answer：Alright, so I have this problem here about finding the range of intercept values of the tangent line of the curve ( y = e^x ) at the point ( (x_0, e^{x_0}) ) on the y-axis. The options are A: ((-1, 1]), B: ((-infty, 1]), C: ((-infty, 0]), and D: ((0, 1]). Hmm, okay, let's break this down step by step.First, I remember that the equation of a tangent line to a curve at a particular point involves the derivative of the function at that point. So, for ( y = e^x ), the derivative is ( y' = e^x ). That means at the point ( (x_0, e^{x_0}) ), the slope of the tangent line is ( e^{x_0} ).Now, using the point-slope form of a line, which is ( y - y_1 = m(x - x_1) ), where ( m ) is the slope and ( (x_1, y_1) ) is the point of tangency, I can plug in the values. So, substituting, we get:[y - e^{x_0} = e^{x_0}(x - x_0)]Simplifying this, it becomes:[y = e^{x_0}x - e^{x_0}x_0 + e^{x_0}]Which simplifies further to:[y = e^{x_0}x + e^{x_0}(1 - x_0)]So, the y-intercept of this tangent line is ( e^{x_0}(1 - x_0) ). Let's denote this y-intercept as ( b ), so:[b = e^{x_0}(1 - x_0)]Now, the question is asking for the range of possible values of ( b ) as ( x_0 ) varies over all real numbers. So, I need to analyze the function ( b(x_0) = e^{x_0}(1 - x_0) ) and determine its range.To find the range, I should probably find the maximum and minimum values of ( b(x_0) ). To do this, I'll take the derivative of ( b(x_0) ) with respect to ( x_0 ) and find its critical points.So, let's compute the derivative ( b'(x_0) ):[b'(x_0) = frac{d}{dx_0} left[ e^{x_0}(1 - x_0) right]]Using the product rule, which states that ( frac{d}{dx}[u cdot v] = u'v + uv' ), where ( u = e^{x_0} ) and ( v = (1 - x_0) ), we get:[u' = e^{x_0} quad text{and} quad v' = -1]So,[b'(x_0) = e^{x_0}(1 - x_0) + e^{x_0}(-1) = e^{x_0}(1 - x_0 - 1) = e^{x_0}(-x_0)]Therefore,[b'(x_0) = -x_0 e^{x_0}]To find the critical points, we set ( b'(x_0) = 0 ):[-x_0 e^{x_0} = 0]Since ( e^{x_0} ) is never zero for any real ( x_0 ), the only solution is ( x_0 = 0 ).Now, we need to determine whether this critical point is a maximum or a minimum. We can do this by analyzing the sign of ( b'(x_0) ) around ( x_0 = 0 ).- For ( x_0 < 0 ), say ( x_0 = -1 ), ( b'(x_0) = -(-1)e^{-1} = e^{-1} > 0 ). So, the function is increasing.- For ( x_0 > 0 ), say ( x_0 = 1 ), ( b'(x_0) = -(1)e^{1} = -e < 0 ). So, the function is decreasing.This means that at ( x_0 = 0 ), the function ( b(x_0) ) reaches a maximum.Now, let's compute ( b(0) ):[b(0) = e^{0}(1 - 0) = 1 times 1 = 1]So, the maximum value of ( b ) is 1.Next, we need to consider the behavior of ( b(x_0) ) as ( x_0 ) approaches positive and negative infinity.- As ( x_0 to infty ), ( e^{x_0} ) grows exponentially, while ( (1 - x_0) ) becomes very negative. So, ( b(x_0) = e^{x_0}(1 - x_0) ) will approach negative infinity because the negative term dominates the exponential growth. - As ( x_0 to -infty ), ( e^{x_0} ) approaches zero, and ( (1 - x_0) ) approaches positive infinity. However, since ( e^{x_0} ) approaches zero faster than ( (1 - x_0) ) grows, the product ( b(x_0) ) will approach zero.Therefore, combining these observations:- The maximum value of ( b ) is 1.- As ( x_0 to infty ), ( b to -infty ).- As ( x_0 to -infty ), ( b to 0 ).Thus, the range of ( b ) is all real numbers less than or equal to 1, which can be written as ( (-infty, 1] ).Looking back at the options:A: ((-1, 1]) – This is incorrect because the range extends to negative infinity, not just -1.B: ((-infty, 1]) – This matches our conclusion.C: ((-infty, 0]) – This is incorrect because the maximum value is 1, not 0.D: ((0, 1]) – This is incorrect because the range includes negative values as well.Therefore, the correct answer is B: ((-infty, 1]).**Final Answer**boxed{B}

answer：Okay, so I need to calculate the expression (frac{tan frac{pi}{6} + i}{tan frac{pi}{6} - i}) and determine if it's a tenth root of unity. If it is, I have to find the integer (n) between 0 and 9 such that the expression equals (cos frac{2n pi}{10} + i sin frac{2n pi}{10}).First, let me recall what a tenth root of unity is. It's a complex number (z) such that (z^{10} = 1). These roots can be expressed in the form (cos frac{2n pi}{10} + i sin frac{2n pi}{10}) for (n = 0, 1, 2, ..., 9). So, if the given expression can be written in this form, it is a tenth root of unity.Let me start by simplifying the given expression. The expression is (frac{tan frac{pi}{6} + i}{tan frac{pi}{6} - i}). I know that (tan frac{pi}{6}) is (frac{1}{sqrt{3}}), so substituting that in, the expression becomes (frac{frac{1}{sqrt{3}} + i}{frac{1}{sqrt{3}} - i}).To simplify this complex fraction, I can multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of (frac{1}{sqrt{3}} - i) is (frac{1}{sqrt{3}} + i). So, multiplying both numerator and denominator by this conjugate:[frac{left(frac{1}{sqrt{3}} + iright)left(frac{1}{sqrt{3}} + iright)}{left(frac{1}{sqrt{3}} - iright)left(frac{1}{sqrt{3}} + iright)}]Let me compute the numerator first. Expanding the numerator:[left(frac{1}{sqrt{3}} + iright)left(frac{1}{sqrt{3}} + iright) = left(frac{1}{sqrt{3}}right)^2 + 2 cdot frac{1}{sqrt{3}} cdot i + i^2][= frac{1}{3} + frac{2i}{sqrt{3}} - 1][= left(frac{1}{3} - 1right) + frac{2i}{sqrt{3}}][= -frac{2}{3} + frac{2i}{sqrt{3}}]Now, the denominator:[left(frac{1}{sqrt{3}} - iright)left(frac{1}{sqrt{3}} + iright) = left(frac{1}{sqrt{3}}right)^2 - (i)^2][= frac{1}{3} - (-1)][= frac{1}{3} + 1][= frac{4}{3}]So, putting the numerator over the denominator:[frac{-frac{2}{3} + frac{2i}{sqrt{3}}}{frac{4}{3}} = frac{-2 + 2isqrt{3}}{4} = frac{-1 + isqrt{3}}{2}]Wait, that doesn't seem right. Let me check my calculations again.Wait, I think I made a mistake in the numerator. Let me recalculate:Numerator:[left(frac{1}{sqrt{3}} + iright)left(frac{1}{sqrt{3}} + iright) = left(frac{1}{sqrt{3}}right)^2 + 2 cdot frac{1}{sqrt{3}} cdot i + i^2][= frac{1}{3} + frac{2i}{sqrt{3}} - 1][= left(frac{1}{3} - 1right) + frac{2i}{sqrt{3}}][= -frac{2}{3} + frac{2i}{sqrt{3}}]Denominator:[left(frac{1}{sqrt{3}} - iright)left(frac{1}{sqrt{3}} + iright) = left(frac{1}{sqrt{3}}right)^2 - (i)^2][= frac{1}{3} - (-1)][= frac{1}{3} + 1 = frac{4}{3}]So, the fraction becomes:[frac{-frac{2}{3} + frac{2i}{sqrt{3}}}{frac{4}{3}} = frac{-2 + 2isqrt{3}}{4} = frac{-1 + isqrt{3}}{2}]Wait, that's correct. So, the simplified form is (frac{-1 + isqrt{3}}{2}).Now, I need to see if this is a tenth root of unity. Let me recall that the tenth roots of unity are given by (e^{2pi i k/10}) for (k = 0, 1, ..., 9), which can be written as (cos frac{2pi k}{10} + i sin frac{2pi k}{10}).Alternatively, I can write (frac{-1 + isqrt{3}}{2}) in polar form. Let me compute its magnitude and argument.The magnitude is:[sqrt{left(frac{-1}{2}right)^2 + left(frac{sqrt{3}}{2}right)^2} = sqrt{frac{1}{4} + frac{3}{4}} = sqrt{1} = 1]So, the magnitude is 1, which is a good sign because roots of unity lie on the unit circle.Now, the argument. The real part is (-frac{1}{2}) and the imaginary part is (frac{sqrt{3}}{2}). So, this corresponds to an angle in the second quadrant.The reference angle can be found by:[theta = arctanleft(frac{sqrt{3}/2}{1/2}right) = arctan(sqrt{3}) = frac{pi}{3}]Since it's in the second quadrant, the argument is (pi - frac{pi}{3} = frac{2pi}{3}).So, the complex number is (e^{i frac{2pi}{3}}), which is indeed a root of unity. Specifically, it's a third root of unity because ((e^{i frac{2pi}{3}})^3 = e^{i 2pi} = 1).But we need to check if it's a tenth root of unity. Let me see if (frac{2pi}{3}) can be expressed as (frac{2pi k}{10}) for some integer (k).Simplify (frac{2pi}{3}):[frac{2pi}{3} = frac{20pi}{30} = frac{2pi cdot 10}{30} = frac{2pi cdot (10)}{30}]Wait, that doesn't directly help. Let me see:We have (e^{i theta}) where (theta = frac{2pi}{3}). We want to see if (theta = frac{2pi k}{10}) for some integer (k).So, set (frac{2pi}{3} = frac{2pi k}{10}). Dividing both sides by (2pi):[frac{1}{3} = frac{k}{10} implies k = frac{10}{3}]But (k) must be an integer between 0 and 9. Since (frac{10}{3}) is approximately 3.333, which is not an integer, this suggests that (frac{-1 + isqrt{3}}{2}) is not a tenth root of unity.Wait, but that contradicts my earlier thought because I know that (e^{i frac{2pi}{3}}) is a third root of unity, not a tenth. So, perhaps the expression is not a tenth root of unity.But wait, let me double-check my calculations because the initial expression was (frac{tan frac{pi}{6} + i}{tan frac{pi}{6} - i}), which I simplified to (frac{-1 + isqrt{3}}{2}). Let me verify that step again.Starting with (frac{frac{1}{sqrt{3}} + i}{frac{1}{sqrt{3}} - i}). Multiply numerator and denominator by (frac{1}{sqrt{3}} + i):Numerator:[left(frac{1}{sqrt{3}} + iright)left(frac{1}{sqrt{3}} + iright) = frac{1}{3} + frac{2i}{sqrt{3}} + i^2 = frac{1}{3} + frac{2i}{sqrt{3}} - 1 = -frac{2}{3} + frac{2i}{sqrt{3}}]Denominator:[left(frac{1}{sqrt{3}} - iright)left(frac{1}{sqrt{3}} + iright) = frac{1}{3} - (i)^2 = frac{1}{3} + 1 = frac{4}{3}]So, the fraction is:[frac{-frac{2}{3} + frac{2i}{sqrt{3}}}{frac{4}{3}} = frac{-2 + 2isqrt{3}}{4} = frac{-1 + isqrt{3}}{2}]Yes, that's correct. So, the expression simplifies to (frac{-1 + isqrt{3}}{2}), which is (e^{i frac{2pi}{3}}), a third root of unity, not a tenth.Wait, but the problem says to determine if it's a tenth root of unity. Maybe I made a mistake in the simplification. Let me try another approach.Alternatively, I can write the original expression as:[frac{tan frac{pi}{6} + i}{tan frac{pi}{6} - i} = frac{frac{sin frac{pi}{6}}{cos frac{pi}{6}} + i}{frac{sin frac{pi}{6}}{cos frac{pi}{6}} - i}]Let me factor out (frac{1}{cos frac{pi}{6}}) from numerator and denominator:[= frac{frac{1}{cos frac{pi}{6}} (sin frac{pi}{6} + i cos frac{pi}{6})}{frac{1}{cos frac{pi}{6}} (sin frac{pi}{6} - i cos frac{pi}{6})} = frac{sin frac{pi}{6} + i cos frac{pi}{6}}{sin frac{pi}{6} - i cos frac{pi}{6}}]Now, let me write this as:[frac{sin theta + i cos theta}{sin theta - i cos theta}]where (theta = frac{pi}{6}).Let me multiply numerator and denominator by (i):[= frac{i sin theta - cos theta}{i sin theta + cos theta}]Wait, that might not help. Alternatively, I can write the numerator and denominator in terms of exponentials.Recall that (e^{iphi} = cos phi + i sin phi). Let me see if I can express the numerator and denominator in terms of exponentials.Wait, the numerator is (sin theta + i cos theta). Let me factor out (i):[sin theta + i cos theta = i (cos theta - i sin theta) = i e^{-itheta}]Similarly, the denominator is (sin theta - i cos theta). Let me factor out (i):[sin theta - i cos theta = -i (cos theta + i sin theta) = -i e^{itheta}]So, the expression becomes:[frac{i e^{-itheta}}{-i e^{itheta}} = frac{i}{-i} cdot frac{e^{-itheta}}{e^{itheta}} = (-1) cdot e^{-i2theta} = -e^{-i2theta}]Substituting (theta = frac{pi}{6}):[-e^{-i2 cdot frac{pi}{6}} = -e^{-ifrac{pi}{3}} = e^{ipi} cdot e^{-ifrac{pi}{3}} = e^{i(pi - frac{pi}{3})} = e^{ifrac{2pi}{3}}]So, again, we get (e^{ifrac{2pi}{3}}), which is a third root of unity, not a tenth.Wait, but the problem says to determine if it's a tenth root of unity. Maybe I'm missing something. Let me check if (e^{ifrac{2pi}{3}}) is a tenth root of unity.A tenth root of unity satisfies (z^{10} = 1). So, let's compute ((e^{ifrac{2pi}{3}})^{10} = e^{ifrac{20pi}{3}}).Simplify (frac{20pi}{3}):[frac{20pi}{3} = 6pi + frac{2pi}{3} = 3 cdot 2pi + frac{2pi}{3}]Since (e^{i2pi k} = 1) for any integer (k), we have:[e^{ifrac{20pi}{3}} = e^{i(6pi + frac{2pi}{3})} = e^{i6pi} cdot e^{ifrac{2pi}{3}} = (e^{i2pi})^3 cdot e^{ifrac{2pi}{3}} = 1^3 cdot e^{ifrac{2pi}{3}} = e^{ifrac{2pi}{3}} neq 1]So, ((e^{ifrac{2pi}{3}})^{10} neq 1), meaning (e^{ifrac{2pi}{3}}) is not a tenth root of unity. Therefore, the given expression is not a tenth root of unity.Wait, but the problem says to determine if it is a tenth root of unity. Maybe I made a mistake in the simplification. Let me try another approach.Alternatively, I can write the original expression as:[frac{tan frac{pi}{6} + i}{tan frac{pi}{6} - i} = frac{frac{1}{sqrt{3}} + i}{frac{1}{sqrt{3}} - i}]Let me represent this as a complex number (z = frac{a + ib}{a - ib}), where (a = frac{1}{sqrt{3}}) and (b = 1).I recall that (frac{a + ib}{a - ib} = frac{(a + ib)^2}{a^2 + b^2}). Let me verify that:[frac{a + ib}{a - ib} = frac{(a + ib)(a + ib)}{(a - ib)(a + ib)} = frac{a^2 - b^2 + 2iab}{a^2 + b^2}]Yes, that's correct. So, applying this to our case:[z = frac{a + ib}{a - ib} = frac{a^2 - b^2 + 2iab}{a^2 + b^2}]Substituting (a = frac{1}{sqrt{3}}) and (b = 1):[a^2 = frac{1}{3}, quad b^2 = 1, quad ab = frac{1}{sqrt{3}}]So,[z = frac{frac{1}{3} - 1 + 2i cdot frac{1}{sqrt{3}}}{frac{1}{3} + 1} = frac{-frac{2}{3} + frac{2i}{sqrt{3}}}{frac{4}{3}} = frac{-2 + 2isqrt{3}}{4} = frac{-1 + isqrt{3}}{2}]Again, we get the same result. So, the expression simplifies to (frac{-1 + isqrt{3}}{2}), which is (e^{ifrac{2pi}{3}}), a third root of unity.Since it's not a tenth root of unity, the answer would be that it is not a tenth root of unity. However, the problem seems to suggest that it is, so perhaps I made a mistake in my approach.Wait, let me think differently. Maybe I can express the original expression in terms of exponentials directly.Let me write (tan frac{pi}{6} = frac{1}{sqrt{3}}), so the expression is:[frac{frac{1}{sqrt{3}} + i}{frac{1}{sqrt{3}} - i}]Let me write this as:[frac{1 + isqrt{3}}{1 - isqrt{3}} cdot frac{1}{sqrt{3}} div frac{1}{sqrt{3}} = frac{1 + isqrt{3}}{1 - isqrt{3}}]Wait, no, that's not correct. Let me see:Actually, (frac{frac{1}{sqrt{3}} + i}{frac{1}{sqrt{3}} - i} = frac{1 + isqrt{3}}{1 - isqrt{3}}) because if I multiply numerator and denominator by (sqrt{3}), I get:[frac{1 + isqrt{3}}{1 - isqrt{3}}]Yes, that's correct.Now, let me compute this fraction:[frac{1 + isqrt{3}}{1 - isqrt{3}} = frac{(1 + isqrt{3})^2}{(1)^2 + (sqrt{3})^2} = frac{1 + 2isqrt{3} - 3}{1 + 3} = frac{-2 + 2isqrt{3}}{4} = frac{-1 + isqrt{3}}{2}]Again, same result. So, it's confirmed that the expression simplifies to (frac{-1 + isqrt{3}}{2}), which is (e^{ifrac{2pi}{3}}), a third root of unity.Therefore, it is not a tenth root of unity because its angle is (frac{2pi}{3}), which is not a multiple of (frac{2pi}{10}).Wait, but let me check if (frac{2pi}{3}) is a multiple of (frac{2pi}{10}). Let me see:[frac{2pi}{3} = frac{20pi}{30} = frac{2pi cdot 10}{30} = frac{2pi cdot (10)}{30}]Wait, that's not helpful. Alternatively, let me see if (frac{2pi}{3}) can be written as (frac{2pi k}{10}) for some integer (k):[frac{2pi}{3} = frac{2pi k}{10} implies frac{1}{3} = frac{k}{10} implies k = frac{10}{3}]Since (k) must be an integer, this is not possible. Therefore, the expression is not a tenth root of unity.But wait, the problem says to determine if it is a tenth root of unity. Maybe I made a mistake in the initial simplification. Let me try another approach.Let me consider the original expression:[frac{tan frac{pi}{6} + i}{tan frac{pi}{6} - i}]Let me write this as:[frac{tan theta + i}{tan theta - i} quad text{where} quad theta = frac{pi}{6}]I recall that (tan theta = frac{sin theta}{cos theta}), so:[frac{frac{sin theta}{cos theta} + i}{frac{sin theta}{cos theta} - i} = frac{sin theta + i cos theta}{sin theta - i cos theta}]Let me write this as:[frac{sin theta + i cos theta}{sin theta - i cos theta} = frac{e^{i(frac{pi}{2} - theta)}}{e^{-i(frac{pi}{2} - theta)}} = e^{i2(frac{pi}{2} - theta)} = e^{i(pi - 2theta)}]Wait, let me verify that step. Let me recall that (sin theta = cos(frac{pi}{2} - theta)) and (cos theta = sin(frac{pi}{2} - theta)). So, the numerator is:[sin theta + i cos theta = cosleft(frac{pi}{2} - thetaright) + i sinleft(frac{pi}{2} - thetaright) = e^{ileft(frac{pi}{2} - thetaright)}]Similarly, the denominator is:[sin theta - i cos theta = cosleft(frac{pi}{2} - thetaright) - i sinleft(frac{pi}{2} - thetaright) = e^{-ileft(frac{pi}{2} - thetaright)}]Therefore, the expression becomes:[frac{e^{ileft(frac{pi}{2} - thetaright)}}{e^{-ileft(frac{pi}{2} - thetaright)}} = e^{i2left(frac{pi}{2} - thetaright)} = e^{i(pi - 2theta)}]Substituting (theta = frac{pi}{6}):[e^{ileft(pi - 2 cdot frac{pi}{6}right)} = e^{ileft(pi - frac{pi}{3}right)} = e^{ifrac{2pi}{3}}]Again, same result. So, it's confirmed that the expression is (e^{ifrac{2pi}{3}}), a third root of unity, not a tenth.Therefore, the answer is that it is not a tenth root of unity. However, the problem seems to suggest that it is, so perhaps I made a mistake in interpreting the problem.Wait, let me read the problem again:"Calculate (frac{tan frac{pi}{6} + i}{tan frac{pi}{6} - i}) and determine if it is a tenth root of unity. If so, find the value of (n) such that it equals (cos frac{2n pi}{10} + i sin frac{2n pi}{10}) for some integer (n) between 0 and 9 inclusive."Hmm, maybe I made a mistake in the initial simplification. Let me try another approach.Let me consider the original expression:[frac{tan frac{pi}{6} + i}{tan frac{pi}{6} - i}]Let me write (tan frac{pi}{6} = frac{1}{sqrt{3}}), so:[frac{frac{1}{sqrt{3}} + i}{frac{1}{sqrt{3}} - i}]Let me multiply numerator and denominator by (sqrt{3}):[frac{1 + isqrt{3}}{1 - isqrt{3}}]Now, let me compute this fraction:[frac{1 + isqrt{3}}{1 - isqrt{3}} = frac{(1 + isqrt{3})^2}{(1)^2 + (sqrt{3})^2} = frac{1 + 2isqrt{3} - 3}{1 + 3} = frac{-2 + 2isqrt{3}}{4} = frac{-1 + isqrt{3}}{2}]Again, same result. So, it's confirmed that the expression simplifies to (frac{-1 + isqrt{3}}{2}), which is (e^{ifrac{2pi}{3}}), a third root of unity.Therefore, it is not a tenth root of unity. However, the problem seems to suggest that it is, so perhaps I made a mistake in the problem statement.Wait, let me check the problem statement again:"Calculate (frac{tan frac{pi}{6} + i}{tan frac{pi}{6} - i}) and determine if it is a tenth root of unity. If so, find the value of (n) such that it equals (cos frac{2n pi}{10} + i sin frac{2n pi}{10}) for some integer (n) between 0 and 9 inclusive."Wait, perhaps I misread the problem. Let me check if it's (tan frac{pi}{6}) or (tan frac{pi}{3}). No, it's (tan frac{pi}{6}).Alternatively, maybe the problem is to calculate (frac{tan frac{pi}{6} + i}{tan frac{pi}{6} - i}) and see if it's a tenth root of unity, and if so, find (n).But from my calculations, it's a third root of unity, not a tenth. Therefore, the answer is that it is not a tenth root of unity.However, the initial assistant's answer suggested it was a tenth root of unity with (n=1). Let me see where that might have come from.Wait, perhaps the assistant made a mistake in the simplification. Let me check their steps:They started with:[frac{tan frac{pi}{6} + i}{tan frac{pi}{6} - i} = frac{frac{sin frac{pi}{6}}{cos frac{pi}{6}} + i}{frac{sin frac{pi}{6}}{cos frac{pi}{6}} - i}]Then, simplified (tan frac{pi}{6} = frac{1}{sqrt{3}}):[frac{frac{1}{sqrt{3}} + i}{frac{1}{sqrt{3}} - i}]Then, multiplied numerator and denominator by (frac{1}{sqrt{3}} + i):[frac{frac{1}{3} - 1 + frac{2i}{sqrt{3}}}{1 - 3} = frac{frac{-2}{3} + frac{2i}{sqrt{3}}}{-2} = frac{1}{sqrt{3}} - i]Wait, that's different from my result. Let me see:They have:Numerator after multiplication:[left(frac{1}{sqrt{3}} + iright)left(frac{1}{sqrt{3}} + iright) = frac{1}{3} + frac{2i}{sqrt{3}} - 1 = -frac{2}{3} + frac{2i}{sqrt{3}}]Denominator:[left(frac{1}{sqrt{3}} - iright)left(frac{1}{sqrt{3}} + iright) = frac{1}{3} + 1 = frac{4}{3}]So, the fraction is:[frac{-frac{2}{3} + frac{2i}{sqrt{3}}}{frac{4}{3}} = frac{-2 + 2isqrt{3}}{4} = frac{-1 + isqrt{3}}{2}]Wait, that's the same as my result. But the assistant then wrote:"Utilizing Euler’s formula and knowing (cos frac{pi}{6} = frac{sqrt{3}}{2}) and (sin frac{pi}{6} = frac{1}{2}), we rewrite it as:[frac{1}{sqrt{3}} - i = cos frac{5pi}{6} - isin frac{5pi}{6}]"Wait, that seems incorrect. Because (frac{1}{sqrt{3}} - i) is not equal to (cos frac{5pi}{6} - isin frac{5pi}{6}).Wait, (cos frac{5pi}{6} = -frac{sqrt{3}}{2}) and (sin frac{5pi}{6} = frac{1}{2}), so (cos frac{5pi}{6} - isin frac{5pi}{6} = -frac{sqrt{3}}{2} - ifrac{1}{2}), which is not equal to (frac{1}{sqrt{3}} - i).So, the assistant made a mistake in that step. Instead, the correct expression is (frac{-1 + isqrt{3}}{2}), which is (e^{ifrac{2pi}{3}}), as I calculated.Therefore, the assistant's conclusion that it's a tenth root of unity with (n=1) is incorrect.So, to summarize, the expression simplifies to (e^{ifrac{2pi}{3}}), which is a third root of unity, not a tenth. Therefore, it is not a tenth root of unity.However, the problem asks to determine if it is a tenth root of unity. If it is, find (n). Since it's not, the answer is that it is not a tenth root of unity.But wait, perhaps I made a mistake in interpreting the problem. Let me check again.Wait, the assistant's answer concluded that it is a tenth root of unity with (n=1). Let me see how they arrived at that.They wrote:"Thus, the expression corresponds to:[cos left(-frac{pi}{6}right) + i sin left(-frac{pi}{6}right) = cos frac{11pi}{6} + i sin frac{11pi}{6}]This aligns with[cos frac{22 pi}{10} + i sin frac{22 pi}{10}]Therefore, (n = boxed{11}), but since (n) must be between 0 and 9, we take (n = 11 - 10 = boxed{1})."Wait, but (cos frac{11pi}{6} + i sin frac{11pi}{6}) is equal to (cos frac{pi}{6} - i sin frac{pi}{6}), which is (frac{sqrt{3}}{2} - ifrac{1}{2}), which is not equal to (frac{-1 + isqrt{3}}{2}).So, the assistant made a mistake in equating the expression to (cos frac{11pi}{6} + i sin frac{11pi}{6}).Therefore, the correct conclusion is that the expression is not a tenth root of unity.But wait, let me double-check. Maybe I can express (frac{-1 + isqrt{3}}{2}) as a tenth root of unity.The tenth roots of unity are at angles (frac{2pi k}{10}) for (k=0,1,...,9). Let me compute the angle of (frac{-1 + isqrt{3}}{2}):The real part is (-frac{1}{2}), the imaginary part is (frac{sqrt{3}}{2}). So, the angle is (frac{2pi}{3}), as I calculated earlier.Now, let me see if (frac{2pi}{3}) is a multiple of (frac{2pi}{10}):[frac{2pi}{3} = frac{2pi}{10} cdot frac{10}{3} = frac{2pi}{10} cdot 3.overline{3}]Since (3.overline{3}) is not an integer, (frac{2pi}{3}) is not a multiple of (frac{2pi}{10}). Therefore, (frac{-1 + isqrt{3}}{2}) is not a tenth root of unity.Therefore, the answer is that it is not a tenth root of unity.But the problem seems to suggest that it is, so perhaps I made a mistake in the initial simplification. Let me try one more time.Wait, perhaps I made a mistake in the initial multiplication. Let me recompute the numerator and denominator.Numerator:[left(frac{1}{sqrt{3}} + iright)left(frac{1}{sqrt{3}} + iright) = frac{1}{3} + frac{2i}{sqrt{3}} + i^2 = frac{1}{3} + frac{2i}{sqrt{3}} - 1 = -frac{2}{3} + frac{2i}{sqrt{3}}]Denominator:[left(frac{1}{sqrt{3}} - iright)left(frac{1}{sqrt{3}} + iright) = frac{1}{3} - (i)^2 = frac{1}{3} + 1 = frac{4}{3}]So, the fraction is:[frac{-frac{2}{3} + frac{2i}{sqrt{3}}}{frac{4}{3}} = frac{-2 + 2isqrt{3}}{4} = frac{-1 + isqrt{3}}{2}]Yes, same result. So, it's confirmed.Therefore, the expression is not a tenth root of unity.But the problem asks to determine if it is a tenth root of unity. If so, find (n). Since it's not, the answer is that it is not a tenth root of unity.However, the assistant's answer suggested it was, so perhaps there's a misunderstanding. Alternatively, maybe the problem was misstated.Alternatively, perhaps the original expression was meant to be (frac{tan frac{pi}{5} + i}{tan frac{pi}{5} - i}), which might result in a tenth root of unity. But as per the problem, it's (frac{pi}{6}).Therefore, based on my calculations, the expression is not a tenth root of unity.

answer：Alright, so I have this problem here where I'm given that ( b > a > 0 ) and ( a + b = 1 ). I need to figure out the correct order of these expressions: ( 2ab ), ( dfrac{a^4 - b^4}{a - b} ), ( dfrac{a + b}{2} ), and ( b ). The options are A, B, C, and D, each suggesting a different order.First, let me understand what each expression represents. 1. ( 2ab ): This is twice the product of ( a ) and ( b ). Since ( a ) and ( b ) are positive and less than 1 (because ( a + b = 1 )), their product will be less than both ( a ) and ( b ). So, ( 2ab ) should be a relatively small number.2. ( dfrac{a^4 - b^4}{a - b} ): This looks like a difference of squares or something that can be factored. Maybe I can simplify this expression.3. ( dfrac{a + b}{2} ): Since ( a + b = 1 ), this simplifies to ( dfrac{1}{2} ). So, this is just 0.5.4. ( b ): This is the larger of the two numbers ( a ) and ( b ), so it should be greater than ( a ) and all the other expressions.Okay, let's tackle ( dfrac{a^4 - b^4}{a - b} ) first. I remember that ( a^4 - b^4 ) can be factored as ( (a^2 + b^2)(a + b)(a - b) ). So, if I factor that, the expression becomes:[dfrac{(a^2 + b^2)(a + b)(a - b)}{a - b}]The ( (a - b) ) terms cancel out, leaving me with:[(a^2 + b^2)(a + b)]Since ( a + b = 1 ), this simplifies further to:[a^2 + b^2]So, ( dfrac{a^4 - b^4}{a - b} = a^2 + b^2 ). That's easier to work with.Now, I need to compare ( 2ab ), ( a^2 + b^2 ), ( dfrac{1}{2} ), and ( b ).I know from the AM-GM inequality that for positive numbers ( a ) and ( b ):[dfrac{a + b}{2} geq sqrt{ab}]But since ( a neq b ) (because ( b > a )), the inequality is strict:[dfrac{a + b}{2} > sqrt{ab}]Squaring both sides:[left( dfrac{a + b}{2} right)^2 > ab]Which simplifies to:[dfrac{(a + b)^2}{4} > ab]Given that ( a + b = 1 ), this becomes:[dfrac{1}{4} > ab]So, ( ab < dfrac{1}{4} ), which means ( 2ab < dfrac{1}{2} ).That's useful because ( dfrac{a + b}{2} = dfrac{1}{2} ), so ( 2ab < dfrac{1}{2} ).Next, let's look at ( a^2 + b^2 ). I know that:[(a + b)^2 = a^2 + 2ab + b^2]Since ( a + b = 1 ), this becomes:[1 = a^2 + 2ab + b^2]So, ( a^2 + b^2 = 1 - 2ab ).From earlier, we have ( 2ab < dfrac{1}{2} ), so ( 1 - 2ab > 1 - dfrac{1}{2} = dfrac{1}{2} ). Therefore, ( a^2 + b^2 > dfrac{1}{2} ).So, ( a^2 + b^2 ) is greater than ( dfrac{1}{2} ).Now, let's compare ( a^2 + b^2 ) and ( b ). Since ( b > a ), ( b ) is greater than ( a ), but how does ( a^2 + b^2 ) compare to ( b )?Let me think. Since ( a + b = 1 ), ( a = 1 - b ). So, ( a^2 + b^2 = (1 - b)^2 + b^2 = 1 - 2b + b^2 + b^2 = 1 - 2b + 2b^2 ).So, ( a^2 + b^2 = 2b^2 - 2b + 1 ).I need to see whether ( 2b^2 - 2b + 1 ) is greater than or less than ( b ).Let's subtract ( b ) from both sides:[2b^2 - 2b + 1 - b = 2b^2 - 3b + 1]So, we need to determine the sign of ( 2b^2 - 3b + 1 ).Let me find the roots of ( 2b^2 - 3b + 1 = 0 ):Using the quadratic formula:[b = dfrac{3 pm sqrt{9 - 8}}{4} = dfrac{3 pm 1}{4}]So, the roots are ( b = 1 ) and ( b = dfrac{1}{2} ).Since ( b > a ) and ( a + b = 1 ), ( b ) must be greater than ( dfrac{1}{2} ) because if ( b = dfrac{1}{2} ), then ( a = dfrac{1}{2} ), but ( b > a ), so ( b > dfrac{1}{2} ).Therefore, ( b ) is between ( dfrac{1}{2} ) and 1.Now, let's analyze the quadratic ( 2b^2 - 3b + 1 ). The coefficient of ( b^2 ) is positive, so the parabola opens upwards. The roots are at ( b = dfrac{1}{2} ) and ( b = 1 ). Therefore, for ( b ) between ( dfrac{1}{2} ) and 1, the quadratic is negative because it's below the x-axis between the roots.So, ( 2b^2 - 3b + 1 < 0 ) when ( dfrac{1}{2} < b < 1 ).Therefore, ( a^2 + b^2 - b < 0 ), which implies ( a^2 + b^2 < b ).So, putting it all together:- ( 2ab < dfrac{1}{2} )- ( dfrac{1}{2} < a^2 + b^2 < b )Therefore, the order is:( 2ab < dfrac{a + b}{2} < dfrac{a^4 - b^4}{a - b} < b )Looking back at the options:A: ( 2ab < dfrac{a^4 - b^4}{a - b} < dfrac{a + b}{2} < b )B: ( 2ab < dfrac{a + b}{2} < dfrac{a^4 - b^4}{a - b} < b )C: ( dfrac{a^4 - b^4}{a - b} < 2ab < dfrac{a + b}{2} < b )D: ( 2ab < dfrac{a + b}{2} < b < dfrac{a^4 - b^4}{a - b} )From my deductions, the correct order is ( 2ab < dfrac{1}{2} < a^2 + b^2 < b ), which corresponds to option B.

answer：Okay, so I have this quadratic inequality problem here. It says that the solution set of the quadratic inequality ( ax^2 + bx + c < 0 ) is ( {x | x < frac{1}{3} ) or ( x > frac{1}{2} } ). I need to determine the solution set of another quadratic inequality ( cx^2 - bx + a > 0 ) in terms of ( x ).Alright, let's start by understanding what the given solution set tells us about the quadratic ( ax^2 + bx + c ). The solution set ( x < frac{1}{3} ) or ( x > frac{1}{2} ) means that the quadratic is negative outside the interval ( [frac{1}{3}, frac{1}{2}] ). So, the graph of the quadratic must open upwards or downwards?Wait, if the quadratic is negative outside the interval, that means it opens upwards because the parabola is below the x-axis outside the roots. But actually, hold on. If the quadratic is negative when ( x < frac{1}{3} ) or ( x > frac{1}{2} ), that suggests that the parabola opens upwards because it's below the x-axis on both ends. Hmm, but actually, no. If a quadratic opens upwards, it's positive outside the roots and negative between them. But here, it's negative outside the roots, which would mean it opens downward. So, ( a ) must be negative.Right, so ( a < 0 ). That's important because it affects the direction of the parabola.Now, the roots of the quadratic ( ax^2 + bx + c = 0 ) are ( x = frac{1}{3} ) and ( x = frac{1}{2} ). So, using Vieta's formulas, the sum of the roots is ( frac{1}{3} + frac{1}{2} = frac{5}{6} ), and the product of the roots is ( frac{1}{3} times frac{1}{2} = frac{1}{6} ).Vieta's formulas tell us that for a quadratic ( ax^2 + bx + c = 0 ), the sum of the roots is ( -frac{b}{a} ) and the product is ( frac{c}{a} ). So, we can set up the following equations:1. ( frac{5}{6} = -frac{b}{a} )2. ( frac{1}{6} = frac{c}{a} )From the first equation, solving for ( b ):( frac{5}{6} = -frac{b}{a} ) implies ( b = -frac{5}{6}a ).From the second equation, solving for ( c ):( frac{1}{6} = frac{c}{a} ) implies ( c = frac{1}{6}a ).So, now we have expressions for ( b ) and ( c ) in terms of ( a ). Let's keep that in mind.Now, we need to solve the inequality ( cx^2 - bx + a > 0 ). Let's substitute ( b ) and ( c ) with the expressions we found.Substituting ( b = -frac{5}{6}a ) and ( c = frac{1}{6}a ) into the inequality:( frac{1}{6}a x^2 - (-frac{5}{6}a)x + a > 0 )Simplify the terms:First, distribute the negative sign in the second term:( frac{1}{6}a x^2 + frac{5}{6}a x + a > 0 )Now, let's factor out ( a ) from each term:( a left( frac{1}{6}x^2 + frac{5}{6}x + 1 right) > 0 )Hmm, that looks a bit messy. Maybe we can factor out ( frac{1}{6} ) from the quadratic expression inside the parentheses:( a times frac{1}{6} (x^2 + 5x + 6) > 0 )So, simplifying:( frac{a}{6} (x^2 + 5x + 6) > 0 )Now, since ( a < 0 ), as we established earlier, ( frac{a}{6} ) is also negative. Let's denote ( k = frac{a}{6} ), which is negative.So, the inequality becomes:( k (x^2 + 5x + 6) > 0 )Since ( k < 0 ), we can divide both sides of the inequality by ( k ), but we have to reverse the inequality sign:( x^2 + 5x + 6 < 0 )Alright, now we have a simpler quadratic inequality: ( x^2 + 5x + 6 < 0 ). Let's factor this quadratic:Looking for two numbers that multiply to 6 and add up to 5. Those numbers are 2 and 3.So, ( x^2 + 5x + 6 = (x + 2)(x + 3) )Therefore, the inequality becomes:( (x + 2)(x + 3) < 0 )Now, we need to find the values of ( x ) for which this product is negative. To do that, let's find the critical points where each factor is zero:( x + 2 = 0 ) => ( x = -2 )( x + 3 = 0 ) => ( x = -3 )These critical points divide the real number line into three intervals:1. ( x < -3 )2. ( -3 < x < -2 )3. ( x > -2 )Now, we can test each interval to determine the sign of the product ( (x + 2)(x + 3) ) in that interval.1. **Interval ( x < -3 ):** Let's pick ( x = -4 ). - ( (-4 + 2) = -2 ) (negative) - ( (-4 + 3) = -1 ) (negative) - Product: ( (-2)(-1) = 2 ) (positive) So, the product is positive in this interval.2. **Interval ( -3 < x < -2 ):** Let's pick ( x = -2.5 ). - ( (-2.5 + 2) = -0.5 ) (negative) - ( (-2.5 + 3) = 0.5 ) (positive) - Product: ( (-0.5)(0.5) = -0.25 ) (negative) So, the product is negative in this interval.3. **Interval ( x > -2 ):** Let's pick ( x = 0 ). - ( (0 + 2) = 2 ) (positive) - ( (0 + 3) = 3 ) (positive) - Product: ( (2)(3) = 6 ) (positive) So, the product is positive in this interval.We are looking for where the product is less than zero, which is in the interval ( -3 < x < -2 ).Therefore, the solution set for the inequality ( (x + 2)(x + 3) < 0 ) is ( (-3, -2) ).But wait, let's double-check our steps to make sure we didn't make any mistakes.1. We started with the given solution set and correctly identified that ( a < 0 ).2. Using Vieta's formulas, we found expressions for ( b ) and ( c ) in terms of ( a ).3. Substituted these into the new inequality and simplified correctly.4. Factored the quadratic correctly into ( (x + 2)(x + 3) ).5. Tested the intervals correctly and found the solution set.Everything seems to check out. So, the solution set for ( cx^2 - bx + a > 0 ) is indeed ( (-3, -2) ).**Final Answer**The solution set is boxed{(-3, -2)}.

answer：Okay, so I have this problem where I need to find the maximum value of the expression (frac{ab}{2a + b + 1}) given that (a) and (b) are positive real numbers satisfying ((2a + b)^2 = 1 + 6ab). Hmm, let me try to figure this out step by step.First, I should probably start by understanding the given equation: ((2a + b)^2 = 1 + 6ab). Maybe I can expand the left side and see if I can manipulate it to express one variable in terms of the other or find some relationship between (a) and (b).Expanding the left side: ((2a + b)^2 = 4a^2 + 4ab + b^2). So, the equation becomes:(4a^2 + 4ab + b^2 = 1 + 6ab).Let me rearrange this equation to bring all terms to one side:(4a^2 + 4ab + b^2 - 6ab - 1 = 0).Simplifying the like terms:(4a^2 - 2ab + b^2 - 1 = 0).Hmm, this is a quadratic in terms of (a) and (b). Maybe I can factor this or find a way to express (ab) in terms of (2a + b). Let me see.Wait, the original equation is ((2a + b)^2 = 1 + 6ab). Maybe I can solve for (ab) directly.Starting from ((2a + b)^2 = 1 + 6ab), subtract 1 from both sides:((2a + b)^2 - 1 = 6ab).So, (ab = frac{(2a + b)^2 - 1}{6}).That's useful because the expression we need to maximize is (frac{ab}{2a + b + 1}). If I substitute (ab) from above, I get:(frac{frac{(2a + b)^2 - 1}{6}}{2a + b + 1}).Simplify the numerator:(frac{(2a + b)^2 - 1}{6(2a + b + 1)}).Notice that the numerator is a difference of squares: ((2a + b)^2 - 1^2 = (2a + b - 1)(2a + b + 1)). So, factoring that:(frac{(2a + b - 1)(2a + b + 1)}{6(2a + b + 1)}).The ((2a + b + 1)) terms cancel out:(frac{2a + b - 1}{6}).So, the expression simplifies to (frac{2a + b - 1}{6}). Therefore, to maximize (frac{ab}{2a + b + 1}), I just need to maximize (frac{2a + b - 1}{6}), which is equivalent to maximizing (2a + b - 1).Given that (a) and (b) are positive real numbers, I need to find the maximum value of (2a + b) under the constraint ((2a + b)^2 = 1 + 6ab).Let me denote (t = 2a + b). Then, from the given equation:(t^2 = 1 + 6ab).But from earlier, we have (ab = frac{t^2 - 1}{6}). So, substituting back into the equation:(t^2 = 1 + 6 times frac{t^2 - 1}{6}).Simplify the right side:(t^2 = 1 + (t^2 - 1)).Which simplifies to:(t^2 = t^2).Hmm, that's an identity, which doesn't give me new information. Maybe I need another approach.Wait, perhaps I can use the AM-GM inequality on the terms (2a) and (b). The AM-GM inequality states that for positive real numbers, the arithmetic mean is greater than or equal to the geometric mean. So,(frac{2a + b}{2} geq sqrt{2a times b}).Squaring both sides:(left(frac{2a + b}{2}right)^2 geq 2ab).Multiply both sides by 4:((2a + b)^2 geq 8ab).But from the given equation, ((2a + b)^2 = 1 + 6ab). So, substituting:(1 + 6ab geq 8ab).Subtract (6ab) from both sides:(1 geq 2ab).So, (ab leq frac{1}{2}).Hmm, but how does this help me? I need to relate this back to (t = 2a + b).Wait, earlier I found that the expression simplifies to (frac{t - 1}{6}), so I need to maximize (t). So, perhaps I can find the maximum value of (t = 2a + b) given the constraint.Let me try to express (t) in terms of (ab). From the given equation:(t^2 = 1 + 6ab).So, (ab = frac{t^2 - 1}{6}).But I also have from AM-GM that (ab leq frac{t^2}{8}), since (ab leq left(frac{2a + b}{2}right)^2 / 2 = frac{t^2}{8}).So, combining these two expressions for (ab):(frac{t^2 - 1}{6} leq frac{t^2}{8}).Multiply both sides by 24 to eliminate denominators:(4(t^2 - 1) leq 3t^2).Expanding:(4t^2 - 4 leq 3t^2).Subtract (3t^2) from both sides:(t^2 - 4 leq 0).So,(t^2 leq 4).Taking square roots (since (t) is positive):(t leq 2).Therefore, the maximum value of (t = 2a + b) is 2. So, going back to the expression we need to maximize:(frac{t - 1}{6} leq frac{2 - 1}{6} = frac{1}{6}).So, the maximum value of (frac{ab}{2a + b + 1}) is (frac{1}{6}).But wait, I should check if this maximum is achievable. When does equality hold? For AM-GM, equality holds when (2a = b). So, if (2a = b), then (t = 2a + b = 2a + 2a = 4a). But we also have (t leq 2), so (4a leq 2) which implies (a leq frac{1}{2}).From the given equation, if (b = 2a), then:((2a + b)^2 = (4a)^2 = 16a^2 = 1 + 6ab = 1 + 6a(2a) = 1 + 12a^2).So,(16a^2 = 1 + 12a^2).Subtract (12a^2):(4a^2 = 1).So,(a^2 = frac{1}{4}).Therefore,(a = frac{1}{2}) (since (a) is positive).Then, (b = 2a = 1).So, when (a = frac{1}{2}) and (b = 1), we have (t = 2a + b = 2 times frac{1}{2} + 1 = 1 + 1 = 2), which satisfies the maximum (t = 2).Therefore, the maximum value of (frac{ab}{2a + b + 1}) is indeed (frac{1}{6}).**Final Answer**The maximum value is boxed{dfrac{1}{6}}.

answer：First, determine the total amount of work required to mow the lawn. This can be calculated by multiplying the number of people by the time they take. So, 6 people mowing for 5 hours equals 30 person-hours.Next, to find out how many people are needed to complete the same amount of work in 3 hours, divide the total work by the new time. Therefore, 30 person-hours divided by 3 hours equals 10 people.Finally, subtract the original number of people from the new number of people to determine how many additional people are needed. 10 people minus 6 people equals 4 more people.