💡Okay, so I have this physics problem about two rockets in space with negligible gravity. Both rockets are initially at rest in the observer's reference frame. The masses are m and 9m. A constant force F acts on the smaller rocket (mass m) for a distance d, and it acquires momentum p. Now, the same force F acts on the larger rocket (mass 9m) for the same distance d, and I need to find out how much momentum the larger rocket acquires. The options are p/9, p/3, p, 3p, or 9p.Alright, let me start by recalling some basic physics concepts. Momentum is mass times velocity, right? So p = mv. When a force is applied over a distance, work is done, which translates into kinetic energy. The work done by the force F over distance d is W = F*d. This work becomes the kinetic energy of the rocket, which is (1/2)mv².So for the first rocket with mass m, the work done is F*d, and that equals (1/2)mv². From this, I can solve for the velocity v. Let me write that down:F*d = (1/2)mv²So, v = sqrt(2Fd/m)Then, the momentum p is mv, so:p = m*sqrt(2Fd/m) = sqrt(2Fd*m)Wait, let me check that. If v = sqrt(2Fd/m), then p = m*sqrt(2Fd/m). Let's simplify that:p = sqrt(m² * 2Fd/m) = sqrt(2Fd*m)Yes, that's correct.Now, for the second rocket with mass 9m, the same force F is applied over the same distance d. So, the work done is still F*d, which equals the kinetic energy of the second rocket:F*d = (1/2)*(9m)*v'²Solving for v':v' = sqrt(2Fd/(9m)) = (1/3)*sqrt(2Fd/m)So, the velocity of the second rocket is one-third of the velocity of the first rocket.Now, the momentum p' for the second rocket is:p' = (9m)*v' = 9m*(1/3)*sqrt(2Fd/m) = 3*sqrt(2Fd*m)But wait, from the first rocket, we know that sqrt(2Fd*m) is equal to p. So, p' = 3p.Hmm, that seems straightforward. But let me double-check because sometimes when dealing with forces and distances, it's easy to mix up time and velocity.Another way to approach this is by considering the work-energy principle. The work done on each rocket is the same, F*d, which equals the kinetic energy gained. For the first rocket, (1/2)mv² = F*d, and for the second rocket, (1/2)*(9m)*v'² = F*d.So, setting them equal:(1/2)mv² = (1/2)*(9m)*v'²Simplifying, v² = 9v'², so v = 3v'Therefore, the velocity of the first rocket is three times that of the second rocket.But momentum is mass times velocity. So for the first rocket, p = mv, and for the second rocket, p' = 9m*v'Since v = 3v', then p = m*3v' = 3mv'But p' = 9m*v', so p' = 3*(3mv') = 3pWait, that seems a bit convoluted. Let me clarify:From v = 3v', we have v' = v/3.Then, p' = 9m*v' = 9m*(v/3) = 3mvBut p = mv, so p' = 3p.Yes, that makes sense.Alternatively, thinking about the relationship between momentum and kinetic energy. Kinetic energy is (p²)/(2m). So, for the first rocket, KE = p²/(2m) = F*d.For the second rocket, KE = (p')²/(2*9m) = F*d.Setting them equal:p²/(2m) = (p')²/(18m)Simplify:p²/m = (p')²/(9m)Multiply both sides by 9m:9p² = (p')²So, p' = 3pYep, that's another way to see it.So, regardless of the approach, whether through velocity, kinetic energy, or directly relating momentum, the result is the same: the momentum of the second rocket is three times that of the first rocket.Therefore, the answer should be 3p, which is option D.

💡Okay, so I have this problem where I need to find all real polynomials ( p(x) ) such that ( 1 + p(x) equiv frac{p(x-1) + p(x+1)}{2} ). Hmm, that looks like a functional equation for polynomials. I remember that polynomials have certain properties, especially when it comes to their degrees and coefficients. Maybe I can use that to figure out what ( p(x) ) must be.First, let me rewrite the equation to make it a bit clearer:[ 1 + p(x) = frac{p(x-1) + p(x+1)}{2} ]I can multiply both sides by 2 to eliminate the fraction:[ 2 + 2p(x) = p(x-1) + p(x+1) ]So, rearranging terms:[ p(x+1) - 2p(x) + p(x-1) = 2 ]Hmm, this looks familiar. It reminds me of the second difference in sequences or the second derivative in calculus. For polynomials, the second difference is related to the second derivative, which for a polynomial of degree ( n ) is a polynomial of degree ( n-2 ). Wait, but in this case, the second difference is equal to 2, which is a constant. That suggests that the second derivative of ( p(x) ) is constant. And if the second derivative is constant, that means ( p(x) ) is a quadratic polynomial. Because the second derivative of a quadratic is a constant, right?So, if ( p(x) ) is a quadratic polynomial, let me assume that ( p(x) = ax^2 + bx + c ). Then, I can compute ( p(x+1) ) and ( p(x-1) ) and plug them into the equation to find the coefficients ( a ), ( b ), and ( c ).Let's compute ( p(x+1) ):[ p(x+1) = a(x+1)^2 + b(x+1) + c = a(x^2 + 2x + 1) + b(x + 1) + c = ax^2 + 2ax + a + bx + b + c ]Similarly, ( p(x-1) ):[ p(x-1) = a(x-1)^2 + b(x-1) + c = a(x^2 - 2x + 1) + b(x - 1) + c = ax^2 - 2ax + a + bx - b + c ]Now, let's compute ( p(x+1) + p(x-1) ):[ p(x+1) + p(x-1) = (ax^2 + 2ax + a + bx + b + c) + (ax^2 - 2ax + a + bx - b + c) ][ = 2ax^2 + 2bx + 2a + 2c ]So, the average of ( p(x+1) ) and ( p(x-1) ) is:[ frac{p(x+1) + p(x-1)}{2} = frac{2ax^2 + 2bx + 2a + 2c}{2} = ax^2 + bx + (a + c) ]According to the original equation, this should equal ( 1 + p(x) ):[ 1 + p(x) = 1 + ax^2 + bx + c ]So, setting them equal:[ ax^2 + bx + (a + c) = 1 + ax^2 + bx + c ]Subtracting ( ax^2 + bx + c ) from both sides:[ a = 1 ]So, ( a = 1 ). That means the coefficient of ( x^2 ) is 1. The coefficients ( b ) and ( c ) can be any real numbers because they cancel out in the equation. Therefore, the general solution is:[ p(x) = x^2 + bx + c ]Where ( b ) and ( c ) are real constants. Wait, let me double-check. If I plug ( p(x) = x^2 + bx + c ) back into the original equation, does it hold?Compute ( p(x+1) ):[ (x+1)^2 + b(x+1) + c = x^2 + 2x + 1 + bx + b + c ]Compute ( p(x-1) ):[ (x-1)^2 + b(x-1) + c = x^2 - 2x + 1 + bx - b + c ]Average of ( p(x+1) ) and ( p(x-1) ):[ frac{(x^2 + 2x + 1 + bx + b + c) + (x^2 - 2x + 1 + bx - b + c)}{2} ][ = frac{2x^2 + 2bx + 2 + 2c}{2} ][ = x^2 + bx + (1 + c) ]Now, ( 1 + p(x) = 1 + x^2 + bx + c = x^2 + bx + (1 + c) )Yes, they are equal. So, the solution holds for any real constants ( b ) and ( c ). Therefore, all real polynomials satisfying the given equation are quadratic polynomials of the form ( p(x) = x^2 + bx + c ), where ( b ) and ( c ) are real numbers.**Final Answer**The real polynomials satisfying the equation are all quadratic polynomials of the form boxed{x^2 + bx + c} where ( b ) and ( c ) are real constants.

💡First, I need to calculate the volume of the original sphere with a radius of 6 inches using the formula for the volume of a sphere, which is ( V = frac{4}{3}pi r^3 ).Substituting the given radius:[V = frac{4}{3}pi (6)^3 = frac{4}{3}pi (216) = 288pi text{ cubic inches}]Next, I need to find the volume of the larger sphere, which is triple the volume of the original sphere:[3V = 3 times 288pi = 864pi text{ cubic inches}]Let ( r ) be the radius of the larger sphere. Using the volume formula again:[frac{4}{3}pi r^3 = 864pi][r^3 = 648]To find ( r ), I take the cube root of 648. Recognizing that ( 648 = 6^3 times 2 ), I can simplify:[r = 6sqrt[3]{2}]The diameter of the sphere is twice the radius:[text{Diameter} = 2r = 12sqrt[3]{2}]Finally, expressing the diameter in the form ( asqrt[3]{b} ), I identify ( a = 12 ) and ( b = 2 ). Therefore, ( a + b = 12 + 2 = 14 ).

💡Okay, so I have this problem with three sets: A, B, and C. Let me try to understand each part step by step.First, set A is defined as all real numbers x such that 3 is less than or equal to x, and x is less than 7. So, in interval notation, that would be [3, 7). Got that.Set B is all real numbers x where x is greater than 2 and less than 10. So, that's (2, 10). Alright.Set C is all real numbers x where x is less than a. So, that's (-∞, a). I need to find two things: first, the complement of A intersected with B, and second, the range of a such that A is a subset of C.Starting with part (1): Find (complement of A) ∩ B.Alright, the complement of A with respect to the real numbers. Since A is [3, 7), its complement would be everything not in A. So, that's all x such that x < 3 or x ≥ 7. So, in interval notation, that's (-∞, 3) ∪ [7, ∞).Now, I need to intersect this complement with set B, which is (2, 10). So, let's visualize this. The complement of A is two intervals: from negative infinity to 3, and from 7 to infinity. But set B is only from 2 to 10.So, intersecting (-∞, 3) with (2, 10) would give me (2, 3). Because that's the overlap between those two intervals. Similarly, intersecting [7, ∞) with (2, 10) gives me [7, 10). So, combining these two results, the intersection is (2, 3) union [7, 10).Wait, let me make sure. So, the complement of A is everything less than 3 and everything greater than or equal to 7. Intersecting that with B, which is between 2 and 10, so for the less than 3 part, it's from 2 to 3, since B starts at 2. And for the greater than or equal to 7 part, it's from 7 to 10, since B ends at 10. So, yeah, that makes sense.So, part (1) is (2, 3) union [7, 10). I think that's correct.Moving on to part (2): If A is a subset of C, find the range of a.Set A is [3, 7), and set C is (-∞, a). So, for A to be a subset of C, every element of A must be in C. That means that the entire interval [3, 7) must be contained within (-∞, a).So, the upper bound of A is 7. For C to include all elements of A, a must be greater than or equal to 7. Because if a were less than 7, then the interval [3, 7) would not be entirely contained in (-∞, a). For example, if a was 6, then 6.5 is in A but not in C, which would mean A is not a subset of C.Therefore, a must be at least 7. So, the range of a is a ≥ 7.Wait, let me think again. If a is exactly 7, then C is (-∞, 7). But A is [3, 7), which doesn't include 7. So, 7 is not in A, so C doesn't need to include 7. Hmm, but actually, since A doesn't include 7, as long as a is greater than or equal to 7, C will include all elements of A. If a is exactly 7, then C is (-∞, 7), which still includes all of A because A doesn't reach 7. So, a can be 7 or any number larger than 7.So, yeah, the range is a ≥ 7.I think that's it. Let me just recap:1. The complement of A is (-∞, 3) ∪ [7, ∞). Intersecting that with B, which is (2, 10), gives (2, 3) ∪ [7, 10).2. For A to be a subset of C, which is (-∞, a), a must be at least 7.Yeah, that makes sense. I don't think I made any mistakes here.

💡Okay, so I've got this problem here about finite sets in three-dimensional space and figuring out the minimum and maximum number of planes needed to separate all the points. Let me try to unpack this step by step.First, the problem defines a function f(S) for a finite set S in ℝ³. This function f(S) is the minimum number of planes needed such that each region created by these planes contains at most one point from S. Then, M(n) is the maximum value of f(S) when the size of S is n, and m(n) is the minimum value of f(S) for the same size n. The goal is to find the product M(200) multiplied by m(200).Alright, so I need to figure out what M(200) and m(200) are. Let's start with understanding what f(S) represents. If I have a set S of points in 3D space, f(S) is the smallest number of planes required to ensure that each point is in its own separate region. That means no two points from S can be in the same region after the planes have divided the space.To find M(n), which is the maximum f(S) over all sets S with |S| = n, I need to consider the worst-case scenario. That is, the configuration of points that requires the most planes to separate them. On the flip side, m(n) is the easiest case, where the points can be separated with the fewest planes.Let me think about how planes divide space. Each plane can split the space into two regions. When you add another plane, it can intersect the first plane along a line, creating more regions. In three dimensions, the maximum number of regions created by k planes is given by the formula:R(k) = (k³ + 5k + 6)/6Wait, is that right? Let me recall. The formula for the maximum number of regions divided by k hyperplanes in n-dimensional space is a known combinatorial problem. In three dimensions, it's:R(k) = C(k, 0) + C(k, 1) + C(k, 2) + C(k, 3)Which simplifies to:R(k) = 1 + k + (k(k - 1))/2 + (k(k - 1)(k - 2))/6So, for k planes, the maximum number of regions is R(k) = (k³ + 5k + 6)/6. Let me verify that with small k:For k=1: R(1) = (1 + 5 + 6)/6 = 12/6 = 2. Correct, one plane divides space into two regions.For k=2: R(2) = (8 + 10 + 6)/6 = 24/6 = 4. Correct, two planes can divide space into four regions.For k=3: R(3) = (27 + 15 + 6)/6 = 48/6 = 8. Correct, three planes can divide space into eight regions.So, the formula holds. Therefore, for a given number of regions needed, say n=200, we need to find the smallest k such that R(k) ≥ 200.So, M(n) is essentially the smallest k where R(k) is at least n. Therefore, M(200) is the smallest k such that (k³ + 5k + 6)/6 ≥ 200.Let me solve for k:(k³ + 5k + 6)/6 ≥ 200Multiply both sides by 6:k³ + 5k + 6 ≥ 1200So, k³ + 5k ≥ 1194Let me try k=10:10³ + 5*10 = 1000 + 50 = 1050 < 1194k=11:11³ + 5*11 = 1331 + 55 = 1386 ≥ 1194So, k=11 is the smallest integer where R(k) ≥ 200. Therefore, M(200)=11.Now, what about m(n)? m(n) is the minimum f(S) over all sets S with |S|=n. That is, the easiest case where the points can be separated with the fewest planes.What's the minimum number of planes needed to separate n points in 3D space? Well, in the best case, if all points are colinear, meaning they lie on a straight line, then you can separate them with n-1 planes, each plane slicing between two consecutive points.Wait, but in 3D, a plane can separate two points, but if they are colinear, you can actually separate them with a single plane if they are not too close. Hmm, maybe I need to think more carefully.Actually, in 3D, if points are colinear, you can separate them with a single plane by rotating the plane around the line. But wait, no, a single plane can only separate the space into two halves. So, to separate multiple points on a line, you would need multiple planes.Wait, perhaps the minimum number of planes needed is n-1, similar to how in 2D, you need n-1 lines to separate n colinear points. But in 3D, maybe you can do better?Wait, no, actually, in 3D, if points are colinear, you can still only separate them with planes, but each plane can only separate one point from the rest. So, to separate each point individually, you might still need n-1 planes.Wait, let me think. If you have points on a line, you can use a plane to separate one point from the rest. Then, for the remaining n-1 points, you can use another plane to separate another point, and so on. So, in total, you would need n-1 planes.Alternatively, is there a way to separate multiple points with a single plane? For example, if points are arranged in some specific configuration, maybe a single plane can separate multiple pairs of points.But in the minimal case, where points are arranged in the easiest possible way, which is colinear, you can't do better than n-1 planes because each plane can only separate one point from the rest. So, m(n) would be n-1.Wait, but in 3D, maybe you can separate more points with fewer planes? For example, if points are arranged in a grid or something, but in the minimal case, the easiest is colinear, so m(n)=n-1.Wait, actually, no. Wait, if points are in general position, meaning no three are colinear, then perhaps you can separate them with fewer planes. But m(n) is the minimum over all possible configurations, so the minimal f(S) would be when the points are arranged in the easiest way to separate, which is colinear, requiring n-1 planes.Wait, but actually, in 3D, if points are colinear, you can separate them with n-1 planes, each plane slicing between two points. So, for 200 points, you would need 199 planes. Therefore, m(200)=199.Wait, but that seems contradictory because in 3D, you can separate multiple points with a single plane if they are arranged in certain ways. But in the minimal case, where the points are arranged to be as easy as possible to separate, which is colinear, you still need n-1 planes.Wait, but actually, in 3D, if points are colinear, you can separate them with a single plane by rotating the plane around the line. But no, because a single plane can only divide the space into two halves. So, if you have points on a line, you can use a plane to separate one point from the rest, but then you still have n-1 points on the line, which would require another plane, and so on.Therefore, yes, for colinear points, you need n-1 planes to separate each point individually. So, m(n)=n-1.Wait, but I'm a bit confused because in 2D, for colinear points, you need n-1 lines to separate them, but in 3D, maybe you can do it with fewer planes because planes can be arranged in different orientations.Wait, no, because each plane can only separate the space into two parts. So, to separate each point individually, you still need n-1 planes, each separating one point from the rest.Therefore, m(n)=n-1, so m(200)=199.Wait, but I'm not entirely sure. Let me think again. If you have points arranged in a line, can you separate them with fewer than n-1 planes? For example, if you have four points on a line, can you separate them with two planes instead of three?Let me visualize. If I have four points on a line, A, B, C, D. If I place a plane between A and B, then another plane between C and D, that would separate A, B, C, D into four regions. Wait, no, because the regions would be A, B, C, D, but actually, the regions would be A, B, C, D, but the space is divided into four regions by two planes. Wait, no, two planes in 3D can divide space into four regions, but if the points are on a line, each plane can only separate one point from the rest.Wait, no, actually, if you have two planes, you can separate the line into three segments, each containing one point. Wait, no, because two planes can intersect along a line, and if the points are on that line, then the planes can't separate them because they all lie on the line of intersection.Wait, that's a problem. If points are on the line of intersection of two planes, then those two planes can't separate the points because they all lie on the line where the planes meet.So, in that case, you can't use two planes to separate four colinear points because the points lie on the line of intersection, making it impossible to separate them with just two planes.Therefore, to separate four colinear points, you need three planes, each separating one point from the rest. So, in general, for n colinear points, you need n-1 planes.Therefore, m(n)=n-1.So, m(200)=199.Therefore, M(200)=11 and m(200)=199, so their product is 11*199=2189.Wait, but let me double-check the calculation for M(200). Earlier, I used the formula R(k)=(k³ +5k +6)/6 and found that for k=11, R(11)=232, which is greater than 200, and for k=10, R(10)=176, which is less than 200. So, M(200)=11.Yes, that seems correct.So, putting it all together, M(200)=11 and m(200)=199, so their product is 11*199=2189.I think that's the answer.

💡Alright, so I've got this quadratic function: ( y = x^2 + bx + b - 1 ). There are two parts to this problem. Let me tackle them one by one.**Part (1): When ( y = 0 ), find the value of ( x ) in terms of ( b ).**Okay, so when ( y = 0 ), the equation becomes:[ x^2 + bx + b - 1 = 0 ]This is a quadratic equation in ( x ). To solve for ( x ), I can try factoring it. Let me see if it factors nicely.Looking at the coefficients, I notice that the constant term is ( b - 1 ). Maybe I can factor this as ( (x + 1)(x + (b - 1)) ). Let me check:[ (x + 1)(x + b - 1) = x^2 + (b - 1)x + x + (b - 1) ]Simplifying the middle terms:[ x^2 + (b - 1 + 1)x + (b - 1) = x^2 + bx + (b - 1) ]Yes, that matches the original equation! So, the factored form is:[ (x + 1)(x + b - 1) = 0 ]Setting each factor equal to zero gives the solutions:[ x + 1 = 0 Rightarrow x = -1 ][ x + b - 1 = 0 Rightarrow x = 1 - b ]So, the solutions are ( x = -1 ) and ( x = 1 - b ).**Part (2): The parabola intersects the x-axis at points ( A ) and ( B ). A line through ( E(4, 2) ) intersects the parabola at ( P ) and ( Q ). Points ( P ) and ( Q ) are in the first and fourth quadrants, respectively. Connect ( AP ) and ( AQ ) to intersect the y-axis at ( M(0, m) ) and ( N(0, n) ).**There are two subparts here:**① When ( b < 2 ), find the abscissa ( x_P ) of point ( P ) in terms of ( m ) and ( b ).**Alright, so ( b < 2 ). From part (1), the roots are ( x = -1 ) and ( x = 1 - b ). Since ( b < 2 ), ( 1 - b > -1 ). So, point ( A ) is ( (-1, 0) ) and point ( B ) is ( (1 - b, 0) ).We need to find the x-coordinate of point ( P ). A line passes through ( E(4, 2) ) and intersects the parabola at ( P ) and ( Q ). Since ( P ) is in the first quadrant, its x-coordinate ( x_P ) is positive, and since ( Q ) is in the fourth quadrant, its x-coordinate ( x_Q ) is positive but y-coordinate is negative.Let me denote the equation of line ( PQ ) as ( y = kx + c ). Since it passes through ( E(4, 2) ), substituting:[ 2 = 4k + c Rightarrow c = 2 - 4k ]So, the equation of the line is ( y = kx + 2 - 4k ).This line intersects the parabola ( y = x^2 + bx + b - 1 ). Setting them equal:[ x^2 + bx + b - 1 = kx + 2 - 4k ]Bring all terms to one side:[ x^2 + (b - k)x + (b - 1 - 2 + 4k) = 0 ]Simplify:[ x^2 + (b - k)x + (b - 3 + 4k) = 0 ]Let me denote this quadratic equation as:[ x^2 + (b - k)x + (b - 3 + 4k) = 0 ]Let the roots be ( x_P ) and ( x_Q ). From Vieta's formulas:[ x_P + x_Q = -(b - k) ][ x_P x_Q = b - 3 + 4k ]But I need to relate this to points ( M ) and ( N ). Points ( M ) and ( N ) are where lines ( AP ) and ( AQ ) intersect the y-axis.Let me find the equations of lines ( AP ) and ( AQ ).First, for line ( AP ):Point ( A ) is ( (-1, 0) ) and point ( P ) is ( (x_P, y_P) ). The slope of ( AP ) is:[ m_{AP} = frac{y_P - 0}{x_P - (-1)} = frac{y_P}{x_P + 1} ]So, the equation of line ( AP ) is:[ y = frac{y_P}{x_P + 1}(x + 1) ]This intersects the y-axis at ( x = 0 ):[ y = frac{y_P}{x_P + 1}(0 + 1) = frac{y_P}{x_P + 1} ]So, point ( M ) is ( (0, frac{y_P}{x_P + 1}) ). Therefore, ( m = frac{y_P}{x_P + 1} ).Similarly, for line ( AQ ):Point ( Q ) is ( (x_Q, y_Q) ). The slope of ( AQ ) is:[ m_{AQ} = frac{y_Q - 0}{x_Q - (-1)} = frac{y_Q}{x_Q + 1} ]Equation of line ( AQ ):[ y = frac{y_Q}{x_Q + 1}(x + 1) ]Intersecting the y-axis at ( x = 0 ):[ y = frac{y_Q}{x_Q + 1} ]So, point ( N ) is ( (0, frac{y_Q}{x_Q + 1}) ). Therefore, ( n = frac{y_Q}{x_Q + 1} ).But I need to express ( x_P ) in terms of ( m ) and ( b ). Let me see.From ( m = frac{y_P}{x_P + 1} ), we can express ( y_P = m(x_P + 1) ).But ( y_P ) is also equal to the value of the quadratic at ( x_P ):[ y_P = x_P^2 + b x_P + b - 1 ]So:[ m(x_P + 1) = x_P^2 + b x_P + b - 1 ]Let me rearrange this:[ x_P^2 + b x_P + b - 1 - m x_P - m = 0 ][ x_P^2 + (b - m) x_P + (b - 1 - m) = 0 ]This is a quadratic in ( x_P ). Let me denote it as:[ x_P^2 + (b - m) x_P + (b - 1 - m) = 0 ]Let me denote this equation as (1).But from earlier, when we set the line equal to the parabola, we had:[ x^2 + (b - k)x + (b - 3 + 4k) = 0 ]Let me denote this as equation (2).Equation (1) and equation (2) are both quadratics with roots ( x_P ) and ( x_Q ). Therefore, their coefficients must be proportional.But equation (1) is:[ x_P^2 + (b - m) x_P + (b - 1 - m) = 0 ]Equation (2) is:[ x^2 + (b - k)x + (b - 3 + 4k) = 0 ]Since both have the same roots, their coefficients must be equal (up to a constant multiple, but since the leading coefficient is 1 in both, they must be identical).Therefore:1. Coefficient of ( x ):[ b - m = b - k ]Which simplifies to:[ -m = -k Rightarrow m = k ]2. Constant term:[ b - 1 - m = b - 3 + 4k ]Substitute ( k = m ):[ b - 1 - m = b - 3 + 4m ]Simplify:[ -1 - m = -3 + 4m ]Bring all terms to one side:[ -1 + 3 = 4m + m ][ 2 = 5m ][ m = frac{2}{5} ]Wait, that's interesting. So, ( m = frac{2}{5} ). But the problem asks for ( x_P ) in terms of ( m ) and ( b ). Hmm, perhaps I made a wrong assumption.Wait, no, actually, since ( m = k ), and from equation (2), the sum of roots is ( x_P + x_Q = -(b - k) = -(b - m) ). From equation (1), the sum of roots is also ( -(b - m) ). So, that's consistent.But from equation (1), the product of roots is ( x_P x_Q = b - 1 - m ). From equation (2), the product is ( b - 3 + 4k = b - 3 + 4m ). Therefore:[ b - 1 - m = b - 3 + 4m ]Simplify:[ -1 - m = -3 + 4m ][ -1 + 3 = 4m + m ][ 2 = 5m ][ m = frac{2}{5} ]So, ( m ) is a constant, not depending on ( b ). But the problem says "find the value of the abscissa ( x_P ) of point ( P ); (express in terms of ( m ) and ( b ))". But if ( m ) is a constant, then ( x_P ) can be expressed in terms of ( m ) and ( b ).Wait, but from equation (1):[ x_P^2 + (b - m) x_P + (b - 1 - m) = 0 ]We can solve for ( x_P ) using quadratic formula:[ x_P = frac{-(b - m) pm sqrt{(b - m)^2 - 4 cdot 1 cdot (b - 1 - m)}}{2} ]Simplify discriminant:[ D = (b - m)^2 - 4(b - 1 - m) ][ D = b^2 - 2bm + m^2 - 4b + 4 + 4m ]But since ( m = frac{2}{5} ), substitute:[ D = b^2 - 2b cdot frac{2}{5} + left(frac{2}{5}right)^2 - 4b + 4 + 4 cdot frac{2}{5} ]Simplify:[ D = b^2 - frac{4}{5}b + frac{4}{25} - 4b + 4 + frac{8}{5} ]Combine like terms:- ( b^2 )- ( -frac{4}{5}b - 4b = -frac{4}{5}b - frac{20}{5}b = -frac{24}{5}b )- ( frac{4}{25} + 4 + frac{8}{5} = frac{4}{25} + frac{100}{25} + frac{40}{25} = frac{144}{25} )So, ( D = b^2 - frac{24}{5}b + frac{144}{25} )This can be written as:[ D = left(b - frac{12}{5}right)^2 ]Therefore, square root of D is ( |b - frac{12}{5}| ). Since ( b < 2 ), and ( frac{12}{5} = 2.4 ), so ( b - frac{12}{5} ) is negative, so ( sqrt{D} = frac{12}{5} - b ).Thus, ( x_P ) is:[ x_P = frac{-(b - m) pm left(frac{12}{5} - bright)}{2} ]But since ( P ) is in the first quadrant, ( x_P ) must be positive. Let's consider the '+' sign:[ x_P = frac{-(b - m) + left(frac{12}{5} - bright)}{2} ]Simplify numerator:[ -b + m + frac{12}{5} - b = -2b + m + frac{12}{5} ]So,[ x_P = frac{-2b + m + frac{12}{5}}{2} = -b + frac{m}{2} + frac{6}{5} ]But since ( m = frac{2}{5} ), substitute:[ x_P = -b + frac{2}{10} + frac{6}{5} = -b + frac{1}{5} + frac{6}{5} = -b + frac{7}{5} ]Wait, but the problem asks to express ( x_P ) in terms of ( m ) and ( b ), not substituting ( m ). So, perhaps I should not substitute ( m = frac{2}{5} ).Let me go back. From equation (1):[ x_P^2 + (b - m) x_P + (b - 1 - m) = 0 ]We can write this as:[ x_P^2 + (b - m) x_P + (b - 1 - m) = 0 ]Let me denote ( x_P = t ). Then:[ t^2 + (b - m) t + (b - 1 - m) = 0 ]We can solve for ( t ):[ t = frac{-(b - m) pm sqrt{(b - m)^2 - 4(b - 1 - m)}}{2} ]But earlier, we found that ( m = frac{2}{5} ), so substituting:[ t = frac{-(b - frac{2}{5}) pm sqrt{(b - frac{2}{5})^2 - 4(b - 1 - frac{2}{5})}}{2} ]Simplify inside the square root:[ (b - frac{2}{5})^2 - 4(b - frac{7}{5}) ][ = b^2 - frac{4}{5}b + frac{4}{25} - 4b + frac{28}{5} ][ = b^2 - frac{24}{5}b + frac{144}{25} ]Which is ( left(b - frac{12}{5}right)^2 ), as before.So, ( t = frac{-(b - frac{2}{5}) pm left(frac{12}{5} - bright)}{2} )Taking the positive root (since ( x_P ) is positive):[ t = frac{-(b - frac{2}{5}) + left(frac{12}{5} - bright)}{2} ][ = frac{-b + frac{2}{5} + frac{12}{5} - b}{2} ][ = frac{-2b + frac{14}{5}}{2} ][ = -b + frac{7}{5} ]But since ( m = frac{2}{5} ), we can write:[ x_P = -b + frac{7}{5} = -b + frac{2}{5} + 1 = m + 1 - b ]Wait, that doesn't seem right. Let me check the algebra.Wait, no. From the previous step:[ x_P = -b + frac{7}{5} ]But ( m = frac{2}{5} ), so ( frac{7}{5} = m + 1 ), since ( m = frac{2}{5} Rightarrow m + 1 = frac{7}{5} ). Therefore:[ x_P = -b + m + 1 ]But the problem asks for ( x_P ) in terms of ( m ) and ( b ). So, perhaps it's better to leave it as:[ x_P = -b + frac{7}{5} ]But since ( m = frac{2}{5} ), we can express ( frac{7}{5} = m + 1 ). So:[ x_P = -b + m + 1 ]But I'm not sure if this is the simplest form. Alternatively, since ( m = frac{2}{5} ), it's a constant, so ( x_P ) is linear in ( b ).Wait, but the problem says "express in terms of ( m ) and ( b )", so perhaps I should keep ( m ) as a variable, not substituting its value. Let me try that.From equation (1):[ x_P^2 + (b - m) x_P + (b - 1 - m) = 0 ]Let me solve for ( x_P ):[ x_P = frac{-(b - m) pm sqrt{(b - m)^2 - 4(b - 1 - m)}}{2} ]Simplify the discriminant:[ D = (b - m)^2 - 4(b - 1 - m) ][ = b^2 - 2bm + m^2 - 4b + 4 + 4m ][ = b^2 - 2bm + m^2 - 4b + 4m + 4 ]This doesn't seem to factor nicely. Maybe I made a mistake earlier.Wait, earlier I found that ( m = frac{2}{5} ), which is a constant, so perhaps ( x_P ) can be expressed as ( x_P = m - b + 1 ). Let me check:From equation (1):[ x_P^2 + (b - m) x_P + (b - 1 - m) = 0 ]If ( x_P = m - b + 1 ), let's plug it in:[ (m - b + 1)^2 + (b - m)(m - b + 1) + (b - 1 - m) = 0 ]Expand ( (m - b + 1)^2 ):[ m^2 - 2bm + b^2 + 2m - 2b + 1 ]Expand ( (b - m)(m - b + 1) ):[ (b - m)(m - b + 1) = -(m - b)(m - b + 1) = -(m - b)^2 - (m - b) ][ = -m^2 + 2bm - b^2 - m + b ]Now, add all terms:[ (m^2 - 2bm + b^2 + 2m - 2b + 1) + (-m^2 + 2bm - b^2 - m + b) + (b - 1 - m) = 0 ]Simplify term by term:- ( m^2 - m^2 = 0 )- ( -2bm + 2bm = 0 )- ( b^2 - b^2 = 0 )- ( 2m - m - m = 0 )- ( -2b + b + b = 0 )- ( 1 - 1 = 0 )So, indeed, it sums to 0. Therefore, ( x_P = m - b + 1 ) is a solution.Since ( x_P ) is positive and in the first quadrant, this must be the correct root. Therefore, the abscissa ( x_P ) is:[ x_P = m - b + 1 ]**② When ( b = -3 ), prove that ( OM cdot ON ) is a constant value.**Alright, set ( b = -3 ). The quadratic function becomes:[ y = x^2 - 3x - 4 ]Find points ( A ) and ( B ). From part (1), the roots are ( x = -1 ) and ( x = 1 - b = 1 - (-3) = 4 ). So, ( A(-1, 0) ) and ( B(4, 0) ).We need to find ( OM ) and ( ON ), which are the y-intercepts of lines ( AP ) and ( AQ ), respectively.From part ①, when ( b = -3 ), ( x_P = m - (-3) + 1 = m + 4 ). So, ( x_P = m + 4 ).Since ( P ) lies on the parabola, its y-coordinate is:[ y_P = (m + 4)^2 - 3(m + 4) - 4 ]Simplify:[ y_P = m^2 + 8m + 16 - 3m - 12 - 4 ][ y_P = m^2 + 5m ]So, point ( P ) is ( (m + 4, m^2 + 5m) ).Similarly, for point ( Q ), since the line passes through ( E(4, 2) ), and the line intersects the parabola at ( P ) and ( Q ), we can find ( x_Q ) in terms of ( n ).But perhaps there's a better way. Since ( OM = m ) and ( ON = n ), we need to find ( OM cdot ON = m cdot n ).From part ①, we have ( x_P = m - b + 1 ). When ( b = -3 ), ( x_P = m + 4 ).Similarly, for point ( Q ), the abscissa ( x_Q ) can be found similarly. Let me denote ( x_Q = n - b + 1 = n + 4 ).But wait, is this correct? Let me think.Actually, from part ①, the abscissa ( x_P ) is ( m - b + 1 ). Similarly, for point ( Q ), if we denote ( x_Q ), it would be ( n - b + 1 ). But since ( Q ) is in the fourth quadrant, its y-coordinate is negative, so ( n ) would be negative.But perhaps there's a relationship between ( m ) and ( n ).Alternatively, since points ( P ) and ( Q ) lie on the line passing through ( E(4, 2) ), we can use the fact that the slope from ( E ) to ( P ) is the same as the slope from ( E ) to ( Q ).Let me denote the slope as ( k ). Then, the equation of line ( PQ ) is:[ y - 2 = k(x - 4) ]This line intersects the parabola ( y = x^2 - 3x - 4 ). Setting them equal:[ x^2 - 3x - 4 = kx - 4k + 2 ]Bring all terms to one side:[ x^2 - (3 + k)x - 4 + 4k - 2 = 0 ]Simplify:[ x^2 - (3 + k)x + (4k - 6) = 0 ]The roots of this quadratic are ( x_P ) and ( x_Q ). From Vieta's formulas:[ x_P + x_Q = 3 + k ][ x_P x_Q = 4k - 6 ]But from part ①, ( x_P = m + 4 ) and ( x_Q = n + 4 ). Wait, no, earlier I thought ( x_P = m + 4 ), but actually, when ( b = -3 ), ( x_P = m - (-3) + 1 = m + 4 ). Similarly, ( x_Q = n - (-3) + 1 = n + 4 ).So, ( x_P = m + 4 ) and ( x_Q = n + 4 ).From Vieta's:[ x_P + x_Q = (m + 4) + (n + 4) = m + n + 8 = 3 + k ][ x_P x_Q = (m + 4)(n + 4) = mn + 4m + 4n + 16 = 4k - 6 ]So, we have two equations:1. ( m + n + 8 = 3 + k Rightarrow k = m + n + 5 )2. ( mn + 4m + 4n + 16 = 4k - 6 )Substitute ( k ) from equation 1 into equation 2:[ mn + 4m + 4n + 16 = 4(m + n + 5) - 6 ]Simplify RHS:[ 4m + 4n + 20 - 6 = 4m + 4n + 14 ]So, equation becomes:[ mn + 4m + 4n + 16 = 4m + 4n + 14 ]Subtract ( 4m + 4n ) from both sides:[ mn + 16 = 14 ][ mn = -2 ]Therefore, ( OM cdot ON = |m cdot n| = | -2 | = 2 ), which is a constant.So, regardless of the value of ( b ), when ( b = -3 ), the product ( OM cdot ON ) is always 2.