💡Alright, so I've got this problem here about an odd function f(x) defined on all real numbers. It says that f(x) equals x squared when x is greater than or equal to zero. Since it's an odd function, I remember that means f(-x) = -f(x). So, for negative x, f(x) should be -x squared. That makes sense.The inequality given is f(x + a) is greater than or equal to 2 times f(x) for any x in the interval [a, a + 2]. I need to find the range of the real number a that satisfies this condition.Okay, let's break this down. Since f(x) is defined differently for positive and negative x, I should consider cases based on the value of x + a. But since x is in [a, a + 2], let's see what that means for x + a. If x is in [a, a + 2], then x + a is in [2a, 2a + 2]. So, depending on the value of a, x + a could be positive or negative.But wait, since f(x) is x squared for x >= 0 and -x squared for x < 0, maybe I can express f(x + a) in terms of x and a. Let me write that out.For x >= 0, f(x) = x^2. For x < 0, f(x) = -x^2. So, f(x + a) would be (x + a)^2 if x + a >= 0, and -(x + a)^2 if x + a < 0.But since x is in [a, a + 2], let's see what x + a is. If a is positive, then x + a is definitely positive because x is at least a, so x + a is at least 2a. If a is negative, x + a could be negative or positive depending on how negative a is.Hmm, this might get complicated. Maybe I should consider different cases for a.Case 1: a >= 0.In this case, since x is in [a, a + 2], x + a is in [2a, 2a + 2]. Since a is non-negative, 2a is non-negative, so x + a is non-negative. Therefore, f(x + a) = (x + a)^2.So, the inequality becomes (x + a)^2 >= 2x^2.Let me expand that: x^2 + 2ax + a^2 >= 2x^2.Subtract 2x^2 from both sides: -x^2 + 2ax + a^2 >= 0.Let me rearrange: -x^2 + 2ax + a^2 >= 0.Multiply both sides by -1 (which reverses the inequality): x^2 - 2ax - a^2 <= 0.So, x^2 - 2ax - a^2 <= 0.This is a quadratic in x. Let's find its roots.The quadratic equation x^2 - 2ax - a^2 = 0.Using the quadratic formula: x = [2a ± sqrt(4a^2 + 4a^2)] / 2 = [2a ± sqrt(8a^2)] / 2 = [2a ± 2a*sqrt(2)] / 2 = a ± a*sqrt(2).So, the roots are x = a(1 + sqrt(2)) and x = a(1 - sqrt(2)).Since sqrt(2) is approximately 1.414, 1 - sqrt(2) is negative. So, the quadratic is less than or equal to zero between the roots x = a(1 - sqrt(2)) and x = a(1 + sqrt(2)).But our x is in [a, a + 2]. So, for the inequality x^2 - 2ax - a^2 <= 0 to hold for all x in [a, a + 2], the entire interval [a, a + 2] must lie within [a(1 - sqrt(2)), a(1 + sqrt(2))].But since a >= 0, a(1 - sqrt(2)) is negative, and a(1 + sqrt(2)) is positive. So, the interval [a, a + 2] is entirely within [a(1 - sqrt(2)), a(1 + sqrt(2))] only if a + 2 <= a(1 + sqrt(2)).Let me write that: a + 2 <= a(1 + sqrt(2)).Subtract a from both sides: 2 <= a*sqrt(2).So, a >= 2 / sqrt(2) = sqrt(2).So, in this case, a must be greater than or equal to sqrt(2).Case 2: a < 0.In this case, since a is negative, x is in [a, a + 2]. Let's see what x + a is. x + a is in [2a, 2a + 2]. Since a is negative, 2a is negative, but 2a + 2 could be positive or negative depending on a.Let me find when 2a + 2 >= 0: 2a + 2 >= 0 => a >= -1.So, if a >= -1, then x + a could be positive or negative. If a < -1, then x + a is always negative because 2a + 2 < 0.Wait, let's see: If a < -1, then 2a + 2 < 0, so x + a is in [2a, 2a + 2], which is entirely negative because 2a + 2 < 0. So, f(x + a) = -(x + a)^2.If a >= -1, then x + a could be positive or negative depending on x.This is getting a bit messy. Maybe I should handle a < -1 and -1 <= a < 0 separately.Subcase 2a: a < -1.Then x + a is in [2a, 2a + 2], which is entirely negative because 2a + 2 < 0 (since a < -1). Therefore, f(x + a) = -(x + a)^2.So, the inequality becomes -(x + a)^2 >= 2x^2.But since -(x + a)^2 is negative and 2x^2 is non-negative, this inequality cannot hold because a negative number cannot be greater than or equal to a non-negative number.Therefore, there are no solutions for a < -1.Subcase 2b: -1 <= a < 0.In this case, x is in [a, a + 2]. Since a >= -1, a + 2 >= 1. So, x ranges from a (which is >= -1) to a + 2 (which is >= 1). Therefore, x can be both negative and positive.So, f(x) is x^2 for x >= 0 and -x^2 for x < 0.Similarly, f(x + a) will depend on whether x + a is positive or negative.But since a is between -1 and 0, x is in [a, a + 2], so x + a is in [2a, 2a + 2].Since a >= -1, 2a >= -2, and 2a + 2 >= 0. So, x + a can be negative or positive.Therefore, f(x + a) is (x + a)^2 if x + a >= 0, and -(x + a)^2 if x + a < 0.Similarly, f(x) is x^2 if x >= 0, and -x^2 if x < 0.So, the inequality f(x + a) >= 2f(x) can have different forms depending on the sign of x and x + a.This seems complicated, but maybe I can consider different intervals within [a, a + 2].Let me find the point where x + a = 0, which is x = -a.Since a is between -1 and 0, -a is between 0 and 1.So, in the interval [a, a + 2], the point x = -a is within [a, a + 2] because a <= -a <= a + 2.Wait, let's check: a <= -a <= a + 2.Since a is negative, -a is positive. Is -a <= a + 2?Yes, because a + 2 >= 1 (since a >= -1), and -a <= 1 (since a >= -1).So, x = -a is within [a, a + 2].Therefore, we can split the interval [a, a + 2] into two parts: [a, -a] and [-a, a + 2].In [a, -a], x is negative, so f(x) = -x^2. Also, x + a is in [2a, 0], which is negative, so f(x + a) = -(x + a)^2.In [-a, a + 2], x is non-negative, so f(x) = x^2. Also, x + a is in [0, 2a + 2], which is non-negative because a >= -1, so 2a + 2 >= 0.Therefore, in [-a, a + 2], f(x + a) = (x + a)^2.So, we have two cases:1. For x in [a, -a]: f(x + a) = -(x + a)^2 and f(x) = -x^2.So, the inequality becomes -(x + a)^2 >= 2*(-x^2).Simplify: -(x + a)^2 >= -2x^2.Multiply both sides by -1 (reverses inequality): (x + a)^2 <= 2x^2.Which is similar to the previous case.Expand: x^2 + 2ax + a^2 <= 2x^2.Subtract 2x^2: -x^2 + 2ax + a^2 <= 0.Multiply by -1: x^2 - 2ax - a^2 >= 0.So, x^2 - 2ax - a^2 >= 0.This quadratic in x is the same as before. Its roots are x = a(1 ± sqrt(2)).So, the quadratic is positive outside the roots. So, x <= a(1 - sqrt(2)) or x >= a(1 + sqrt(2)).But in this subcase, x is in [a, -a]. Since a is negative, -a is positive.So, we need x in [a, -a] to satisfy x <= a(1 - sqrt(2)) or x >= a(1 + sqrt(2)).But a(1 - sqrt(2)) is more negative than a because sqrt(2) > 1, so 1 - sqrt(2) < 0.Similarly, a(1 + sqrt(2)) is more negative than a because a is negative.Wait, let's compute a(1 - sqrt(2)) and a(1 + sqrt(2)).Since a is negative, a(1 - sqrt(2)) is a positive number because 1 - sqrt(2) is negative, and a is negative, so negative times negative is positive.Similarly, a(1 + sqrt(2)) is negative because 1 + sqrt(2) is positive, and a is negative.So, the quadratic is positive when x <= a(1 - sqrt(2)) or x >= a(1 + sqrt(2)).But in our interval [a, -a], x ranges from a (negative) to -a (positive). So, the inequality x^2 - 2ax - a^2 >= 0 holds when x <= a(1 - sqrt(2)) or x >= a(1 + sqrt(2)).But in [a, -a], x is from a to -a. So, the inequality holds when x <= a(1 - sqrt(2)) or x >= a(1 + sqrt(2)).But since a(1 - sqrt(2)) is positive and a(1 + sqrt(2)) is negative, and x is in [a, -a], which is from a (negative) to -a (positive), we need to see if the entire interval [a, -a] is covered by x <= a(1 - sqrt(2)) or x >= a(1 + sqrt(2)).But a(1 - sqrt(2)) is positive, and a(1 + sqrt(2)) is negative.So, in the interval [a, -a], the inequality holds when x <= a(1 - sqrt(2)) or x >= a(1 + sqrt(2)).But since x is in [a, -a], which includes both negative and positive values, we need to check if the inequality holds for all x in [a, -a].But this seems tricky. Maybe instead of trying to handle both cases, I should consider the maximum and minimum of the inequality.Alternatively, maybe I can find the minimum of the expression f(x + a) - 2f(x) over x in [a, a + 2] and ensure it's non-negative.But this might be complicated.Alternatively, since in the first case when a >= sqrt(2), the inequality holds, and when a < sqrt(2), it might not. But we need to check for a < 0.Wait, but in the case when a < 0, we saw that for a < -1, the inequality cannot hold because f(x + a) is negative and 2f(x) is non-negative, so the inequality can't be satisfied.For a between -1 and 0, we have to ensure that the inequality holds for both parts of the interval [a, -a] and [-a, a + 2].In the interval [a, -a], the inequality reduces to (x + a)^2 <= 2x^2, which is similar to the a >= 0 case.But in this case, x is negative, so x = -|x|.Wait, maybe I can substitute x = -y where y is positive in [a, -a].Let me try that. Let x = -y, where y is in [0, -a].Then, the inequality becomes ( -y + a )^2 <= 2*(-y)^2.Simplify: (a - y)^2 <= 2y^2.Expand: a^2 - 2ay + y^2 <= 2y^2.Subtract 2y^2: a^2 - 2ay - y^2 <= 0.Which is the same as y^2 + 2ay - a^2 >= 0.This is a quadratic in y: y^2 + 2ay - a^2 >= 0.Find its roots: y = [-2a ± sqrt(4a^2 + 4a^2)] / 2 = [-2a ± sqrt(8a^2)] / 2 = [-2a ± 2|a|sqrt(2)] / 2.Since a is negative, |a| = -a.So, y = [-2a ± 2*(-a)sqrt(2)] / 2 = [-2a ∓ 2a sqrt(2)] / 2 = -a ∓ a sqrt(2).So, y = -a(1 + sqrt(2)) or y = -a(1 - sqrt(2)).But y is in [0, -a], since x = -y is in [a, 0].So, the quadratic y^2 + 2ay - a^2 is positive outside the roots y = -a(1 - sqrt(2)) and y = -a(1 + sqrt(2)).But since y is in [0, -a], and a is negative, let's see:-a(1 - sqrt(2)) is positive because a is negative, so 1 - sqrt(2) is negative, so negative times negative is positive.Similarly, -a(1 + sqrt(2)) is positive.So, the quadratic is positive when y <= -a(1 - sqrt(2)) or y >= -a(1 + sqrt(2)).But y is in [0, -a]. So, the inequality y^2 + 2ay - a^2 >= 0 holds when y <= -a(1 - sqrt(2)) or y >= -a(1 + sqrt(2)).But since y is in [0, -a], and -a(1 + sqrt(2)) > -a (because 1 + sqrt(2) > 1), the inequality holds when y <= -a(1 - sqrt(2)).But y is in [0, -a], so we need 0 <= y <= -a(1 - sqrt(2)).But -a(1 - sqrt(2)) must be greater than or equal to 0, which it is because a is negative.But also, -a(1 - sqrt(2)) must be less than or equal to -a.So, -a(1 - sqrt(2)) <= -a.Divide both sides by -a (which is positive, so inequality remains the same):1 - sqrt(2) >= 1.But 1 - sqrt(2) is approximately -0.414, which is less than 1. So, this inequality is not true.Therefore, the quadratic y^2 + 2ay - a^2 >= 0 does not hold for all y in [0, -a]. Hence, the inequality f(x + a) >= 2f(x) does not hold for all x in [a, -a] when a is between -1 and 0.Therefore, there are no solutions for a < 0.So, combining both cases, the only solutions are when a >= sqrt(2).Therefore, the range of a is [sqrt(2), +infty).

💡Okay, so I have this problem here about a parabola and a circle, and I need to find the minimum value of m + |AB|, where m is the distance from point A on the parabola to the y-axis, and B is any point on the circle. Hmm, let me try to break this down step by step.First, let's understand the given equations. The parabola is y² = 4x. I remember that the standard form of a parabola that opens to the right is y² = 4ax, where (a, 0) is the focus. So in this case, 4a = 4, which means a = 1. Therefore, the focus F of the parabola is at (1, 0). Got that.Next, the circle is given by (x + 3)² + (y + 3)² = 4. That's the standard equation of a circle with center at (-3, -3) and radius 2. So, the circle is centered at (-3, -3) and has a radius of 2 units. I should probably note that down.Now, point A is on the parabola y² = 4x. The distance from A to the y-axis is m. Since the y-axis is the line x = 0, the distance from any point (x, y) to the y-axis is just the absolute value of its x-coordinate, right? So, if A is (x, y), then m = |x|. But since the parabola y² = 4x only exists for x ≥ 0 (because y² is always non-negative), m is just x. So, m = x for point A on the parabola.So, we need to minimize m + |AB|, which is x + |AB|. Now, |AB| is the distance between point A on the parabola and point B on the circle. Since B can be any point on the circle, we need to find the minimum value of x + |AB| over all possible points A on the parabola and B on the circle.Hmm, this seems a bit abstract. Maybe I can visualize it. The parabola opens to the right, and the circle is located at (-3, -3) with radius 2. So, the circle is in the third quadrant, and the parabola is in the first and fourth quadrants. I need to find a point A on the parabola such that when I add the distance from A to the y-axis (which is just its x-coordinate) and the distance from A to some point B on the circle, the total is minimized.I wonder if there's a way to use the properties of the parabola here. I remember that for a parabola, the distance from any point on the parabola to the focus is equal to the distance from that point to the directrix. The directrix of the parabola y² = 4x is the line x = -1. So, for any point A on the parabola, the distance from A to F (the focus at (1, 0)) is equal to the distance from A to the directrix x = -1.Wait, that might be useful. Since m is the distance from A to the y-axis, which is x, and the distance from A to the directrix is x + 1 (since the directrix is at x = -1, the distance from (x, y) to x = -1 is |x - (-1)| = x + 1). But since A is on the parabola, the distance from A to F is equal to the distance from A to the directrix, so |AF| = x + 1.Therefore, x = |AF| - 1. So, m = x = |AF| - 1. Therefore, m + |AB| = |AF| - 1 + |AB|. So, if I can minimize |AF| + |AB|, then subtracting 1 will give me the minimum of m + |AB|.So, the problem reduces to minimizing |AF| + |AB|, where A is on the parabola and B is on the circle. Hmm, that seems more manageable.Now, let's think about |AF| + |AB|. Since F is fixed at (1, 0), and B is any point on the circle, maybe we can consider the triangle inequality or some geometric property here.Wait, if I fix A, then |AF| + |AB| is the sum of distances from A to F and from A to B. But since B can be any point on the circle, maybe for each A, the minimal |AB| is the distance from A to the center of the circle minus the radius? Or is it the distance from A to the center minus the radius?Wait, actually, the minimal distance from A to any point on the circle is |AO| - r, where O is the center of the circle and r is its radius, provided that A is outside the circle. If A is inside the circle, then the minimal distance would be zero, but in this case, since the circle is at (-3, -3) and the parabola is in the first and fourth quadrants, point A is likely outside the circle.So, the minimal |AB| for a given A is |AO| - r, where O is (-3, -3) and r = 2. Therefore, |AB| ≥ |AO| - 2. So, m + |AB| = |AF| - 1 + |AB| ≥ |AF| - 1 + |AO| - 2 = |AF| + |AO| - 3.Therefore, to minimize m + |AB|, we need to minimize |AF| + |AO| - 3. So, the problem reduces to minimizing |AF| + |AO|, and then subtracting 3.But |AF| + |AO| is the sum of distances from A to F and from A to O. Hmm, I'm not sure if there's a direct geometric interpretation for this sum. Maybe I can think of reflecting a point or something.Wait, in optimization problems involving distances, sometimes reflecting a point across a line or a curve can help. For example, in the shortest path problems, reflecting a point across a mirror gives the optimal path.In this case, since A is on the parabola, which has a focus F and directrix x = -1, maybe reflecting O across the directrix or something like that could help.Let me try reflecting the center of the circle O(-3, -3) across the directrix x = -1. The reflection of a point (x, y) across the vertical line x = a is (2a - x, y). So, reflecting (-3, -3) across x = -1 would give (2*(-1) - (-3), -3) = (-2 + 3, -3) = (1, -3). So, the reflection point is (1, -3).Hmm, interesting. So, the reflection of O across the directrix is (1, -3). Let's call this point O'.Now, since A is on the parabola, |AF| = |A directrix|, which is x + 1. So, |AF| = x + 1. Also, |AO| is the distance from A to O, which is sqrt[(x + 3)² + (y + 3)²]. But since A is on the parabola, y² = 4x, so y can be expressed in terms of x.Alternatively, maybe I can express |AO| in terms of |AF| and |O'O| or something.Wait, let's think about triangle inequality. |AF| + |AO| ≥ |FO|, but F is (1, 0) and O is (-3, -3). So, |FO| is the distance between (1, 0) and (-3, -3), which is sqrt[(1 + 3)² + (0 + 3)²] = sqrt[16 + 9] = sqrt[25] = 5.So, |AF| + |AO| ≥ 5. Therefore, the minimal value of |AF| + |AO| is 5, achieved when A lies on the line segment connecting F and O. But wait, A is on the parabola, so does the line segment FO intersect the parabola?Let me check. The line segment from F(1, 0) to O(-3, -3). Let's parametrize this line. The parametric equations can be written as x = 1 - 4t, y = 0 - 3t, where t ranges from 0 to 1.We can find if this line intersects the parabola y² = 4x. Substitute x and y from the parametric equations into the parabola equation:( -3t )² = 4*(1 - 4t )So, 9t² = 4 - 16tBring all terms to one side:9t² + 16t - 4 = 0Solve for t:t = [-16 ± sqrt(256 + 144)] / (2*9) = [-16 ± sqrt(400)] / 18 = [-16 ± 20]/18So, two solutions:t = (-16 + 20)/18 = 4/18 = 2/9t = (-16 - 20)/18 = -36/18 = -2Since t ranges from 0 to 1, only t = 2/9 is valid.So, the line FO intersects the parabola at t = 2/9. Therefore, the point A is at x = 1 - 4*(2/9) = 1 - 8/9 = 1/9, y = -3*(2/9) = -2/3.So, point A is (1/9, -2/3). Therefore, when A is at (1/9, -2/3), |AF| + |AO| = |FO| = 5.Therefore, the minimal value of |AF| + |AO| is 5, achieved at A = (1/9, -2/3). Therefore, the minimal value of m + |AB| is 5 - 3 = 2.Wait, hold on. Earlier, I had m + |AB| ≥ |AF| + |AO| - 3, and since |AF| + |AO| ≥ 5, then m + |AB| ≥ 5 - 3 = 2. So, the minimal value is 2.But let me verify this. If A is at (1/9, -2/3), then m = x = 1/9. Then, |AB| is the minimal distance from A to the circle. Since the minimal |AB| is |AO| - r, where O is (-3, -3) and r = 2.Compute |AO|: distance from (1/9, -2/3) to (-3, -3):sqrt[(1/9 + 3)² + (-2/3 + 3)²] = sqrt[(28/9)² + (7/3)²] = sqrt[(784/81) + (49/9)] = sqrt[(784 + 441)/81] = sqrt[1225/81] = 35/9 ≈ 3.888...So, |AB| ≥ |AO| - r = 35/9 - 2 = 35/9 - 18/9 = 17/9 ≈ 1.888...Therefore, m + |AB| ≥ 1/9 + 17/9 = 18/9 = 2.So, the minimal value is indeed 2, achieved when A is at (1/9, -2/3) and B is the point on the circle closest to A, which lies along the line connecting A and O.Therefore, the minimal value of m + |AB| is 2.**Final Answer**The minimum value of ( m + |AB| ) is boxed{2}.

💡First, I need to determine the speed of the train during the second part of the journey. Let's denote this speed as ( v ) km/h.The train covers the first ( x ) km at a speed of 30 km/h. The time taken for this part of the journey is:[text{Time}_1 = frac{x}{30} text{ hours}]Next, the train covers the second ( 2x ) km at the unknown speed ( v ). The time taken for this part is:[text{Time}_2 = frac{2x}{v} text{ hours}]The total distance covered by the train is ( 3x ) km, and the average speed for the entire journey is 22.5 km/h. Therefore, the total time taken for the entire journey is:[text{Total Time} = frac{3x}{22.5} text{ hours}]The sum of the individual times should equal the total time:[frac{x}{30} + frac{2x}{v} = frac{3x}{22.5}]To simplify the equation, I can multiply both sides by ( 30v ) to eliminate the denominators:[v cdot x + 60x = frac{90v cdot x}{22.5}]Dividing both sides by ( x ) to eliminate ( x ):[v + 60 = frac{90v}{22.5}]Simplifying the right side of the equation:[v + 60 = 4v]Subtracting ( v ) from both sides to isolate ( v ):[60 = 3v]Finally, dividing both sides by 3 to solve for ( v ):[v = 20 text{ km/h}]

💡To determine how many bushes Natalie needs to pick, we'll first calculate the total number of containers required for both the zucchinis and the containers she wants to keep. Each zucchini requires 4 containers of blueberries, so for 60 zucchinis, she needs:60 zucchinis × 4 containers/zucchini = 240 containers.Adding the 20 containers she wants to keep:240 containers + 20 containers = 260 containers.Since each bush yields 10 containers, the number of bushes needed is:260 containers ÷ 10 containers/bush = 26 bushes.Therefore, Natalie needs to pick 26 bushes.

💡Alright, so I have this problem here where I need to use mathematical induction to prove an inequality. The inequality is given as:1 + 1/2 + 1/3 + 1/4 + … + 1/(2^{n-1} + 1) < f(n) for n ≥ 2, where n is a positive integer.And the question is specifically asking about what happens when we go from n = k to n = k+1 in the induction step. It wants to know how many items are added to the left side of the inequality. The options are:A: 1 itemB: k itemsC: 2^{k-1} itemsD: 2^{k} itemsOkay, so first, I need to recall how mathematical induction works. It has two main steps: the base case and the induction step. The base case is usually proving the statement for the initial value, which here is n = 2. Then, assuming it's true for some arbitrary n = k, we need to show it's true for n = k+1.But the question isn't asking me to prove the entire inequality; it's specifically about the change in the number of terms when moving from k to k+1. So, I need to figure out how many new terms are added to the left side when n increases by 1.Let me write down the left side for n = k and n = k+1 to see the difference.For n = k:Left side = 1 + 1/2 + 1/3 + … + 1/(2^{k-1} + 1)For n = k+1:Left side = 1 + 1/2 + 1/3 + … + 1/(2^{(k+1)-1} + 1) = 1 + 1/2 + 1/3 + … + 1/(2^{k} + 1)So, the difference between the two is the number of terms from 1/(2^{k-1} + 2) up to 1/(2^{k} + 1). I need to count how many terms that is.Let me denote the number of terms added as T. So,T = (2^{k} + 1) - (2^{k-1} + 1) = 2^{k} + 1 - 2^{k-1} - 1 = 2^{k} - 2^{k-1}Simplify that:2^{k} - 2^{k-1} = 2^{k-1}(2 - 1) = 2^{k-1}So, T = 2^{k-1} items are added when moving from n = k to n = k+1.Let me double-check this calculation to make sure I didn't make a mistake.Starting with n = k, the last term is 1/(2^{k-1} + 1). For n = k+1, the last term is 1/(2^{k} + 1). So, the number of new terms is the number of integers from (2^{k-1} + 2) to (2^{k} + 1), inclusive.The number of integers from a to b inclusive is b - a + 1. So,Number of terms = (2^{k} + 1) - (2^{k-1} + 2) + 1 = 2^{k} + 1 - 2^{k-1} - 2 + 1 = 2^{k} - 2^{k-1} - 0 = 2^{k-1}Yes, that confirms it. So, the number of items added is 2^{k-1}, which is option C.I think that makes sense because each time n increases by 1, the upper limit of the summation doubles and adds 1, which would mean the number of new terms added is exponential in terms of k.So, the answer should be C: 2^{k-1} items.**Final Answer**boxed{C}

💡Okay, so I have this trigonometry problem here: I need to find the value of sin 15 degrees multiplied by sin 75 degrees, and then choose the correct option from A to D. The options are 1/2, sqrt(3)/2, 1/4, and sqrt(3)/4. Hmm, let me think about how to approach this.First, I remember that there are some trigonometric identities that can help simplify products of sine functions. Maybe I can use the product-to-sum formulas? Let me recall what those are. I think one of them is sin A sin B equals something like cos(A - B) minus cos(A + B) all over 2. Let me write that down:sin A sin B = [cos(A - B) - cos(A + B)] / 2Okay, so if I let A be 15 degrees and B be 75 degrees, then:sin 15° sin 75° = [cos(15° - 75°) - cos(15° + 75°)] / 2Let me compute the angles inside the cosines:15° - 75° = -60°, and 15° + 75° = 90°So, substituting these back in:= [cos(-60°) - cos(90°)] / 2I know that cosine is an even function, so cos(-60°) is the same as cos(60°). And cos(90°) is 0. So:= [cos(60°) - 0] / 2= cos(60°) / 2Now, cos(60°) is 0.5 or 1/2. So:= (1/2) / 2= 1/4Wait, so that would make the value 1/4, which is option C. Hmm, but let me double-check in case I made a mistake.Alternatively, maybe I can use another identity or approach. I remember that sin(75°) can be written as sin(90° - 15°), which is cos(15°). So, sin 75° = cos 15°. Therefore, the original expression becomes:sin 15° sin 75° = sin 15° cos 15°Now, I recall that sin A cos A is equal to (1/2) sin 2A. So, applying that here:sin 15° cos 15° = (1/2) sin(30°)Sin 30° is 1/2, so:= (1/2) * (1/2)= 1/4Okay, that's the same result as before. So, it seems consistent. Therefore, the value is 1/4, which is option C.But just to make sure, let me think if there's another way to approach this, maybe using exact values of sine and cosine for these angles.I know that sin 15° can be expressed using the sine of a difference formula:sin(45° - 30°) = sin 45° cos 30° - cos 45° sin 30°Similarly, sin 75° can be expressed as sin(45° + 30°):sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°Let me compute both of these.First, sin 15°:sin 15° = sin(45° - 30°) = sin 45° cos 30° - cos 45° sin 30°I know that sin 45° = sqrt(2)/2, cos 30° = sqrt(3)/2, cos 45° = sqrt(2)/2, and sin 30° = 1/2.So plugging these in:= (sqrt(2)/2)(sqrt(3)/2) - (sqrt(2)/2)(1/2)= (sqrt(6)/4) - (sqrt(2)/4)= (sqrt(6) - sqrt(2)) / 4Similarly, sin 75°:sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°Again, plugging in the known values:= (sqrt(2)/2)(sqrt(3)/2) + (sqrt(2)/2)(1/2)= (sqrt(6)/4) + (sqrt(2)/4)= (sqrt(6) + sqrt(2)) / 4Now, let's multiply sin 15° and sin 75°:sin 15° sin 75° = [(sqrt(6) - sqrt(2))/4] * [(sqrt(6) + sqrt(2))/4]This looks like the product of a difference and a sum, which is a difference of squares:= [(sqrt(6))^2 - (sqrt(2))^2] / (4 * 4)= [6 - 2] / 16= 4 / 16= 1/4Okay, so that's the same answer again. So, regardless of the method I use—whether it's the product-to-sum formula, using the identity sin A cos A, or expanding the sines using angle addition formulas—I end up with 1/4.Therefore, I'm confident that the correct answer is 1/4, which is option C.**Final Answer**boxed{C}