💡Alright, so I've got this problem here about a sine function, and I need to figure out its analytical expression. Let me try to break it down step by step.First, the function is given as ( y = 2sin(omega x + varphi) ), where ( omega > 0 ) and ( 0 < varphi < pi ). They mention that the distance between two adjacent highest points on its graph is ( pi ). Hmm, I remember that the distance between two highest points on a sine graph is the period of the function. So, the period ( T ) is ( pi ).I recall that the period of a sine function ( y = Asin(omega x + varphi) ) is given by ( T = frac{2pi}{omega} ). So, if the period is ( pi ), I can set up the equation:[frac{2pi}{omega} = pi]Solving for ( omega ), I divide both sides by ( pi ):[frac{2}{omega} = 1 implies omega = 2]Okay, so now I know that ( omega = 2 ). So the function becomes ( y = 2sin(2x + varphi) ).Next, the problem says that the graph of the function ( f(x) ) is shifted to the left by ( frac{pi}{6} ) units. Shifting a function left by ( a ) units involves replacing ( x ) with ( x + a ) in the function. So, shifting ( f(x) ) left by ( frac{pi}{6} ) would give us:[fleft(x + frac{pi}{6}right)]But wait, actually, the function ( f(x) ) is being shifted, so if the original function is ( y = 2sin(2x + varphi) ), then shifting it left by ( frac{pi}{6} ) would result in:[y = 2sinleft(2left(x + frac{pi}{6}right) + varphiright)]Let me simplify that:[y = 2sinleft(2x + frac{pi}{3} + varphiright)]So, the shifted function is ( y = 2sinleft(2x + frac{pi}{3} + varphiright) ).Now, the resulting graph is symmetric about the axis ( x = frac{pi}{4} ). Hmm, symmetry about a vertical line for a sine function usually means that the line is either a peak, a trough, or a point of inflection. Since it's a sine function, which is symmetric about its midline, but in this case, it's shifted, so the axis of symmetry is ( x = frac{pi}{4} ).I think that for the graph to be symmetric about ( x = frac{pi}{4} ), the function must satisfy the condition:[fleft(frac{pi}{4} + aright) = fleft(frac{pi}{4} - aright)]for any ( a ). This is the definition of symmetry about the line ( x = frac{pi}{4} ).So, applying this to our shifted function:[2sinleft(2left(frac{pi}{4} + aright) + frac{pi}{3} + varphiright) = 2sinleft(2left(frac{pi}{4} - aright) + frac{pi}{3} + varphiright)]Simplify both sides:Left side:[2sinleft(frac{pi}{2} + 2a + frac{pi}{3} + varphiright)]Right side:[2sinleft(frac{pi}{2} - 2a + frac{pi}{3} + varphiright)]So, we have:[sinleft(frac{pi}{2} + 2a + frac{pi}{3} + varphiright) = sinleft(frac{pi}{2} - 2a + frac{pi}{3} + varphiright)]I know that ( sin(theta) = sin(pi - theta) ), so for the above equality to hold for all ( a ), the arguments must either be equal or supplementary angles. But since this has to hold for any ( a ), the only way this can be true is if the coefficients of ( a ) are zero. Let me think about that.Wait, actually, if we consider that the equality must hold for all ( a ), then the functions inside the sine must be equal or supplementary. Let me write the equation again:[sinleft(frac{pi}{2} + frac{pi}{3} + varphi + 2aright) = sinleft(frac{pi}{2} + frac{pi}{3} + varphi - 2aright)]Let me denote ( C = frac{pi}{2} + frac{pi}{3} + varphi ). Then the equation becomes:[sin(C + 2a) = sin(C - 2a)]Using the sine identity ( sin(A) = sin(B) ) implies ( A = B + 2kpi ) or ( A = pi - B + 2kpi ) for some integer ( k ).So, applying this:Case 1:[C + 2a = C - 2a + 2kpi]Simplify:[4a = 2kpi implies a = frac{kpi}{2}]But this has to hold for all ( a ), which is only possible if ( k = 0 ) and ( a = 0 ), which is trivial and doesn't give us any new information.Case 2:[C + 2a = pi - (C - 2a) + 2kpi]Simplify:[C + 2a = pi - C + 2a + 2kpi]Subtract ( 2a ) from both sides:[C = pi - C + 2kpi]Bring ( C ) to the right:[2C = pi + 2kpi]So,[C = frac{pi}{2} + kpi]But ( C = frac{pi}{2} + frac{pi}{3} + varphi ), so:[frac{pi}{2} + frac{pi}{3} + varphi = frac{pi}{2} + kpi]Subtract ( frac{pi}{2} ) from both sides:[frac{pi}{3} + varphi = kpi]Therefore,[varphi = kpi - frac{pi}{3}]Given that ( 0 < varphi < pi ), let's find possible values for ( k ).If ( k = 1 ):[varphi = pi - frac{pi}{3} = frac{2pi}{3}]Which is within the range ( 0 < varphi < pi ).If ( k = 0 ):[varphi = -frac{pi}{3}]Which is negative, so it doesn't fit.If ( k = 2 ):[varphi = 2pi - frac{pi}{3} = frac{5pi}{3}]Which is greater than ( pi ), so it's out of range.Therefore, the only valid solution is ( varphi = frac{2pi}{3} ).So, going back to the function ( f(x) ), which was shifted left by ( frac{pi}{6} ) to get the symmetric graph. Wait, actually, let me clarify: the original function is ( y = 2sin(2x + varphi) ), and shifting it left by ( frac{pi}{6} ) gives us ( y = 2sin(2(x + frac{pi}{6}) + varphi) = 2sin(2x + frac{pi}{3} + varphi) ). We found that ( varphi = frac{2pi}{3} ), so plugging that in:[y = 2sinleft(2x + frac{pi}{3} + frac{2pi}{3}right) = 2sin(2x + pi)]Wait, ( frac{pi}{3} + frac{2pi}{3} = pi ), so the function becomes ( 2sin(2x + pi) ). But ( sin(2x + pi) = -sin(2x) ), so this simplifies to ( -2sin(2x) ). Hmm, that seems a bit odd, but let's check.Wait, maybe I made a mistake here. Let me retrace:We had ( f(x) = 2sin(2x + varphi) ). Then, shifting left by ( frac{pi}{6} ) gives us ( 2sin(2(x + frac{pi}{6}) + varphi) = 2sin(2x + frac{pi}{3} + varphi) ). Then, we determined that ( varphi = frac{2pi}{3} ), so substituting back:[2sinleft(2x + frac{pi}{3} + frac{2pi}{3}right) = 2sin(2x + pi)]Yes, that's correct. So, ( 2sin(2x + pi) ) is indeed ( -2sin(2x) ). But looking at the options, none of them are ( -2sin(2x) ). Wait, let me check the options again.The options are:A: ( f(x) = 2sin(x + frac{pi}{6}) )B: ( f(x) = 2sin(2x + frac{2pi}{3}) )C: ( f(x) = 2sin(x + frac{pi}{3}) )D: ( f(x) = 2sin(2x + frac{5pi}{6}) )Hmm, none of these are ( -2sin(2x) ). Did I make a mistake somewhere?Wait, maybe I misinterpreted the shifting. Let me think again. The function ( f(x) ) is shifted left by ( frac{pi}{6} ) to get the symmetric graph. So, actually, the original function is ( f(x) = 2sin(2x + varphi) ), and after shifting left by ( frac{pi}{6} ), it becomes ( 2sin(2(x + frac{pi}{6}) + varphi) = 2sin(2x + frac{pi}{3} + varphi) ). This shifted function is symmetric about ( x = frac{pi}{4} ).But when I solved for ( varphi ), I got ( varphi = frac{2pi}{3} ). So, plugging that back into ( f(x) ), which is the original function before shifting, we have:[f(x) = 2sin(2x + frac{2pi}{3})]Which is option B. Wait, but earlier when I substituted into the shifted function, I got ( 2sin(2x + pi) ). So, perhaps I confused the original function with the shifted one.Let me clarify:- Original function: ( f(x) = 2sin(2x + varphi) )- Shifted function: ( f(x + frac{pi}{6}) = 2sin(2(x + frac{pi}{6}) + varphi) = 2sin(2x + frac{pi}{3} + varphi) )This shifted function is symmetric about ( x = frac{pi}{4} ). So, we set up the equation for symmetry and found ( varphi = frac{2pi}{3} ). Therefore, the original function ( f(x) ) is:[f(x) = 2sin(2x + frac{2pi}{3})]Which is exactly option B. So, I think I was confused earlier when I thought the shifted function was ( -2sin(2x) ), but actually, the original function is ( 2sin(2x + frac{2pi}{3}) ), which is option B.Let me just verify that shifting this function left by ( frac{pi}{6} ) gives a function symmetric about ( x = frac{pi}{4} ).Shifting ( f(x) = 2sin(2x + frac{2pi}{3}) ) left by ( frac{pi}{6} ) gives:[fleft(x + frac{pi}{6}right) = 2sinleft(2left(x + frac{pi}{6}right) + frac{2pi}{3}right) = 2sinleft(2x + frac{pi}{3} + frac{2pi}{3}right) = 2sin(2x + pi)]As I had before, which is ( -2sin(2x) ). Now, is this function symmetric about ( x = frac{pi}{4} )?Let me check if ( fleft(frac{pi}{4} + aright) = fleft(frac{pi}{4} - aright) ).Compute ( fleft(frac{pi}{4} + aright) = -2sinleft(2left(frac{pi}{4} + aright)right) = -2sinleft(frac{pi}{2} + 2aright) )Similarly, ( fleft(frac{pi}{4} - aright) = -2sinleft(2left(frac{pi}{4} - aright)right) = -2sinleft(frac{pi}{2} - 2aright) )Now, ( sinleft(frac{pi}{2} + 2aright) = cos(2a) ) and ( sinleft(frac{pi}{2} - 2aright) = cos(2a) ). Therefore,[fleft(frac{pi}{4} + aright) = -2cos(2a) quad text{and} quad fleft(frac{pi}{4} - aright) = -2cos(2a)]So, they are equal. Therefore, the shifted function is indeed symmetric about ( x = frac{pi}{4} ). That checks out.Therefore, the original function ( f(x) ) is ( 2sin(2x + frac{2pi}{3}) ), which is option B.I think that makes sense now. Initially, I was confused about whether I was looking at the original function or the shifted one, but clarifying that helped me see that option B is indeed the correct answer.**Final Answer**boxed{B}

💡Okay, so I have this problem here: (1 + x)^n is expanded as a0 + a1x + a2x² + ... + anx^n. They tell me that a2 equals a3, and I need to find n. Hmm, okay, let me think about how to approach this.First, I remember that when you expand (1 + x)^n using the binomial theorem, each coefficient ak is equal to the combination C(n, k). So, a2 is C(n, 2) and a3 is C(n, 3). The problem says these two are equal, so I can set them equal to each other: C(n, 2) = C(n, 3).Now, I need to recall the formula for combinations. C(n, k) is n! / (k!(n - k)!). So, plugging in the values, C(n, 2) is n! / (2!(n - 2)!) and C(n, 3) is n! / (3!(n - 3)!). Setting these equal gives me:n! / (2!(n - 2)!) = n! / (3!(n - 3)!).Okay, so I can cancel out the n! on both sides since they're the same and non-zero. That leaves me with:1 / (2!(n - 2)!) = 1 / (3!(n - 3)!).Simplifying the factorials, 2! is 2 and 3! is 6. So, substituting those in:1 / (2(n - 2)!) = 1 / (6(n - 3)!).Hmm, now I need to solve for n. Let's cross-multiply to get rid of the fractions:6(n - 3)! = 2(n - 2)!.Divide both sides by 2 to simplify:3(n - 3)! = (n - 2)!.Now, I know that (n - 2)! is equal to (n - 2)(n - 3)!. So, substituting that in:3(n - 3)! = (n - 2)(n - 3)!.Oh, look at that! Both sides have (n - 3)! so I can divide both sides by (n - 3)! as long as (n - 3)! is not zero. Since factorials are defined for non-negative integers, n must be at least 3, so (n - 3)! is fine.Dividing both sides by (n - 3)! gives:3 = n - 2.So, solving for n:n = 3 + 2 = 5.Wait, let me double-check that. If n is 5, then C(5, 2) is 10 and C(5, 3) is also 10. Yeah, that works. So n must be 5.I think that's it. I went through the steps, set the combinations equal, simplified, and solved for n. It makes sense because when n is 5, the coefficients for x² and x³ are both 10, which satisfies the condition a2 = a3.

💡First, I need to determine the original number that was mistakenly divided by 1.8. Given that the quotient was 21 and the remainder was 2, I can set up the equation:Number = Quotient × Divisor + RemainderNumber = 21 × 1.8 + 2Calculating this gives:Number = 37.8 + 2Number = 39.8Now that I have the original number, I can find the correct value by dividing it by 18:Correct Quotient = Number / 18Correct Quotient = 39.8 / 18Performing the division:Correct Quotient ≈ 2.21111111111...Finally, rounding this to two decimal places:Correct Quotient ≈ 2.21

💡First, I recognize that (80) can be expressed as (2^4 cdot 5). Using the properties of logarithms, I can break down (log_{10}80) into (log_{10}(2^4 cdot 5)), which simplifies to (4log_{10}2 + log_{10}5). Given that (log_{10}2 = x) and (log_{10}5 = y), this becomes (4x + y).Next, I need to find (log_{8}80). Since (8 = 2^3), I can express (log_{10}8) as (3log_{10}2), which is (3x).Using the change of base formula, (log_{8}80 = frac{log_{10}80}{log_{10}8}), I substitute the values I found:[log_{8}80 = frac{4x + y}{3x}]This matches option B.

💡Okay, so I have this problem: "How many y-intercepts does the graph of the parabola x = 3y² - 5y + 1 have?" Hmm, let me think about how to approach this.First, I remember that a y-intercept occurs where the graph crosses the y-axis. And on the y-axis, the value of x is always 0. So, to find the y-intercepts, I need to set x equal to 0 and solve for y. That makes sense.So, starting with the equation of the parabola:x = 3y² - 5y + 1Since we're looking for y-intercepts, I'll set x to 0:0 = 3y² - 5y + 1Now, this is a quadratic equation in terms of y. The general form of a quadratic equation is ay² + by + c = 0, where a, b, and c are constants. In this case, a is 3, b is -5, and c is 1.I recall that the number of real solutions to a quadratic equation can be determined by the discriminant, which is given by the formula:Discriminant (Δ) = b² - 4acLet me calculate the discriminant for this equation:Δ = (-5)² - 4 * 3 * 1Δ = 25 - 12Δ = 13Okay, so the discriminant is 13. Now, I remember that if the discriminant is positive, there are two distinct real solutions; if it's zero, there's exactly one real solution; and if it's negative, there are no real solutions.In this case, since Δ = 13, which is positive, there are two distinct real solutions. That means the quadratic equation 3y² - 5y + 1 = 0 has two real roots. Therefore, the graph of the parabola x = 3y² - 5y + 1 will cross the y-axis at two distinct points.Wait, just to make sure I'm not making a mistake here. Let me double-check my calculations.Starting with the discriminant:Δ = b² - 4acΔ = (-5)² - 4 * 3 * 1Δ = 25 - 12Δ = 13Yes, that's correct. So, the discriminant is indeed positive, meaning two real solutions. Therefore, there are two y-intercepts.I think that's it. So, the graph of the given parabola has two y-intercepts.

💡Okay, so I have this problem where I need to find the maximum value of the minimum among four expressions: 1/a, 2/b, 4/c, and the cube root of abc. All a, b, c are positive numbers. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, let me write down what I need to find. I need to maximize M, where M is the minimum of those four expressions. So, M is the smallest among 1/a, 2/b, 4/c, and cube root of abc. My goal is to make this smallest value as large as possible.Maybe I can start by setting all these expressions equal to each other because if I can make them all the same, that might give me the maximum M. Let me assume that 1/a = 2/b = 4/c = cube root(abc) = M. Then, I can express a, b, c in terms of M.So, if 1/a = M, then a = 1/M. Similarly, 2/b = M implies b = 2/M, and 4/c = M implies c = 4/M. Now, let's plug these into the cube root expression.Cube root of abc would be cube root of (1/M * 2/M * 4/M) = cube root of (8/M^3). So, cube root(8/M^3) = 2/M. But we also have cube root(abc) = M, so 2/M = M. Solving for M, we get M^2 = 2, so M = sqrt(2). Since M is positive, we take the positive root.Wait, so if I set all four expressions equal to sqrt(2), then a = 1/sqrt(2), b = 2/sqrt(2) = sqrt(2), and c = 4/sqrt(2) = 2*sqrt(2). Let me check if this works.Calculating 1/a: 1/(1/sqrt(2)) = sqrt(2). 2/b: 2/sqrt(2) = sqrt(2). 4/c: 4/(2*sqrt(2)) = 2/sqrt(2) = sqrt(2). Cube root of abc: cube root of (1/sqrt(2) * sqrt(2) * 2*sqrt(2)). Let's compute that:1/sqrt(2) * sqrt(2) = 1. Then, 1 * 2*sqrt(2) = 2*sqrt(2). Cube root of 2*sqrt(2). Hmm, 2*sqrt(2) is 2^(3/2), so cube root of 2^(3/2) is 2^(1/2) = sqrt(2). Perfect, so all four expressions equal sqrt(2). So, M is indeed sqrt(2).But wait, is this the maximum possible? Could there be a higher M? Let me think. Suppose I try to make M larger than sqrt(2). Then, 1/a, 2/b, 4/c would have to be at least M, which is larger than sqrt(2). But then, the cube root of abc would have to be at least M as well. Let's see if that's possible.If M is larger than sqrt(2), say M = sqrt(2) + ε for some ε > 0, then a would be less than 1/sqrt(2), b less than 2/sqrt(2), and c less than 4/sqrt(2). Then, abc would be less than (1/sqrt(2))*(2/sqrt(2))*(4/sqrt(2)) = (1*2*4)/(sqrt(2)*sqrt(2)*sqrt(2)) = 8/(2*sqrt(2)) = 4/sqrt(2) = 2*sqrt(2). So, cube root of abc would be less than cube root(2*sqrt(2)) = sqrt(2). But we wanted cube root(abc) to be at least M, which is larger than sqrt(2). That's a contradiction. Therefore, M cannot be larger than sqrt(2).So, sqrt(2) is indeed the maximum value of M. Therefore, the maximum value of the minimum of those four expressions is sqrt(2).**Final Answer**The maximum value is boxed{sqrt{2}}.