💡Okay, so I have this problem where a father and his son are skating around a circular rink. Initially, they're both going in the same direction, and the father overtakes the son from time to time. Then, the son starts skating in the opposite direction, and now they meet five times more often. I need to find the ratio of their skating speeds.Hmm, let me think. So, when they're moving in the same direction, the father is faster, so he overtakes the son. The time between overtakes depends on their relative speed. If they're moving in the same direction, the relative speed is the difference between their speeds. When the son starts moving in the opposite direction, their relative speed becomes the sum of their speeds because they're moving towards each other.The problem says they meet five times more often after the son changes direction. So, the frequency of meetings increases by a factor of five. Since frequency is inversely proportional to the time between meetings, and the time between meetings is related to the relative speed, I can set up a ratio of their relative speeds before and after the son changes direction.Let me denote the son's speed as ( v_s ) and the father's speed as ( v_f ). When they're moving in the same direction, the relative speed is ( v_f - v_s ). When they're moving in opposite directions, the relative speed is ( v_f + v_s ). The frequency of meetings is proportional to the relative speed. So, if they meet five times more often when moving towards each other, the ratio of their relative speeds should be 5. That gives me the equation:[frac{v_f + v_s}{v_f - v_s} = 5]Now, I need to solve for the ratio ( frac{v_f}{v_s} ). Let me set ( frac{v_f}{v_s} = k ), so ( v_f = k v_s ). Substituting this into the equation:[frac{k v_s + v_s}{k v_s - v_s} = 5]Factor out ( v_s ) from numerator and denominator:[frac{(k + 1) v_s}{(k - 1) v_s} = 5]The ( v_s ) terms cancel out:[frac{k + 1}{k - 1} = 5]Now, solve for ( k ):Multiply both sides by ( k - 1 ):[k + 1 = 5(k - 1)]Expand the right side:[k + 1 = 5k - 5]Bring all terms to one side:[k + 1 - 5k + 5 = 0]Combine like terms:[-4k + 6 = 0]Solve for ( k ):[-4k = -6 k = frac{6}{4} = frac{3}{2}]So, the ratio of the father's speed to the son's speed is ( frac{3}{2} ).Wait, let me double-check my steps. I set up the relative speeds correctly, right? When moving in the same direction, the relative speed is the difference, and when moving opposite, it's the sum. The frequency is proportional to the relative speed, so the ratio of frequencies is equal to the ratio of relative speeds. That makes sense because if you're moving towards each other, you cover the distance between each other faster, so you meet more often.Plugging in ( k = frac{3}{2} ) back into the original equation:[frac{frac{3}{2} + 1}{frac{3}{2} - 1} = frac{frac{5}{2}}{frac{1}{2}} = 5]Yes, that checks out. So, the ratio is indeed ( frac{3}{2} ).I think I've got it right. The father is 1.5 times faster than the son.

💡First, identify the total amount of bread served, which is 0.9 loaves.Next, note the amount of wheat bread served, which is 0.5 loaves.To find the amount of white bread served, subtract the wheat bread from the total bread:0.9 - 0.5 = 0.4.Therefore, 0.4 loaves of white bread were served.

💡Alright, so I have this problem here where I need to find the number of intersection points between two functions: f(x) = 2 ln x and g(x) = x² - 4x + 5. The options are A: 3, B: 2, C: 1, D: 0. Hmm, okay, let's break this down step by step.First off, to find the intersection points of two functions, I know I need to set them equal to each other and solve for x. So, that would mean solving the equation 2 ln x = x² - 4x + 5. The number of real solutions to this equation will tell me how many times the two graphs intersect.Now, solving this equation algebraically might be tricky because it involves both a logarithmic function and a quadratic function. These types of equations usually don't have straightforward algebraic solutions, especially when they're mixed like this. So, maybe I should think about the behavior of each function separately and see how they might intersect.Let's start with f(x) = 2 ln x. I remember that the natural logarithm function, ln x, is only defined for x > 0. It increases as x increases, but it does so very slowly. Multiplying it by 2 just makes it increase a bit faster, but it's still a logarithmic function, which means it grows without bound as x approaches infinity, but it does so more and more slowly.On the other hand, g(x) = x² - 4x + 5 is a quadratic function. Quadratic functions are parabolas, and since the coefficient of x² is positive (1 in this case), it opens upwards. The vertex of this parabola will be its minimum point. To find the vertex, I can use the formula for the x-coordinate of the vertex, which is -b/(2a). Here, a = 1 and b = -4, so the x-coordinate is -(-4)/(2*1) = 4/2 = 2. Plugging x = 2 back into g(x), we get g(2) = (2)² - 4*(2) + 5 = 4 - 8 + 5 = 1. So, the vertex is at (2, 1).Now, since the parabola opens upwards and its vertex is at (2, 1), the function g(x) will increase as we move away from x = 2 in both directions. That means as x approaches positive infinity, g(x) will go to infinity, and as x approaches negative infinity, g(x) will also go to infinity. However, since f(x) is only defined for x > 0, we only need to consider x values greater than 0.Looking at f(x) = 2 ln x, as x approaches 0 from the right, ln x approaches negative infinity, so f(x) approaches negative infinity. As x increases, f(x) increases, but as I mentioned earlier, it does so very slowly. On the other hand, g(x) at x = 0 is g(0) = 0 - 0 + 5 = 5. So, at x = 0, g(x) is 5, while f(x) is approaching negative infinity. That means near x = 0, g(x) is much higher than f(x).At x = 2, g(x) is 1, and f(2) = 2 ln 2 ≈ 2 * 0.693 ≈ 1.386. So, at x = 2, f(x) is approximately 1.386, which is higher than g(x)'s value of 1. That suggests that somewhere between x = 0 and x = 2, the two functions cross each other because f(x) starts below g(x) and ends up above g(x) at x = 2.Now, let's check what happens as x increases beyond 2. At x = 3, g(3) = 9 - 12 + 5 = 2, and f(3) = 2 ln 3 ≈ 2 * 1.0986 ≈ 2.197. So, f(x) is still above g(x) at x = 3. At x = 4, g(4) = 16 - 16 + 5 = 5, and f(4) = 2 ln 4 ≈ 2 * 1.386 ≈ 2.772. Now, g(x) is 5, which is higher than f(x)'s 2.772. So, between x = 3 and x = 4, g(x) overtakes f(x) again.This suggests that there's another intersection point between x = 3 and x = 4. So, we have one intersection between x = 0 and x = 2, and another between x = 3 and x = 4. That would make two intersection points in total.But wait, let's double-check to make sure there aren't more intersection points. For that, I should analyze the behavior of the functions beyond x = 4. As x increases further, g(x) = x² - 4x + 5 will grow much faster than f(x) = 2 ln x because quadratic functions grow much faster than logarithmic functions. So, beyond x = 4, g(x) will continue to outpace f(x), meaning they won't intersect again.What about between x = 2 and x = 3? At x = 2, f(x) ≈ 1.386 and g(x) = 1. At x = 3, f(x) ≈ 2.197 and g(x) = 2. So, f(x) is increasing faster than g(x) in this interval, but since f(x) is already above g(x) at x = 2 and remains above at x = 3, there's no crossing in this interval.To be thorough, maybe I should consider the derivatives of both functions to understand their rates of increase. The derivative of f(x) is f'(x) = 2/x, which decreases as x increases. The derivative of g(x) is g'(x) = 2x - 4, which increases as x increases. At x = 2, f'(2) = 2/2 = 1, and g'(2) = 4 - 4 = 0. So, at x = 2, f(x) is increasing at a rate of 1, while g(x) is at its minimum slope of 0.As x increases beyond 2, f'(x) decreases, while g'(x) increases. This means that g(x) will eventually overtake f(x) in terms of growth rate, which aligns with our earlier observation that g(x) surpasses f(x) around x = 4.Additionally, let's consider the second derivatives to understand the concavity. The second derivative of f(x) is f''(x) = -2/x², which is always negative for x > 0, meaning f(x) is concave down everywhere in its domain. The second derivative of g(x) is g''(x) = 2, which is always positive, meaning g(x) is concave up everywhere.This difference in concavity might also influence the number of intersection points. Since f(x) is concave down and g(x) is concave up, they can intersect at most twice. This is because a concave down function can intersect a concave up function at two points, but not more, due to their respective curvatures.To further confirm, let's analyze the function h(x) = g(x) - f(x) = x² - 4x + 5 - 2 ln x. We can look for the number of zeros of h(x), which correspond to the intersection points of f(x) and g(x).Calculating h(x) at various points:- At x approaching 0 from the right: h(x) approaches 5 - (-∞) = ∞.- At x = 1: h(1) = 1 - 4 + 5 - 0 = 2.- At x = 2: h(2) = 4 - 8 + 5 - 2 ln 2 ≈ 1 - 1.386 ≈ -0.386.- At x = 3: h(3) = 9 - 12 + 5 - 2 ln 3 ≈ 2 - 2.197 ≈ -0.197.- At x = 4: h(4) = 16 - 16 + 5 - 2 ln 4 ≈ 5 - 2.772 ≈ 2.228.So, h(x) starts at positive infinity at x approaching 0, decreases to h(2) ≈ -0.386, then slightly increases to h(3) ≈ -0.197, and then increases further to h(4) ≈ 2.228.This means h(x) crosses the x-axis somewhere between x = 0 and x = 2 (from positive to negative), and then again between x = 3 and x = 4 (from negative to positive). Therefore, there are two zeros, meaning two intersection points.To ensure there are no more zeros, let's check the behavior as x approaches infinity. As x becomes very large, g(x) = x² - 4x + 5 grows like x², while f(x) = 2 ln x grows much slower. Therefore, h(x) = g(x) - f(x) will approach infinity as x approaches infinity, meaning h(x) remains positive beyond x = 4 and doesn't cross the x-axis again.Additionally, since h(x) is continuous for x > 0 (as both f(x) and g(x) are continuous there), and we've established that it crosses the x-axis twice, there are exactly two intersection points.Just to be extra cautious, let's consider if there could be any more intersection points. Suppose we check x = 5: h(5) = 25 - 20 + 5 - 2 ln 5 ≈ 10 - 3.218 ≈ 6.782, which is positive. So, h(x) continues to increase beyond x = 4, confirming no further crossings.In summary, by analyzing the behavior of both functions, their derivatives, and evaluating h(x) at key points, it's clear that the two functions intersect exactly twice. Therefore, the correct answer should be B: 2.

💡Okay, so I have this problem here with complex numbers. Let me try to figure it out step by step. Alright, the problem says: Given ( z_1 = m + i ) and ( z_2 = 1 - 2i ), if ( frac{z_1}{z_2} = -frac{1}{2} ), find the value of the real number ( m ). The options are A: 2, B: -2, C: ( frac{1}{2} ), D: ( -frac{1}{2} ).Hmm, okay. So, I need to find the value of ( m ) such that when I divide ( z_1 ) by ( z_2 ), I get ( -frac{1}{2} ). Since both ( z_1 ) and ( z_2 ) are complex numbers, dividing them will involve some complex number arithmetic.First, let me recall how to divide complex numbers. I think you multiply the numerator and the denominator by the conjugate of the denominator to make the denominator real. The conjugate of ( z_2 = 1 - 2i ) is ( 1 + 2i ). So, I can multiply both the numerator ( z_1 ) and the denominator ( z_2 ) by ( 1 + 2i ) to simplify the expression.Let me write that down:[frac{z_1}{z_2} = frac{m + i}{1 - 2i}]Multiplying numerator and denominator by the conjugate of the denominator:[frac{(m + i)(1 + 2i)}{(1 - 2i)(1 + 2i)}]Okay, now I need to multiply out the numerator and the denominator.Starting with the denominator:[(1 - 2i)(1 + 2i) = 1 cdot 1 + 1 cdot 2i - 2i cdot 1 - 2i cdot 2i]Simplify each term:[1 + 2i - 2i - 4i^2]Wait, ( i^2 = -1 ), so ( -4i^2 = -4(-1) = 4 ). Also, the middle terms ( 2i - 2i ) cancel out.So, denominator simplifies to:[1 + 4 = 5]Alright, that's straightforward. Now, the numerator:[(m + i)(1 + 2i)]Let me expand this:[m cdot 1 + m cdot 2i + i cdot 1 + i cdot 2i]Simplify each term:[m + 2mi + i + 2i^2]Again, ( i^2 = -1 ), so ( 2i^2 = 2(-1) = -2 ).Combine like terms:Real parts: ( m - 2 )Imaginary parts: ( 2mi + i = (2m + 1)i )So, numerator becomes:[(m - 2) + (2m + 1)i]Putting it all together, the fraction becomes:[frac{(m - 2) + (2m + 1)i}{5}]Which can be written as:[frac{m - 2}{5} + frac{2m + 1}{5}i]Now, according to the problem, this fraction equals ( -frac{1}{2} ). But ( -frac{1}{2} ) is a real number, which means the imaginary part of the fraction must be zero. So, we can set up two equations: one for the real part and one for the imaginary part.Real part equation:[frac{m - 2}{5} = -frac{1}{2}]Imaginary part equation:[frac{2m + 1}{5} = 0]Let me solve the imaginary part equation first because it seems simpler.Imaginary part:[frac{2m + 1}{5} = 0]Multiply both sides by 5:[2m + 1 = 0]Subtract 1:[2m = -1]Divide by 2:[m = -frac{1}{2}]Hmm, so ( m = -frac{1}{2} ). Let me check if this also satisfies the real part equation.Real part equation:[frac{m - 2}{5} = -frac{1}{2}]Substitute ( m = -frac{1}{2} ):[frac{(-frac{1}{2}) - 2}{5} = frac{-frac{1}{2} - 2}{5} = frac{-frac{5}{2}}{5} = -frac{5}{2} times frac{1}{5} = -frac{1}{2}]Yes, that works out. So, both the real and imaginary parts are satisfied when ( m = -frac{1}{2} ).Let me just recap to make sure I didn't make any mistakes. I started by multiplying numerator and denominator by the conjugate of the denominator to make the denominator real. Then I expanded both the numerator and denominator, simplified them, and separated into real and imaginary parts. Since the result is a real number, the imaginary part must be zero, which gave me an equation to solve for ( m ). Then I checked if this value also satisfies the real part equation, and it did. So, I think my answer is correct.Looking back at the options, D is ( -frac{1}{2} ), so that should be the correct answer.**Final Answer**boxed{D}

💡Okay, so I have this problem about skew lines and two propositions. I need to figure out if each proposition is correct or not and then choose the right option from A to D. Let me try to break this down step by step.First, let me recall what skew lines are. Skew lines are lines in three-dimensional space that are neither parallel nor intersecting. So, they don't lie in the same plane, and they don't cross each other. Got it.Now, let's look at Proposition I. It says: If line ( a ) is on plane ( alpha ) and line ( b ) is on plane ( beta ), and they are skew lines, then the intersection line ( c ) of planes ( alpha ) and ( beta ) intersects at most one of ( a ) or ( b ).Hmm, okay. So, ( a ) and ( b ) are skew, meaning they don't intersect and aren't parallel. They lie on different planes ( alpha ) and ( beta ). The intersection of ( alpha ) and ( beta ) is line ( c ). The proposition claims that ( c ) can intersect at most one of ( a ) or ( b ).Let me visualize this. If ( a ) is on ( alpha ) and ( c ) is the intersection of ( alpha ) and ( beta ), then ( c ) must lie on both ( alpha ) and ( beta ). So, ( a ) is on ( alpha ), and ( c ) is also on ( alpha ). Therefore, ( a ) and ( c ) are both on ( alpha ). Since ( a ) and ( c ) are both on the same plane, they must either intersect or be parallel.But ( a ) and ( b ) are skew, so ( a ) can't be parallel to ( b ). Wait, but ( c ) is the intersection of ( alpha ) and ( beta ), so ( c ) is on both planes. If ( a ) is on ( alpha ), then ( a ) and ( c ) must intersect because they are on the same plane. Similarly, ( b ) is on ( beta ), so ( b ) and ( c ) must intersect as well because they are on the same plane.Wait a minute, that means ( c ) intersects both ( a ) and ( b ). But Proposition I says ( c ) intersects at most one of them. So, that would mean Proposition I is incorrect because ( c ) actually intersects both.Hmm, maybe I'm misunderstanding something. Let me think again. If ( a ) and ( b ) are skew, they don't lie on the same plane, so ( alpha ) and ( beta ) must be different planes. Their intersection is line ( c ). Since ( a ) is on ( alpha ), ( a ) and ( c ) must intersect at some point on ( alpha ). Similarly, ( b ) is on ( beta ), so ( b ) and ( c ) must intersect at some point on ( beta ). So, ( c ) intersects both ( a ) and ( b ). Therefore, Proposition I is incorrect.Okay, moving on to Proposition II. It says: There does not exist such an infinite number of lines, any two of which are skew lines.So, the proposition is claiming that you can't have infinitely many lines where every pair is skew. I need to determine if this is true or not.Let me think about how lines can be arranged in space. If I have multiple lines, they can be parallel, intersecting, or skew. To have all pairs skew, they must not be parallel and not intersecting.Is it possible to have infinitely many such lines? Let me consider an example. Imagine taking a set of lines that are all parallel to each other but not lying on the same plane. Wait, no, if they are parallel, they are not skew. Skew lines are not parallel and don't intersect.Alternatively, think about lines that lie on different planes, each pair not intersecting and not parallel. Can we have infinitely many such lines?Actually, yes. For example, consider lines that are all parallel to the z-axis but shifted along the x and y axes. Each line would be of the form ( x = a, y = b, z = t ) where ( a ) and ( b ) vary. These lines are all parallel, so they are not skew. Hmm, that doesn't work.Wait, maybe another approach. Consider lines that lie on different planes, each pair not intersecting and not parallel. For instance, take lines that are generators of a hyperboloid. A hyperboloid of one sheet has two families of skew lines. Each line from one family is skew with every line from the other family, but lines within the same family are parallel. So, that's not quite what we need.Wait, maybe if we take lines that are all non-parallel and non-intersecting. For example, take lines that are all parallel to different directions and arranged such that no two intersect. But in three-dimensional space, it's tricky because lines can either intersect, be parallel, or be skew.But actually, in three-dimensional space, you can have infinitely many lines where each pair is skew. For example, take lines that are all parallel to the z-axis but shifted along the x and y axes in a way that no two are parallel or intersecting. Wait, but if they are all parallel to the z-axis, they are parallel to each other, so they are not skew.Hmm, maybe another way. Consider lines that lie on different planes, each plane arranged such that any two lines from different planes are skew. For example, take planes that are all different and arranged in a way that their intersection lines don't cause any two lines to intersect or be parallel.Wait, actually, in three-dimensional space, it's possible to have infinitely many lines where each pair is skew. For example, take lines that are all parallel to different directions and arranged such that no two are coplanar. Since they are not coplanar, they can't intersect, and since they are not parallel, they are not parallel. Therefore, they are skew.So, Proposition II is saying that such an infinite number of lines does not exist, but in reality, it does exist. Therefore, Proposition II is incorrect.Wait, but I'm a bit confused. Let me think again. If I have infinitely many lines, each pair being skew, is that possible?I think yes. For example, consider lines that are all parallel to the z-axis but shifted along the x-axis. Wait, no, those are parallel. So, that's not skew.Alternatively, consider lines that are all parallel to the x-axis but shifted along the y and z axes. Again, they are parallel, not skew.Wait, maybe another approach. Consider lines that are all non-parallel and non-intersecting. For example, take lines that are all generators of a hyperboloid, but not just one family. Wait, no, the hyperboloid has two families, each family's lines are parallel, so not skew.Wait, maybe take lines that are all skew to each other by having different directions and positions. For example, take lines that are all skew to a given line, but arranged such that they are also skew to each other.I think it's possible. For example, take a line ( l ) and then take infinitely many lines that are all skew to ( l ) and also skew to each other. This can be done by ensuring that each new line is not parallel to ( l ) and doesn't intersect ( l ), and also doesn't intersect or is parallel to any of the previously added lines.Therefore, Proposition II is incorrect because such an infinite number of lines does exist.Wait, but I'm not entirely sure. Maybe there's a limit to how many skew lines you can have. Let me think about it more carefully.In three-dimensional space, given any line, you can have infinitely many lines that are skew to it. For example, take a line ( l ) and then consider all lines that are not parallel and do not intersect ( l ). There are infinitely many such lines.Now, if I want to have infinitely many lines where every pair is skew, I need to ensure that not only each line is skew to ( l ), but also each pair among themselves is skew.Is that possible? Let me consider.Suppose I have line ( l_1 ). Then, I can have line ( l_2 ) skew to ( l_1 ). Then, line ( l_3 ) skew to both ( l_1 ) and ( l_2 ). And so on.But as I add more lines, ensuring that each new line is skew to all previous ones becomes more complex. However, in three-dimensional space, it's possible to have infinitely many such lines.For example, consider lines that are all parallel to different directions and arranged such that no two are coplanar. Since they are not coplanar, they can't intersect, and since they are not parallel, they are skew.Therefore, Proposition II is incorrect because such an infinite set of lines does exist.Wait, but I'm still a bit unsure. Maybe there's a mathematical result that says you can't have infinitely many mutually skew lines. Let me recall.I think in three-dimensional space, you can have infinitely many mutually skew lines. For example, consider the set of all lines that are parallel to the z-axis but shifted along the x and y axes. Wait, no, those are parallel, not skew.Wait, another example: take lines that are all parallel to the line ( x = y = z ), but shifted in different directions. Wait, no, if they are parallel, they are not skew.Hmm, maybe another approach. Consider lines that are all skew to each other by having different directions and positions. For example, take lines that are all skew to the x-axis, y-axis, and z-axis.Wait, no, the axes themselves are not skew. The x-axis and y-axis intersect, so they are not skew. Similarly, x-axis and z-axis intersect.Wait, maybe take lines that are all skew to each other by having different directions and positions. For example, take lines that are all generators of a hyperboloid, but not just one family.Wait, a hyperboloid of one sheet has two families of skew lines, each family being a set of lines where each line is skew to all others in the opposite family, but within the same family, they are parallel. So, that's not mutually skew.Wait, maybe consider lines that are all skew to each other by having different directions and positions. For example, take lines that are all skew to each other by having different directions and positions.I think it's possible. For example, take lines that are all skew to each other by having different directions and positions. Since in three-dimensional space, given any line, you can have infinitely many lines skew to it, and you can arrange them such that they are also skew to each other.Therefore, Proposition II is incorrect because such an infinite number of lines does exist.Wait, but I'm still not entirely confident. Maybe I should look for a mathematical proof or a known result.Upon reflection, I recall that in three-dimensional space, it's possible to have infinitely many mutually skew lines. For example, consider the set of all lines that are parallel to the z-axis but shifted along the x and y axes. Wait, no, those are parallel, not skew.Wait, another example: consider lines that are all skew to each other by having different directions and positions. For example, take lines that are all skew to each other by having different directions and positions.I think it's possible. For example, take lines that are all skew to each other by having different directions and positions. Since in three-dimensional space, given any line, you can have infinitely many lines skew to it, and you can arrange them such that they are also skew to each other.Therefore, Proposition II is incorrect because such an infinite number of lines does exist.Wait, but I'm still a bit confused. Maybe I should think about it differently. Let's consider the number of degrees of freedom.In three-dimensional space, a line is determined by four parameters (since it's the intersection of two planes). To have two lines be skew, they must not intersect and not be parallel. So, for each pair of lines, there are certain conditions they must satisfy.But if I want infinitely many lines where each pair is skew, I need to ensure that for every pair, they don't intersect and aren't parallel. This seems challenging, but I think it's possible.For example, consider lines that are all skew to each other by having different directions and positions. Since in three-dimensional space, given any line, you can have infinitely many lines skew to it, and you can arrange them such that they are also skew to each other.Therefore, Proposition II is incorrect because such an infinite number of lines does exist.Okay, so after all that, I think:- Proposition I is incorrect because the intersection line ( c ) actually intersects both ( a ) and ( b ).- Proposition II is incorrect because it's possible to have infinitely many lines where each pair is skew.Therefore, the correct answer is D: Both propositions are incorrect.

💡Alright, so I'm trying to solve this problem about a hyperbola and some inequalities involving vectors. Let me go through it step by step because it seems a bit tricky, but I think I can figure it out.First, the problem says: "The line ( x = 2 ) intersects the asymptotes of the hyperbola ( frac{x^2}{4} - y^2 = 1 ) at points ( A ) and ( B ). Let ( P ) be any point on the hyperbola, and suppose ( overrightarrow{OP} = aoverrightarrow{OA} + boverrightarrow{OB} ) where ( a, b in mathbb{R} ) and ( O ) is the origin. Then which of the following inequalities always holds? The options are A: ( a^2 + b^2 geq 1 ), B: ( |ab| geq 1 ), C: ( |a + b| geq 1 ), D: ( |a - b| geq 2 )."Okay, so I need to find which inequality is always true for any point ( P ) on the hyperbola. Let me break it down.**Step 1: Find the asymptotes of the hyperbola.**The given hyperbola is ( frac{x^2}{4} - y^2 = 1 ). I remember that the standard form of a hyperbola centered at the origin is ( frac{x^2}{a^2} - frac{y^2}{b^2} = 1 ), and its asymptotes are ( y = pm frac{b}{a}x ).In this case, ( a^2 = 4 ) so ( a = 2 ), and ( b^2 = 1 ) so ( b = 1 ). Therefore, the asymptotes are ( y = pm frac{1}{2}x ).**Step 2: Find points ( A ) and ( B ) where ( x = 2 ) intersects the asymptotes.**The line ( x = 2 ) is a vertical line. To find the intersection points with the asymptotes, substitute ( x = 2 ) into the equations of the asymptotes.For the first asymptote ( y = frac{1}{2}x ):- Substitute ( x = 2 ): ( y = frac{1}{2} times 2 = 1 ). So, one point is ( (2, 1) ).For the second asymptote ( y = -frac{1}{2}x ):- Substitute ( x = 2 ): ( y = -frac{1}{2} times 2 = -1 ). So, the other point is ( (2, -1) ).Therefore, points ( A ) and ( B ) are ( (2, 1) ) and ( (2, -1) ) respectively.**Step 3: Express vector ( overrightarrow{OP} ) in terms of ( overrightarrow{OA} ) and ( overrightarrow{OB} ).**Given that ( overrightarrow{OP} = aoverrightarrow{OA} + boverrightarrow{OB} ), let's write this in coordinates.- ( overrightarrow{OA} ) is the vector from the origin to ( A ), which is ( (2, 1) ).- ( overrightarrow{OB} ) is the vector from the origin to ( B ), which is ( (2, -1) ).So, ( overrightarrow{OP} = a(2, 1) + b(2, -1) = (2a + 2b, a - b) ).Therefore, the coordinates of point ( P ) are ( (2a + 2b, a - b) ).**Step 4: Substitute ( P ) into the hyperbola equation.**Since ( P ) lies on the hyperbola ( frac{x^2}{4} - y^2 = 1 ), substitute ( x = 2a + 2b ) and ( y = a - b ) into the equation.Let's compute each term:First, ( x^2 ):( (2a + 2b)^2 = 4a^2 + 8ab + 4b^2 ).Then, ( frac{x^2}{4} = frac{4a^2 + 8ab + 4b^2}{4} = a^2 + 2ab + b^2 ).Next, ( y^2 ):( (a - b)^2 = a^2 - 2ab + b^2 ).Now, plug these into the hyperbola equation:( a^2 + 2ab + b^2 - (a^2 - 2ab + b^2) = 1 ).Simplify the left side:( a^2 + 2ab + b^2 - a^2 + 2ab - b^2 = 1 ).Simplify term by term:- ( a^2 - a^2 = 0 )- ( 2ab + 2ab = 4ab )- ( b^2 - b^2 = 0 )So, the equation becomes:( 4ab = 1 ).Therefore, ( ab = frac{1}{4} ).**Step 5: Analyze the inequalities given the condition ( ab = frac{1}{4} ).**We have ( ab = frac{1}{4} ). Let's check each option to see which inequality always holds.**Option A: ( a^2 + b^2 geq 1 )**Hmm, I know that ( (a + b)^2 = a^2 + 2ab + b^2 ) and ( (a - b)^2 = a^2 - 2ab + b^2 ). Since ( ab = frac{1}{4} ), maybe I can relate ( a^2 + b^2 ) to something.But let's test with specific values. Suppose ( a = b = frac{1}{2} ). Then ( ab = frac{1}{4} ), which satisfies the condition. Then ( a^2 + b^2 = frac{1}{4} + frac{1}{4} = frac{1}{2} ), which is less than 1. So, ( a^2 + b^2 ) can be less than 1, meaning Option A is not necessarily true.**Option B: ( |ab| geq 1 )**We already have ( ab = frac{1}{4} ), so ( |ab| = frac{1}{4} ), which is less than 1. Therefore, this inequality does not hold. So, Option B is false.**Option C: ( |a + b| geq 1 )**Let me see. From ( ab = frac{1}{4} ), can I find a relationship involving ( a + b )?I recall the inequality that relates arithmetic mean and geometric mean (AM-GM). For real numbers, ( |a + b| geq 2sqrt{|ab|} ). But wait, is that always true?Wait, actually, the AM-GM inequality states that for non-negative real numbers, ( frac{a + b}{2} geq sqrt{ab} ). But here, ( a ) and ( b ) can be positive or negative because they are real numbers. So, I need to be careful.But since ( ab = frac{1}{4} ), which is positive, that means ( a ) and ( b ) must have the same sign. So, either both positive or both negative.Let me consider ( a ) and ( b ) positive first. Then, by AM-GM:( frac{a + b}{2} geq sqrt{ab} = sqrt{frac{1}{4}} = frac{1}{2} ).Multiplying both sides by 2:( a + b geq 1 ).Similarly, if ( a ) and ( b ) are both negative, then ( a + b leq -1 ). Therefore, ( |a + b| geq 1 ).So, regardless of the signs, ( |a + b| geq 1 ). Therefore, Option C is always true.**Option D: ( |a - b| geq 2 )**Hmm, let's test this. Let me take specific values. Suppose ( a = 1 ) and ( b = frac{1}{4} ). Then ( ab = frac{1}{4} ), which satisfies the condition. Then ( |a - b| = |1 - frac{1}{4}| = frac{3}{4} ), which is less than 2. So, this inequality does not hold. Therefore, Option D is false.**Conclusion:**After checking each option, only Option C: ( |a + b| geq 1 ) always holds true given the condition ( ab = frac{1}{4} ).**Final Answer**boxed{C}