(x-4).Therefore, the zeros of F(x) are the zeros of f(x+4) and the zeros of g(x-4). So, if I can find the intervals where the zeros of f(x+4) and g(x-4) lie, I can figure out the intervals [a-1, a] and [b-1, b].Let me first analyze f(x). It looks like f(x) is an alternating series. Maybe it's related to the expansion of ln(1 + x) or something similar? Wait, the expansion of ln(1 + x) is x - x²/2 + x³/3 - x⁴/4 + …, which is similar to f(x) but starts at x instead of 1. So, f(x) is actually 1 + ln(1 + x) because if I add 1 to the expansion of ln(1 + x), I get 1 + x - x²/2 + x³/3 - x⁴/4 + …, which is exactly f(x). So, f(x) = 1 + ln(1 + x).Similarly, let's look at g(x). The expansion is 1 - x + x²/2 - x³/3 + x⁴/4 - … - x²⁰¹⁷/2017. That looks like the expansion of ln(1 + x) but with alternating signs. Wait, actually, the expansion of ln(1 - x) is -x - x²/2 - x³/3 - x⁴/4 - …, but that's not exactly what we have here. However, if I factor out a negative sign, it becomes similar to f(-x). Let me check:f(-x) = 1 + (-x) - (-x)²/2 + (-x)³/3 - (-x)⁴/4 + … + (-x)²⁰¹⁷/2017Simplify that:f(-x) = 1 - x - x²/2 - x³/3 - x⁴/4 - … - x²⁰¹⁷/2017Hmm, that's not quite g(x). g(x) is 1 - x + x²/2 - x³/3 + x⁴/4 - … - x²⁰¹⁷/2017. So, the signs alternate differently. Wait, maybe g(x) is 1 + ln(1 - x)? Let's see:The expansion of ln(1 - x) is -x - x²/2 - x³/3 - x⁴/4 - …, so 1 + ln(1 - x) would be 1 - x - x²/2 - x³/3 - x⁴/4 - …, which is similar to f(-x). But g(x) alternates signs differently. So, maybe g(x) is related to 1 - ln(1 + x)? Let me check:1 - ln(1 + x) = 1 - (x - x²/2 + x³/3 - x⁴/4 + …) = 1 - x + x²/2 - x³/3 + x⁴/4 - …, which is exactly g(x). So, g(x) = 1 - ln(1 + x).Wait, that seems right. So, f(x) = 1 + ln(1 + x) and g(x) = 1 - ln(1 + x). So, that simplifies things a bit.Therefore, f(x) = 1 + ln(1 + x) and g(x) = 1 - ln(1 + x). So, f(x+4) = 1 + ln(1 + x + 4) = 1 + ln(x + 5). Similarly, g(x - 4) = 1 - ln(1 + x - 4) = 1 - ln(x - 3).So, F(x) = f(x+4)·g(x-4) = [1 + ln(x + 5)]·[1 - ln(x - 3)].So, the zeros of F(x) are the solutions to [1 + ln(x + 5)]·[1 - ln(x - 3)] = 0. Therefore, either 1 + ln(x + 5) = 0 or 1 - ln(x - 3) = 0.Let me solve these equations:1. 1 + ln(x + 5) = 0 => ln(x + 5) = -1 => x + 5 = e^{-1} => x = e^{-1} - 5 ≈ (0.3679) - 5 ≈ -4.63212. 1 - ln(x - 3) = 0 => ln(x - 3) = 1 => x - 3 = e^{1} => x = e + 3 ≈ 2.7183 + 3 ≈ 5.7183So, the zeros of F(x) are approximately at x ≈ -4.6321 and x ≈ 5.7183.Now, the problem states that all zeros of F(x) are in the intervals [a - 1, a] ∪ [b - 1, b], where a < b and b is an integer. So, I need to find integers a and b such that:- The zero at ≈ -4.6321 is in [a - 1, a]- The zero at ≈ 5.7183 is in [b - 1, b]So, let's find a such that -4.6321 is in [a - 1, a]. Since a must be an integer, let's see:If a = -4, then [a - 1, a] = [-5, -4]. But -4.6321 is between -5 and -4, so it is in [-5, -4]. So, a = -4.Similarly, for the zero at ≈5.7183, we need to find b such that 5.7183 is in [b - 1, b]. Let's see:If b = 6, then [b - 1, b] = [5, 6]. But 5.7183 is between 5 and 6, so it is in [5, 6]. So, b = 6.Therefore, a = -4 and b = 6. So, a + b = -4 + 6 = 2.Wait, but let me double-check. The problem says "all the zeros of the function F(x) are in the interval [a-1, a] ∪ [b-1,b] (a < b, b∈ℤ)". So, the zeros are approximately -4.6321 and 5.7183. So, -4.6321 is in [-5, -4], which is [a - 1, a] with a = -4. Similarly, 5.7183 is in [5, 6], which is [b - 1, b] with b = 6. So, a = -4, b = 6, so a + b = 2.Yes, that seems correct.

answer：Okay, so I need to find the domain of this function: ( y = dfrac{ sqrt{ log_{ frac{1}{2}}(x-1) } }{ |x| - z } ). Hmm, let me break this down step by step. First, I remember that the domain of a function is all the real numbers for which the function is defined. So, I need to look at each part of this function and figure out what restrictions there are on ( x ).Starting with the numerator: ( sqrt{ log_{ frac{1}{2}}(x-1) } ). For a square root to be defined, the expression inside the square root must be greater than or equal to zero. So, I need:( log_{ frac{1}{2}}(x-1) geq 0 )Okay, logarithms can be tricky, especially with bases less than 1. I recall that ( log_b(a) ) where ( 0 < b < 1 ) is a decreasing function. That means as ( a ) increases, the logarithm decreases. So, ( log_{ frac{1}{2}}(x-1) geq 0 ) implies that ( x-1 ) must be less than or equal to 1 because the logarithm is decreasing. Wait, let me think about that again.Actually, since the base is less than 1, ( log_{ frac{1}{2}}(x-1) geq 0 ) when ( x-1 leq 1 ). Because when the argument of the logarithm is less than or equal to 1, the logarithm is non-negative. So, solving ( x - 1 leq 1 ) gives ( x leq 2 ).But also, the argument of the logarithm must be positive. So, ( x - 1 > 0 ) which means ( x > 1 ). So combining these two, ( x ) must be greater than 1 and less than or equal to 2. So, ( 1 < x leq 2 ).Now, moving on to the denominator: ( |x| - z ). For the function to be defined, the denominator cannot be zero. So, ( |x| - z neq 0 ), which means ( |x| neq z ). But wait, the problem doesn't specify what ( z ) is. Is ( z ) a constant or a variable? If ( z ) is a constant, then ( |x| neq z ) would impose another restriction on ( x ). However, since ( z ) isn't defined here, maybe it's a typo or perhaps it's supposed to be another variable. Hmm, this is confusing.Assuming ( z ) is a constant, then ( |x| neq z ) would mean ( x neq z ) and ( x neq -z ). But since from the numerator we already have ( x > 1 ), ( x ) is positive, so ( x neq z ) if ( z ) is positive. If ( z ) is negative, ( x ) is already greater than 1, so ( x ) can't be equal to a negative number anyway. So, maybe the only restriction is ( x neq z ) if ( z ) is positive.But without knowing the value of ( z ), it's hard to specify exactly. Maybe the problem expects ( z ) to be a specific value, but since it's not given, perhaps we can ignore it or assume it doesn't affect the domain? Or maybe it's a variable, but then the function would have two variables, ( x ) and ( z ), which complicates things.Wait, looking back at the problem, it's written as ( |x| - z ). If ( z ) is a variable, then the function is actually a function of two variables, ( x ) and ( z ). But the question is about the domain of ( y ) as a function of ( x ), so perhaps ( z ) is a constant. Given that, I think the safest assumption is that ( z ) is a constant, and we need to exclude any ( x ) such that ( |x| = z ). But since ( x > 1 ), we only need to consider ( x = z ) if ( z > 1 ). If ( z leq 1 ), then ( x ) can't be equal to ( z ) anyway because ( x > 1 ). So, if ( z > 1 ), we have to exclude ( x = z ) from the domain. If ( z leq 1 ), there's no additional restriction.But since ( z ) isn't specified, maybe the problem expects us to consider ( z ) as a constant and state the domain in terms of ( z ). Alternatively, perhaps it's a typo and should be ( |x - z| ) or something else. But I'll proceed with the given.So, combining the two conditions:1. From the numerator: ( 1 < x leq 2 )2. From the denominator: ( x neq z ) if ( z > 1 )Therefore, the domain is ( (1, 2] ) excluding ( x = z ) if ( z ) is in that interval. But since ( z ) isn't specified, maybe the answer is just ( (1, 2) ) assuming ( z ) isn't in that interval or it's negligible.Wait, actually, in the original function, the denominator is ( |x| - z ). If ( z ) is a constant, and ( x ) is in ( (1, 2] ), then ( |x| = x ) because ( x ) is positive. So, the denominator is ( x - z ). Therefore, ( x - z neq 0 ) implies ( x neq z ).So, if ( z ) is in ( (1, 2] ), then ( x ) cannot be ( z ). If ( z ) is outside this interval, then there's no restriction from the denominator.But since ( z ) isn't given, perhaps the domain is ( (1, 2) ) because at ( x = 2 ), the logarithm becomes zero, so the numerator is zero, making the entire expression zero, which is defined. Wait, no, the numerator is zero, but the denominator at ( x = 2 ) is ( |2| - z = 2 - z ). So, unless ( z = 2 ), the function is defined at ( x = 2 ).But again, without knowing ( z ), it's hard to say. Maybe the problem expects us to assume ( z ) is such that ( 2 - z neq 0 ), so ( z neq 2 ). But that's speculative.Alternatively, maybe the problem has a typo, and it's supposed to be ( |x - 1| - z ) or something else. But I'll stick with the given.So, to summarize:- From the numerator: ( 1 < x leq 2 )- From the denominator: ( x neq z )Therefore, the domain is ( (1, 2] ) excluding ( x = z ) if ( z ) is in ( (1, 2] ).But since ( z ) isn't specified, perhaps the safest answer is ( (1, 2) ), assuming ( z ) isn't in that interval or that the function is defined at ( x = 2 ) unless ( z = 2 ).Wait, actually, at ( x = 2 ), the numerator is zero, so the function is zero, which is defined as long as the denominator isn't zero. So, if ( z neq 2 ), ( x = 2 ) is included. If ( z = 2 ), then ( x = 2 ) is excluded.But since ( z ) isn't given, maybe the answer is ( (1, 2) ) to be safe, or ( (1, 2] ) if we assume ( z neq 2 ).Hmm, I'm a bit confused here because of the ( z ). Maybe I should look back at the problem.Wait, the function is ( y = dfrac{ sqrt{ log_{ frac{1}{2}}(x-1) } }{ |x| - z } ). So, ( z ) is in the denominator. If ( z ) is a constant, then the domain depends on ( z ). If ( z ) is a variable, then it's a function of two variables, but the problem is asking for the domain of ( y ) as a function of ( x ), so ( z ) must be a constant.Therefore, the domain is ( (1, 2] ) excluding ( x = z ) if ( z ) is in ( (1, 2] ). But since ( z ) isn't specified, perhaps the answer is ( (1, 2) ) assuming ( z ) isn't in that interval, or ( (1, 2] ) if ( z ) is outside.But I think the standard approach is to consider ( z ) as a constant and state the domain in terms of ( z ). So, the domain is ( (1, 2] ) excluding ( x = z ) if ( z ) is in ( (1, 2] ).Alternatively, if ( z ) is a variable, the domain would be more complex, but I think it's safe to assume ( z ) is a constant here.So, putting it all together, the domain is ( x ) such that ( 1 < x leq 2 ) and ( x neq z ) if ( z ) is in ( (1, 2] ). But since ( z ) isn't specified, maybe the answer is ( (1, 2) ).Wait, no, at ( x = 2 ), the numerator is zero, so the function is defined as long as the denominator isn't zero. So, if ( z neq 2 ), ( x = 2 ) is included. If ( z = 2 ), then ( x = 2 ) is excluded.But since ( z ) isn't given, perhaps the answer is ( (1, 2) ) to be safe, or ( (1, 2] ) if we assume ( z neq 2 ).I think the most precise answer is ( (1, 2) ) because if ( z = 2 ), then ( x = 2 ) would make the denominator zero, which is undefined. But since ( z ) isn't specified, we can't be sure, so it's safer to exclude ( x = 2 ) just in case.Wait, no, at ( x = 2 ), the numerator is zero, so even if the denominator is non-zero, the function is defined. So, if ( z neq 2 ), ( x = 2 ) is included. If ( z = 2 ), it's excluded. But since ( z ) isn't specified, maybe the answer is ( (1, 2] ) with the caveat that if ( z = 2 ), it's excluded.But I think the problem expects a specific interval, so probably ( (1, 2) ).Wait, let me check again.From the numerator: ( log_{1/2}(x-1) geq 0 ) implies ( x - 1 leq 1 ) because the base is less than 1, so the inequality flips. So, ( x leq 2 ). Also, ( x - 1 > 0 ) implies ( x > 1 ). So, ( 1 < x leq 2 ).From the denominator: ( |x| - z neq 0 ). Since ( x > 1 ), ( |x| = x ), so ( x - z neq 0 ) implies ( x neq z ).Therefore, the domain is ( (1, 2] ) excluding ( x = z ) if ( z ) is in ( (1, 2] ). But since ( z ) isn't specified, we can't exclude a specific point, so the domain is ( (1, 2] ).Wait, but if ( z ) is in ( (1, 2] ), then ( x = z ) would make the denominator zero, so we have to exclude it. But since ( z ) isn't given, maybe the answer is ( (1, 2) ) to avoid that point.I'm a bit torn here. I think the safest answer is ( (1, 2) ) because at ( x = 2 ), the numerator is zero, but the denominator is ( 2 - z ). If ( z = 2 ), then it's undefined, but if ( z neq 2 ), it's defined. Since ( z ) isn't specified, it's safer to exclude ( x = 2 ) to ensure the function is defined for all ( x ) in the interval.Wait, no, because at ( x = 2 ), the function is zero divided by ( 2 - z ). If ( 2 - z neq 0 ), it's defined. So, unless ( z = 2 ), ( x = 2 ) is included. But since ( z ) isn't specified, we can't assume ( z = 2 ). So, the domain should be ( (1, 2] ).But I'm not entirely sure. Maybe the problem expects ( (1, 2) ) because of the denominator, but I think ( (1, 2] ) is correct because at ( x = 2 ), the function is defined unless ( z = 2 ).I think I'll go with ( (1, 2] ) as the domain, but I'm a bit uncertain because of the ( z ). Maybe the answer is ( (1, 2) ) to be safe.Wait, let me think about it again. The function is ( y = dfrac{ sqrt{ log_{ frac{1}{2}}(x-1) } }{ |x| - z } ). So, the numerator is zero at ( x = 2 ), making the entire expression zero, which is fine as long as the denominator isn't zero. So, if ( z neq 2 ), ( x = 2 ) is included. If ( z = 2 ), it's excluded. But since ( z ) isn't specified, we can't exclude ( x = 2 ) without knowing ( z ). Therefore, the domain is ( (1, 2] ).Yes, I think that's the correct approach. So, the domain is ( (1, 2] ).

answer：Okay, so I've got this problem here where I need to compare three different expressions: a, b, and c. The expressions are given as:- ( a = 2ln 1.01 )- ( b = ln 1.02 )- ( c = sqrt{1.04} - 1 )And I need to figure out the order of these three values from smallest to largest. The options are:A: ( a < b < c )B: ( b < c < a )C: ( b < a < c )D: ( c < a < b )Alright, let's start by understanding each of these expressions individually.First, ( a = 2ln 1.01 ). I know that the natural logarithm function, ( ln x ), grows slowly for values of x close to 1. So, ( ln 1.01 ) is a small positive number. Multiplying it by 2 will just make it a bit larger, but still small.Next, ( b = ln 1.02 ). Similar to ( a ), this is also a natural logarithm of a number slightly larger than 1, so it's another small positive number. But since 1.02 is larger than 1.01, ( ln 1.02 ) should be larger than ( ln 1.01 ). However, ( a ) is twice ( ln 1.01 ), so I need to see whether ( 2ln 1.01 ) is larger or smaller than ( ln 1.02 ).Then, ( c = sqrt{1.04} - 1 ). The square root of 1.04 is a number slightly larger than 1, so subtracting 1 will give a small positive number. I need to compare this with the logarithmic expressions.To compare these, I think it might help to approximate each value numerically. Let me try that.Starting with ( a = 2ln 1.01 ). I remember that ( ln(1 + x) ) can be approximated by the Taylor series expansion: ( ln(1 + x) approx x - frac{x^2}{2} + frac{x^3}{3} - dots ) for small x. Since 0.01 is a small number, this approximation should be pretty good.So, ( ln 1.01 = ln(1 + 0.01) approx 0.01 - frac{(0.01)^2}{2} + frac{(0.01)^3}{3} ). Let's compute that:( 0.01 = 0.01 )( frac{(0.01)^2}{2} = frac{0.0001}{2} = 0.00005 )( frac{(0.01)^3}{3} = frac{0.000001}{3} approx 0.000000333 )So, ( ln 1.01 approx 0.01 - 0.00005 + 0.000000333 approx 0.009950333 )Therefore, ( a = 2 times 0.009950333 approx 0.019900666 )Now, ( b = ln 1.02 ). Similarly, using the Taylor series:( ln 1.02 = ln(1 + 0.02) approx 0.02 - frac{(0.02)^2}{2} + frac{(0.02)^3}{3} )Calculating each term:( 0.02 = 0.02 )( frac{(0.02)^2}{2} = frac{0.0004}{2} = 0.0002 )( frac{(0.02)^3}{3} = frac{0.000008}{3} approx 0.000002666 )So, ( ln 1.02 approx 0.02 - 0.0002 + 0.000002666 approx 0.019802666 )Therefore, ( b approx 0.019802666 )Comparing ( a ) and ( b ):( a approx 0.019900666 )( b approx 0.019802666 )So, ( a > b )Now, let's compute ( c = sqrt{1.04} - 1 ). Again, using a Taylor series approximation for the square root function around 1. The expansion is ( sqrt{1 + x} approx 1 + frac{x}{2} - frac{x^2}{8} + frac{x^3}{16} - dots ) for small x.Here, ( x = 0.04 ), which is a bit larger, so the approximation might not be as accurate, but let's try:( sqrt{1.04} = sqrt{1 + 0.04} approx 1 + frac{0.04}{2} - frac{(0.04)^2}{8} + frac{(0.04)^3}{16} )Calculating each term:( 1 = 1 )( frac{0.04}{2} = 0.02 )( frac{(0.04)^2}{8} = frac{0.0016}{8} = 0.0002 )( frac{(0.04)^3}{16} = frac{0.000064}{16} = 0.000004 )So, ( sqrt{1.04} approx 1 + 0.02 - 0.0002 + 0.000004 = 1.019804 )Therefore, ( c = sqrt{1.04} - 1 approx 1.019804 - 1 = 0.019804 )So, ( c approx 0.019804 )Now, let's compare all three:- ( a approx 0.019900666 )- ( b approx 0.019802666 )- ( c approx 0.019804 )So, ordering them from smallest to largest:( b approx 0.019802666 ) is the smallest,Then ( c approx 0.019804 ) is slightly larger,And ( a approx 0.019900666 ) is the largest.Therefore, the order is ( b < c < a ), which corresponds to option B.Wait, but let me double-check because sometimes approximations can be misleading.Alternatively, maybe I can use exact values or better approximations.Let me compute ( ln 1.01 ) more accurately.Using a calculator:( ln 1.01 approx 0.00995033 )So, ( a = 2 times 0.00995033 approx 0.01990066 )( ln 1.02 approx 0.0198026 )( sqrt{1.04} approx 1.0198039 ), so ( c = 1.0198039 - 1 = 0.0198039 )So, now:- ( a approx 0.01990066 )- ( c approx 0.0198039 )- ( b approx 0.0198026 )So, ( b approx 0.0198026 ) is the smallest,Then ( c approx 0.0198039 ),Then ( a approx 0.01990066 )So, indeed, ( b < c < a ), which is option B.But wait, in the initial problem statement, the options are:A: ( a < b < c )B: ( b < c < a )C: ( b < a < c )D: ( c < a < b )So, according to my calculations, the correct order is ( b < c < a ), which is option B.But let me think again. Maybe I made a mistake in approximating ( sqrt{1.04} ). Let me compute ( sqrt{1.04} ) more accurately.Using a better approximation method or a calculator:( sqrt{1.04} approx 1.0198039 )So, ( c = 1.0198039 - 1 = 0.0198039 )Comparing with ( b = ln 1.02 approx 0.0198026 )So, ( c ) is approximately 0.0198039, which is slightly larger than ( b approx 0.0198026 )Therefore, ( b < c )And ( a approx 0.01990066 ), which is larger than both ( b ) and ( c )So, the order is ( b < c < a ), which is option B.Wait, but in the initial problem, the options are:A: a < b < cB: b < c < aC: b < a < cD: c < a < bSo, according to my calculations, the correct answer is B.But let me cross-verify with another approach.Another way is to consider the functions involved.We know that ( ln(1 + x) ) is concave, so ( 2ln(1 + x) ) is also concave.Similarly, ( sqrt{1 + 4x} - 1 ) is a concave function as well.But perhaps comparing derivatives or using inequalities.Alternatively, consider that for small x, ( ln(1 + x) approx x - x^2/2 ), and ( sqrt{1 + x} approx 1 + x/2 - x^2/8 ).So, let's express all in terms of x.Let me set x = 0.01.Then:- ( a = 2ln(1 + x) approx 2(x - x^2/2) = 2x - x^2 )- ( b = ln(1 + 2x) approx 2x - (2x)^2/2 = 2x - 2x^2 )- ( c = sqrt{1 + 4x} - 1 approx (1 + 2x - (4x)^2/8) - 1 = 2x - (16x^2)/8 = 2x - 2x^2 )Wait, so according to this approximation:- ( a approx 2x - x^2 )- ( b approx 2x - 2x^2 )- ( c approx 2x - 2x^2 )So, ( a approx 2x - x^2 ), which is larger than ( b ) and ( c ) because ( 2x - x^2 > 2x - 2x^2 )And ( b ) and ( c ) are approximately equal in this approximation.But in reality, when we computed numerically, ( c ) was slightly larger than ( b ).So, perhaps the next term in the expansion would show that ( c > b ).Let me include the next term in the expansion.For ( ln(1 + x) approx x - x^2/2 + x^3/3 )So, ( a = 2ln(1 + x) approx 2x - x^2 + (2x^3)/3 )Similarly, ( b = ln(1 + 2x) approx 2x - (4x^2)/2 + (8x^3)/3 = 2x - 2x^2 + (8x^3)/3 )And ( c = sqrt{1 + 4x} - 1 approx 2x - (16x^2)/8 + (3*(4x)^3)/(16*1) = 2x - 2x^2 + (3*64x^3)/16 = 2x - 2x^2 + 12x^3 )Wait, that seems off. Let me correct that.The expansion for ( sqrt{1 + y} ) is ( 1 + y/2 - y^2/8 + y^3/16 - dots )So, for ( y = 4x ), ( sqrt{1 + 4x} = 1 + (4x)/2 - (4x)^2/8 + (4x)^3/16 - dots = 1 + 2x - 2x^2 + 4x^3 - dots )Therefore, ( c = sqrt{1 + 4x} - 1 approx 2x - 2x^2 + 4x^3 )So, now, let's write all three expressions with the cubic term:- ( a approx 2x - x^2 + (2x^3)/3 )- ( b approx 2x - 2x^2 + (8x^3)/3 )- ( c approx 2x - 2x^2 + 4x^3 )Now, let's plug in x = 0.01:Compute each term:For ( a ):- ( 2x = 0.02 )- ( -x^2 = -0.0001 )- ( (2x^3)/3 = (2*0.000001)/3 ≈ 0.000000666 )- So, ( a ≈ 0.02 - 0.0001 + 0.000000666 ≈ 0.019900666 )For ( b ):- ( 2x = 0.02 )- ( -2x^2 = -0.0002 )- ( (8x^3)/3 ≈ (8*0.000001)/3 ≈ 0.000002666 )- So, ( b ≈ 0.02 - 0.0002 + 0.000002666 ≈ 0.019802666 )For ( c ):- ( 2x = 0.02 )- ( -2x^2 = -0.0002 )- ( 4x^3 = 4*0.000001 = 0.000004 )- So, ( c ≈ 0.02 - 0.0002 + 0.000004 ≈ 0.019804 )So, with the cubic terms, we have:- ( a ≈ 0.019900666 )- ( b ≈ 0.019802666 )- ( c ≈ 0.019804 )So, indeed, ( b < c < a )Therefore, the correct order is ( b < c < a ), which is option B.I think this confirms my earlier conclusion. So, the answer should be option B.

answer：First, calculate the total amount of money Sam has initially by converting her dimes and quarters into cents.Sam has 19 dimes and 6 quarters:- 19 dimes × 10 cents per dime = 190 cents- 6 quarters × 25 cents per quarter = 150 centsTotal amount = 190 cents + 150 cents = 340 centsNext, determine the cost of the 4 candy bars. Each candy bar costs 3 dimes:- 3 dimes × 10 cents per dime = 30 cents per candy bar- 4 candy bars × 30 cents per candy bar = 120 centsSubtract the cost of the candy bars from the total amount to find out how much money Sam has left before buying lollipops:- 340 cents - 120 cents = 220 centsSam has 195 cents left after buying lollipops. Calculate the amount spent on lollipops:- 220 cents - 195 cents = 25 centsSince each lollipop costs 1 quarter (25 cents), determine the number of lollipops bought:- 25 cents ÷ 25 cents per lollipop = 1 lollipop

answer：Okay, so I need to find the tens digit of 36 raised to the 12th power. Hmm, that sounds a bit tricky at first, but maybe I can break it down step by step.First, I remember that when dealing with powers, especially large exponents like 12, it's helpful to look for patterns in the last few digits. Since I'm interested in the tens digit, I should probably focus on the last two digits of 36^12. If I can figure out what those two digits are, then the tens digit will just be the second-to-last digit of that number.Alright, let's start by calculating the first few powers of 36 and see if there's a pattern in the last two digits.36^1 is just 36. So, the last two digits are 36.36^2 is 36 multiplied by 36. Let me do that multiplication:36x36----216 (which is 36 x 6)1080 (which is 36 x 30, shifted one position to the left)----1296So, 36^2 is 1296, and the last two digits are 96.Next, 36^3 would be 36^2 multiplied by 36, which is 1296 x 36. That's a bit more complex, but I can do it step by step.1296 x 36:First, multiply 1296 by 6:1296 x 6 = 7776Then, multiply 1296 by 30:1296 x 30 = 38,880Now, add those two results together:7776 + 38,880 = 46,656So, 36^3 is 46,656, and the last two digits are 56.Moving on to 36^4, which is 36^3 multiplied by 36, so 46,656 x 36. Again, breaking it down:46,656 x 6 = 279,93646,656 x 30 = 1,399,680Adding those together:279,936 + 1,399,680 = 1,679,616So, 36^4 is 1,679,616, and the last two digits are 16.Now, 36^5 would be 36^4 multiplied by 36, which is 1,679,616 x 36. Let's compute that:1,679,616 x 6 = 10,077,6961,679,616 x 30 = 50,388,480Adding those:10,077,696 + 50,388,480 = 60,466,176So, 36^5 is 60,466,176, and the last two digits are 76.Continuing to 36^6, which is 36^5 multiplied by 36, so 60,466,176 x 36. Let's calculate:60,466,176 x 6 = 362,797,05660,466,176 x 30 = 1,813,985,280Adding those:362,797,056 + 1,813,985,280 = 2,176,782,336So, 36^6 is 2,176,782,336, and the last two digits are 36.Wait a minute, that's interesting. The last two digits of 36^6 are 36, which is the same as the last two digits of 36^1. Does this mean there's a repeating cycle every 5 powers? Let's check the next few to confirm.36^7 would be 36^6 multiplied by 36, which is 2,176,782,336 x 36. But since we're only interested in the last two digits, and 36^6 ends with 36, multiplying by 36 again would give the same last two digits as 36^2, which was 96. So, 36^7 should end with 96.Similarly, 36^8 would end with the same last two digits as 36^3, which were 56.36^9 would end with 16, like 36^4.36^10 would end with 76, like 36^5.36^11 would end with 36, like 36^6.36^12 would end with 96, like 36^7.Wait, but earlier I thought the cycle was every 5 powers, but now it seems like it's every 5 powers starting from 36^1. So, 36^1 ends with 36, 36^2 with 96, 36^3 with 56, 36^4 with 16, 36^5 with 76, and then 36^6 back to 36, and so on.So, the cycle is indeed every 5 powers: 36, 96, 56, 16, 76, and then repeats.Given that, to find the last two digits of 36^12, I can determine where 12 falls in the cycle.Since the cycle length is 5, I can divide 12 by 5 to find the remainder.12 divided by 5 is 2 with a remainder of 2.So, the remainder is 2, which means 36^12 will have the same last two digits as 36^2, which were 96.Therefore, the last two digits of 36^12 are 96, and the tens digit is 9.Wait, but earlier I thought the cycle was every 5 powers, but when I calculated 36^6, it ended with 36, which is the same as 36^1. So, the cycle is indeed every 5 powers.But when I calculated 36^12, I thought it would end with 96, which is the same as 36^2. But according to the cycle, since 12 mod 5 is 2, it should correspond to the second position in the cycle, which is 96. So, the last two digits are 96, and the tens digit is 9.But wait, in the initial calculation, I thought the cycle was every 5 powers, but when I calculated 36^6, it ended with 36, which is the same as 36^1. So, the cycle is indeed every 5 powers.But when I calculated 36^12, I thought it would end with 96, which is the same as 36^2. But according to the cycle, since 12 mod 5 is 2, it should correspond to the second position in the cycle, which is 96. So, the last two digits are 96, and the tens digit is 9.Wait, but in the initial problem, the assistant concluded that the tens digit is 3, but according to my calculations, it should be 9. Did I make a mistake somewhere?Let me double-check.36^1: 36, last two digits 3636^2: 1296, last two digits 9636^3: 46656, last two digits 5636^4: 1679616, last two digits 1636^5: 60466176, last two digits 7636^6: 2176782336, last two digits 36So, the cycle is indeed 36, 96, 56, 16, 76, and then repeats.So, the cycle length is 5.Therefore, to find 36^12, we can find 12 mod 5, which is 2 (since 5*2=10, 12-10=2).So, 36^12 corresponds to the second position in the cycle, which is 96.Therefore, the last two digits are 96, and the tens digit is 9.But the assistant initially concluded that the tens digit is 3, which contradicts my calculations.Wait, perhaps the assistant made a mistake in identifying the cycle.Let me check the assistant's thought process.The assistant said:"Notice that the last two digits of 36^6 are the same as the last two digits of 36^1. This suggests that there is a cycle of length 5 in the last two digits of powers of 36. Since 12 is a multiple of 6 (which is the cycle length), the last two digits of 36^12 will be the same as the last two digits of 36^6, which are 36.Therefore, the tens digit of 36^12 is boxed{3} ."Wait, the assistant said the cycle length is 5, but then said 12 is a multiple of 6, which is incorrect because 12 is not a multiple of 6 in terms of the cycle length. The cycle length is 5, not 6.So, the assistant incorrectly identified the cycle length as 6 instead of 5, leading to the wrong conclusion.Therefore, the correct cycle length is 5, and 12 mod 5 is 2, so 36^12 ends with 96, making the tens digit 9.So, the correct answer should be 9, not 3.I think I need to clarify this to ensure the correct answer is provided.