💡Okay, so I have this function f(x) defined in two parts. If x is less than 1, f(x) is 2 raised to the power of x. But if x is greater than or equal to 1, f(x) is equal to f(x minus 1). Hmm, that seems a bit recursive. I need to find f(log₂9). Alright, let's break this down step by step.First, I should figure out what log₂9 is. I know that log₂9 is the exponent to which I have to raise 2 to get 9. Since 2^3 is 8 and 2^4 is 16, log₂9 must be somewhere between 3 and 4. Specifically, log₂9 is approximately 3.169925001. So, it's definitely greater than 1. That means I have to use the second part of the function definition, which is f(x) = f(x - 1).So, f(log₂9) = f(log₂9 - 1). Let me compute log₂9 - 1. Since log₂9 is approximately 3.1699, subtracting 1 gives me approximately 2.1699. Hmm, that's still greater than 1, so I have to apply the function definition again. Therefore, f(2.1699) = f(2.1699 - 1) = f(1.1699).Now, 1.1699 is still greater than 1, so I need to subtract 1 once more. f(1.1699) = f(1.1699 - 1) = f(0.1699). Okay, 0.1699 is less than 1, so now I can use the first part of the function definition, which is f(x) = 2^x.So, f(0.1699) = 2^0.1699. Let me calculate that. I know that 2^0.1699 is approximately equal to 9/8 because 9 divided by 8 is 1.125, and 2^0.1699 is roughly 1.125. Let me verify that.Wait, 2^0.1699. Let me compute it more accurately. Since 2^0.1699 is the same as e^(0.1699 * ln2). Let's compute 0.1699 * ln2. ln2 is approximately 0.6931. So, 0.1699 * 0.6931 ≈ 0.1176. Then, e^0.1176 ≈ 1.1247, which is approximately 9/8 because 9 divided by 8 is 1.125. So, yes, 2^0.1699 is roughly 9/8.Therefore, f(log₂9) = 9/8. Let me check if I did everything correctly. Starting from log₂9, which is about 3.1699, subtracting 1 three times to get to 0.1699, then computing 2^0.1699 as approximately 9/8. That seems right.Wait, but let me think again. Is there a more straightforward way to see this without approximating? Maybe using properties of logarithms and exponents.I know that log₂9 is equal to log₂(9/8 * 8) = log₂(9/8) + log₂8. Since log₂8 is 3, log₂9 = log₂(9/8) + 3. So, log₂9 - 3 = log₂(9/8). Therefore, f(log₂9) = f(log₂(9/8)). Since log₂(9/8) is less than 1, f(log₂(9/8)) = 2^{log₂(9/8)} = 9/8. That's a cleaner way to see it without approximating.So, yes, f(log₂9) is 9/8, which is option A. I think that's the correct answer.**Final Answer**boxed{A}

💡Alright, so I have this problem here about a circle and a point inside it. Let me try to parse it step by step.First, the circle C is given by the equation ( x^2 + y^2 - 2x + 2my + m^2 - 7 = 0 ). They mention that ( m > 0 ), so I know m is a positive real number. The point P is (2, 0), and it's inside this circle. There's a line AB passing through P that intersects the circle at points A and B. The maximum area of triangle ABC is 4, and I need to find the range of m.Hmm, okay. Let me recall that the general equation of a circle is ( (x - h)^2 + (y - k)^2 = r^2 ), where (h, k) is the center and r is the radius. So, maybe I should rewrite the given equation in this standard form.Starting with the given equation:( x^2 + y^2 - 2x + 2my + m^2 - 7 = 0 ).Let me group the x terms and y terms:( (x^2 - 2x) + (y^2 + 2my) + (m^2 - 7) = 0 ).Now, I'll complete the square for the x terms and the y terms.For the x terms: ( x^2 - 2x ). To complete the square, take half of -2, which is -1, square it to get 1. So, add and subtract 1:( (x^2 - 2x + 1 - 1) = (x - 1)^2 - 1 ).For the y terms: ( y^2 + 2my ). Half of 2m is m, square it to get ( m^2 ). So, add and subtract ( m^2 ):( (y^2 + 2my + m^2 - m^2) = (y + m)^2 - m^2 ).Putting it all back into the equation:( (x - 1)^2 - 1 + (y + m)^2 - m^2 + m^2 - 7 = 0 ).Simplify the constants:-1 - 7 = -8, and the ( -m^2 + m^2 ) cancels out.So, the equation becomes:( (x - 1)^2 + (y + m)^2 - 8 = 0 ),which simplifies to:( (x - 1)^2 + (y + m)^2 = 8 ).Okay, so the circle has center at (1, -m) and radius ( sqrt{8} = 2sqrt{2} ).Now, point P is (2, 0), and it's inside the circle. So, the distance from P to the center should be less than the radius.Let me compute the distance between (2, 0) and (1, -m):Distance formula: ( sqrt{(2 - 1)^2 + (0 - (-m))^2} = sqrt{1 + m^2} ).Since P is inside the circle, this distance must be less than the radius:( sqrt{1 + m^2} < 2sqrt{2} ).Squaring both sides:( 1 + m^2 < 8 ),so ( m^2 < 7 ),which gives ( m < sqrt{7} ).Since m is positive, ( 0 < m < sqrt{7} ).But wait, the problem mentions the maximum area of triangle ABC is 4. So, I need to consider that as well.Hmm, triangle ABC. Wait, in the problem statement, it says "the line AB passing through point P intersects circle C at points A and B." So, AB is a chord passing through P, and C is the center of the circle, right?So, triangle ABC is formed by points A, B, and C, where C is the center (1, -m). So, the triangle is formed by two points on the circle and the center.I need to find the maximum area of triangle ABC, which is given as 4. So, I need to express the area of triangle ABC in terms of m and find the range of m such that the maximum area is 4.Let me recall that the area of a triangle can be given by ( frac{1}{2}ab sin theta ), where a and b are sides, and θ is the angle between them.In this case, for triangle ABC, sides AC and BC are both radii of the circle, so they are equal to ( 2sqrt{2} ). So, the area would be ( frac{1}{2} times 2sqrt{2} times 2sqrt{2} times sin theta ), where θ is the angle at point C.Simplifying that, ( frac{1}{2} times 8 times sin theta = 4 sin theta ).So, the area is ( 4 sin theta ). The maximum area occurs when ( sin theta ) is maximum, which is 1. So, the maximum area is 4, which is given. So, that tells me that the maximum area is achieved when θ is 90 degrees, making triangle ABC a right-angled triangle.Wait, but in the problem, it's given that the maximum area is 4. So, that must mean that the maximum possible area is 4, which occurs when θ is 90 degrees.But how does this relate to the position of point P?Hmm, because the line AB passes through P, which is inside the circle. So, the chord AB passes through P, and the triangle ABC has maximum area when AB is such that angle ACB is 90 degrees.But how does the position of P affect this? Maybe the length of AB is related to the distance from P to the center?Wait, perhaps I need to consider the length of AB. Since AB is a chord passing through P, the length of AB depends on the distance from the center to the chord.Wait, the formula for the length of a chord is ( 2sqrt{r^2 - d^2} ), where d is the distance from the center to the chord.In this case, the chord AB passes through P, so the distance from the center to AB is less than or equal to the distance from the center to P, which is ( sqrt{1 + m^2} ).Wait, but if AB passes through P, then the distance from the center to AB is less than or equal to the distance from the center to P.But in our case, when the area is maximized, the angle ACB is 90 degrees, which would mean that AB is such that triangle ABC is right-angled at C.In that case, AB would be the hypotenuse of the right-angled triangle, so AB would be ( sqrt{AC^2 + BC^2} = sqrt{(2sqrt{2})^2 + (2sqrt{2})^2} = sqrt{8 + 8} = sqrt{16} = 4 ).So, the length of AB when the area is maximized is 4.But AB is a chord of the circle. The length of a chord is related to the distance from the center to the chord. The formula is ( AB = 2sqrt{r^2 - d^2} ), where d is the distance from the center to the chord.So, in this case, when AB = 4, we have:( 4 = 2sqrt{(2sqrt{2})^2 - d^2} ),which simplifies to:( 2 = sqrt{8 - d^2} ),squaring both sides:( 4 = 8 - d^2 ),so ( d^2 = 4 ),thus ( d = 2 ).So, the distance from the center to the chord AB is 2 when the area is maximized.But since AB passes through P, the distance from the center to AB is less than or equal to the distance from the center to P.Wait, but in this case, when the area is maximized, the distance from the center to AB is exactly 2. So, that must mean that the distance from the center to P is greater than or equal to 2, because the chord AB passes through P, so the minimal distance from the center to AB is less than or equal to the distance from the center to P.Wait, actually, the distance from the center to AB is the minimal distance from the center to the chord, which is less than or equal to the distance from the center to any point on the chord, including P.So, in this case, since the distance from the center to AB is 2, and P lies on AB, the distance from the center to P must be greater than or equal to 2.But earlier, we found that the distance from the center to P is ( sqrt{1 + m^2} ), which must be greater than or equal to 2.So, ( sqrt{1 + m^2} geq 2 ),squaring both sides:( 1 + m^2 geq 4 ),so ( m^2 geq 3 ),which gives ( m geq sqrt{3} ).But we also had earlier that ( m < sqrt{7} ) because P is inside the circle.So, combining these two results, ( sqrt{3} leq m < sqrt{7} ).Therefore, the range of m is from ( sqrt{3} ) to ( sqrt{7} ), not including ( sqrt{7} ).Let me just double-check my reasoning.1. I converted the circle equation to standard form, found center (1, -m) and radius ( 2sqrt{2} ).2. Calculated the distance from P(2,0) to the center: ( sqrt{1 + m^2} ).3. Since P is inside the circle, this distance must be less than the radius: ( sqrt{1 + m^2} < 2sqrt{2} ), leading to ( m < sqrt{7} ).4. The maximum area of triangle ABC is 4, which occurs when angle ACB is 90 degrees, making AB = 4.5. The length of AB is related to the distance from the center to AB: ( AB = 2sqrt{r^2 - d^2} ). Plugging in AB=4 and r=2√2, we found d=2.6. Since AB passes through P, the distance from the center to AB (which is 2) must be less than or equal to the distance from the center to P. Thus, ( sqrt{1 + m^2} geq 2 ), leading to ( m geq sqrt{3} ).7. Combining both inequalities, ( sqrt{3} leq m < sqrt{7} ).Yes, that seems consistent. I think that's the correct range for m.

💡Okay, so I need to find the range of the function ( g(x) = frac{1}{x^2 + x} ). Hmm, let's see. I remember that the range of a function is all the possible output values it can produce. So, I need to figure out all the possible values that ( g(x) ) can take as ( x ) varies over its domain.First, I should probably figure out the domain of ( g(x) ). The function is a fraction, so the denominator can't be zero because division by zero is undefined. So, I need to find where ( x^2 + x ) is not zero.Let me solve ( x^2 + x = 0 ). Factoring, I get ( x(x + 1) = 0 ). So, the solutions are ( x = 0 ) and ( x = -1 ). That means ( g(x) ) is undefined at ( x = 0 ) and ( x = -1 ). Therefore, the domain of ( g(x) ) is all real numbers except ( x = 0 ) and ( x = -1 ).Now, to find the range, I need to consider the possible values of ( g(x) ). Since ( g(x) = frac{1}{x^2 + x} ), the output depends on the values of ( x^2 + x ). Let me denote ( y = x^2 + x ). Then, ( g(x) = frac{1}{y} ). So, the range of ( g(x) ) depends on the range of ( y ).Let's analyze ( y = x^2 + x ). This is a quadratic function. I can rewrite it in vertex form to understand its behavior better. Completing the square:( y = x^2 + x = x^2 + x + frac{1}{4} - frac{1}{4} = left(x + frac{1}{2}right)^2 - frac{1}{4} ).So, the vertex form is ( y = left(x + frac{1}{2}right)^2 - frac{1}{4} ). This tells me that the parabola opens upwards (since the coefficient of the squared term is positive) and has its vertex at ( left(-frac{1}{2}, -frac{1}{4}right) ).Since the parabola opens upwards, the minimum value of ( y ) is ( -frac{1}{4} ), and it can increase without bound as ( x ) moves away from the vertex in either direction. However, we need to remember that ( y ) cannot be zero because ( g(x) ) is undefined at ( x = 0 ) and ( x = -1 ).So, the range of ( y = x^2 + x ) is ( [-frac{1}{4}, 0) cup (0, infty) ). But wait, ( y ) can't be zero, so it's actually ( [-frac{1}{4}, 0) cup (0, infty) ).Now, since ( g(x) = frac{1}{y} ), the range of ( g(x) ) will depend on the range of ( y ). Let's consider the two intervals separately:1. When ( y ) is in ( (0, infty) ): - As ( y ) approaches zero from the positive side, ( frac{1}{y} ) approaches positive infinity. - As ( y ) increases without bound, ( frac{1}{y} ) approaches zero. - Therefore, ( frac{1}{y} ) can take any positive value from just above zero to positive infinity.2. When ( y ) is in ( [-frac{1}{4}, 0) ): - Here, ( y ) is negative. So, ( frac{1}{y} ) will also be negative. - As ( y ) approaches zero from the negative side, ( frac{1}{y} ) approaches negative infinity. - The minimum value of ( y ) is ( -frac{1}{4} ), so the maximum value of ( frac{1}{y} ) in this interval is ( frac{1}{-frac{1}{4}} = -4 ). - Therefore, ( frac{1}{y} ) can take any negative value from just below zero to negative infinity, but it's bounded above by -4.Wait a minute, that seems conflicting with my initial thought that the range is ( (0, infty) ). Maybe I made a mistake earlier.Let me double-check. The function ( y = x^2 + x ) can take negative values between ( -frac{1}{4} ) and 0, and positive values greater than 0. So, ( frac{1}{y} ) can be negative when ( y ) is negative and positive when ( y ) is positive.But looking back at the original function ( g(x) = frac{1}{x^2 + x} ), I need to consider both cases where ( y ) is positive and where ( y ) is negative.However, when ( y ) is negative, ( frac{1}{y} ) is negative. So, does that mean the range includes negative numbers as well?Wait, but when I initially thought about it, I considered only the positive outputs because ( x^2 ) is always positive, but I forgot that ( x ) can make the denominator negative.So, perhaps the range is not just ( (0, infty) ), but also includes negative values.But let's think about it again. When ( y ) is negative, ( frac{1}{y} ) is negative. So, the range should include both positive and negative values.But wait, let's test specific values.When ( x = 1 ), ( y = 1 + 1 = 2 ), so ( g(1) = frac{1}{2} ).When ( x = -2 ), ( y = 4 - 2 = 2 ), so ( g(-2) = frac{1}{2} ).When ( x = -0.5 ), which is the vertex, ( y = (-0.5)^2 + (-0.5) = 0.25 - 0.5 = -0.25 ), so ( g(-0.5) = frac{1}{-0.25} = -4 ).When ( x ) approaches 0 from the positive side, ( y ) approaches 0 from the positive side, so ( g(x) ) approaches positive infinity.When ( x ) approaches 0 from the negative side, ( y ) approaches 0 from the negative side, so ( g(x) ) approaches negative infinity.Similarly, when ( x ) approaches -1 from the positive side, ( y ) approaches 0 from the positive side, so ( g(x) ) approaches positive infinity.When ( x ) approaches -1 from the negative side, ( y ) approaches 0 from the negative side, so ( g(x) ) approaches negative infinity.So, putting this all together, ( g(x) ) can take any positive value greater than 0, and any negative value less than 0, but it cannot take values between -4 and 0.Wait, no. When ( y ) is in ( [-frac{1}{4}, 0) ), ( frac{1}{y} ) is in ( (-infty, -4] ).And when ( y ) is in ( (0, infty) ), ( frac{1}{y} ) is in ( (0, infty) ).So, the range of ( g(x) ) is ( (-infty, -4] cup (0, infty) ).But wait, earlier I thought the range was ( (0, infty) ), but now I'm seeing that it also includes negative values less than or equal to -4.So, I need to reconcile this.Let me think again about the function ( y = x^2 + x ). Its minimum value is ( -frac{1}{4} ), so ( y ) can be as low as ( -frac{1}{4} ), but not lower. Therefore, ( frac{1}{y} ) can be as low as ( -4 ) (since ( frac{1}{-frac{1}{4}} = -4 )), but it can't be higher than that in the negative side.So, the negative part of the range is ( (-infty, -4] ), and the positive part is ( (0, infty) ).Therefore, the range of ( g(x) ) is all real numbers except those between ( -4 ) and 0, excluding 0.So, in interval notation, that's ( (-infty, -4] cup (0, infty) ).But wait, let me verify this with some test points.If ( x = -1.5 ), then ( y = (-1.5)^2 + (-1.5) = 2.25 - 1.5 = 0.75 ), so ( g(-1.5) = frac{1}{0.75} approx 1.333 ), which is positive.If ( x = -0.25 ), then ( y = (-0.25)^2 + (-0.25) = 0.0625 - 0.25 = -0.1875 ), so ( g(-0.25) = frac{1}{-0.1875} approx -5.333 ), which is less than -4.If ( x = -0.75 ), then ( y = (-0.75)^2 + (-0.75) = 0.5625 - 0.75 = -0.1875 ), so ( g(-0.75) = frac{1}{-0.1875} approx -5.333 ), same as above.Wait, but earlier I thought the minimum value of ( y ) is ( -frac{1}{4} ), which is -0.25, so ( frac{1}{y} ) would be -4. But in these test points, I'm getting values less than -4.Hmm, that seems contradictory.Wait, let's calculate ( y ) at ( x = -0.5 ), which is the vertex. ( y = (-0.5)^2 + (-0.5) = 0.25 - 0.5 = -0.25 ). So, ( g(-0.5) = frac{1}{-0.25} = -4 ).So, at ( x = -0.5 ), ( g(x) = -4 ).But when I took ( x = -0.25 ) and ( x = -0.75 ), I got ( g(x) approx -5.333 ), which is less than -4.Wait, that suggests that ( g(x) ) can be less than -4, which contradicts the idea that the minimum value of ( y ) is -0.25, leading to ( g(x) ) being -4.But actually, ( y ) can be less than -0.25? Wait, no. The vertex is at ( y = -0.25 ), which is the minimum value of ( y ). So, ( y ) can't be less than -0.25. Therefore, ( frac{1}{y} ) can't be less than -4.But in my test points, I got ( g(x) = -5.333 ), which is less than -4. That must mean I made a mistake in my calculations.Wait, let me recalculate ( y ) at ( x = -0.25 ):( y = (-0.25)^2 + (-0.25) = 0.0625 - 0.25 = -0.1875 ).Wait, that's correct. So, ( y = -0.1875 ), which is greater than -0.25. So, ( frac{1}{y} = frac{1}{-0.1875} approx -5.333 ), which is less than -4.But wait, if ( y ) can be as low as -0.25, then ( frac{1}{y} ) can be as low as -4. But in this case, ( y = -0.1875 ) is greater than -0.25, so ( frac{1}{y} ) is less than -4.Wait, that doesn't make sense. If ( y ) is greater than -0.25 (closer to zero), then ( frac{1}{y} ) is less than -4 (more negative). But if ( y ) is exactly -0.25, ( frac{1}{y} = -4 ). So, as ( y ) approaches -0.25 from above (i.e., from values closer to zero), ( frac{1}{y} ) approaches -4 from below (i.e., more negative).Wait, that means that the range of ( g(x) ) when ( y ) is negative is ( (-infty, -4] ).Because as ( y ) approaches -0.25 from above (i.e., from values like -0.1875, -0.2, etc.), ( frac{1}{y} ) approaches -4 from below (i.e., becomes more negative). So, the maximum value of ( frac{1}{y} ) in the negative range is -4, and it can go to negative infinity as ( y ) approaches 0 from the negative side.Therefore, the range of ( g(x) ) is ( (-infty, -4] cup (0, infty) ).But wait, earlier I thought the range was ( (0, infty) ), but now I'm seeing that it also includes negative values less than or equal to -4.So, to confirm, let's consider the behavior of ( g(x) ):- As ( x ) approaches 0 from the positive side, ( y = x^2 + x ) approaches 0 from the positive side, so ( g(x) ) approaches positive infinity.- As ( x ) approaches 0 from the negative side, ( y = x^2 + x ) approaches 0 from the negative side, so ( g(x) ) approaches negative infinity.- As ( x ) approaches -1 from the positive side, ( y = x^2 + x ) approaches 0 from the positive side, so ( g(x) ) approaches positive infinity.- As ( x ) approaches -1 from the negative side, ( y = x^2 + x ) approaches 0 from the negative side, so ( g(x) ) approaches negative infinity.- At ( x = -0.5 ), ( y = -0.25 ), so ( g(x) = -4 ).Therefore, the function ( g(x) ) can take any positive value greater than 0, and any negative value less than or equal to -4.So, the range of ( g(x) ) is ( (-infty, -4] cup (0, infty) ).But wait, in my initial analysis, I thought the range was ( (0, infty) ), but now I'm seeing that it also includes negative values. So, I need to make sure I'm not missing anything.Let me consider the function ( g(x) = frac{1}{x^2 + x} ) and see if it can take any value between -4 and 0.Suppose ( g(x) = c ), where ( c ) is a constant. Then, ( frac{1}{x^2 + x} = c ), which implies ( x^2 + x = frac{1}{c} ).So, ( x^2 + x - frac{1}{c} = 0 ).This is a quadratic equation in ( x ). For real solutions to exist, the discriminant must be non-negative.The discriminant ( D = 1^2 - 4 times 1 times (-frac{1}{c}) = 1 + frac{4}{c} ).For real solutions, ( D geq 0 ), so ( 1 + frac{4}{c} geq 0 ).Let's solve this inequality:( 1 + frac{4}{c} geq 0 )Multiply both sides by ( c ). But we have to be careful about the sign of ( c ).Case 1: ( c > 0 )Then, multiplying both sides by ( c ) (positive, so inequality sign remains):( c + 4 geq 0 )Since ( c > 0 ), ( c + 4 > 4 > 0 ). So, this inequality is always true for ( c > 0 ). Therefore, for any positive ( c ), there exists a real solution ( x ), meaning ( g(x) ) can take any positive value.Case 2: ( c < 0 )Then, multiplying both sides by ( c ) (negative, so inequality sign reverses):( c + 4 leq 0 )So, ( c leq -4 )Therefore, for ( c < 0 ), real solutions exist only if ( c leq -4 ).So, combining both cases, the range of ( g(x) ) is ( (-infty, -4] cup (0, infty) ).Therefore, the function ( g(x) = frac{1}{x^2 + x} ) can take any positive real value and any real value less than or equal to -4.So, the range is all real numbers except those between -4 and 0, excluding 0.In interval notation, that's ( (-infty, -4] cup (0, infty) ).I think that's the correct range.

💡First, identify the cost of the computer game, which is 66.Next, calculate the total cost of the three movie tickets by multiplying the number of tickets by the price per ticket: 3 tickets * 12 per ticket = 36.Finally, add the cost of the computer game and the total cost of the movie tickets to find the total amount spent on entertainment: 66 + 36 = 102.

💡Okay, so I have this problem here, and I need to find the value of P and then use that to find Q. Let me try to break it down step by step.First, the problem gives me an expression involving a fraction with a numerator and a denominator. The numerator is a sum of terms like 1×2×4, 2×4×8, 3×6×12, and so on up to 1999×3998×7996. The denominator is the sum of cubes from 1³ to 1999³. Then, I need to take the cube root of that fraction to get P.Alright, let me write that down:P = [(1×2×4 + 2×4×8 + 3×6×12 + ... + 1999×3998×7996) / (1³ + 2³ + 3³ + ... + 1999³)]^(1/3)Hmm, okay. So, the numerator is a sum of products, each term being three numbers multiplied together. Let me see if I can find a pattern here.Looking at the first few terms:1×2×4 = 82×4×8 = 643×6×12 = 216Wait a second, these look like multiples of 8. Let me check:1×2×4 = 8 = 8×1³2×4×8 = 64 = 8×2³3×6×12 = 216 = 8×3³Oh! So each term in the numerator is 8 times n³, where n is the term number. So, the numerator is the sum from n=1 to n=1999 of 8n³.So, numerator = 8(1³ + 2³ + 3³ + ... + 1999³)And the denominator is just the sum of cubes from 1³ to 1999³.So, the fraction becomes:[8(1³ + 2³ + 3³ + ... + 1999³)] / (1³ + 2³ + 3³ + ... + 1999³) = 8So, the fraction simplifies to 8, and then taking the cube root of 8 gives P.Cube root of 8 is 2, so P = 2.Okay, that wasn't too bad. Now, moving on to the second part.The problem says: If (x - P)(x - 2Q) - 1 = 0 has two integral roots, find the value of Q.Alright, so I need to find Q such that this quadratic equation has two integer roots. I already know that P is 2, so let me substitute that in.So, the equation becomes:(x - 2)(x - 2Q) - 1 = 0Let me expand this:First, multiply (x - 2)(x - 2Q):= x² - 2Qx - 2x + 4QThen, subtract 1:= x² - (2Q + 2)x + 4Q - 1 = 0So, the quadratic equation is:x² - (2Q + 2)x + (4Q - 1) = 0Now, for this quadratic to have two integral roots, the discriminant must be a perfect square. The discriminant D is given by:D = [-(2Q + 2)]² - 4×1×(4Q - 1)Let me compute that:D = (2Q + 2)² - 4(4Q - 1)First, expand (2Q + 2)²:= 4Q² + 8Q + 4Then, expand 4(4Q - 1):= 16Q - 4So, D = (4Q² + 8Q + 4) - (16Q - 4)Simplify:= 4Q² + 8Q + 4 - 16Q + 4Combine like terms:= 4Q² - 8Q + 8Hmm, so D = 4Q² - 8Q + 8For the quadratic to have integer roots, D must be a perfect square. Let me factor out a 4:D = 4(Q² - 2Q + 2)So, D = 4(Q² - 2Q + 2)Since 4 is a perfect square, Q² - 2Q + 2 must also be a perfect square.Let me set Q² - 2Q + 2 = k², where k is an integer.So, Q² - 2Q + 2 = k²Let me rearrange this equation:Q² - 2Q + (2 - k²) = 0This is a quadratic in Q. For Q to be an integer, the discriminant of this quadratic must be a perfect square.Wait, but this might get complicated. Maybe there's another approach.Alternatively, since D = 4(Q² - 2Q + 2) must be a perfect square, let me denote D = m², where m is an integer.So, 4(Q² - 2Q + 2) = m²Divide both sides by 4:Q² - 2Q + 2 = (m/2)²But since Q is an integer, m must be even. Let me set m = 2n, where n is an integer.Then, (m/2)² = n²So, Q² - 2Q + 2 = n²So, Q² - 2Q + 2 - n² = 0Let me rearrange this:Q² - 2Q + (2 - n²) = 0Again, this is a quadratic in Q. For Q to be integer, the discriminant must be a perfect square.The discriminant for this quadratic in Q is:Δ = (-2)² - 4×1×(2 - n²) = 4 - 8 + 4n² = 4n² - 4 = 4(n² - 1)So, Δ = 4(n² - 1)For Q to be integer, Δ must be a perfect square. So, 4(n² - 1) must be a perfect square.Let me set 4(n² - 1) = k², where k is an integer.So, 4(n² - 1) = k²Divide both sides by 4:n² - 1 = (k/2)²Since n is integer, k must be even. Let me set k = 2m, where m is an integer.Then, (k/2)² = m²So, n² - 1 = m²Rearranged:n² - m² = 1This factors as:(n - m)(n + m) = 1Since n and m are integers, the only solutions are when both factors are 1 or both are -1.Case 1:n - m = 1n + m = 1Adding both equations:2n = 2 => n = 1Subtracting:2m = 0 => m = 0Case 2:n - m = -1n + m = -1Adding both equations:2n = -2 => n = -1Subtracting:2m = 0 => m = 0So, the only solutions are n = 1, m = 0 or n = -1, m = 0.Let me check n = 1:From earlier, Q² - 2Q + 2 = n² = 1So, Q² - 2Q + 2 = 1Q² - 2Q + 1 = 0This factors as (Q - 1)² = 0, so Q = 1Similarly, for n = -1:Q² - 2Q + 2 = (-1)² = 1Same equation: Q² - 2Q + 1 = 0 => Q = 1So, the only integer solution is Q = 1.Let me verify this.If Q = 1, then the quadratic equation is:(x - 2)(x - 2×1) - 1 = 0 => (x - 2)(x - 2) - 1 = 0Which simplifies to:(x - 2)² - 1 = 0 => (x - 2 - 1)(x - 2 + 1) = 0 => (x - 3)(x - 1) = 0So, the roots are x = 3 and x = 1, which are integers. Perfect.Therefore, Q = 1.So, to recap:First, I found that P = 2 by simplifying the given fraction.Then, I substituted P = 2 into the quadratic equation and found that Q must be 1 for the equation to have integer roots.I think that's it. Let me just double-check my steps to make sure I didn't make any mistakes.1. Simplified the numerator as 8 times the sum of cubes, which cancels with the denominator, giving 8. Cube root of 8 is 2, so P = 2. That seems correct.2. Substituted P = 2 into the quadratic equation, expanded it, and found the discriminant must be a perfect square. Through some algebra, I found that Q must be 1. Checked by plugging back in, and it worked.Yeah, I think that's solid.

💡Let's break down the problem step by step to find the ratio of the volume of the second container to the volume of the third container.1. **Define Variables:** - Let ( C ) be the volume of the first container. - Let ( D ) be the volume of the second container. - Let ( E ) be the volume of the third container.2. **First Pour:** - The first container is filled to (frac{3}{5}) of its capacity, so the amount of water is (frac{3}{5}C). - This water is poured into the second container, making it (frac{2}{3}) full. Therefore, (frac{3}{5}C = frac{2}{3}D).3. **Solve for ( C ) in terms of ( D ):** [ frac{3}{5}C = frac{2}{3}D implies C = frac{2}{3} times frac{5}{3}D = frac{10}{9}D ]4. **Second Pour:** - Water is transferred from the second container to the third container until the third container is (frac{1}{2}) full. - The amount of water transferred is (frac{1}{2}E).5. **Determine the Remaining Water in the Second Container:** - The second container was initially (frac{2}{3}D) full. - After transferring (frac{1}{2}E), the remaining water in the second container is (frac{2}{3}D - frac{1}{2}E).6. **Express ( E ) in terms of ( D ):** - Since the amount transferred is (frac{1}{2}E), we have: [ frac{1}{2}E = frac{2}{3}D - left(frac{2}{3}D - frac{1}{2}Eright) ] - Simplifying, we find that ( E = frac{3}{2}D ).7. **Find the Ratio ( frac{D}{E} ):** [ frac{D}{E} = frac{D}{frac{3}{2}D} = frac{2}{3} ]8. **Conclusion:** - The ratio of the volume of the second container to the volume of the third container is (frac{2}{3}).The final answer is (boxed{textbf{(B) } frac{2}{3}}).