💡Alright, so I've got this problem about a 1000-meter race involving three runners: A, B, and C. The problem says that when runner A finishes the race, runner B is 50 meters away from the finish line. Then, when runner B finishes the race, runner C is 100 meters away from the finish line. The question is asking how many meters away runner C is from the finish line when runner A finishes.Okay, let me break this down step by step. First, I need to understand the relationships between the runners' speeds and distances. Since all runners are running at constant speeds, the distances they cover in the same amount of time should be proportional to their speeds.So, when runner A finishes the 1000 meters, runner B is 50 meters behind. That means runner B has run 950 meters in the same time it took runner A to run 1000 meters. From this, I can figure out the ratio of their speeds. If I let the speed of runner A be ( v_A ) and the speed of runner B be ( v_B ), then the time it takes runner A to finish is ( t = frac{1000}{v_A} ). In this same time, runner B runs 950 meters, so ( 950 = v_B times t ). Substituting ( t ) from the first equation into the second gives ( 950 = v_B times frac{1000}{v_A} ), which simplifies to ( frac{v_B}{v_A} = frac{950}{1000} = 0.95 ). So, runner B's speed is 95% of runner A's speed.Next, when runner B finishes the race, runner C is 100 meters away from the finish line. That means runner C has run 900 meters in the same time it took runner B to run 1000 meters. Let's denote runner C's speed as ( v_C ). The time it takes runner B to finish is ( t' = frac{1000}{v_B} ). In this same time, runner C runs 900 meters, so ( 900 = v_C times t' ). Substituting ( t' ) gives ( 900 = v_C times frac{1000}{v_B} ), which simplifies to ( frac{v_C}{v_B} = frac{900}{1000} = 0.9 ). Therefore, runner C's speed is 90% of runner B's speed.Now, I need to find out how far runner C is from the finish line when runner A finishes. Since runner A finishes in time ( t = frac{1000}{v_A} ), I need to find out how much distance runner C covers in this time. Runner C's speed is 0.9 times runner B's speed, and runner B's speed is 0.95 times runner A's speed. So, runner C's speed is ( v_C = 0.9 times v_B = 0.9 times 0.95 times v_A = 0.855 times v_A ).Therefore, the distance runner C covers in time ( t ) is ( d_C = v_C times t = 0.855 times v_A times frac{1000}{v_A} = 0.855 times 1000 = 855 ) meters. Since the race is 1000 meters, runner C is ( 1000 - 855 = 145 ) meters away from the finish line when runner A finishes.Wait, let me double-check that. If runner C is running at 85.5% the speed of runner A, then in the time runner A runs 1000 meters, runner C runs 855 meters, leaving 145 meters. That seems consistent.Alternatively, I can think about the ratios directly. Since runner B runs 950 meters when runner A runs 1000 meters, and runner C runs 900 meters when runner B runs 1000 meters, I can set up a proportion to find out how far runner C runs when runner A runs 1000 meters.Let me denote the distance runner C runs when runner A runs 1000 meters as ( d ). The ratio of runner B's distance to runner A's distance is ( frac{950}{1000} ), and the ratio of runner C's distance to runner B's distance is ( frac{900}{1000} ). So, combining these ratios, the ratio of runner C's distance to runner A's distance is ( frac{950}{1000} times frac{900}{1000} = frac{950 times 900}{1000 times 1000} = frac{855000}{1000000} = 0.855 ). Therefore, ( d = 0.855 times 1000 = 855 ) meters, which means runner C is 145 meters away from the finish line.Another way to look at it is by considering the time it takes for each runner to finish. Let's say the time it takes runner A to finish is ( t ). Then, runner B's time to finish 1000 meters is ( frac{1000}{v_B} ). But when runner A finishes, runner B has only run 950 meters, so the time ( t ) is also equal to ( frac{950}{v_B} ). Therefore, ( frac{1000}{v_B} = frac{950}{v_B} + Delta t ), where ( Delta t ) is the additional time runner B takes to finish the remaining 50 meters. But this might complicate things, so I think the ratio approach is more straightforward.I could also use relative speeds, but since all runners start at the same time and run at constant speeds, the ratios of their distances should suffice. I think I've covered the necessary steps and verified the calculations, so I'm confident that the answer is 145 meters.

💡Alright, so I've got this problem here about a function f(x). Let me try to understand what it's asking. The domain of f(x) is all real numbers, which means it's defined everywhere. They also tell me that f(-1) equals 2. Okay, so when x is -1, the function value is 2. Then, it says that for any real number x, the derivative f'(x) is greater than 2. Hmm, so the slope of the function is always steeper than 2. The question is asking for the solution set where f(x) is greater than 2x + 4. So, I need to find all the x-values where f(x) is above the line 2x + 4. Let me think about how to approach this. Since we have information about the derivative, maybe I can use calculus here. I remember that if the derivative of a function is always positive, the function is increasing. In this case, f'(x) is greater than 2, which is positive, so f(x) is strictly increasing. That means as x increases, f(x) increases as well.But how does that help me compare f(x) to 2x + 4? Maybe I can define a new function that represents the difference between f(x) and 2x + 4. Let's try that. Let me define F(x) = f(x) - (2x + 4). So, F(x) is the difference between f(x) and the linear function 2x + 4. Now, let's see what F(-1) is. Since f(-1) is 2, plugging in x = -1 into F(x) gives F(-1) = f(-1) - (2*(-1) + 4) = 2 - (-2 + 4) = 2 - 2 = 0. So, F(-1) is 0. That means at x = -1, f(x) equals 2x + 4.Next, let's find the derivative of F(x). Since F(x) = f(x) - (2x + 4), the derivative F'(x) would be f'(x) - 2. We know that f'(x) is greater than 2, so F'(x) = f'(x) - 2 > 0. This tells me that F(x) is also strictly increasing because its derivative is always positive.Since F(x) is strictly increasing and F(-1) = 0, what does that tell me about F(x) for x values greater than -1? Well, if the function is increasing and it's 0 at x = -1, then for any x greater than -1, F(x) must be greater than 0. That means f(x) - (2x + 4) > 0, or f(x) > 2x + 4.On the other hand, for x values less than -1, since F(x) is increasing, F(x) would be less than 0. So, f(x) < 2x + 4 for x < -1.Therefore, the solution set where f(x) > 2x + 4 is all real numbers greater than -1. In interval notation, that would be (-1, ∞).Wait, let me double-check. If F(x) is increasing and F(-1) = 0, then to the right of x = -1, F(x) should be positive, meaning f(x) > 2x + 4. To the left of x = -1, F(x) is negative, so f(x) < 2x + 4. That makes sense because the function f(x) is growing faster than the linear function 2x + 4 due to its derivative being greater than 2.Is there any other point I need to consider? Maybe the behavior as x approaches infinity or negative infinity. But since f'(x) is always greater than 2, f(x) will outpace any linear function as x increases. As x decreases, since f(x) is increasing, it might not necessarily outpace 2x + 4, but in this case, we've already determined that F(x) is negative for x < -1, so f(x) is less than 2x + 4 there.I think that covers it. So, the solution set is all real numbers greater than -1.

💡Okay, so I have this problem about a batsman who scored 1 run in his 18th innings, and that makes his average 18. I need to find his average up to the 17th innings, but there's also this condition that he must maintain a minimum required run rate of 5 runs per over in each innings. Hmm, okay, let's break this down step by step.First, I need to understand what the average means in cricket. The average is the total number of runs scored divided by the number of innings played. So, if the batsman's average after 18 innings is 18, that means his total runs after 18 innings is 18 times 18, which is 324 runs. Okay, that makes sense.Now, he scored 1 run in his 18th innings. So, to find out how many runs he had after 17 innings, I just subtract that 1 run from the total. So, 324 minus 1 is 323 runs after 17 innings. Therefore, his average after 17 innings would be 323 divided by 17, which is 19. So, his average up to the 17th innings is 19.But wait, there's this other condition about maintaining a minimum required run rate of 5 runs per over in each innings. I'm not entirely sure how that affects the problem. Run rate is usually runs per over, so 5 runs per over means he needs to score at least 5 runs every over he faces in each innings. But the problem doesn't specify how many overs he faced in each innings, so I'm not sure how to apply this condition.Maybe it's just additional information that isn't necessary for calculating the average? Or perhaps it's a constraint that affects the total runs he could have scored? I'm not sure. Since the problem doesn't provide details about the number of overs or how the run rate impacts the total runs, I think I can proceed with the average calculation as I did before.So, summarizing my thoughts: The batsman's average after 18 innings is 18, meaning he scored 324 runs in total. After scoring 1 run in the 18th innings, his total after 17 innings is 323 runs. Dividing that by 17 gives an average of 19. The run rate condition might be extra information that doesn't affect the calculation since we don't have specifics on overs faced.I think that's the solution. His average up to the 17th innings is 19.

💡Okay, so I have this polar equation: (4rho cdot sin^2 frac{theta}{2} = 5). I need to figure out what kind of curve this represents. The options are Circle, Ellipse, One branch of a hyperbola, or Parabola. Hmm, let me think about how to approach this.First, I remember that in polar coordinates, (rho) is the radius and (theta) is the angle. The equation involves (sin^2 frac{theta}{2}), which makes me think of some trigonometric identities. Maybe I can simplify this equation using an identity.I recall that (sin^2 x = frac{1 - cos 2x}{2}). So, if I let (x = frac{theta}{2}), then (sin^2 frac{theta}{2} = frac{1 - cos theta}{2}). Let me substitute that into the equation.So, replacing (sin^2 frac{theta}{2}) with (frac{1 - cos theta}{2}), the equation becomes:[4rho cdot frac{1 - cos theta}{2} = 5]Simplifying that, the 4 and the 2 cancel out to 2:[2rho (1 - cos theta) = 5]Okay, so now I have (2rho (1 - cos theta) = 5). I need to convert this into Cartesian coordinates to see what kind of curve it is. I remember that in polar coordinates, (rho = sqrt{x^2 + y^2}) and (cos theta = frac{x}{rho}).Let me substitute these into the equation. So, replacing (rho) with (sqrt{x^2 + y^2}) and (cos theta) with (frac{x}{sqrt{x^2 + y^2}}), the equation becomes:[2sqrt{x^2 + y^2} left(1 - frac{x}{sqrt{x^2 + y^2}}right) = 5]Let me simplify the expression inside the parentheses first:[1 - frac{x}{sqrt{x^2 + y^2}} = frac{sqrt{x^2 + y^2} - x}{sqrt{x^2 + y^2}}]So, substituting back into the equation:[2sqrt{x^2 + y^2} cdot frac{sqrt{x^2 + y^2} - x}{sqrt{x^2 + y^2}} = 5]The (sqrt{x^2 + y^2}) in the numerator and denominator cancel out, leaving:[2(sqrt{x^2 + y^2} - x) = 5]Let me distribute the 2:[2sqrt{x^2 + y^2} - 2x = 5]Hmm, so I have (2sqrt{x^2 + y^2} - 2x = 5). I need to get rid of the square root to see the equation more clearly. Let me isolate the square root term first.Adding (2x) to both sides:[2sqrt{x^2 + y^2} = 5 + 2x]Now, I can divide both sides by 2 to simplify:[sqrt{x^2 + y^2} = frac{5}{2} + x]But maybe it's better to square both sides to eliminate the square root. Let's do that.Squaring both sides:[(sqrt{x^2 + y^2})^2 = left(frac{5}{2} + xright)^2]Simplifying both sides:[x^2 + y^2 = left(frac{5}{2}right)^2 + 2 cdot frac{5}{2} cdot x + x^2][x^2 + y^2 = frac{25}{4} + 5x + x^2]Now, subtract (x^2) from both sides to simplify:[y^2 = frac{25}{4} + 5x]Hmm, so this simplifies to:[y^2 = 5x + frac{25}{4}]I can write this as:[y^2 = 5left(x + frac{5}{4}right)]This looks familiar. It resembles the standard form of a parabola. The standard form of a parabola that opens to the right is (y^2 = 4p(x - h)), where ((h, k)) is the vertex and (p) is the focal length.Comparing this to my equation, (y^2 = 5left(x + frac{5}{4}right)), it's similar but with 5 instead of 4p. So, in this case, (4p = 5), which means (p = frac{5}{4}). The vertex is at (left(-frac{5}{4}, 0right)).Therefore, this equation represents a parabola. So, the correct answer should be D: Parabola.Let me just double-check my steps to make sure I didn't make any mistakes. Starting from the polar equation, I used the identity for (sin^2 frac{theta}{2}), converted to Cartesian coordinates, and simplified step by step. Each step seems logical, and the final equation is indeed a parabola. So, I think I'm confident with this answer.**Final Answer**boxed{D}

💡Okay, so I'm trying to figure out if the function ( f(x) = frac{1}{x} ) has a "drifting point." From the problem statement, a "drifting point" ( x_0 ) is a real number in the domain of ( f ) such that ( f(x_0 + 1) = f(x_0) + f(1) ). First, let me write down what ( f(1) ) is. Since ( f(x) = frac{1}{x} ), then ( f(1) = frac{1}{1} = 1 ). So, the equation we need to satisfy is ( f(x_0 + 1) = f(x_0) + 1 ).Substituting ( f(x) ) into this equation, we get:[frac{1}{x_0 + 1} = frac{1}{x_0} + 1]Hmm, let me solve this equation for ( x_0 ). First, subtract ( frac{1}{x_0} ) from both sides:[frac{1}{x_0 + 1} - frac{1}{x_0} = 1]To combine the fractions on the left, I need a common denominator, which is ( x_0(x_0 + 1) ). So:[frac{x_0 - (x_0 + 1)}{x_0(x_0 + 1)} = 1]Simplify the numerator:[frac{x_0 - x_0 - 1}{x_0(x_0 + 1)} = 1 implies frac{-1}{x_0(x_0 + 1)} = 1]So, we have:[-1 = x_0(x_0 + 1)]Which simplifies to:[x_0^2 + x_0 + 1 = 0]Now, I need to solve this quadratic equation. Let me compute the discriminant to see if there are real solutions. The discriminant ( D ) is given by:[D = b^2 - 4ac = (1)^2 - 4(1)(1) = 1 - 4 = -3]Since the discriminant is negative, there are no real solutions. This means there is no real number ( x_0 ) that satisfies the equation ( f(x_0 + 1) = f(x_0) + f(1) ). Therefore, the function ( f(x) = frac{1}{x} ) does not have a "drifting point."**Final Answer**The function ( f(x) = frac{1}{x} ) does not have a "drifting point." The final answer is boxed{text{No}}.---Now, moving on to the second problem: proving that the function ( f(x) = x^2 + 2^x ) has a "drifting point" in the interval ( (0, 1) ).Again, a "drifting point" ( x_0 ) satisfies ( f(x_0 + 1) = f(x_0) + f(1) ). Let's compute ( f(1) ):[f(1) = (1)^2 + 2^1 = 1 + 2 = 3]So, the equation becomes:[f(x_0 + 1) = f(x_0) + 3]Substituting ( f(x) ) into this equation:[(x_0 + 1)^2 + 2^{x_0 + 1} = x_0^2 + 2^{x_0} + 3]Let me expand and simplify both sides. Starting with the left side:[(x_0 + 1)^2 = x_0^2 + 2x_0 + 1]And ( 2^{x_0 + 1} = 2 cdot 2^{x_0} ). So, the left side becomes:[x_0^2 + 2x_0 + 1 + 2 cdot 2^{x_0}]The right side is:[x_0^2 + 2^{x_0} + 3]Subtracting the right side from both sides to set the equation to zero:[(x_0^2 + 2x_0 + 1 + 2 cdot 2^{x_0}) - (x_0^2 + 2^{x_0} + 3) = 0]Simplify:[2x_0 + 1 + 2^{x_0} - 3 = 0 implies 2x_0 + 2^{x_0} - 2 = 0]So, we have:[2x_0 + 2^{x_0} - 2 = 0]Let me define a function ( h(x) = 2x + 2^x - 2 ). We need to show that ( h(x) = 0 ) has a solution in ( (0, 1) ).I'll evaluate ( h(x) ) at the endpoints of the interval.First, at ( x = 0 ):[h(0) = 2(0) + 2^0 - 2 = 0 + 1 - 2 = -1]At ( x = 1 ):[h(1) = 2(1) + 2^1 - 2 = 2 + 2 - 2 = 2]So, ( h(0) = -1 ) and ( h(1) = 2 ). Since ( h(x) ) is continuous (as it's composed of continuous functions) and changes sign from negative to positive over the interval ( (0, 1) ), by the Intermediate Value Theorem, there exists some ( x_0 in (0, 1) ) such that ( h(x_0) = 0 ).Therefore, the function ( f(x) = x^2 + 2^x ) has a "drifting point" in the interval ( (0, 1) ).**Final Answer**The function ( f(x) = x^2 + 2^x ) has a "drifting point" in the interval ( (0, 1) ). The final answer is boxed{text{Yes}}.---Finally, the third problem: determining the range of values for the real number ( a ) such that the function ( f(x) = lgleft(frac{a}{x^2 + 1}right) ) has a "drifting point" in the interval ( (0, +infty) ).First, let's recall that a "drifting point" ( x_0 ) satisfies ( f(x_0 + 1) = f(x_0) + f(1) ). Let's compute ( f(1) ):[f(1) = lgleft(frac{a}{1^2 + 1}right) = lgleft(frac{a}{2}right)]So, the equation becomes:[lgleft(frac{a}{(x_0 + 1)^2 + 1}right) = lgleft(frac{a}{x_0^2 + 1}right) + lgleft(frac{a}{2}right)]Using the logarithm property ( lg A + lg B = lg(AB) ), the right side becomes:[lgleft(frac{a}{x_0^2 + 1} cdot frac{a}{2}right) = lgleft(frac{a^2}{2(x_0^2 + 1)}right)]So, we have:[lgleft(frac{a}{(x_0 + 1)^2 + 1}right) = lgleft(frac{a^2}{2(x_0^2 + 1)}right)]Since the logarithm function is injective (one-to-one), the arguments must be equal:[frac{a}{(x_0 + 1)^2 + 1} = frac{a^2}{2(x_0^2 + 1)}]Assuming ( a neq 0 ) (since ( lg ) of zero is undefined), we can divide both sides by ( a ):[frac{1}{(x_0 + 1)^2 + 1} = frac{a}{2(x_0^2 + 1)}]Cross-multiplying:[2(x_0^2 + 1) = a[(x_0 + 1)^2 + 1]]Let me expand ( (x_0 + 1)^2 ):[(x_0 + 1)^2 = x_0^2 + 2x_0 + 1]So, substituting back:[2(x_0^2 + 1) = a(x_0^2 + 2x_0 + 1 + 1) = a(x_0^2 + 2x_0 + 2)]Expanding the left side:[2x_0^2 + 2 = a x_0^2 + 2a x_0 + 2a]Bring all terms to one side:[2x_0^2 + 2 - a x_0^2 - 2a x_0 - 2a = 0]Factor like terms:[(2 - a)x_0^2 - 2a x_0 + (2 - 2a) = 0]So, we have a quadratic equation in ( x_0 ):[(2 - a)x_0^2 - 2a x_0 + (2 - 2a) = 0]Let me denote this quadratic as ( g(x) = (2 - a)x^2 - 2a x + (2 - 2a) ). For there to be a real solution ( x_0 ) in ( (0, +infty) ), the quadratic must have at least one positive real root.First, let's consider the discriminant ( D ) of the quadratic:[D = (-2a)^2 - 4(2 - a)(2 - 2a) = 4a^2 - 4(2 - a)(2 - 2a)]Let me compute ( (2 - a)(2 - 2a) ):[(2 - a)(2 - 2a) = 4 - 4a - 2a + 2a^2 = 4 - 6a + 2a^2]So, the discriminant becomes:[D = 4a^2 - 4(4 - 6a + 2a^2) = 4a^2 - 16 + 24a - 8a^2 = -4a^2 + 24a - 16]For the quadratic to have real roots, ( D geq 0 ):[-4a^2 + 24a - 16 geq 0]Multiply both sides by (-1) (which reverses the inequality):[4a^2 - 24a + 16 leq 0]Divide both sides by 4:[a^2 - 6a + 4 leq 0]Solve the quadratic inequality ( a^2 - 6a + 4 leq 0 ). First, find the roots:[a = frac{6 pm sqrt{36 - 16}}{2} = frac{6 pm sqrt{20}}{2} = frac{6 pm 2sqrt{5}}{2} = 3 pm sqrt{5}]So, the quadratic ( a^2 - 6a + 4 ) is less than or equal to zero between its roots:[3 - sqrt{5} leq a leq 3 + sqrt{5}]But we also need to ensure that the quadratic ( g(x) ) has at least one positive real root. Let's analyze the quadratic ( g(x) = (2 - a)x^2 - 2a x + (2 - 2a) ).Case 1: ( 2 - a > 0 ) (i.e., ( a < 2 ))In this case, the parabola opens upwards. For it to have at least one positive root, either:- The quadratic has a root at ( x = 0 ), but ( g(0) = 2 - 2a ). For ( a neq 1 ), ( g(0) neq 0 ). So, not helpful.- The quadratic has a positive root, which can be ensured if ( g(0) ) and ( g(infty) ) have opposite signs or if the vertex is below the x-axis.But since the parabola opens upwards, if the minimum is below the x-axis, it will have two real roots. But we need at least one positive root.Alternatively, since we already have the discriminant condition, and ( a ) is between ( 3 - sqrt{5} ) and ( 3 + sqrt{5} ), but ( 3 - sqrt{5} approx 0.764 ) and ( 3 + sqrt{5} approx 5.236 ). So, ( a ) is between approximately 0.764 and 5.236.But since we have ( a < 2 ) in this case, the upper bound is 2. So, ( 3 - sqrt{5} leq a < 2 ).Case 2: ( 2 - a < 0 ) (i.e., ( a > 2 ))In this case, the parabola opens downwards. For it to have at least one positive root, the maximum must be above the x-axis, which is already considered in the discriminant condition. However, since ( a > 2 ), but our discriminant condition only allows up to ( a leq 3 + sqrt{5} approx 5.236 ). But we need to check if the roots are positive. Let's denote the roots as ( x_1 ) and ( x_2 ). The product of the roots is ( frac{2 - 2a}{2 - a} ). If ( a > 2 ), then ( 2 - a < 0 ), so the product is ( frac{2 - 2a}{2 - a} = frac{-2(a - 1)}{-(a - 2)} = frac{2(a - 1)}{a - 2} ). Since ( a > 2 ), both numerator and denominator are positive, so the product is positive. The sum of the roots is ( frac{2a}{2 - a} ). Since ( a > 2 ), ( 2 - a < 0 ), so the sum is negative. Therefore, both roots are negative because their sum is negative and their product is positive. Hence, in this case, there are no positive roots. Therefore, ( a > 2 ) does not yield a positive root.Case 3: ( 2 - a = 0 ) (i.e., ( a = 2 ))Then the equation becomes linear:[0 cdot x^2 - 4x + (2 - 4) = -4x - 2 = 0 implies x = -frac{1}{2}]Which is negative, so not in ( (0, +infty) ).Therefore, combining these cases, the only valid range for ( a ) is ( 3 - sqrt{5} leq a < 2 ).But let's verify if ( a = 3 - sqrt{5} ) is valid. Compute ( 3 - sqrt{5} approx 3 - 2.236 = 0.764 ). Since ( a ) must be positive (as the argument of the logarithm must be positive for all ( x ) in ( (0, +infty) )), ( a > 0 ). So, ( 3 - sqrt{5} approx 0.764 > 0 ), which is acceptable.Also, when ( a = 3 - sqrt{5} ), the discriminant is zero, so there is exactly one real root. Let's check if this root is positive.The quadratic equation becomes:[(2 - a)x^2 - 2a x + (2 - 2a) = 0]At ( a = 3 - sqrt{5} ), ( 2 - a = 2 - (3 - sqrt{5}) = -1 + sqrt{5} approx -1 + 2.236 = 1.236 > 0 ). So, the parabola opens upwards. The single root is at the vertex. The x-coordinate of the vertex is ( frac{2a}{2(2 - a)} = frac{a}{2 - a} ). Since ( a = 3 - sqrt{5} ) and ( 2 - a = -1 + sqrt{5} ), we have:[x = frac{3 - sqrt{5}}{-1 + sqrt{5}} = frac{(3 - sqrt{5})}{(sqrt{5} - 1)} ]Multiply numerator and denominator by ( sqrt{5} + 1 ):[x = frac{(3 - sqrt{5})(sqrt{5} + 1)}{(sqrt{5} - 1)(sqrt{5} + 1)} = frac{3sqrt{5} + 3 - 5 - sqrt{5}}{5 - 1} = frac{2sqrt{5} - 2}{4} = frac{sqrt{5} - 1}{2} approx frac{2.236 - 1}{2} = frac{1.236}{2} approx 0.618]Which is positive. So, ( a = 3 - sqrt{5} ) is valid.Similarly, for ( a ) approaching 2 from below, the quadratic equation has two roots, but as ( a ) approaches 2, one root approaches zero and the other approaches negative infinity. Wait, no, let me check.Wait, when ( a ) approaches 2 from below, ( 2 - a ) approaches 0 from above. The quadratic equation becomes:[(2 - a)x^2 - 2a x + (2 - 2a) = 0]As ( a to 2^- ), ( 2 - a to 0^+ ), ( -2a to -4 ), ( 2 - 2a to -2 ). So, the equation becomes approximately:[0 cdot x^2 - 4x - 2 = 0 implies -4x - 2 = 0 implies x = -frac{1}{2}]Which is negative. So, as ( a ) approaches 2 from below, one root approaches negative infinity and the other approaches a finite negative value. Wait, that contradicts my earlier thought.Wait, actually, let's consider the quadratic equation as ( a ) approaches 2 from below. Let me denote ( a = 2 - epsilon ), where ( epsilon to 0^+ ).Then:[(2 - a) = epsilon][-2a = -4 + 2epsilon][2 - 2a = 2 - 4 + 2epsilon = -2 + 2epsilon]So, the quadratic equation becomes:[epsilon x^2 + (-4 + 2epsilon)x + (-2 + 2epsilon) = 0]Divide through by ( epsilon ) (since ( epsilon neq 0 )):[x^2 + left( frac{-4 + 2epsilon}{epsilon} right)x + left( frac{-2 + 2epsilon}{epsilon} right) = 0]Simplify:[x^2 + left( -frac{4}{epsilon} + 2 right)x + left( -frac{2}{epsilon} + 2 right) = 0]As ( epsilon to 0^+ ), the coefficients become:[x^2 - infty x - infty = 0]Which suggests that one root goes to negative infinity and the other approaches a finite value. But since ( x_0 ) must be positive, we need to ensure that at least one root is positive.Wait, perhaps my approach is complicating things. Let's instead consider the quadratic equation:[(2 - a)x^2 - 2a x + (2 - 2a) = 0]We can analyze the roots using the quadratic formula:[x = frac{2a pm sqrt{D}}{2(2 - a)} = frac{2a pm sqrt{-4a^2 + 24a - 16}}{2(2 - a)}]But since we already have the discriminant condition, and we know ( D geq 0 ) when ( 3 - sqrt{5} leq a leq 3 + sqrt{5} ), and we've established that for ( a < 2 ), the quadratic opens upwards and has a positive root, while for ( a > 2 ), it opens downwards and has negative roots.Therefore, the valid range for ( a ) is ( 3 - sqrt{5} leq a < 2 ).But let's also ensure that the argument of the logarithm is positive for all ( x ) in ( (0, +infty) ). The argument is ( frac{a}{x^2 + 1} ). Since ( x^2 + 1 > 0 ) for all real ( x ), we just need ( a > 0 ). Our range ( 3 - sqrt{5} approx 0.764 ) is positive, so it's acceptable.Hence, the range of ( a ) is ( [3 - sqrt{5}, 2) ).**Final Answer**The range of values for ( a ) is boxed{[3 - sqrt{5}, 2)}.

💡Alright, so I have this problem that says if a bounded figure has an axis of symmetry and a center of symmetry, then the center of symmetry lies on the axis of symmetry. Hmm, okay, let me try to wrap my head around this.First, let me recall what these terms mean. An axis of symmetry is a line where if you reflect the figure over that line, it looks the same. So, for example, if I have a figure that's symmetric about the x-axis, then for every point (x, y) in the figure, there's also a point (x, -y). Got it.A center of symmetry is a point where if you rotate the figure 180 degrees around that point, it looks the same. So, for any point (x, y), there's another point (-x, -y) in the figure. That makes sense.Now, the figure is bounded, which means all its points lie within some fixed distance from the origin or some central point. So, it doesn't go off to infinity in any direction.Okay, so the problem is saying that if a figure has both an axis of symmetry and a center of symmetry, then that center must lie on the axis. I need to prove that.Let me try to visualize this. Suppose the figure is symmetric about the x-axis, so that's our axis of symmetry. And it also has a center of symmetry at some point, say (h, k). I need to show that k must be zero, meaning the center lies on the x-axis.Wait, is that necessarily true? Let me think. If the center is at (h, k), then reflecting over the x-axis would take (h, k) to (h, -k). But since the figure is symmetric about the x-axis, reflecting the center over the x-axis should still be a center of symmetry, right? Because the figure looks the same after reflection.But if (h, k) is a center of symmetry, then so is (h, -k). Hmm, but a figure can't have two different centers of symmetry unless they are the same point. Otherwise, it would imply some kind of periodicity or something, which might not be compatible with being bounded.Wait, is that always true? If a figure has two centers of symmetry, does that mean it's unbounded? Maybe. Because if you have two centers, you can keep translating the figure by the vector connecting those centers, and it would keep repeating, which would make it unbounded.But our figure is bounded, so it can't have two distinct centers of symmetry. Therefore, the only way for both (h, k) and (h, -k) to be centers of symmetry is if they are the same point. That would mean k = 0, so the center is on the x-axis.Okay, that seems like a good line of reasoning. Let me try to formalize it.Suppose the figure has an axis of symmetry, say the x-axis, and a center of symmetry at (h, k). Since the figure is symmetric about the x-axis, reflecting the center (h, k) over the x-axis gives another center of symmetry at (h, -k). But a bounded figure can't have two distinct centers of symmetry because that would imply an infinite sequence of centers, contradicting boundedness. Therefore, (h, k) must equal (h, -k), which implies k = 0. Hence, the center of symmetry lies on the x-axis.Wait, does this hold in general? What if the axis of symmetry isn't the x-axis? Suppose it's some arbitrary line. Then, reflecting the center over that line should give another center, and for the figure to be bounded, those two centers must coincide, meaning the center lies on the axis.Yeah, that makes sense. So, regardless of the axis of symmetry, reflecting the center over that axis must give the same center, meaning it lies on the axis.Let me think of an example. Take a circle. It has infinitely many axes of symmetry and a center of symmetry at its center. The center lies on all those axes, which is consistent with the statement.Another example: a rectangle. It has two axes of symmetry (the vertical and horizontal lines through its center) and a center of symmetry at its center. The center lies on both axes, so again, consistent.What about a figure that's symmetric about the x-axis and has a center of symmetry not on the x-axis? Wait, according to our reasoning, that can't happen because it would imply two centers, making the figure unbounded. So, such a figure can't exist if it's bounded.Is there a way to make a bounded figure with an axis of symmetry and a center of symmetry not on the axis? I don't think so. Because reflecting the center over the axis would give another center, and the figure would have to repeat infinitely in that direction, making it unbounded.Therefore, the conclusion seems solid: if a bounded figure has both an axis of symmetry and a center of symmetry, the center must lie on the axis.I think I've got it. So, to summarize:1. Assume the figure has an axis of symmetry and a center of symmetry.2. Reflecting the center over the axis gives another center of symmetry.3. For the figure to be bounded, these two centers must coincide.4. Therefore, the center lies on the axis.Yeah, that seems like a valid proof. I don't see any flaws in this reasoning. It relies on the fact that a bounded figure can't have multiple centers of symmetry unless they're the same point. And reflecting the center over the axis gives another center, so they must be the same, meaning the center is on the axis.I think that's a solid argument. Maybe I should write it out more formally to make sure it's clear.