answer：First, I need to determine the total cost for each cycle by adding the purchase price to the repairs and accessories.For Cycle A:- Purchase Price: Rs. 1800- Repairs and Accessories: Rs. 200- Total Cost: Rs. 1800 + Rs. 200 = Rs. 2000For Cycle B:- Purchase Price: Rs. 2400- Repairs and Accessories: Rs. 300- Total Cost: Rs. 2400 + Rs. 300 = Rs. 2700For Cycle C:- Purchase Price: Rs. 3200- Repairs and Accessories: Rs. 400- Total Cost: Rs. 3200 + Rs. 400 = Rs. 3600Next, I'll calculate the discounted price for each cycle based on the given discount rates.For Cycle A:- Discount: 10% of Rs. 2000 = 0.10 × Rs. 2000 = Rs. 200- Discounted Price: Rs. 2000 - Rs. 200 = Rs. 1800For Cycle B:- Discount: 12% of Rs. 2700 = 0.12 × Rs. 2700 = Rs. 324- Discounted Price: Rs. 2700 - Rs. 324 = Rs. 2376For Cycle C:- Discount: 15% of Rs. 3600 = 0.15 × Rs. 3600 = Rs. 540- Discounted Price: Rs. 3600 - Rs. 540 = Rs. 3060After applying the discounts, I'll calculate the sales tax on each discounted price.For Cycle A:- Sales Tax: 5% of Rs. 1800 = 0.05 × Rs. 1800 = Rs. 90- Final Selling Price: Rs. 1800 + Rs. 90 = Rs. 1890For Cycle B:- Sales Tax: 6% of Rs. 2376 = 0.06 × Rs. 2376 = Rs. 142.56- Final Selling Price: Rs. 2376 + Rs. 142.56 = Rs. 2518.56For Cycle C:- Sales Tax: 7% of Rs. 3060 = 0.07 × Rs. 3060 = Rs. 214.20- Final Selling Price: Rs. 3060 + Rs. 214.20 = Rs. 3274.20Finally, I'll sum up the final selling prices of all three cycles to get the combined final selling price.Combined Final Selling Price:- Rs. 1890 (Cycle A) + Rs. 2518.56 (Cycle B) + Rs. 3274.20 (Cycle C) = Rs. 7682.76Therefore, the combined final selling price of all three cycles is Rs. 7682.76.

answer：First, I need to determine the total distance the car travels. The car goes 100 km uphill and 50 km downhill, so the total distance is 150 km.Next, I'll calculate the time taken for each part of the journey. For the uphill section, the car travels at 30 km/hr, so the time taken is 100 km divided by 30 km/hr, which equals 10/3 hours or approximately 3.33 hours.For the downhill section, the car travels at 70 km/hr, so the time taken is 50 km divided by 70 km/hr, which equals 5/7 hours or approximately 0.71 hours.Adding these two times together gives the total time for the journey: 10/3 hours plus 5/7 hours. To add these fractions, I'll find a common denominator, which is 21. Converting both fractions, I get 70/21 plus 15/21, which equals 85/21 hours or approximately 4.05 hours.Finally, to find the average speed, I'll divide the total distance by the total time: 150 km divided by 4.05 hours, which is approximately 37.04 km/hr.Therefore, the average speed of the car for the entire journey is approximately 37.04 km/hr.

answer：Okay, so I have this function f(x) = e^x sin x, and I need to answer three questions about it. Let me take them one by one.**Problem (1): Find the intervals of monotonicity for f(x).**Alright, monotonicity refers to where the function is increasing or decreasing. To find that, I need to compute the derivative of f(x) and analyze its sign.So, f(x) = e^x sin x. The derivative f’(x) will be the derivative of e^x times sin x plus e^x times the derivative of sin x. That is:f’(x) = e^x sin x + e^x cos x.I can factor out e^x:f’(x) = e^x (sin x + cos x).Hmm, e^x is always positive, so the sign of f’(x) depends on sin x + cos x.I remember that sin x + cos x can be rewritten using a trigonometric identity. Specifically, sin x + cos x = √2 sin(x + π/4). Let me verify that:Using the identity: a sin x + b cos x = √(a² + b²) sin(x + φ), where φ = arctan(b/a). Here, a = 1 and b = 1, so √(1 + 1) = √2, and φ = arctan(1) = π/4. So yes, sin x + cos x = √2 sin(x + π/4).Therefore, f’(x) = √2 e^x sin(x + π/4).Since √2 e^x is always positive, the sign of f’(x) is determined by sin(x + π/4).So, f’(x) > 0 when sin(x + π/4) > 0, and f’(x) < 0 when sin(x + π/4) < 0.The sine function is positive in intervals where its argument is between 0 and π, and negative between π and 2π, and this repeats every 2π.So, sin(x + π/4) > 0 when:0 < x + π/4 < π ⇒ -π/4 < x < 3π/4.Similarly, sin(x + π/4) < 0 when:π < x + π/4 < 2π ⇒ 3π/4 < x < 7π/4.But since sine is periodic with period 2π, this pattern repeats every 2π. So, in general, for any integer k, the intervals where f’(x) > 0 are:[2kπ - π/4, 2kπ + 3π/4].And the intervals where f’(x) < 0 are:[2kπ + 3π/4, 2kπ + 7π/4].Therefore, f(x) is increasing on [2kπ - π/4, 2kπ + 3π/4] and decreasing on [2kπ + 3π/4, 2kπ + 7π/4] for all integers k.**Problem (2): If for any x ∈ [0, π/2], f(x) ≥ kx always holds, find the range of the real number k.**Okay, so we need to find all real numbers k such that e^x sin x ≥ kx for all x in [0, π/2].Let me define a new function g(x) = f(x) - kx = e^x sin x - kx.We need g(x) ≥ 0 for all x in [0, π/2].To ensure this, the minimum value of g(x) on [0, π/2] should be greater than or equal to zero.So, first, let's find the critical points of g(x) by taking its derivative:g’(x) = f’(x) - k = e^x (sin x + cos x) - k.Let me denote h(x) = e^x (sin x + cos x). Then, g’(x) = h(x) - k.So, the critical points occur when h(x) = k.We need to analyze h(x) on [0, π/2].Compute h’(x):h’(x) = derivative of e^x (sin x + cos x) = e^x (sin x + cos x) + e^x (cos x - sin x) = e^x (2 cos x).So, h’(x) = 2 e^x cos x.Since cos x is non-negative on [0, π/2], h’(x) ≥ 0 on [0, π/2]. Therefore, h(x) is non-decreasing on [0, π/2].Thus, the minimum value of h(x) is at x = 0, and the maximum is at x = π/2.Compute h(0):h(0) = e^0 (sin 0 + cos 0) = 1*(0 + 1) = 1.Compute h(π/2):h(π/2) = e^{π/2} (sin(π/2) + cos(π/2)) = e^{π/2} (1 + 0) = e^{π/2}.So, h(x) ∈ [1, e^{π/2}] on [0, π/2].Now, let's analyze g’(x):Case 1: If k ≤ 1.Since h(x) ≥ 1, and k ≤ 1, then h(x) - k ≥ 0. Therefore, g’(x) ≥ 0 on [0, π/2]. So, g(x) is non-decreasing on [0, π/2].Thus, the minimum of g(x) is at x = 0:g(0) = e^0 sin 0 - k*0 = 0.So, g(x) ≥ 0 for all x in [0, π/2] when k ≤ 1.Case 2: If 1 < k < e^{π/2}.Here, h(x) is increasing from 1 to e^{π/2}. So, h(x) = k will have exactly one solution x_0 in (0, π/2). For x < x_0, h(x) < k, so g’(x) < 0, meaning g(x) is decreasing. For x > x_0, h(x) > k, so g’(x) > 0, meaning g(x) is increasing.Therefore, g(x) has a minimum at x = x_0. We need to check if g(x_0) ≥ 0.But since x_0 is in (0, π/2), and g(0) = 0, if g(x) is decreasing before x_0, then g(x_0) < g(0) = 0, which violates the condition. Therefore, such k does not satisfy the requirement.Case 3: If k ≥ e^{π/2}.Then, h(x) ≤ e^{π/2} < k. So, h(x) - k < 0 for all x in [0, π/2]. Thus, g’(x) < 0, meaning g(x) is decreasing on [0, π/2].Therefore, the minimum of g(x) is at x = π/2:g(π/2) = e^{π/2} sin(π/2) - k*(π/2) = e^{π/2} - (k π)/2.But since k ≥ e^{π/2}, we have g(π/2) ≤ e^{π/2} - (e^{π/2} * π)/2. Since π/2 > 1, this is negative. Therefore, g(x) < 0 at x = π/2, which violates the condition.Therefore, only Case 1 is valid, so k must be ≤ 1.**Problem (3): Let F(x) = f(x) + e^x cos x, where x ∈ [-2015π/2, 2017π/2]. Draw all the tangent lines to the graph of F(x) at point M((π - 1)/2, 0), and let the abscissas of the points of tangency form the sequence {x_n}. Find the sum S of all terms of the sequence {x_n}.**Okay, this seems a bit more involved. Let's break it down.First, F(x) = e^x sin x + e^x cos x = e^x (sin x + cos x).We need to find all tangent lines to F(x) that pass through the point M((π - 1)/2, 0). The points of tangency will have abscissas x_n, and we need to find the sum of all x_n in the given interval.Let me denote the point of tangency as (x_0, F(x_0)). The tangent line at this point has slope F’(x_0).Compute F’(x):F’(x) = derivative of e^x (sin x + cos x) = e^x (sin x + cos x) + e^x (cos x - sin x) = 2 e^x cos x.So, F’(x) = 2 e^x cos x.Therefore, the equation of the tangent line at (x_0, F(x_0)) is:y - F(x_0) = F’(x_0)(x - x_0).We know this tangent line passes through M((π - 1)/2, 0). So, substituting x = (π - 1)/2 and y = 0 into the equation:0 - F(x_0) = F’(x_0) left( frac{pi - 1}{2} - x_0 right).So,- F(x_0) = F’(x_0) left( frac{pi - 1}{2} - x_0 right).Substitute F(x_0) and F’(x_0):- e^{x_0} (sin x_0 + cos x_0) = 2 e^{x_0} cos x_0 left( frac{pi - 1}{2} - x_0 right).We can divide both sides by e^{x_0} (since e^{x_0} ≠ 0):- (sin x_0 + cos x_0) = 2 cos x_0 left( frac{pi - 1}{2} - x_0 right).Let me rearrange this:- sin x_0 - cos x_0 = 2 cos x_0 left( frac{pi - 1}{2} - x_0 right).Let me bring all terms to one side:- sin x_0 - cos x_0 - 2 cos x_0 left( frac{pi - 1}{2} - x_0 right) = 0.Factor out cos x_0:- sin x_0 - cos x_0 - 2 cos x_0 cdot frac{pi - 1}{2} + 2 cos x_0 x_0 = 0.Simplify:- sin x_0 - cos x_0 - (π - 1) cos x_0 + 2 x_0 cos x_0 = 0.Combine like terms:- sin x_0 - [1 + (π - 1)] cos x_0 + 2 x_0 cos x_0 = 0.Simplify inside the brackets:- sin x_0 - π cos x_0 + 2 x_0 cos x_0 = 0.Factor cos x_0:- sin x_0 + cos x_0 (-π + 2 x_0) = 0.Let me write this as:sin x_0 = cos x_0 (2 x_0 - π).Divide both sides by cos x_0 (assuming cos x_0 ≠ 0):tan x_0 = 2 x_0 - π.So, the equation we need to solve is:tan x_0 = 2 x_0 - π.Let me denote y = x_0. Then, the equation is:tan y = 2 y - π.We need to find all solutions y in the interval [-2015π/2, 2017π/2].Hmm, this seems challenging. Let me think about the behavior of the functions tan y and 2 y - π.First, tan y has vertical asymptotes at y = (2k + 1)π/2 for integers k. Between each pair of asymptotes, tan y goes from -∞ to +∞.The function 2 y - π is a straight line with slope 2 and y-intercept -π.So, in each interval between two consecutive asymptotes of tan y, the line 2 y - π will intersect tan y exactly once. Because tan y is increasing from -∞ to +∞, and the line is also increasing, so they cross once per interval.Therefore, the number of solutions y in the interval [-2015π/2, 2017π/2] is equal to the number of intervals between the asymptotes within that range.Let me compute how many intervals we have.First, the interval is from -2015π/2 to 2017π/2.Note that 2017π/2 = 1008π + π/2, and -2015π/2 = -1007π - π/2.So, the interval spans from just below -1007π to just above 1008π.Each interval between asymptotes is of length π, starting at (2k - 1)π/2 to (2k + 1)π/2.Wait, actually, the asymptotes are at y = (2k + 1)π/2 for integer k. So, between each pair of consecutive asymptotes, the interval is of length π.So, the number of intervals within [-2015π/2, 2017π/2] is approximately the total length divided by π.Total length is 2017π/2 - (-2015π/2) = (2017 + 2015)π/2 = 4032π/2 = 2016π.Number of intervals is 2016π / π = 2016. But since each interval is between two asymptotes, the number of solutions is equal to the number of intervals, which is 2016.Wait, but actually, each interval between asymptotes will have exactly one solution, so the number of solutions is equal to the number of intervals, which is 2016.But let me verify.Wait, the interval from -2015π/2 to 2017π/2 is symmetric around π/2? Wait, no.Wait, -2015π/2 is equal to -1007.5π, and 2017π/2 is 1008.5π.So, the total number of intervals is the number of π-length intervals between -1007.5π and 1008.5π.Each interval is centered at kπ, from (k - 0.5)π to (k + 0.5)π.So, from -1007.5π to 1008.5π, the number of intervals is 1008 - (-1007) = 2015? Wait, not quite.Wait, each interval is of length π, starting at (2k - 1)π/2 to (2k + 1)π/2.Wait, perhaps it's better to count the number of asymptotes within the interval.The asymptotes are at y = (2k + 1)π/2 for integer k.So, the left endpoint is -2015π/2 = -1007.5π, and the right endpoint is 2017π/2 = 1008.5π.So, the asymptotes within this interval are from k = -1007 to k = 1008.Wait, let me compute:The asymptotes are at y = (2k + 1)π/2.We need to find all integers k such that (2k + 1)π/2 is within [-2015π/2, 2017π/2].So,-2015π/2 ≤ (2k + 1)π/2 ≤ 2017π/2.Multiply all parts by 2/π:-2015 ≤ 2k + 1 ≤ 2017.Subtract 1:-2016 ≤ 2k ≤ 2016.Divide by 2:-1008 ≤ k ≤ 1008.So, k ranges from -1008 to 1008, inclusive. That's 2017 values.But each asymptote is at y = (2k + 1)π/2 for k from -1008 to 1008.But the number of intervals between asymptotes is one less than the number of asymptotes. So, from k = -1008 to k = 1008, there are 2017 asymptotes, hence 2016 intervals.Therefore, there are 2016 solutions y in the interval [-2015π/2, 2017π/2].Each solution y corresponds to an x_n, which is the abscissa of the point of tangency.Now, we need to find the sum S of all x_n.Given that the equation is tan y = 2 y - π, and the solutions are symmetric in some way?Wait, let me think about the equation tan y = 2 y - π.If we let y = π/2 - z, then tan(π/2 - z) = cot z = 2(π/2 - z) - π = π - 2 z - π = -2 z.So, cot z = -2 z.Which is equivalent to tan z = -1/(2 z).Hmm, not sure if that helps.Alternatively, perhaps the solutions are symmetric around π/2.Wait, let's consider that for each solution y, there might be another solution symmetric around π/2.Let me test this.Suppose y is a solution, so tan y = 2 y - π.Let me see if π - y is also a solution.Compute tan(π - y) = -tan y.Compute 2(π - y) - π = 2π - 2 y - π = π - 2 y.So, tan(π - y) = -tan y = -(2 y - π) = -2 y + π.Compare to 2(π - y) - π = π - 2 y.So, tan(π - y) = π - 2 y = 2(π - y) - π.Therefore, if y is a solution, then π - y is also a solution.Thus, the solutions come in pairs symmetric about π/2.Therefore, for each solution y, there is another solution π - y.Therefore, the sum of each pair is y + (π - y) = π.Since there are 2016 solutions, which is an even number, the number of such pairs is 2016 / 2 = 1008.Therefore, the sum S is 1008 * π.Hence, S = 1008π.**Final Answer**- (1) The intervals of increase are boxed{left[2kpi - frac{pi}{4}, 2kpi + frac{3pi}{4}right]} and the intervals of decrease are boxed{left[2kpi + frac{3pi}{4}, 2kpi + frac{7pi}{4}right]} for all integers k.- (2) The range of k is boxed{(-infty, 1]}.- (3) The sum S is boxed{1008pi}.

answer：First, I recognize that the product of the HCF and LCM of two numbers equals the product of the numbers themselves. Given that one number is 24 and the other is ( x ), I can set up the equation:[text{HCF} times text{LCM} = 24 times x]From the problem, the reciprocal of the HCF is ( frac{1}{17} ), so the HCF is 17. Similarly, the reciprocal of the LCM is ( frac{1}{312} ), so the LCM is 312. Substituting these values into the equation:[17 times 312 = 24 times x]To solve for ( x ), I divide both sides by 24:[x = frac{17 times 312}{24}]Simplifying the equation, I notice that 24 and 312 are both divisible by 8:[frac{24}{8} = 3 quad text{and} quad frac{312}{8} = 39]This reduces the equation to:[x = frac{17 times 39}{3}]Multiplying 17 and 39 gives:[x = frac{663}{3}]Finally, dividing 663 by 3:[x = 221]Therefore, the other number is 221.

answer：Okay, so I'm trying to solve this problem about two circles, Γ₁ and Γ₂, that have the same radius and are tangent at point P. There's a line ℓ passing through the center of Γ₁, which is O₁, and it's tangent to Γ₂ at point A. This line ℓ intersects Γ₁ again at point X, which is between A and O₁. Then, M is the midpoint of AX, and Y is the other intersection point of PM with Γ₂. I need to prove that XY is parallel to the line connecting the centers, O₁O₂.Alright, let me start by drawing a diagram to visualize this setup. I have two circles of equal radius touching at point P. The line ℓ goes through O₁ and is tangent to Γ₂ at A. So, ℓ is a tangent to Γ₂, which means that at point A, the radius O₂A is perpendicular to ℓ. That gives me a right angle at A between O₂A and ℓ.Since ℓ passes through O₁ and is tangent to Γ₂ at A, I can note that O₁A is the tangent line, and O₂A is perpendicular to it. So, triangle O₁AO₂ is a right triangle at A. Also, since both circles have the same radius, let's denote the radius as r. So, O₁P = O₂P = r, and since the circles are tangent at P, the distance between the centers O₁O₂ is 2r.Now, the line ℓ intersects Γ₁ again at X. Since ℓ passes through O₁ and intersects Γ₁, X must be diametrically opposite to the point where ℓ enters Γ₁. Wait, no, because ℓ is tangent to Γ₂, so it's not necessarily passing through the entire circle Γ₁. Hmm, maybe I need to think differently.Let me consider the power of point O₁ with respect to Γ₂. The power of O₁ is equal to the square of the length of the tangent from O₁ to Γ₂, which is O₁A². Since Γ₁ and Γ₂ are tangent at P, the distance between their centers is 2r, so O₁O₂ = 2r. Therefore, the power of O₁ with respect to Γ₂ is O₁O₂² - r² = (2r)² - r² = 4r² - r² = 3r². So, O₁A² = 3r², which means O₁A = r√3.Okay, so the length of the tangent from O₁ to Γ₂ is r√3. That might be useful later.Now, looking at line ℓ, which is tangent to Γ₂ at A and passes through O₁. It intersects Γ₁ at X. Since ℓ passes through O₁, which is the center of Γ₁, and intersects Γ₁ at X, the segment O₁X is a radius of Γ₁, so O₁X = r. But wait, O₁A is r√3, and O₁X is r, so point X is closer to O₁ than point A is.Let me try to find the coordinates of these points to make things clearer. Maybe coordinate geometry can help here. Let's place the centers O₁ and O₂ on the x-axis for simplicity. Let me set O₁ at (-r, 0) and O₂ at (r, 0), so that the distance between them is 2r, as required.Since the circles are tangent at P, which lies on the line connecting the centers, so P is at the midpoint between O₁ and O₂, which is (0, 0). Wait, no, because both circles have radius r, so if O₁ is at (-r, 0) and O₂ is at (r, 0), then the point of tangency P would be at (0, 0), right? Because each circle extends r units from their centers, so they meet at the origin.Now, line ℓ passes through O₁ (-r, 0) and is tangent to Γ₂ at A. Let me find the coordinates of point A. Since ℓ is tangent to Γ₂ at A, and Γ₂ is centered at (r, 0) with radius r, the tangent from O₁ (-r, 0) to Γ₂ will touch Γ₂ at some point A.The equation of Γ₂ is (x - r)² + y² = r². The tangent from O₁ (-r, 0) to Γ₂ can be found using the formula for the tangent from an external point to a circle. The condition is that the distance from O₁ to the tangent line is equal to the radius.Alternatively, since we know that the tangent from O₁ to Γ₂ will form a right angle with the radius O₂A. So, triangle O₁O₂A is a right triangle at A.Given that O₁O₂ = 2r, and O₂A = r, we can find O₁A using the Pythagorean theorem:O₁A² + O₂A² = O₁O₂²O₁A² + r² = (2r)²O₁A² = 4r² - r² = 3r²So, O₁A = r√3, which matches what I found earlier.Now, to find the coordinates of A, let's consider the line ℓ. Since ℓ passes through O₁ (-r, 0) and is tangent to Γ₂ at A, which is at a distance r√3 from O₁. Let me parametrize line ℓ.Let me denote the slope of line ℓ as m. The equation of ℓ is y = m(x + r). Since it's tangent to Γ₂ at A, the distance from O₂ (r, 0) to line ℓ must be equal to the radius r.The distance from a point (x₀, y₀) to the line ax + by + c = 0 is |ax₀ + by₀ + c| / sqrt(a² + b²). So, rewriting ℓ's equation in standard form: mx - y + mr = 0.The distance from O₂ (r, 0) to ℓ is |m*r - 0 + mr| / sqrt(m² + 1) = |2mr| / sqrt(m² + 1). This distance must equal r, the radius.So, |2mr| / sqrt(m² + 1) = rDivide both sides by r (assuming r ≠ 0):|2m| / sqrt(m² + 1) = 1Square both sides:(4m²) / (m² + 1) = 1Multiply both sides by (m² + 1):4m² = m² + 1Subtract m²:3m² = 1So, m² = 1/3, which means m = ±1/√3.Therefore, the slope of line ℓ is either 1/√3 or -1/√3. Let's choose m = 1/√3 for simplicity; the other case would be symmetric.So, the equation of ℓ is y = (1/√3)(x + r).Now, let's find the coordinates of point A, which is the point of tangency on Γ₂. Since ℓ is tangent to Γ₂ at A, and we know the slope of ℓ is 1/√3, the radius O₂A is perpendicular to ℓ, so its slope is -√3.The line O₂A passes through O₂ (r, 0) and has slope -√3, so its equation is y = -√3(x - r).Now, find the intersection point A between ℓ and O₂A.Set the two equations equal:(1/√3)(x + r) = -√3(x - r)Multiply both sides by √3 to eliminate the denominator:(x + r) = -3(x - r)Expand:x + r = -3x + 3rBring all terms to one side:x + r + 3x - 3r = 04x - 2r = 04x = 2rx = (2r)/4 = r/2Now, substitute x = r/2 into the equation of ℓ:y = (1/√3)(r/2 + r) = (1/√3)(3r/2) = (3r)/(2√3) = (r√3)/2So, point A is at (r/2, (r√3)/2).Now, let's find point X, which is the other intersection of line ℓ with Γ₁. Γ₁ is centered at O₁ (-r, 0) with radius r, so its equation is (x + r)² + y² = r².We already know that ℓ passes through O₁ (-r, 0) and point A (r/2, (r√3)/2). Let's find the other intersection point X.Parametrize line ℓ. Let me use a parameter t such that when t = 0, we are at O₁ (-r, 0), and when t = 1, we are at A (r/2, (r√3)/2). So, the parametric equations are:x = -r + t*(r/2 + r) = -r + t*(3r/2)y = 0 + t*((r√3)/2 - 0) = t*(r√3)/2So, x = -r + (3r/2)ty = (r√3/2)tNow, substitute these into the equation of Γ₁:(x + r)² + y² = r²Substitute x and y:(-r + (3r/2)t + r)² + ((r√3/2)t)² = r²Simplify:((3r/2)t)² + ( (r√3/2)t )² = r²Compute each term:(9r²/4)t² + (3r²/4)t² = r²Combine like terms:(9r²/4 + 3r²/4)t² = r²(12r²/4)t² = r²3r² t² = r²Divide both sides by r² (assuming r ≠ 0):3t² = 1t² = 1/3t = ±1/√3We already have t = 1 corresponding to point A, so the other intersection point X corresponds to t = -1/√3.So, substituting t = -1/√3 into the parametric equations:x = -r + (3r/2)*(-1/√3) = -r - (3r)/(2√3) = -r - (r√3)/2y = (r√3/2)*(-1/√3) = -r/2So, point X is at (-r - (r√3)/2, -r/2).Now, let's find the midpoint M of AX. Point A is at (r/2, (r√3)/2), and point X is at (-r - (r√3)/2, -r/2).The midpoint M has coordinates:M_x = (r/2 + (-r - (r√3)/2)) / 2 = (r/2 - r - (r√3)/2) / 2 = (-r/2 - (r√3)/2) / 2 = (-r - r√3)/4M_y = ((r√3)/2 + (-r/2)) / 2 = (r√3/2 - r/2) / 2 = (r(√3 - 1)/2) / 2 = r(√3 - 1)/4So, M is at ((-r - r√3)/4, r(√3 - 1)/4).Now, we need to find point Y, which is the other intersection of line PM with Γ₂. First, let's find the equation of line PM.Point P is the point of tangency between Γ₁ and Γ₂, which we placed at (0, 0). So, P is at (0, 0), and M is at ((-r - r√3)/4, r(√3 - 1)/4).So, the line PM goes from (0, 0) to ((-r - r√3)/4, r(√3 - 1)/4). Let's find its parametric equations.Let parameter s vary from 0 to 1 to go from P to M. So,x = 0 + s*((-r - r√3)/4 - 0) = s*(-r(1 + √3)/4)y = 0 + s*(r(√3 - 1)/4 - 0) = s*r(√3 - 1)/4So, the parametric equations are:x = -s*r(1 + √3)/4y = s*r(√3 - 1)/4We need to find the other intersection point Y of this line with Γ₂. Γ₂ is centered at (r, 0) with radius r, so its equation is (x - r)² + y² = r².Substitute x and y from the parametric equations into Γ₂'s equation:(-s*r(1 + √3)/4 - r)² + (s*r(√3 - 1)/4)² = r²Simplify the first term:(-s*r(1 + √3)/4 - r) = -r*(s(1 + √3)/4 + 1)Let me factor out -r:= -r*(s(1 + √3)/4 + 1)So, the first term squared is [ -r*(s(1 + √3)/4 + 1) ]² = r²*(s(1 + √3)/4 + 1)²The second term is [s*r(√3 - 1)/4]² = r²*s²*(√3 - 1)²/16So, putting it all together:r²*(s(1 + √3)/4 + 1)² + r²*s²*(√3 - 1)²/16 = r²Divide both sides by r²:(s(1 + √3)/4 + 1)² + s²*(√3 - 1)²/16 = 1Let me expand the first term:Let me denote a = s(1 + √3)/4 + 1a = 1 + s(1 + √3)/4So, a² = [1 + s(1 + √3)/4]² = 1 + 2*s(1 + √3)/4 + s²(1 + √3)²/16Simplify:= 1 + (s(1 + √3))/2 + s²(1 + 2√3 + 3)/16= 1 + (s(1 + √3))/2 + s²(4 + 2√3)/16= 1 + (s(1 + √3))/2 + s²(2 + √3)/8Now, the second term is s²*(√3 - 1)²/16Compute (√3 - 1)² = 3 - 2√3 + 1 = 4 - 2√3So, the second term is s²*(4 - 2√3)/16 = s²*(2 - √3)/8Now, putting it all together:1 + (s(1 + √3))/2 + s²(2 + √3)/8 + s²(2 - √3)/8 = 1Combine the s² terms:s²[(2 + √3) + (2 - √3)]/8 = s²*(4)/8 = s²/2So, the equation becomes:1 + (s(1 + √3))/2 + s²/2 = 1Subtract 1 from both sides:(s(1 + √3))/2 + s²/2 = 0Multiply both sides by 2:s(1 + √3) + s² = 0Factor out s:s(s + 1 + √3) = 0So, s = 0 or s = -1 - √3s = 0 corresponds to point P (0, 0), so the other intersection point Y corresponds to s = -1 - √3.Now, let's find the coordinates of Y using s = -1 - √3.x = -s*r(1 + √3)/4 = -(-1 - √3)*r(1 + √3)/4 = (1 + √3)*r(1 + √3)/4Multiply out:(1 + √3)(1 + √3) = 1 + 2√3 + 3 = 4 + 2√3So, x = r*(4 + 2√3)/4 = r*(2 + √3)/2Similarly, y = s*r(√3 - 1)/4 = (-1 - √3)*r(√3 - 1)/4Multiply out:(-1 - √3)(√3 - 1) = (-1)(√3 - 1) - √3(√3 - 1) = -√3 + 1 - 3 + √3 = (-√3 + √3) + (1 - 3) = 0 - 2 = -2So, y = (-2)r/4 = -r/2Therefore, point Y is at (r*(2 + √3)/2, -r/2)Now, we have points X and Y. Let's write down their coordinates:Point X: (-r - (r√3)/2, -r/2)Point Y: (r*(2 + √3)/2, -r/2)I need to find the slope of line XY and compare it to the slope of O₁O₂.First, let's compute the coordinates more neatly.Point X:x = -r - (r√3)/2 = -r(1 + √3/2)y = -r/2Point Y:x = r*(2 + √3)/2 = r(1 + √3/2)y = -r/2So, both points X and Y have the same y-coordinate, -r/2. That means the line XY is horizontal, right? Because the y-coordinates are the same.Wait, but O₁O₂ is along the x-axis from (-r, 0) to (r, 0), so it's also horizontal. Therefore, if XY is horizontal and O₁O₂ is horizontal, they are parallel.But wait, let me double-check. The slope of XY is (y_Y - y_X)/(x_Y - x_X). Since y_Y = y_X = -r/2, the numerator is 0, so the slope is 0. Similarly, O₁O₂ is along the x-axis, so its slope is also 0. Therefore, XY is parallel to O₁O₂.Wait, that seems too straightforward. Did I make a mistake somewhere?Let me recap:- I placed the centers O₁ at (-r, 0) and O₂ at (r, 0), with P at (0, 0).- Found point A at (r/2, (r√3)/2) on Γ₂.- Found point X at (-r - (r√3)/2, -r/2) on Γ₁.- Midpoint M of AX is at ((-r - r√3)/4, r(√3 - 1)/4).- Found point Y by extending PM to intersect Γ₂ again, resulting in Y at (r*(2 + √3)/2, -r/2).- Observed that both X and Y have y-coordinate -r/2, so XY is horizontal, hence parallel to O₁O₂, which is also horizontal.Yes, that seems correct. Therefore, XY is parallel to O₁O₂.I think this approach works. By assigning coordinates and computing the positions, I was able to show that both XY and O₁O₂ are horizontal lines, hence parallel. This satisfies the problem's requirement.

answer：Okay, so I have this problem here: Given that 4^a equals 2^(a+2), I need to find the solution set for the inequality a^(2x+1) > a^(x-1). Hmm, let me try to figure this out step by step.First, I need to solve for 'a' from the equation 4^a = 2^(a+2). I remember that 4 is the same as 2 squared, so maybe I can rewrite 4^a in terms of base 2. Let me try that.So, 4^a is equal to (2^2)^a, which simplifies to 2^(2a). So now, the equation becomes 2^(2a) = 2^(a+2). Since the bases are the same, I can set the exponents equal to each other. That gives me 2a = a + 2. Hmm, okay, solving for 'a', I subtract 'a' from both sides, which gives me a = 2. So, a is 2. Got that part.Now, moving on to the inequality: a^(2x+1) > a^(x-1). Since I found that a is 2, I can substitute that into the inequality. So, it becomes 2^(2x+1) > 2^(x-1). Again, the bases are the same, so I can set the exponents in an inequality. That means 2x + 1 > x - 1.Let me solve this inequality. Subtract 'x' from both sides: 2x + 1 - x > -1, which simplifies to x + 1 > -1. Then, subtract 1 from both sides: x > -2. So, the solution is x greater than -2.Wait, but hold on a second. I should make sure that the base 'a' is positive because when dealing with exponential functions, if the base is between 0 and 1, the inequality sign flips when you take logarithms or compare exponents. But in this case, a is 2, which is greater than 1, so the inequality sign stays the same. So, my solution x > -2 should be correct.Let me double-check by plugging in a value greater than -2, say x = 0. Then, the left side is 2^(2*0 +1) = 2^1 = 2, and the right side is 2^(0 -1) = 2^(-1) = 0.5. So, 2 > 0.5 is true. What about a value less than -2, say x = -3. Then, left side is 2^(2*(-3)+1) = 2^(-5) = 1/32, and the right side is 2^(-3 -1) = 2^(-4) = 1/16. So, 1/32 > 1/16 is false. That confirms that x > -2 is correct.I think that's all. So, the solution set is all real numbers x such that x is greater than -2.