💡Alright, so I have this problem about two circles and their positional relationship. The question is asking whether they are externally tangent, internally tangent, disjoint, or contained. The options are A, B, C, D respectively. Let me try to figure this out step by step.First, I need to understand the equations of the circles given. The first circle is given by the equation ( x^2 + y^2 = 1 ). That looks familiar—it's the standard equation of a circle centered at the origin (0,0) with a radius of 1. So, center at (0,0) and radius 1. Got that.The second circle has the equation ( x^2 + y^2 - 6y + 5 = 0 ). Hmm, this one isn't in the standard form, so I need to rearrange it to find its center and radius. Let me recall that the standard form of a circle is ( (x - h)^2 + (y - k)^2 = r^2 ), where (h,k) is the center and r is the radius.So, starting with ( x^2 + y^2 - 6y + 5 = 0 ), I can try completing the square for the y-terms. Let's see:First, group the x and y terms:( x^2 + (y^2 - 6y) + 5 = 0 )Now, to complete the square for ( y^2 - 6y ), I take half of the coefficient of y, which is -6, so half is -3, and then square it, getting 9. So, I add and subtract 9 inside the equation to maintain equality.So, rewriting:( x^2 + (y^2 - 6y + 9 - 9) + 5 = 0 )Simplify:( x^2 + (y - 3)^2 - 9 + 5 = 0 )Combine constants:( x^2 + (y - 3)^2 - 4 = 0 )Move the constant to the other side:( x^2 + (y - 3)^2 = 4 )Okay, so the second circle is centered at (0, 3) with a radius of 2 because ( r^2 = 4 ) implies ( r = 2 ).Now, I have both circles in standard form:1. First circle: Center at (0,0), radius 1.2. Second circle: Center at (0,3), radius 2.Next, I need to find the distance between the centers of these two circles. Since both centers are on the y-axis (their x-coordinates are 0), the distance between them is simply the difference in their y-coordinates.So, the distance between centers is |3 - 0| = 3 units.Now, to determine the positional relationship, I need to compare this distance with the sum and difference of their radii.The radii are 1 and 2, so:- Sum of radii: 1 + 2 = 3- Difference of radii: |2 - 1| = 1Now, the distance between centers is 3, which is exactly equal to the sum of the radii. When the distance between centers equals the sum of the radii, the circles are externally tangent to each other. That means they touch at exactly one point, and they don't intersect at two points or one inside the other.Let me just visualize this. The first circle is small, radius 1, at the origin. The second circle is a bit larger, radius 2, centered at (0,3). So, the distance between centers is 3, which is the same as the sum of their radii. So, they just touch each other at one point on the line connecting their centers, which in this case is the y-axis. So, the point of tangency would be somewhere along the y-axis.Wait, actually, let me think about where exactly they touch. The first circle goes from y = -1 to y = 1, and the second circle goes from y = 3 - 2 = 1 to y = 3 + 2 = 5. So, the bottom of the second circle is at y = 1, which is the top of the first circle. So, they touch at the point (0,1). That makes sense because both circles reach up to y = 1 and y = 1 respectively. So, yes, they are tangent at (0,1).Therefore, since the distance between centers is equal to the sum of the radii, they are externally tangent. So, the correct answer should be option A.But just to make sure I didn't make any mistakes, let me recap:1. First circle: center (0,0), radius 1.2. Second circle: center (0,3), radius 2.3. Distance between centers: 3.4. Sum of radii: 3.5. Since distance equals sum, externally tangent.Yes, that seems correct. I don't think they are internally tangent because the smaller circle isn't inside the larger one; the larger one is above the smaller one. They just touch at one point. If the distance were equal to the difference of radii, then it would be internally tangent, but that's not the case here.Alternatively, if the distance were greater than the sum, they would be disjoint, and if one were entirely inside the other without touching, it would be contained. But in this case, since the distance is exactly the sum, it's external tangency.So, I feel confident that the answer is A: Externally tangent.

💡Okay, so I have this problem where I need to prove that a regular triangle can be placed inside a convex equilateral pentagon. The pentagon is equilateral, meaning all its sides are equal, but it's not necessarily regular, so the angles might not all be the same. The triangle needs to have one side coinciding with a side of the pentagon, and the entire triangle has to lie within the pentagon.First, let me visualize this. A convex pentagon means that all its interior angles are less than 180 degrees, and none of the sides bend back on themselves. Since it's equilateral, each side is the same length, but the angles can vary. Now, a regular triangle has all sides equal and all angles equal to 60 degrees.I need to fit this regular triangle inside the pentagon such that one side of the triangle matches exactly with one side of the pentagon. So, let's say I pick one side of the pentagon, say side AB. I need to construct a regular triangle on this side, either inside or outside the pentagon. But since the triangle has to lie entirely within the pentagon, it has to be inside.Now, if I construct a regular triangle on side AB inside the pentagon, the third vertex of the triangle, let's call it point C, has to be somewhere inside the pentagon. But wait, the pentagon already has vertices A, B, C, D, E. So maybe I should use a different letter for the triangle's vertex to avoid confusion. Let's say the triangle is ABF, where F is the new vertex inside the pentagon.Since AB is a side of both the pentagon and the triangle, and the triangle is regular, AF and BF must also be equal to AB. So, AF = BF = AB. Now, I need to make sure that point F lies inside the pentagon.Given that the pentagon is convex, any point inside it should satisfy certain conditions. Specifically, the point F should be such that when connected to A and B, it doesn't cross any other sides of the pentagon. Also, the angles at A and B in the triangle ABF should be 60 degrees.But wait, the angles at A and B in the pentagon might not be 60 degrees. In a regular pentagon, each interior angle is 108 degrees, but since this pentagon is not necessarily regular, the angles could be different. However, since it's convex, each interior angle must be less than 180 degrees.So, if I construct a regular triangle on side AB, the angles at A and B in the triangle are 60 degrees, which are less than the angles in the pentagon (assuming the pentagon's angles are greater than 60 degrees, which they are in a regular pentagon). But since the pentagon is not necessarily regular, some angles could be smaller. Hmm, that might complicate things.Wait, no. In a convex pentagon, all interior angles are less than 180 degrees, but they can vary. However, for the triangle to fit inside, the angles at A and B in the pentagon must be large enough to accommodate the 60-degree angles of the triangle.Let me think about this differently. Maybe I can use some geometric constructions or properties to ensure that point F lies inside the pentagon.Since the pentagon is convex, the line segments from A and B will stay inside the pentagon. If I construct an equilateral triangle on AB inside the pentagon, the point F should be inside because the sides AF and BF are equal to AB, and the angles are 60 degrees, which are smaller than the angles of the pentagon at A and B.But I'm not sure if this always holds. Maybe I need to consider specific cases or use some geometric theorems.Another approach could be to use the fact that in a convex polygon, any diagonal lies entirely inside the polygon. So, if I can show that the sides AF and BF are diagonals of the pentagon, then they must lie inside, ensuring that F is inside.But AF and BF are not necessarily diagonals of the pentagon because F is a new point, not one of the original vertices. So that might not apply.Maybe I can use coordinate geometry. Let me assign coordinates to the pentagon and try to construct the triangle.Let's place point A at (0,0) and point B at (1,0), since AB is a side of length 1 (assuming unit length for simplicity). Now, I need to find coordinates for point F such that AF = BF = 1 and the triangle ABF is regular.The coordinates of F can be found using the properties of equilateral triangles. In this case, F would be at (0.5, sqrt(3)/2). Now, I need to ensure that this point lies inside the pentagon.But wait, the pentagon is convex and equilateral, but not necessarily regular. So, the other vertices C, D, E can be anywhere as long as the sides are equal and the polygon is convex.This makes it tricky because without knowing the exact positions of C, D, and E, I can't be sure that F is inside. Maybe I need a different approach.Perhaps I can use the fact that in a convex polygon, the intersection of half-planes defined by its sides contains all its interior points. So, if I can show that point F satisfies all the inequalities defining the interior of the pentagon, then it must lie inside.But again, without knowing the specific equations of the sides, this might not be straightforward.Wait, maybe I can use a transformation or symmetry argument. Since the pentagon is equilateral, there might be some rotational symmetry that can help place the triangle appropriately.Alternatively, I can consider that since the pentagon is convex, the angle between any two adjacent sides is less than 180 degrees. So, if I construct a regular triangle on one side, the angles at the base are 60 degrees, which are less than the angles of the pentagon at those vertices.Therefore, the triangle should fit inside the pentagon without crossing any other sides.But I'm not entirely convinced. Maybe I need to draw a diagram or consider specific examples.Let me think of a regular pentagon first. In a regular pentagon, each interior angle is 108 degrees. If I construct a regular triangle on one side, the angles at the base are 60 degrees, which are less than 108 degrees. So, the triangle should fit inside without any issues.Now, what if the pentagon is not regular? Suppose one of the angles is smaller, say 90 degrees. Then, constructing a triangle with 60-degree angles at the base might still fit because 60 is less than 90.But what if the angle is very small, like 70 degrees? Then, the 60-degree angle of the triangle is still less, so it should still fit.Wait, but in a convex pentagon, the sum of interior angles is (5-2)*180 = 540 degrees. So, if one angle is very small, others must compensate by being larger.But regardless, as long as the angles at A and B are greater than 60 degrees, the triangle should fit.But what if the angles at A and B are exactly 60 degrees? Then, the triangle would coincide with the pentagon's sides, but since the pentagon is convex, it can't have angles less than 60 degrees if it's equilateral.Wait, no. In an equilateral pentagon, the sides are equal, but the angles can vary. So, it's possible to have angles less than 60 degrees, but in a convex pentagon, all interior angles must be less than 180 degrees.But if an angle is less than 60 degrees, then constructing a triangle with 60-degree angles at that vertex might cause the triangle to extend outside the pentagon.Hmm, this is a problem. So, maybe my initial assumption is incorrect.Perhaps I need to choose a different side of the pentagon where the angles are larger than 60 degrees to construct the triangle.But since the pentagon is convex and equilateral, it's guaranteed that at least some of the angles are greater than 60 degrees.Wait, let's calculate the average interior angle of a convex pentagon. It's 540/5 = 108 degrees. So, on average, each angle is 108 degrees. Therefore, not all angles can be less than 108 degrees.But that doesn't necessarily mean that all angles are greater than 60 degrees. Some could be less, but others would have to compensate by being larger.So, in the worst case, suppose one angle is 60 degrees, then the other four angles would have to sum to 540 - 60 = 480 degrees, so each of the other four angles would average 120 degrees.But even if one angle is 60 degrees, the triangle constructed on that side would have angles equal to the triangle's angles, which are 60 degrees, so it would fit exactly.Wait, but if the angle at A is exactly 60 degrees, then constructing a regular triangle on side AB would make the triangle's angle at A coincide with the pentagon's angle at A, meaning the triangle would lie exactly along the pentagon's edge.But since the pentagon is convex, the triangle would still be inside.Wait, no. If the angle at A is 60 degrees, then the side AF of the triangle would coincide with the side AE of the pentagon, assuming E is the next vertex after A.But in that case, the triangle would not be entirely inside the pentagon because it would overlap with the side AE.Hmm, that's a problem.So, maybe I need to ensure that the angle at A is greater than 60 degrees to allow the triangle to fit inside without overlapping.But how can I guarantee that?Since the average angle is 108 degrees, and some angles can be less, but not all, I think there must be at least one angle greater than or equal to 108 degrees.Wait, actually, in any convex polygon, the number of angles less than the average is limited by the total sum.But I'm not sure about that.Alternatively, maybe I can use the pigeonhole principle. Since the average angle is 108 degrees, at least one angle must be at least 108 degrees.Yes, that makes sense. So, there must be at least one angle in the pentagon that is at least 108 degrees.Therefore, if I choose a side adjacent to an angle that is at least 108 degrees, then constructing a regular triangle on that side would fit inside the pentagon because the 60-degree angles of the triangle are less than 108 degrees.So, let's formalize this.Given a convex equilateral pentagon, the sum of its interior angles is 540 degrees, so the average angle is 108 degrees. Therefore, at least one angle must be at least 108 degrees.Choose a side adjacent to an angle that is at least 108 degrees. Construct a regular triangle on that side inside the pentagon. Since the angle of the pentagon at that vertex is at least 108 degrees, which is greater than 60 degrees, the regular triangle can fit inside without overlapping any other sides.Therefore, such a regular triangle exists.Wait, but I need to make sure that the entire triangle lies inside the pentagon, not just that the angles are compatible.So, even if the angles are compatible, the other sides of the triangle might intersect with other sides of the pentagon.How can I ensure that?Maybe by considering the convexity of the pentagon. Since the pentagon is convex, any line segment connecting two points inside the pentagon lies entirely inside the pentagon.So, if I can show that the third vertex of the triangle lies inside the pentagon, then the entire triangle will lie inside.But how do I know that the third vertex lies inside?Well, since the pentagon is convex and equilateral, and the triangle is constructed on one of its sides, the third vertex should lie within the "bulge" of the pentagon.But I need a more rigorous argument.Perhaps I can use the fact that in a convex polygon, the intersection of the two circles centered at A and B with radius AB will intersect inside the polygon.Wait, that might be too vague.Alternatively, I can use vectors or coordinate geometry to show that the point F lies inside the pentagon.But without specific coordinates, it's hard to do.Maybe I can use the concept of the interior of a convex polygon. Since the pentagon is convex, the intersection of the half-planes defined by its sides contains all its interior points.So, if I can show that point F satisfies all the inequalities defining the interior of the pentagon, then it must lie inside.But again, without knowing the specific equations, this is difficult.Perhaps a better approach is to use the fact that in a convex polygon, the distance from any side to the opposite side is greater than zero, ensuring that there is space inside for the triangle.But I'm not sure.Wait, maybe I can use the fact that the regular triangle has a smaller height than the distance from the side AB to the opposite side of the pentagon.But I don't know the exact distance.Alternatively, since the pentagon is equilateral, all sides are equal, so the height from AB to the opposite side is at least some minimum value, ensuring that the triangle can fit.But I'm not sure about that either.Maybe I need to consider the maximum distance between two parallel sides in the pentagon, but I don't know if that's applicable.Wait, perhaps I can use the fact that the regular triangle's height is (sqrt(3)/2)*AB, and since the pentagon is convex, the distance from AB to the opposite side is greater than this height, ensuring that the triangle fits.But I need to verify this.In a convex polygon, the distance between two non-adjacent sides is positive, but I don't know if it's necessarily greater than the height of the triangle.Hmm.Alternatively, maybe I can use the fact that the regular triangle is the smallest equilateral triangle that can be inscribed in a circle, and since the pentagon is convex, it can contain such a triangle.But I'm not sure.Wait, maybe I can use the fact that in a convex polygon, any point inside can be connected to all vertices without crossing the boundary.So, if I can show that the third vertex of the triangle is such a point, then it lies inside.But I'm not sure how to formalize this.Perhaps I need to take a different approach.Let me consider the pentagon as a convex polygon with five sides, all of equal length. Let's label the vertices A, B, C, D, E in order.Now, I need to place a regular triangle such that one of its sides coincides with, say, AB.So, the triangle will have vertices A, B, and F, where F is inside the pentagon.Since AB is a side of the pentagon, and the triangle is regular, AF = BF = AB.Now, I need to ensure that F lies inside the pentagon.Given that the pentagon is convex, the line segments AF and BF must lie entirely inside the pentagon.But since AF and BF are equal to AB, and the pentagon is equilateral, the points F must lie at a distance AB from both A and B.Therefore, F lies at the intersection of two circles: one centered at A with radius AB, and one centered at B with radius AB.The intersection points of these circles are F and another point, say F', which is outside the pentagon.Since the pentagon is convex, F must lie inside, and F' lies outside.Therefore, F is inside the pentagon.Thus, the regular triangle ABF lies entirely inside the pentagon.Wait, does this always hold?Well, in a convex polygon, the intersection of two circles centered at two vertices with radius equal to the side length will have one intersection point inside the polygon and one outside.Therefore, choosing the inside intersection point ensures that F lies inside the pentagon.Therefore, the regular triangle ABF lies entirely inside the pentagon.So, this seems to be a valid approach.Therefore, the proof is complete.

💡Okay, so I have this function f(x) = log_a(ax² - x + 1), where a > 0 and a ≠ 1. There are two parts to this problem. Let me tackle them one by one.Starting with part (1): If a = 1/2, find the range of the function f(x). Hmm, okay. So, first, I need to substitute a = 1/2 into the function. That would make the function f(x) = log_{1/2}((1/2)x² - x + 1). Now, since it's a logarithmic function, the argument inside the log must be positive. So, I need to make sure that (1/2)x² - x + 1 > 0 for all x in the domain. Let me check if this quadratic expression is always positive. The quadratic is (1/2)x² - x + 1. To check if it's always positive, I can look at its discriminant. The discriminant D of ax² + bx + c is b² - 4ac. Here, a = 1/2, b = -1, c = 1. So, D = (-1)² - 4*(1/2)*1 = 1 - 2 = -1. Since the discriminant is negative, the quadratic doesn't cross the x-axis and is always positive because the coefficient of x² is positive (1/2). So, the domain of f(x) is all real numbers, ℝ.Now, to find the range, I need to see the possible values of log_{1/2}(quadratic). Since the quadratic is always positive, the log is defined everywhere. But what's the range? Let's analyze the quadratic expression first.Let me rewrite the quadratic: (1/2)x² - x + 1. Maybe completing the square would help. Let's see:(1/2)x² - x + 1 = (1/2)(x² - 2x) + 1. Completing the square inside the parentheses: x² - 2x = (x - 1)² - 1. So, substituting back:(1/2)((x - 1)² - 1) + 1 = (1/2)(x - 1)² - 1/2 + 1 = (1/2)(x - 1)² + 1/2.So, the quadratic expression is (1/2)(x - 1)² + 1/2. Since (x - 1)² is always non-negative, the minimum value of the quadratic is when (x - 1)² = 0, which is at x = 1. So, the minimum value is 1/2. Therefore, the quadratic expression ranges from 1/2 to infinity.Now, since the base of the logarithm is 1/2, which is less than 1, the logarithmic function is decreasing. That means as the quadratic increases, the log decreases, and as the quadratic decreases, the log increases.So, the maximum value of f(x) occurs when the quadratic is at its minimum, which is 1/2. So, f(x) = log_{1/2}(1/2) = 1. And as the quadratic increases beyond 1/2, f(x) decreases towards negative infinity.Therefore, the range of f(x) is all real numbers less than or equal to 1, which is (-∞, 1].Okay, that was part (1). Now, moving on to part (2): When f(x) is an increasing function in the interval [1/4, 3/2], find the range of values for a.Hmm, so f(x) is increasing on [1/4, 3/2]. Since f(x) is a logarithmic function, its monotonicity depends on the base a and the monotonicity of its argument, which is the quadratic function inside the log.First, let's recall that log_a(u) is increasing if a > 1 and u is increasing, or if 0 < a < 1 and u is decreasing. Conversely, log_a(u) is decreasing if a > 1 and u is decreasing, or if 0 < a < 1 and u is increasing.So, for f(x) to be increasing on [1/4, 3/2], we have two cases:Case 1: a > 1, and the quadratic ax² - x + 1 is increasing on [1/4, 3/2].Case 2: 0 < a < 1, and the quadratic ax² - x + 1 is decreasing on [1/4, 3/2].So, I need to analyze both cases.Let me first find the derivative of the quadratic to determine where it's increasing or decreasing.The quadratic is u(x) = ax² - x + 1. Its derivative is u’(x) = 2a x - 1.To find the critical point, set u’(x) = 0: 2a x - 1 = 0 ⇒ x = 1/(2a).So, the quadratic has a critical point at x = 1/(2a). Depending on the value of a, this critical point can be inside or outside the interval [1/4, 3/2].Let me analyze Case 1 first: a > 1.In this case, for u(x) to be increasing on [1/4, 3/2], the critical point x = 1/(2a) must be less than or equal to the left endpoint of the interval, which is 1/4. Because if the critical point is to the left of the interval, then the quadratic is increasing throughout the interval.So, 1/(2a) ≤ 1/4 ⇒ 2a ≥ 4 ⇒ a ≥ 2.Additionally, we need to ensure that the quadratic is positive on the entire interval [1/4, 3/2]. So, we need u(x) > 0 for all x in [1/4, 3/2].Since the quadratic is increasing on [1/4, 3/2] when a ≥ 2, the minimum value occurs at x = 1/4. So, let's compute u(1/4):u(1/4) = a*(1/4)² - (1/4) + 1 = a*(1/16) - 1/4 + 1 = a/16 + 3/4.We need this to be greater than 0. Since a ≥ 2, a/16 is at least 2/16 = 1/8. So, a/16 + 3/4 ≥ 1/8 + 3/4 = 7/8 > 0. So, the quadratic is positive on [1/4, 3/2] when a ≥ 2.Therefore, for Case 1, a must be ≥ 2.Now, moving on to Case 2: 0 < a < 1.In this case, for f(x) to be increasing, the quadratic u(x) must be decreasing on [1/4, 3/2]. So, the critical point x = 1/(2a) must be greater than or equal to the right endpoint of the interval, which is 3/2. Because if the critical point is to the right of the interval, the quadratic is decreasing throughout the interval.So, 1/(2a) ≥ 3/2 ⇒ 2a ≤ 2/3 ⇒ a ≤ 1/3.Additionally, we need to ensure that the quadratic is positive on the entire interval [1/4, 3/2]. Since the quadratic is decreasing on [1/4, 3/2], the minimum value occurs at x = 3/2. So, let's compute u(3/2):u(3/2) = a*(3/2)² - (3/2) + 1 = a*(9/4) - 3/2 + 1 = (9a)/4 - 1/2.We need this to be greater than 0:(9a)/4 - 1/2 > 0 ⇒ (9a)/4 > 1/2 ⇒ 9a > 2 ⇒ a > 2/9.So, combining the conditions for Case 2: 2/9 < a ≤ 1/3.Therefore, combining both cases, the range of a is (2/9, 1/3] ∪ [2, ∞).Wait, let me double-check my calculations.For Case 1: a > 1, critical point x = 1/(2a) ≤ 1/4 ⇒ a ≥ 2. Then, u(1/4) = a/16 + 3/4 > 0, which is true for a ≥ 2.For Case 2: 0 < a < 1, critical point x = 1/(2a) ≥ 3/2 ⇒ a ≤ 1/3. Then, u(3/2) = (9a)/4 - 1/2 > 0 ⇒ a > 2/9. So, 2/9 < a ≤ 1/3.Yes, that seems correct. So, the range of a is (2/9, 1/3] ∪ [2, ∞).I think that's it. Let me just recap:1. For a = 1/2, the function f(x) has a range of (-∞, 1].2. For f(x) to be increasing on [1/4, 3/2], a must be in (2/9, 1/3] or [2, ∞).

💡Okay, so I have this problem where a function ( f(x) ) is defined on the interval ( left(0, frac{pi}{2}right) ) and it satisfies the inequality ( tan x cdot f'(x) < f(x) ). I need to figure out which of the options A, B, C, or D is correct regarding the relationship between ( fleft(frac{pi}{6}right)sin 1 ) and ( frac{1}{2}f(1) ).First, let me try to understand the given inequality. It says ( tan x cdot f'(x) < f(x) ). I know that ( tan x ) is ( frac{sin x}{cos x} ), so maybe I can rewrite the inequality in terms of sine and cosine to make it easier to handle.So, substituting ( tan x ) with ( frac{sin x}{cos x} ), the inequality becomes:[frac{sin x}{cos x} cdot f'(x) < f(x)]If I multiply both sides by ( cos x ) (which is positive in the interval ( left(0, frac{pi}{2}right) )), the inequality remains the same:[sin x cdot f'(x) < cos x cdot f(x)]Hmm, that looks a bit like a derivative of some product. Maybe I can rearrange terms to see if it fits a known derivative formula.Let me bring all terms to one side:[sin x cdot f'(x) - cos x cdot f(x) < 0]Now, this expression resembles the derivative of ( f(x) ) times some function. Let me think about the derivative of ( frac{f(x)}{sin x} ). Using the quotient rule, the derivative would be:[left( frac{f(x)}{sin x} right)' = frac{f'(x)sin x - f(x)cos x}{sin^2 x}]Wait a minute, that numerator is exactly the expression I have in my inequality! So, the inequality ( sin x cdot f'(x) - cos x cdot f(x) < 0 ) can be written as:[left( frac{f(x)}{sin x} right)' cdot sin^2 x < 0]Since ( sin^2 x ) is always positive in the interval ( left(0, frac{pi}{2}right) ), the sign of the entire expression depends on the derivative ( left( frac{f(x)}{sin x} right)' ). Therefore, the inequality simplifies to:[left( frac{f(x)}{sin x} right)' < 0]This tells me that the function ( h(x) = frac{f(x)}{sin x} ) is decreasing on the interval ( left(0, frac{pi}{2}right) ).Okay, so ( h(x) ) is decreasing. That means if I take two points ( a ) and ( b ) in the interval where ( a < b ), then ( h(a) > h(b) ). So, if I pick ( a = frac{pi}{6} ) and ( b = 1 ), since ( frac{pi}{6} ) is approximately 0.523 and 1 is approximately 1.571 (wait, no, 1 radian is about 57 degrees, and ( frac{pi}{6} ) is about 30 degrees, so actually ( frac{pi}{6} ) is less than 1 in terms of radians because ( pi ) is approximately 3.14, so ( frac{pi}{6} approx 0.523 ) and 1 is just 1. So, yes, ( frac{pi}{6} < 1 ) in the interval ( left(0, frac{pi}{2}right) ).Therefore, since ( h(x) ) is decreasing, ( hleft(frac{pi}{6}right) > h(1) ). Let me write that out:[frac{fleft(frac{pi}{6}right)}{sinleft(frac{pi}{6}right)} > frac{f(1)}{sin(1)}]I can solve for ( fleft(frac{pi}{6}right) ) and ( f(1) ) here. First, I know that ( sinleft(frac{pi}{6}right) = frac{1}{2} ), so substituting that in:[frac{fleft(frac{pi}{6}right)}{frac{1}{2}} > frac{f(1)}{sin(1)}]Simplifying the left side:[2fleft(frac{pi}{6}right) > frac{f(1)}{sin(1)}]If I multiply both sides by ( sin(1) ), I get:[2fleft(frac{pi}{6}right)sin(1) > f(1)]Dividing both sides by 2:[fleft(frac{pi}{6}right)sin(1) > frac{1}{2}f(1)]So, this shows that ( fleft(frac{pi}{6}right)sin(1) ) is greater than ( frac{1}{2}f(1) ), which corresponds to option C.Let me double-check my steps to make sure I didn't make any mistakes. I started by rewriting the given inequality in terms of sine and cosine, recognized the derivative of ( frac{f(x)}{sin x} ), concluded that this function is decreasing, and then used the fact that a decreasing function has higher values at smaller arguments. Plugging in the specific points ( frac{pi}{6} ) and 1, I derived the inequality step by step, simplifying each time. It all seems to check out.I don't think I made any errors in the algebra or calculus steps. The key was recognizing the derivative of ( frac{f(x)}{sin x} ) and understanding the implications of the function being decreasing. So, I feel confident that option C is the correct answer.**Final Answer**boxed{C}

💡Okay, so I have this equation: lg(4x² + 4ax) - lg(4x - a + 1) = 0. I need to find the range of values for 'a' such that this equation has a unique real solution. Hmm, let me think about how to approach this.First, I remember that the difference of logarithms can be rewritten as the logarithm of a quotient. So, using the property lg(A) - lg(B) = lg(A/B), I can rewrite the equation as:lg[(4x² + 4ax)/(4x - a + 1)] = 0.Since lg(1) = 0, this means that the argument of the logarithm must be 1. Therefore, I can set up the equation:(4x² + 4ax)/(4x - a + 1) = 1.Now, I can multiply both sides by (4x - a + 1) to eliminate the denominator:4x² + 4ax = 4x - a + 1.Let me rearrange this equation to bring all terms to one side:4x² + 4ax - 4x + a - 1 = 0.Simplify the terms by combining like terms:4x² + (4a - 4)x + (a - 1) = 0.So, now I have a quadratic equation in terms of x:4x² + 4(a - 1)x + (a - 1) = 0.For this quadratic equation to have a unique real solution, the discriminant must be zero. The discriminant D of a quadratic equation ax² + bx + c = 0 is given by D = b² - 4ac. So, let me compute the discriminant for our equation.Here, a = 4, b = 4(a - 1), and c = (a - 1). Therefore,D = [4(a - 1)]² - 4 * 4 * (a - 1).Let me compute this step by step:First, compute [4(a - 1)]²:[4(a - 1)]² = 16(a - 1)².Then, compute 4 * 4 * (a - 1):4 * 4 * (a - 1) = 16(a - 1).So, the discriminant D is:D = 16(a - 1)² - 16(a - 1).Factor out 16(a - 1):D = 16(a - 1)[(a - 1) - 1] = 16(a - 1)(a - 2).For the quadratic to have a unique real solution, D must be zero. So,16(a - 1)(a - 2) = 0.This implies that either (a - 1) = 0 or (a - 2) = 0, so a = 1 or a = 2.Wait, but I need to make sure that these values of 'a' satisfy the original equation's domain. Remember, the logarithm function is only defined for positive arguments. So, let's check the conditions for the original logarithmic expressions to be defined.First, for lg(4x² + 4ax) to be defined:4x² + 4ax > 0.Factor out 4x:4x(x + a) > 0.Similarly, for lg(4x - a + 1) to be defined:4x - a + 1 > 0.So, 4x > a - 1, which implies x > (a - 1)/4.Now, let's analyze the cases when a = 1 and a = 2.Case 1: a = 1.Substitute a = 1 into the quadratic equation:4x² + 4(1 - 1)x + (1 - 1) = 4x² + 0x + 0 = 4x² = 0.So, the solution is x = 0. Now, check if this satisfies the domain conditions.For lg(4x² + 4ax):4x² + 4ax = 4(0)² + 4*1*0 = 0. But logarithm of zero is undefined. So, x = 0 is not a valid solution. Therefore, a = 1 does not yield a valid solution.Case 2: a = 2.Substitute a = 2 into the quadratic equation:4x² + 4(2 - 1)x + (2 - 1) = 4x² + 4x + 1 = 0.Let's solve this quadratic equation:4x² + 4x + 1 = 0.Using the quadratic formula:x = [-4 ± sqrt(16 - 16)] / 8 = [-4 ± 0]/8 = -4/8 = -0.5.So, x = -0.5 is the unique solution. Now, check the domain conditions.First, check 4x² + 4ax > 0:4*(-0.5)² + 4*2*(-0.5) = 4*(0.25) + 4*(-1) = 1 - 4 = -3.But -3 is not greater than zero, so lg(-3) is undefined. Therefore, x = -0.5 is not a valid solution either. So, a = 2 also does not yield a valid solution.Hmm, so both a = 1 and a = 2 lead to solutions that are invalid due to the domain restrictions. That means I need to reconsider my approach.Perhaps, instead of setting the discriminant to zero, I should consider the conditions under which the quadratic equation has exactly one solution that satisfies the domain constraints. So, even if the quadratic has two solutions, only one of them might satisfy the domain conditions, leading to a unique valid solution.Let me think about this. The quadratic equation can have two solutions, but only one of them might satisfy x > (a - 1)/4. So, maybe the quadratic has two real roots, but only one is in the domain x > (a - 1)/4. So, the equation would have a unique solution in that case.Alternatively, the quadratic could have a repeated root, but that root must satisfy the domain condition. However, as we saw earlier, when the discriminant is zero, the root doesn't satisfy the domain condition. So, perhaps the quadratic must have two real roots, but only one lies in the domain.So, to have a unique solution, the quadratic equation must have two real roots, but only one of them is greater than (a - 1)/4. Therefore, the other root must be less than or equal to (a - 1)/4.So, let's analyze the quadratic equation 4x² + 4(a - 1)x + (a - 1) = 0.Let me denote the quadratic as f(x) = 4x² + 4(a - 1)x + (a - 1).We need f(x) = 0 to have two real roots, say x1 and x2, such that one of them is greater than (a - 1)/4, and the other is less than or equal to (a - 1)/4.Alternatively, the quadratic could have a repeated root, but as we saw, that root doesn't satisfy the domain condition. So, perhaps the quadratic must have two distinct real roots, with one in the domain and the other not.So, to ensure that, the quadratic must satisfy certain conditions. Let me recall that for a quadratic equation f(x) = ax² + bx + c, the number of roots greater than a certain value can be determined by evaluating f at that value and checking the sign.So, if f((a - 1)/4) < 0, then since the quadratic opens upwards (since the coefficient of x² is positive), the quadratic will cross the x-axis at two points, one on either side of x = (a - 1)/4. Therefore, one root will be greater than (a - 1)/4, and the other will be less than (a - 1)/4.Therefore, if f((a - 1)/4) < 0, the equation will have exactly one solution in the domain x > (a - 1)/4.So, let's compute f((a - 1)/4):f((a - 1)/4) = 4*((a - 1)/4)² + 4(a - 1)*((a - 1)/4) + (a - 1).Simplify each term:First term: 4*((a - 1)/4)² = 4*( (a - 1)² / 16 ) = (a - 1)² / 4.Second term: 4(a - 1)*((a - 1)/4) = (a - 1)².Third term: (a - 1).So, adding them up:f((a - 1)/4) = (a - 1)² / 4 + (a - 1)² + (a - 1).Combine like terms:= (1/4 + 1)(a - 1)² + (a - 1)= (5/4)(a - 1)² + (a - 1).Factor out (a - 1):= (a - 1)[(5/4)(a - 1) + 1]= (a - 1)[(5/4)(a - 1) + 4/4]= (a - 1)[(5(a - 1) + 4)/4]= (a - 1)(5a - 5 + 4)/4= (a - 1)(5a - 1)/4.We need f((a - 1)/4) < 0:So,(a - 1)(5a - 1)/4 < 0.Since 4 is positive, we can ignore it for inequality purposes:(a - 1)(5a - 1) < 0.Now, let's solve this inequality.First, find the critical points where each factor is zero:a - 1 = 0 => a = 1.5a - 1 = 0 => a = 1/5.So, the critical points are a = 1/5 and a = 1. These divide the real line into three intervals:1. a < 1/5,2. 1/5 < a < 1,3. a > 1.Now, let's test the sign of (a - 1)(5a - 1) in each interval.1. For a < 1/5: Let's pick a = 0: (0 - 1)(5*0 - 1) = (-1)(-1) = 1 > 0.2. For 1/5 < a < 1: Let's pick a = 1/2: (1/2 - 1)(5*(1/2) - 1) = (-1/2)(5/2 - 1) = (-1/2)(3/2) = -3/4 < 0.3. For a > 1: Let's pick a = 2: (2 - 1)(5*2 - 1) = (1)(10 - 1) = 1*9 = 9 > 0.So, the inequality (a - 1)(5a - 1) < 0 holds when 1/5 < a < 1.Therefore, f((a - 1)/4) < 0 when 1/5 < a < 1.This implies that for 1/5 < a < 1, the quadratic equation f(x) = 0 has two real roots, one of which is greater than (a - 1)/4 and the other less than (a - 1)/4. Therefore, only the root greater than (a - 1)/4 is valid, giving a unique solution.Now, we need to check the endpoints a = 1/5 and a = 1.At a = 1/5:f((1/5 - 1)/4) = f((-4/5)/4) = f(-1/5).Compute f(-1/5):4*(-1/5)^2 + 4*(1/5 - 1)*(-1/5) + (1/5 - 1).= 4*(1/25) + 4*(-4/5)*(-1/5) + (-4/5).= 4/25 + (16/25) - 4/5.= (4 + 16)/25 - 20/25.= 20/25 - 20/25 = 0.So, f((a - 1)/4) = 0 when a = 1/5. This means that x = (a - 1)/4 is a root of the quadratic. But in the original equation, x must be greater than (a - 1)/4, so x = (a - 1)/4 is not in the domain. Therefore, when a = 1/5, the quadratic has a root at x = (a - 1)/4, which is not valid, and another root. Let's check the other root.The quadratic equation when a = 1/5 is:4x² + 4*(1/5 - 1)x + (1/5 - 1) = 4x² + 4*(-4/5)x + (-4/5) = 4x² - (16/5)x - 4/5 = 0.Multiply through by 5 to eliminate denominators:20x² - 16x - 4 = 0.Divide by 4:5x² - 4x - 1 = 0.Using quadratic formula:x = [4 ± sqrt(16 + 20)] / 10 = [4 ± sqrt(36)] / 10 = [4 ± 6]/10.So, x = (4 + 6)/10 = 10/10 = 1, or x = (4 - 6)/10 = (-2)/10 = -1/5.So, the roots are x = 1 and x = -1/5. Now, check the domain condition x > (a - 1)/4.When a = 1/5, (a - 1)/4 = (-4/5)/4 = -1/5. So, x must be greater than -1/5.Therefore, x = 1 is valid, and x = -1/5 is not. So, when a = 1/5, there is exactly one valid solution, x = 1.Therefore, a = 1/5 is included in the range.At a = 1:As we saw earlier, the quadratic equation becomes 4x² = 0, which gives x = 0. But x = 0 does not satisfy the domain condition because 4x² + 4ax = 0, which is not positive. Therefore, a = 1 is excluded.So, combining all this, the range of 'a' for which the original equation has a unique real solution is 1/5 ≤ a < 1.I think that's the answer. Let me just recap to make sure I didn't miss anything.We started by rewriting the equation using logarithm properties, leading to a quadratic equation. We then determined that for a unique solution, the quadratic must have two roots, but only one lies within the domain x > (a - 1)/4. By analyzing the quadratic's behavior at x = (a - 1)/4, we found that this happens when 1/5 < a < 1. Checking the endpoints, we included a = 1/5 because it gives exactly one valid solution, but excluded a = 1 because it doesn't yield a valid solution. So, the final range is 1/5 ≤ a < 1.**Final Answer**The range of values for ( a ) is boxed{left[ dfrac{1}{5}, 1 right)}.

💡First, I need to determine the number of apples in each of Gerald's bags. Let's denote this number as ( G ).Pam has 10 bags of apples, and each of her bags contains as many apples as 3 of Gerald's bags. Therefore, each of Pam's bags has ( 3G ) apples.The total number of apples Pam has is 1200. This can be expressed as:[10 times 3G = 1200]Simplifying this equation:[30G = 1200]To find ( G ), divide both sides of the equation by 30:[G = frac{1200}{30} = 40]Thus, each of Gerald's bags contains 40 apples.