🧠:[<think>Okay, let's see. I need to find all values of the parameter ( a ) such that the given equation has two distinct solutions on the interval ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ). The equation is quite complicated, involving seventh powers and parameters. Let me start by carefully going through the problem and the provided solution to understand each step.First, the original equation is:[left(left(1 - x^{2}right)^{2} + 2a^{2} + 5aright)^{7} - left((3a + 2)left(1 - x^{2}right) + 3right)^{7} = 5 - 2a - (3a + 2)x^{2} - 2a^{2} - left(1 - x^{2}right)^{2}]The solution starts by rearranging both sides and introducing the function ( f(t) = t^{7} + t ). They note that this function is strictly increasing, which makes sense because the derivative ( f'(t) = 7t^{6} + 1 ) is always positive. Therefore, if ( f(u) = f(v) ), then ( u = v ).So, they set ( u = left(1 - x^{2}right)^{2} + 2a^{2} + 5a ) and ( v = (3a + 2)left(1 - x^{2}right) + 3 ), leading to the equation ( u = v ):[left(1 - x^{2}right)^{2} + 2a^{2} + 5a = (3a + 2)left(1 - x^{2}right) + 3]Simplifying this gives a quadratic in terms of ( y = 1 - x^{2} ):[y^{2} - (3a + 2)y + 2a^{2} + 5a - 3 = 0]They compute the discriminant ( D = (3a + 2)^2 - 4(2a^{2} + 5a - 3) ), which simplifies to ( (a - 4)^2 ). Therefore, except when ( a = 4 ), the quadratic has two distinct roots ( y_1 = 2a - 1 ) and ( y_2 = a + 3 ). When ( a = 4 ), the quadratic has a repeated root ( y = 7 ).Now, since ( y = 1 - x^{2} ), and ( x ) is in ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ), we need to find the possible values of ( y ). Let's compute the range of ( y ).Given ( x in left[-frac{sqrt{6}}{2}, sqrt{2}right] ), then ( x^2 ) ranges from 0 to ( left(frac{sqrt{6}}{2}right)^2 = frac{6}{4} = 1.5 ). Therefore, ( y = 1 - x^2 ) ranges from ( 1 - 1.5 = -0.5 ) up to ( 1 - 0 = 1 ). So ( y in [-0.5, 1] ).The problem states that the original equation must have two distinct solutions in ( x ). Since ( y = 1 - x^2 ), each ( y ) in ( (-0.5, 1) ) corresponds to two distinct ( x ) values (positive and negative square roots), except when ( y = 1 ), which corresponds to ( x = 0 ), a single solution, and ( y = -0.5 ), which would correspond to ( x = pm sqrt{1.5} ), but since 1.5 is ( frac{3}{2} ), which is within the interval ( x leq sqrt{2} approx 1.414 ). Wait, ( sqrt{1.5} approx 1.2247 ), which is less than ( sqrt{2} approx 1.414 ). So ( x = pm sqrt{1.5} ) is within the interval. But ( y = -0.5 ) would be achieved when ( x^2 = 1.5 ), which is ( x = pm sqrt{1.5} ). So, the endpoints of ( y in [-0.5, 1] ).So, the equation ( y^2 - (3a + 2)y + 2a^2 + 5a - 3 = 0 ) must have solutions ( y ) such that when we solve ( 1 - x^2 = y ), the resulting ( x ) values are distinct and lie within ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ).The solution mentions two cases:1) One root ( y_1 in [-0.5, 1) ), and the other root ( y_2 notin [-1, 1] ), or2) Two distinct roots in ( [-1, -0.5) cup {1} ).But the solution also states that the equation cannot have two distinct solutions belonging to the set ( [-1, -0.5) cup {1} ), so only the first case is possible. Then, they analyze the roots ( y_1 = 2a -1 ) and ( y_2 = a + 3 ).First, they check when ( y_1 = 2a -1 ) is in [-0.5, 1). Let's solve:[-0.5 leq 2a -1 < 1]Adding 1:[0.5 leq 2a < 2 implies 0.25 leq a < 1]So ( a in [0.25, 1) ).In this case, ( y_1 = 2a -1 in [-0.5, 1) ), which would correspond to ( x = pm sqrt{1 - y_1} = pm sqrt{1 - (2a -1)} = pm sqrt{2 - 2a} ). For these x-values to be real, the expression under the square root must be non-negative:[2 - 2a geq 0 implies a leq 1]Which is already satisfied since ( a < 1 ). Also, the x-values ( pm sqrt{2 - 2a} ) must lie within the interval ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ).Since ( sqrt{2 - 2a} ) is positive, we need to ensure ( sqrt{2 - 2a} leq sqrt{2} ), which is always true because ( 2 - 2a leq 2 implies -2a leq 0 implies a geq 0 ). But since ( a geq 0.25 ), this is satisfied. The lower bound: ( sqrt{2 - 2a} geq 0 ), which is always true. So for ( a in [0.25, 1) ), we have two solutions ( x = pm sqrt{2 - 2a} ).Now, the other root ( y_2 = a + 3 ). For the original equation to have only two solutions, ( y_2 ) must not be in the interval ( [-0.5, 1] ). Let's check:If ( y_2 geq 1 ), then ( a + 3 geq 1 implies a geq -2 ).If ( y_2 leq -0.5 ), then ( a + 3 leq -0.5 implies a leq -3.5 ).But since in the first case ( a in [0.25, 1) ), ( y_2 = a + 3 ) will be ( geq 3.25 ), which is outside of [-0.5, 1], so that's okay. So in this case, only ( y_1 ) gives solutions, and since ( y_1 in [-0.5, 1) ), and gives two x-values, which are distinct as long as ( y_1 neq 1 ). Since ( a < 1 ), ( y_1 = 2a -1 < 1 ), so we have two distinct solutions.Second case: ( y_2 = a + 3 in [-0.5, 1) ). Let's see when that happens:[-0.5 leq a + 3 < 1 implies -3.5 leq a < -2]So ( a in [-3.5, -2) ).In this case, ( y_2 = a + 3 in [-0.5, 1) ), leading to ( x = pm sqrt{1 - y_2} = pm sqrt{1 - (a + 3)} = pm sqrt{-a - 2} ). For the square root to be real:[-a - 2 geq 0 implies a leq -2]Which is satisfied since ( a < -2 ). Also, these x-values must be in the interval ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ).Compute ( sqrt{-a - 2} ). Since ( a in [-3.5, -2) ), ( -a in (2, 3.5] ), so ( -a - 2 in (0, 1.5] ). Therefore, ( sqrt{-a - 2} in (0, sqrt{1.5}] approx (0, 1.2247) ). The upper bound ( sqrt{1.5} approx 1.2247 ) is less than ( sqrt{2} approx 1.414 ), so the positive roots are within the interval. The negative roots are greater than ( -sqrt{1.5} approx -1.2247 ), which is greater than ( -sqrt{6}/2 approx -1.2247 ). Wait, ( sqrt{6}/2 approx 1.2247 ), so ( -sqrt{6}/2 approx -1.2247 ). Therefore, ( sqrt{-a - 2} leq sqrt{1.5} approx 1.2247 ), so the x-values ( pm sqrt{-a - 2} ) are within the interval ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ).However, in this case, ( y_1 = 2a -1 ). Since ( a in [-3.5, -2) ), ( y_1 = 2a -1 ) would be ( 2(-3.5) -1 = -8 ) at ( a = -3.5 ), and ( 2(-2) -1 = -5 ) at ( a = -2 ). So ( y_1 in [-8, -5) ), which is outside of [-0.5, 1], so the other root ( y_1 ) doesn't contribute any solutions. Therefore, in this case, only ( y_2 = a + 3 in [-0.5, 1) ) gives two solutions ( x = pm sqrt{-a - 2} ).But we need to ensure that these are distinct solutions. Since ( a in [-3.5, -2) ), ( -a - 2 in (0, 1.5] ), so ( sqrt{-a - 2} ) is positive and non-zero, hence ( x = pm sqrt{-a - 2} ) are two distinct solutions. Additionally, they must lie within the given interval. As calculated before, they do.Therefore, combining both cases, the parameter ( a ) can be in [0.25, 1) or [-3.5, -2). However, the solution mentions:"1) (-0.5 leq 2 a -1 <1 Rightarrow a in[0.25 ; 1)), (1 - x^{2}=2 a -1), (x_{1}=sqrt{2 - 2 a}), (x_{2}=-sqrt{2 - 2 a}). 2) (-0.5 leq a +3 <1 Rightarrow a in[-3.5 ; -2)), (1 - x^{2}=a +3), (x_{1}=sqrt{-a -2}), (x_{2}=-sqrt{-a -2}). The equation cannot have two distinct solutions belonging to the set ([-1 ; -0.5) cup {1})."So according to this, these are the two intervals for ( a ).But the problem says "two distinct solutions on the interval (left[-frac{sqrt{6}}{2}, sqrt{2}right])". Therefore, these two cases give two distinct solutions each. But wait, the problem requires exactly two distinct solutions. However, in each case, we have two solutions. However, we need to check whether in these intervals for ( a ), the equation only has these two solutions. Since the quadratic equation in ( y ) can have two roots, but only one of them is in [-0.5, 1), leading to two x-solutions. The other root is outside, so no x-solutions from that. Therefore, each case gives exactly two x-solutions. But since the original equation was reduced to ( y ) equation, and due to the function ( f(t) ) being strictly increasing, there are no other solutions. Hence, for ( a in [0.25, 1) cup [-3.5, -2) ), the equation has exactly two distinct solutions on the given interval.But wait, the solution says "the equation cannot have two distinct solutions belonging to the set ([-1 ; -0.5) cup {1})". That is, the case where both roots ( y ) are in ( [-1 ; -0.5) cup {1} ), which is outside of [-0.5,1), so even if they are in that set, they don't contribute solutions. But if a quadratic equation had two roots in that set, then ( y ) values would not be in [-0.5,1), so the equation would have no solutions. Therefore, the only possible cases are when one root is in [-0.5,1) and the other is outside [-1,1], which are the two cases considered.But wait, in the problem statement's solution, after substitution, they say:"For the condition of the problem to be satisfied, one root of this equation must belong to the interval ([-0.5 ; 1)), and the second, if it exists, must not belong to the interval ([-1 ; 1]), or this equation must have two distinct solutions belonging to the set ([-1 ; -0.5) cup{1})."But then they conclude that the equation cannot have two distinct solutions in that set. Therefore, only the first case is possible. So the total solutions are when either:- One root is in [-0.5,1), and the other is outside [-1,1]. Since when discriminant is positive (which is always except a=4), there are two roots. But when a=4, only one root y=7.But since D=(a-4)^2, discriminant is zero when a=4, so y=(3a+2)/2. If a=4, then y=(14)/2=7, so y=7, which is outside [-0.5,1), so when a=4, the equation has no solutions. So for a=4, no solutions. For other a, two roots y1=2a-1 and y2=a+3.Therefore, to have solutions, either y1 or y2 must be in [-0.5,1). The other root can be anything, but as per the problem's requirement, to have exactly two distinct x solutions, we need exactly one y in [-0.5,1) leading to two x solutions. If two y's are in [-0.5,1), that would lead to four x solutions (if both y's are in (-0.5,1)), or two x solutions if one y=1 (but y=1 gives one x=0). However, since in our case, the roots are y1=2a-1 and y2=a+3, they cannot both be in [-0.5,1). Let's check.Suppose both y1 and y2 are in [-0.5,1). Then:For y1: -0.5 ≤ 2a -1 <1 ⇒ 0.25 ≤ a <1For y2: -0.5 ≤ a +3 <1 ⇒ -3.5 ≤ a < -2But these intervals for a are disjoint. Therefore, it's impossible for both y1 and y2 to be in [-0.5,1) simultaneously. Hence, only one root can be in [-0.5,1), leading to two x-solutions.Therefore, the possible a values are [0.25,1) and [-3.5,-2). But we need to verify that in each interval, the other root is outside [-1,1].For a ∈ [0.25,1):- y2 = a +3. Since a ≥0.25, y2 ≥ 3.25, which is outside [-1,1].For a ∈ [-3.5,-2):- y1 = 2a -1. Since a ≥-3.5, y1 ≥ -7 -1 = -8, but a < -2 ⇒ y1 < -5. So y1 ∈ [-8, -5), which is outside [-1,1].Hence, in both cases, the other root is outside [-1,1], so only one root contributes to solutions, giving two x-values.Therefore, the values of a are a ∈ [-3.5, -2) ∪ [0.25,1).Now, let's check the endpoints:For a=0.25:y1=2*0.25 -1=0.5 -1=-0.5. So y=-0.5. Then x^2=1 - (-0.5)=1.5, so x=±√1.5≈±1.2247, which is within the interval [-√6/2≈-1.2247, √2≈1.414]. So exactly at the lower endpoint of x. These are two distinct solutions.For a approaching 1 from below:y1 approaches 2*1 -1=1. So y approaches 1 from below. x approaches ±√(2 - 2*1)=0. So x approaches 0. So as a approaches 1, the solutions approach x=0 from both sides. However, at a=1, y1=1, which would give x=0, but a=1 is not included.For a=-3.5:y2= -3.5 +3= -0.5. So y=-0.5, leading to x=±√(1 - (-0.5))=±√1.5, same as above.For a approaching -2 from below:y2 approaches -2 +3=1. So y2 approaches 1 from below. So x approaches ±√( -(-2) -2 )=±√(2 -2)=0. So similar to above.Therefore, at the endpoints of the intervals for a, the solutions are at the edges of the x-interval.Thus, the answer should be a ∈ [-3.5, -2) ∪ [0.25,1), and for each a in these intervals, the solutions are:For a ∈ [0.25,1):x=±√(2 - 2a)For a ∈ [-3.5, -2):x=±√(-a -2)But we need to check if these solutions are indeed within the interval [-√6/2, √2].For a ∈ [0.25,1):x=±√(2 - 2a). Since a ∈ [0.25,1), 2 - 2a ∈ (0, 1.5]. Therefore, √(2 - 2a) ∈ (0, √1.5] ≈ (0, 1.2247]. Since √1.5 ≈1.2247 is exactly √6/2 ≈1.2247. Wait, √6/2 ≈1.2247. So √(2 - 2a) ≤ √6/2. Therefore, x=±√(2 - 2a) ∈ [-√6/2, √6/2]. But the interval is [-√6/2, √2]. Since √6/2 ≈1.2247 and √2≈1.414, so √(2 - 2a) ≤√6/2 < √2. Therefore, the solutions x=±√(2 - 2a) are within the interval [-√6/2, √2].For a ∈ [-3.5, -2):x=±√(-a -2). Since a ∈ [-3.5, -2), -a ∈ (2, 3.5], so -a -2 ∈ (0, 1.5]. Therefore, √(-a -2) ∈ (0, √1.5]≈(0,1.2247]. So x=±√(-a -2) ∈ [-√6/2, √6/2] as before, which is within [-√6/2, √2].Therefore, in both cases, the solutions lie within the given interval. Hence, the final answer is a ∈ [-3.5, -2) ∪ [0.25,1), with corresponding solutions:For a ∈ [0.25,1): x=±√(2 - 2a)For a ∈ [-3.5, -2): x=±√(-a -2)The problem also says "Specify these solutions for each found a."But in the original solution presented, they conclude:"The only solution will be at a=4, this solution is y=7. In all other cases, we have two distinct solutions y1=2a−1 and y2=a+3."Wait, but in the beginning, they said when a=4, discriminant is zero, so only one root y=7, which is outside the interval, so no solutions. Then they proceed to analyze y1 and y2. So their conclusion is:1) For a ∈ [0.25,1): solutions x=±√(2−2a)2) For a ∈ [−3.5,−2): solutions x=±√(−a−2)But in the problem statement, the assistant's solution mentions:"Solution: ... It is necessary to find all values of the parameter ( a )... Let's make the substitution: ( y = 1 - x^{2} ). We arrive at the equation ( y^{2} - (3a + 2)y + 2a^{2} + 5a - 3 = 0 ). For the condition of the problem to be satisfied, one root of this equation must belong to the interval ([-0.5;1)), and the second, if it exists, must not belong to the interval ([-1;1]), or this equation must have two distinct solutions belonging to the set ([-1;-0.5) cup {1}). The discriminant of the equation ( D = (a - 4)^2 ). The only solution will be at ( a = 4 ), this solution is ( y = 7 ). In all other cases, we have two distinct solutions ( y_{1} = 2a -1 ) and ( y_{2} = a + 3 )."Wait, so the assistant's solution then proceeds:1) For ( a in [0.25;1) ), solutions ( x_{1} = sqrt{2 - 2a} ), ( x_{2} = -sqrt{2 - 2a} ).2) For ( a in [-3.5;-2) ), solutions ( x_{1} = sqrt{-a -2} ), ( x_{2} = -sqrt{-a -2} ).And mentions that the equation cannot have two solutions in ([-1;-0.5) cup {1}).Therefore, the final answer is ( a in [-3.5;-2) cup [0.25;1) ), with solutions as above.I need to check if there are any mistakes here. Let's verify with sample a values.Take a=0.5 (which is in [0.25,1)):Then y1=2*0.5 -1=0, which is in [-0.5,1). So x=±√(2 - 2*0.5)=±√1=±1. Check if x=±1 are in [-√6/2≈-1.2247, √2≈1.414]. Yes, they are. So two solutions.Take a=-3 (which is in [-3.5,-2)):y2=-3 +3=0, which is in [-0.5,1). So x=±√(-(-3) -2)=±√(3 -2)=±1. So x=±1, same as above. Wait, but according to the solution, for a=-3, x=±√(-a -2)=±√(3 -2)=±1. Wait, but if a=-3, then y2=a+3=0. Then 1 -x²=0 ⇒x=±1. But since a=-3 is in [-3.5, -2), that's correct. So x=±1 are within the interval.Another test: a=0.25:y1=2*0.25 -1= -0.5. So x²=1 - (-0.5)=1.5, x=±√1.5≈±1.2247, which is exactly the endpoints of the interval [-√6/2, √2] since √6/2≈1.2247. So these are valid.a=-3.5:y2=-3.5 +3= -0.5. So x²=1 - (-0.5)=1.5, x=±√1.5≈±1.2247, again the endpoints.Another test: a=0. Let's see if a=0 is in [0.25,1). No, it's not. So a=0 should not be a solution. For a=0:The equation reduces to y² - (0 +2)y +0 +0 -3= y² -2y -3=0. Solutions y=(2 ±√(4 +12))/2=(2 ±√16)/2=(2 ±4)/2=3 or -1. So y=3 or y=-1. y=3 is outside [-0.5,1), y=-1 is outside. So no solutions. Correct.Take a=-2.5 (in [-3.5,-2)):Then y2=a+3=0.5. So x=±√(-(-2.5)-2)=±√(2.5 -2)=±√0.5≈±0.707. These are within the interval.Take a=-3.5:x=±√1.5≈±1.2247, which is the lower endpoint.Take a approaching -2 from the left, say a=-2.1:y2=-2.1 +3=0.9. x=±√(-(-2.1)-2)=±√(2.1 -2)=±√0.1≈±0.316. Within the interval.Take a=0.25:x=±√(2 -0.5)=±√1.5≈±1.2247, which is the lower endpoint.Take a approaching 1 from below, say a=0.99:x=±√(2 -1.98)=±√0.02≈±0.141, which is within the interval.So all test cases check out. Therefore, the solution provided is correct.Final AnswerThe parameter ( a ) can take values in the intervals ([-3.5, -2)) and ([0.25, 1)). For each ( a ) in ([-3.5, -2)), the solutions are ( x = pm sqrt{-a - 2} ). For each ( a ) in ([0.25, 1)), the solutions are ( x = pm sqrt{2 - 2a} ). Thus, the answer is (boxed{a in [-3.5, -2) cup [0.25, 1)}) with the corresponding solutions as specified.</think>To find all values of the parameter ( a ) for which the given equation has two distinct solutions on the interval (left[-frac{sqrt{6}}{2}, sqrt{2}right]), we start by simplifying the equation using the substitution ( y = 1 - x^2 ). This leads to the quadratic equation:[y^2 - (3a + 2)y + 2a^2 + 5a - 3 = 0]The discriminant of this quadratic equation is ( D = (a - 4)^2 ), which gives the roots ( y_1 = 2a - 1 ) and ( y_2 = a + 3 ) for ( a neq 4 ). When ( a = 4 ), the equation has a repeated root ( y = 7 ), which is outside the interval ([-0.5, 1]).For the original equation to have two distinct solutions on the given interval, one root of the quadratic equation must belong to ([-0.5, 1)) and the other root must not belong to ([-1, 1]). This leads to the following cases:1. Case 1: ( y_1 = 2a - 1 in [-0.5, 1) ) - Solving ( -0.5 leq 2a - 1 < 1 ), we get ( 0.25 leq a < 1 ). - For ( a in [0.25, 1) ), the solutions for ( x ) are ( pm sqrt{2 - 2a} ).2. Case 2: ( y_2 = a + 3 in [-0.5, 1) ) - Solving ( -0.5 leq a + 3 < 1 ), we get ( -3.5 leq a < -2 ). - For ( a in [-3.5, -2) ), the solutions for ( x ) are ( pm sqrt{-a - 2} ).The equation cannot have two distinct solutions belonging to the set ([-1, -0.5) cup {1}), so only the above cases are valid. Thus, the values of ( a ) are ( a in [-3.5, -2) cup [0.25, 1) ). The corresponding solutions for each interval are:- For ( a in [0.25, 1) ): ( x = pm sqrt{2 - 2a} )- For ( a in [-3.5, -2) ): ( x = pm sqrt{-a - 2} )[boxed{a in [-3.5, -2) cup [0.25, 1)}],<think>Okay, so I need to find all values of parameter ( a ) such that the given equation has two distinct solutions on the interval ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ). Let me start by understanding the problem step by step.First, the equation provided is quite complex, but the solution process is outlined. Let me verify each step to make sure I follow.The original equation is:[left(left(1 - x^{2}right)^{2} + 2a^{2} + 5aright)^{7} - left((3a + 2)left(1 - x^{2}right) + 3right)^{7} = 5 - 2a - (3a + 2)x^{2} - 2a^{2} - left(1 - x^{2}right)^{2}]The solution mentions that this can be rewritten by moving all terms to the left side except for the right-hand side expression. Then, they factor it in a way that introduces the function ( f(t) = t^{7} + t ), which is strictly increasing. This is a crucial step because if ( f(u) = f(v) ), then ( u = v ) since ( f ) is injective (one-to-one) due to its strict monotonicity.So, substituting ( u = left(1 - x^{2}right)^{2} + 2a^{2} + 5a ) and ( v = (3a + 2)left(1 - x^{2}right) + 3 ), the equation simplifies to ( u = v ). That gives:[left(1 - x^{2}right)^{2} + 2a^{2} + 5a = (3a + 2)left(1 - x^{2}right) + 3]Rearranging terms, we get a quadratic in ( y = 1 - x^{2} ):[y^{2} - (3a + 2)y + 2a^{2} + 5a - 3 = 0]Let me compute the discriminant ( D ) of this quadratic equation to find the roots. The discriminant is:[D = (3a + 2)^2 - 4 times 1 times (2a^2 + 5a - 3)]Calculating that:[D = 9a^2 + 12a + 4 - 8a^2 - 20a + 12 = (9a^2 - 8a^2) + (12a - 20a) + (4 + 12) = a^2 - 8a + 16 = (a - 4)^2]So the discriminant is a perfect square, which means the roots are real and equal when ( a = 4 ), and distinct otherwise. The roots are:[y_{1,2} = frac{(3a + 2) pm |a - 4|}{2}]But since ( D = (a - 4)^2 ), the absolute value simplifies to ( |a - 4| = |4 - a| ). Therefore, the roots are:[y_1 = frac{(3a + 2) + (4 - a)}{2} = frac{2a + 6}{2} = a + 3]and[y_2 = frac{(3a + 2) - (4 - a)}{2} = frac{4a - 2}{2} = 2a - 1]So, unless ( a = 4 ), we have two distinct roots: ( y_1 = a + 3 ) and ( y_2 = 2a - 1 ). If ( a = 4 ), both roots coincide at ( y = 7 ).Now, since ( y = 1 - x^2 ), we need to consider the possible values of ( y ). Let me think about the domain of ( x ). The interval given is ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ). Let me compute the range of ( x^2 ) over this interval.The maximum value of ( |x| ) in the interval is ( sqrt{2} ), so ( x^2 ) ranges from 0 to ( (sqrt{2})^2 = 2 ). Therefore, ( 1 - x^2 ) ranges from ( 1 - 2 = -1 ) to ( 1 - 0 = 1 ). Hence, ( y in [-1, 1] ).But in the solution explanation, it says that one root must belong to ( [-0.5, 1) ) and the other must not belong to ( [-1, 1] ), or both roots must be in ( [-1, -0.5) cup {1} ). Hmm, I need to verify why that is the case.The original equation after substitution becomes ( y^2 - (3a + 2)y + 2a^2 + 5a - 3 = 0 ), and each ( y ) corresponds to ( x ) via ( y = 1 - x^2 ). Since ( x ) is in ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ), the ( x^2 ) ranges from 0 to 2 (as before), so ( y = 1 - x^2 in [-1, 1] ).But the problem requires the original equation to have two distinct solutions ( x ). Each solution ( y ) can correspond to 0, 1, or 2 solutions ( x ), depending on the value of ( y ):- If ( y = 1 - x^2 ), then ( x^2 = 1 - y ). So ( x = pm sqrt{1 - y} ), provided that ( 1 - y geq 0 ), i.e., ( y leq 1 ).But since ( y in [-1, 1] ), ( 1 - y in [0, 2] ). However, ( sqrt{1 - y} ) must be real, so ( 1 - y geq 0 implies y leq 1 ). So even if ( y ) is in ( [-1, 1] ), ( x ) can be real. However, since ( x ) is restricted to ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ), we need to check if the corresponding ( x ) values lie within this interval.Wait, but if ( x^2 leq 2 ), then ( sqrt{1 - y} leq sqrt{2} ), which is already satisfied since ( x in [-sqrt{2}, sqrt{2}] ). Wait, but ( sqrt{2} ) is the upper limit. So actually, the possible values of ( x ) are already bounded, so ( x^2 leq 2 implies y = 1 - x^2 geq -1 ). Therefore, ( y in [-1, 1] ).Therefore, each solution ( y in [-1, 1] ) will correspond to ( x = pm sqrt{1 - y} ), but we need to check whether these ( x ) values are within the interval ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ). Wait, but ( sqrt{2} approx 1.414 ), and ( sqrt{6}/2 approx 1.224 ). So the interval is from approximately -1.224 to 1.414. So if ( x = sqrt{1 - y} ), then ( sqrt{1 - y} leq sqrt{2} ), which is always true because ( 1 - y leq 2 implies y geq -1 ). Similarly, ( sqrt{1 - y} leq sqrt{6}/2 approx 1.224 implies 1 - y leq 6/4 = 1.5 implies y geq -0.5 ).Wait, hold on. Let me think:The interval for ( x ) is ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ). So ( x ) can be as large as ( sqrt{2} ) and as low as ( -sqrt{6}/2 approx -1.2247 ). Therefore, when we have ( x = pm sqrt{1 - y} ), these values must lie within ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ).Therefore, ( sqrt{1 - y} leq sqrt{2} implies 1 - y leq 2 implies y geq -1 ), which is already part of the domain. But also, the lower bound on ( x ): ( -sqrt{6}/2 leq -sqrt{1 - y} implies sqrt{1 - y} leq sqrt{6}/2 implies 1 - y leq 6/4 = 1.5 implies y geq -0.5 ). Therefore, if ( y geq -0.5 ), then ( sqrt{1 - y} leq sqrt{1.5} approx 1.2247 ), which is exactly ( sqrt{6}/2 ). Therefore, if ( y geq -0.5 ), then ( x = pm sqrt{1 - y} ) will be within the interval ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ). However, if ( y < -0.5 ), then ( sqrt{1 - y} > sqrt{1.5} approx 1.2247 ), which is beyond the lower bound of the interval (since the interval goes down to -1.2247, but the positive square root would exceed the upper limit of the interval? Wait, no. Wait, if ( y < -0.5 ), then ( 1 - y > 1.5 ), so ( sqrt{1 - y} > sqrt{1.5} approx 1.2247 ). But the upper limit of the interval is ( sqrt{2} approx 1.414 ). So ( sqrt{1 - y} ) would be between approximately 1.2247 and 1.414 when ( y ) is between -1 and -0.5. But since ( x ) is allowed up to ( sqrt{2} approx 1.414 ), these are still within the upper limit. Wait, but ( the lower limit is ( -sqrt{6}/2 approx -1.2247 ), so ( x = -sqrt{1 - y} ) would be between -1.414 and -1.2247 when ( y in [-1, -0.5] ). But the interval ( left[-sqrt{6}/2, sqrt{2}right] ) does not include values less than ( -sqrt{6}/2 approx -1.2247 ). Therefore, for ( y in [-1, -0.5) ), ( x = -sqrt{1 - y} ) would be less than ( -sqrt{6}/2 ), hence outside the interval.Therefore, for ( y in [-1, -0.5) ), the corresponding ( x = -sqrt{1 - y} ) is outside the interval, while ( x = sqrt{1 - y} ) is inside the interval if ( sqrt{1 - y} leq sqrt{2} ). But ( 1 - y leq 2 implies y geq -1 ), which is already satisfied. Wait, but ( sqrt{1 - y} leq sqrt{2} ) is always true because ( y geq -1 implies 1 - y leq 2 ). Therefore, for ( y in [-1, -0.5) ), only the positive root ( x = sqrt{1 - y} ) is within the interval, since ( sqrt{1 - y} in (sqrt{1.5}, sqrt{2}] ), which is within ( [sqrt{1.5}, sqrt{2}] subset [1.2247, 1.414] subset [-sqrt{6}/2, sqrt{2}] ). Wait, but the positive root is within the interval, but the negative root ( x = -sqrt{1 - y} ) would be less than ( -sqrt{6}/2 approx -1.2247 ), which is outside the interval. So for ( y in [-1, -0.5) ), there is only one solution ( x = sqrt{1 - y} ).For ( y in [-0.5, 1) ), ( sqrt{1 - y} leq sqrt{1.5} approx 1.2247 ), so both ( x = sqrt{1 - y} ) and ( x = -sqrt{1 - y} ) are within the interval ( left[-sqrt{6}/2, sqrt{2}right] ), since ( sqrt{1 - y} leq sqrt{1.5} approx 1.2247 leq sqrt{2} approx 1.414 ), and similarly for the negative side. So for ( y in [-0.5, 1) ), each ( y ) gives two solutions ( x ).If ( y = 1 ), then ( x^2 = 0 implies x = 0 ), which is a single solution.Therefore, summarizing:- If ( y in [-1, -0.5) ): one solution ( x = sqrt{1 - y} )- If ( y in [-0.5, 1) ): two solutions ( x = pm sqrt{1 - y} )- If ( y = 1 ): one solution ( x = 0 )But since the problem requires two distinct solutions on the interval, we need to have either:1. Two distinct ( y ) values each giving one solution, but since each ( y ) can give either one or two solutions, but to have two distinct solutions in total, maybe:But no, more precisely, each ( y ) can lead to one or two ( x ) solutions. To have two distinct ( x ) solutions in total on the interval, we need:Either:- One ( y in [-0.5, 1) ), which gives two ( x ), but since we need two distinct solutions, this would suffice if there is exactly one such ( y ). However, if there are two different ( y ) values each giving one ( x ), then total two solutions. But according to the problem statement, the equation must have two distinct solutions. So possible cases:Case 1: The quadratic equation in ( y ) has one root ( y_1 in [-0.5, 1) ), which gives two ( x ) solutions. Then total two solutions.Case 2: The quadratic equation has two roots ( y_1, y_2 ). If one of them is in ( [-0.5, 1) ) and the other is not in ( [-1, 1] ), then the first gives two ( x ) solutions, and the second gives none (if outside ( [-1, 1] )) or one (if in ( [-1, -0.5) )), but we need total two. Wait, let's parse the solution explanation:"In the condition of the problem to be satisfied, one root of this equation must belong to the interval ( [-0.5; 1) ), and the second, if it exists, must not belong to the interval ( [-1; 1] ), or this equation must have two distinct solutions belonging to the set ( [-1; -0.5) cup {1} )."Wait, the original solution says that either:- One root is in ( [-0.5, 1) ), which gives two ( x ) solutions, and the other root (if exists) is outside ( [-1, 1] ), so it doesn't contribute.OR- The equation has two roots in ( [-1, -0.5) cup {1} ). But for two roots, each in ( [-1, -0.5) ) or at 1, but since ( y in [-1, 1] ), two roots in ( [-1, -0.5) cup {1} ). However, if there are two roots in ( [-1, -0.5) ), each would give one solution ( x = sqrt{1 - y} ). If there is a root at ( y = 1 ), it gives one solution ( x = 0 ). But if two roots are in ( [-1, -0.5) ), then each gives one solution, totaling two. Similarly, if one root is in ( [-1, -0.5) ) and the other is at 1, then total two solutions. But according to the solution explanation, this case is not possible? The solution says:"The equation cannot have two distinct solutions belonging to the set ( [-1; -0.5) cup {1} )."Therefore, only the first case is possible: one root in ( [-0.5, 1) ) (giving two ( x )), and the other root outside ( [-1, 1] ).But let me check. Suppose there are two roots ( y_1 ) and ( y_2 ). For each root:- If ( y_1 in [-0.5, 1) ), gives two ( x ).- If ( y_2 notin [-1, 1] ), then no solutions from ( y_2 ).- If ( y_2 in [-1, -0.5) cup {1} ), gives one solution.But to have total two solutions, we can have:Either:1. One root ( y_1 in [-0.5, 1) ) (two solutions) and the other root ( y_2 notin [-1, 1] ) (no solutions). Total two solutions.OR2. Two roots ( y_1, y_2 in [-1, -0.5) cup {1} ), each giving one solution. But since each root in ( [-1, -0.5) ) gives one solution and ( y = 1 ) gives one solution, having two such roots would give two solutions. But the solution explanation claims that this case is impossible. Let me see why.Looking back at the quadratic equation in ( y ): ( y^2 - (3a + 2)y + 2a^2 + 5a - 3 = 0 ). The roots are ( y_1 = 2a - 1 ) and ( y_2 = a + 3 ), except when ( a = 4 ), where both roots coincide at ( y = 7 ). So, the roots are ( y = 2a - 1 ) and ( y = a + 3 ). Therefore, for these roots to be in ( [-1, -0.5) cup {1} ), we need:Either both roots in ( [-1, -0.5) ), or one in ( [-1, -0.5) ) and one at 1. But let's check if that's possible.First, when is ( y = 1 ) a root? Substitute ( y = 1 ):( 1 - (3a + 2) + 2a^2 + 5a - 3 = 0 implies 2a^2 + 5a - 3 - 3a - 2 + 1 - 3 = 0 ). Wait, maybe better to substitute ( y = 1 ):( 1^2 - (3a + 2)(1) + 2a^2 + 5a - 3 = 0 implies 1 - 3a - 2 + 2a^2 + 5a - 3 = 0 implies 2a^2 + 2a - 4 = 0 implies a^2 + a - 2 = 0 implies (a + 2)(a - 1) = 0 implies a = -2 ) or ( a = 1 ). So, when ( a = -2 ) or ( a = 1 ), ( y = 1 ) is a root.But according to our earlier roots, for ( a = -2 ), ( y_1 = 2(-2) - 1 = -5 ), ( y_2 = -2 + 3 = 1 ). Similarly, for ( a = 1 ), ( y_1 = 2(1) - 1 = 1 ), ( y_2 = 1 + 3 = 4 ). So in these cases, one root is ( y = 1 ), which gives ( x = 0 ), and the other roots are ( y = -5 ) (for ( a = -2 )) which is outside ( [-1, 1] ), and ( y = 4 ) (for ( a = 1 )), also outside. So in these cases, only ( y = 1 ) gives ( x = 0 ), which is one solution. But the problem requires two distinct solutions, so ( a = -2 ) and ( a = 1 ) would not satisfy the condition unless there's another root contributing.But according to the solution explanation, the only solution is at ( a = 4 ), which gives ( y = 7 ), but this is outside ( [-1, 1] ), so no solutions. Wait, perhaps I need to re-examine the solution.Wait, in the original solution explanation, it says:"The discriminant of the equation ( D=(3 a+2)^{2}-8 a^{2}-20 a+12=(a-4)^{2} ). The only solution will be at ( a=4 ), this solution is ( y=7 ). In all other cases, we have two distinct solutions ( y_{1}=2 a-1 ) and ( y_{2}=a+3 ).1) ( -0.5 leq 2 a-1 <1 Rightarrow a in[0.25 ; 1) ), ( 1 - x^{2}=2 a-1 ), ( x_{1}=sqrt{2-2 a} ), ( x_{2}=-sqrt{2-2 a} ).2) ( -0.5 leq a+3 <1 Rightarrow a in[-3.5 ; -2) ), ( 1 - x^{2}=a+3 ), ( x_{1}=sqrt{-a-2} ), ( x_{2}=-sqrt{-a-2} ).The equation cannot have two distinct solutions belonging to the set ( [-1 ; -0.5) cup{1} )."Wait, but in the original problem statement, the interval for ( x ) is ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ). Therefore, when they consider the two cases for ( a ):1) ( a in [0.25; 1) ): Here, ( y = 2a - 1 ). So ( -0.5 leq 2a - 1 < 1 implies 0.25 leq a < 1 ). Then, ( 1 - x^2 = 2a - 1 implies x^2 = 2 - 2a ). Therefore, ( x = pm sqrt{2 - 2a} ). These solutions must lie within the interval ( left[-frac{sqrt{6}}{2}, sqrt{2}right] ). Let's check:Since ( a in [0.25, 1) ), ( 2 - 2a in (0, 1.5] ). Therefore, ( sqrt{2 - 2a} in (0, sqrt{1.5}] approx (0, 1.2247] ). Since ( sqrt{1.5} approx 1.2247 leq sqrt{6}/2 approx 1.2247 ). Wait, actually ( sqrt{6}/2 approx 1.2247 ), so ( sqrt{2 - 2a} leq sqrt{1.5} = sqrt{6}/2 ). Therefore, ( x = pm sqrt{2 - 2a} ) will be within ( left[-sqrt{6}/2, sqrt{2}right] ). Therefore, these solutions are valid, giving two distinct ( x ).2) ( a in [-3.5; -2) ): Here, ( y = a + 3 ). The condition is ( -0.5 leq a + 3 < 1 implies -3.5 leq a < -2 ). Then, ( 1 - x^2 = a + 3 implies x^2 = 1 - (a + 3) = -a - 2 ). Therefore, ( x = pm sqrt{-a - 2} ). Since ( a in [-3.5, -2) ), ( -a - 2 in (0, 1.5] ). Therefore, ( sqrt{-a - 2} in (0, sqrt{1.5}] approx (0, 1.2247] ). So, similar to the previous case, ( x = pm sqrt{-a - 2} ) will be within the interval, giving two distinct solutions.Therefore, the solution claims that these two intervals for ( a ) are the valid ones where the original equation has two distinct solutions. Additionally, it states that the case where two roots are in ( [-1, -0.5) cup {1} ) is impossible.But why is that impossible? Let's see. Suppose both roots ( y_1 ) and ( y_2 ) are in ( [-1, -0.5) cup {1} ). For that to happen, both ( y_1 ) and ( y_2 ) must be either in ( [-1, -0.5) ) or equal to 1. But since the quadratic equation in ( y ) can have at most two roots, and since the roots are ( y_1 = 2a - 1 ) and ( y_2 = a + 3 ), unless ( a = 4 ), which gives a repeated root.But for both roots to be in ( [-1, -0.5) cup {1} ), we need:Either:- Both ( y_1 = 2a - 1 ) and ( y_2 = a + 3 ) are in ( [-1, -0.5) ).OR- One of them is 1 and the other is in ( [-1, -0.5) ).Let's check if these are possible.Case 1: Both ( y_1, y_2 in [-1, -0.5) ).So:For ( y_1 = 2a - 1 in [-1, -0.5) implies -1 leq 2a - 1 < -0.5 implies 0 leq 2a < 0.5 implies 0 leq a < 0.25 ).For ( y_2 = a + 3 in [-1, -0.5) implies -1 leq a + 3 < -0.5 implies -4 leq a < -3.5 ).But these intervals for ( a ) (0 ≤ a < 0.25 and -4 ≤ a < -3.5) do not overlap. Therefore, there is no ( a ) for which both roots are in ( [-1, -0.5) ).Case 2: One root is 1 and the other is in ( [-1, -0.5) ).As before, if ( y = 1 ) is a root, then ( a = -2 ) or ( a = 1 ).For ( a = -2 ):- ( y_1 = 2(-2) - 1 = -5 ) (outside [-1, 1])- ( y_2 = -2 + 3 = 1 )So ( y = 1 ) gives ( x = 0 ), and ( y = -5 ) gives no solution. Total solutions: 1 (x = 0). Not two.For ( a = 1 ):- ( y_1 = 2(1) - 1 = 1 )- ( y_2 = 1 + 3 = 4 )Similarly, ( y = 1 ) gives x = 0, and y = 4 gives no solution. Total solutions: 1.Therefore, even if one root is 1, the other root is outside the interval, leading to only one solution. Therefore, the case where two roots are in ( [-1, -0.5) cup {1} ) is impossible, as it would require overlapping intervals for ( a ) which do not exist, or leading to only one solution.Therefore, the only possible cases are when one root is in ( [-0.5, 1) ) (giving two solutions) and the other is outside ( [-1, 1] ), which doesn't contribute. Thus, the values of ( a ) are those for which either:1. ( y_1 = 2a - 1 in [-0.5, 1) ) and ( y_2 = a + 3 notin [-1, 1] ), or2. ( y_2 = a + 3 in [-0.5, 1) ) and ( y_1 = 2a - 1 notin [-1, 1] ).But according to the solution explanation, only the first two cases are considered. Let me verify:For case 1: ( y_1 = 2a - 1 in [-0.5, 1) implies -0.5 leq 2a - 1 < 1 implies 0.25 leq a < 1 ). Then ( y_2 = a + 3 ). For ( a < 1 ), ( a + 3 < 4 ). But we need ( y_2 notin [-1, 1] ). Since ( a + 3 ) is linear in ( a ):When ( a geq -4 ), ( y_2 = a + 3 geq -1 ). But for ( a in [0.25, 1) ), ( y_2 in [3.25, 4) ), which is outside ( [-1, 1] ). Therefore, ( y_2 notin [-1, 1] ), so case 1 is valid.For case 2: ( y_2 = a + 3 in [-0.5, 1) implies -0.5 leq a + 3 < 1 implies -3.5 leq a < -2 ). Then ( y_1 = 2a - 1 ). For ( a < -2 ), ( 2a - 1 < -5 ), which is outside ( [-1, 1] ). Therefore, ( y_1 notin [-1, 1] ), so case 2 is valid.Therefore, these are the two intervals where ( a in [0.25, 1) ) and ( a in [-3.5, -2) ).Wait, but the solution explanation mentions:1) ( a in [0.25; 1) )2) ( a in [-3.5; -2) )Therefore, these are the two intervals where the equation has two distinct solutions on the given interval. The problem also asks to specify these solutions for each found ( a ).For case 1: ( a in [0.25, 1) )Equation: ( 1 - x^2 = 2a - 1 implies x^2 = 2 - 2a implies x = pm sqrt{2 - 2a} ).Since ( a in [0.25, 1) ), ( 2 - 2a in (0, 1.5] ). Therefore, ( sqrt{2 - 2a} in (0, sqrt{1.5}] approx (0, 1.2247] ). Since ( sqrt{6}/2 approx 1.2247 ), the solutions ( x = pm sqrt{2 - 2a} ) are within the interval ( left[-sqrt{6}/2, sqrt{2}right] ).For case 2: ( a in [-3.5, -2) )Equation: ( 1 - x^2 = a + 3 implies x^2 = -a - 2 implies x = pm sqrt{-a - 2} ).Since ( a in [-3.5, -2) ), ( -a - 2 in (0, 1.5] ), so ( sqrt{-a - 2} in (0, sqrt{1.5}] approx (0, 1.2247] ). Similarly, the solutions ( x = pm sqrt{-a - 2} ) lie within the interval.Therefore, the solutions for each ( a ) are:- For ( a in [0.25, 1) ): ( x = pm sqrt{2 - 2a} )- For ( a in [-3.5, -2) ): ( x = pm sqrt{-a - 2} )But the problem states "two distinct solutions on the interval". Since both cases give two solutions, which are distinct (positive and negative roots), as long as ( sqrt{2 - 2a} ) and ( sqrt{-a - 2} ) are non-zero and distinct.Wait, but when ( a = 0.25 ), ( x = pm sqrt{2 - 0.5} = pm sqrt{1.5} approx pm 1.2247 ). The lower bound of the interval is ( -sqrt{6}/2 approx -1.2247 ), so ( x = -sqrt{1.5} ) is exactly at the lower bound, which is included in the interval. Similarly, ( x = sqrt{1.5} ) is within the upper bound ( sqrt{2} approx 1.414 ).For ( a ) approaching 1 from below, ( x = pm sqrt{2 - 2a} ) approaches ( x = 0 ), so the solutions converge towards 0, but remain distinct.Similarly, for ( a in [-3.5, -2) ), ( sqrt{-a - 2} ) ranges from ( sqrt{1.5} approx 1.2247 ) (when ( a = -3.5 )) down to ( sqrt{0} = 0 ) (when ( a = -2 )). But since ( a ) approaches -2 from the left, ( sqrt{-a - 2} ) approaches ( sqrt{0} ), so the solutions approach ( x = 0 ), but remain distinct.However, when ( a = -3.5 ), ( x = pm sqrt{-(-3.5) - 2} = pm sqrt{1.5} approx pm 1.2247 ), which is again at the lower bound, but since it's included, it's acceptable.Therefore, both intervals for ( a ) produce two distinct solutions within the given interval.Now, checking if there's any overlap or other cases. For example, when ( a = 4 ), the quadratic equation has a repeated root ( y = 7 ), which is outside ( [-1, 1] ), so no solutions. When ( a ) is outside both intervals, the roots are either both outside ( [-1, 1] ) or one is in ( [-1, 1] ) but not in the required subinterval.Therefore, the final answer should be ( a in [-3.5, -2) cup [0.25, 1) ), and for each ( a ) in these intervals, the solutions are ( pm sqrt{2 - 2a} ) and ( pm sqrt{-a - 2} ) respectively.But let me check for ( a = -3.5 ):( a = -3.5 ):Equation: ( x = pm sqrt{-(-3.5) - 2} = pm sqrt{3.5 - 2} = pm sqrt{1.5} approx pm 1.2247 ), which is exactly the endpoints ( pm sqrt{6}/2 ), included in the interval.For ( a = -2 ):But ( a in [-3.5, -2) ), so ( a = -2 ) is not included. If ( a = -2 ), then ( y = a + 3 = 1 ), which gives ( x = 0 ), only one solution.Similarly, for ( a = 0.25 ):( x = pm sqrt{2 - 2*0.25} = pm sqrt{1.5} approx pm 1.2247 ), included in the interval.For ( a = 1 ):Not included, but approaching 1, ( x ) approaches 0.Therefore, the intervals for ( a ) are correctly identified as ( a in [-3.5, -2) cup [0.25, 1) ).Final AnswerThe values of ( a ) are ( boxed{[-3.5, -2) cup [0.25, 1)} ). For each ( a ) in ( [-3.5, -2) ), the solutions are ( pm sqrt{-a - 2} ), and for each ( a ) in ( [0.25, 1) ), the solutions are ( pm sqrt{2 - 2a} ).</think>To find all values of the parameter ( a ) for which the given equation has two distinct solutions on the interval (left[-frac{sqrt{6}}{2}, sqrt{2}right]), we start by simplifying the equation using the function ( f(t) = t^7 + t ), which is strictly increasing. This allows us to equate the arguments of the function, leading to the quadratic equation in ( y = 1 - x^2 ):[y^2 - (3a + 2)y + 2a^2 + 5a - 3 = 0]The discriminant of this quadratic equation is ( D = (a - 4)^2 ), giving roots ( y_1 = 2a - 1 ) and ( y_2 = a + 3 ). We need these roots to correspond to valid ( x ) values within the given interval. Key Steps:1. Substitution and Quadratic Roots: - The roots ( y_1 = 2a - 1 ) and ( y_2 = a + 3 ) must be analyzed for their contribution to valid ( x ) solutions. - For ( y = 1 - x^2 ), valid ( x ) solutions must lie within (left[-frac{sqrt{6}}{2}, sqrt{2}right]).2. Interval Analysis: - Case 1: ( y_1 = 2a - 1 in [-0.5, 1) ) - This gives ( 0.25 leq a < 1 ), leading to solutions ( x = pm sqrt{2 - 2a} ). - Case 2: ( y_2 = a + 3 in [-0.5, 1) ) - This gives ( -3.5 leq a < -2 ), leading to solutions ( x = pm sqrt{-a - 2} ).3. Verification: - For ( a in [0.25, 1) ), ( sqrt{2 - 2a} ) yields valid solutions within the interval. - For ( a in [-3.5, -2) ), ( sqrt{-a - 2} ) yields valid solutions within the interval. Conclusion:The values of ( a ) for which the equation has two distinct solutions on the interval (left[-frac{sqrt{6}}{2}, sqrt{2}right]) are:- For ( a in [0.25, 1) ), the solutions are ( pm sqrt{2 - 2a} ).- For ( a in [-3.5, -2) ), the solutions are ( pm sqrt{-a - 2} ).Thus, the final answer is:[boxed{[-3.5, -2) cup [0.25, 1)}]For each ( a ) in ( [-3.5, -2) ), the solutions are ( pm sqrt{-a - 2} ), and for each ( a ) in ( [0.25, 1) ), the solutions are ( pm sqrt{2 - 2a} ).]

🔑:Answer: a in[0.25 ; 1), x_{1}=sqrt{2-2 a}, x_{2}=-sqrt{2-2 a}.a in[-3.5 ;-2), x_{1}=sqrt{-a-2}, x_{2}=-sqrt{-a-2}

🧠:[<think>Okay, let's tackle this probability problem. So, the question is about identifying which pair of events are mutually exclusive but not contradictory. The options are A, B, C, D. Hmm, first, I need to remember what mutually exclusive and contradictory mean in probability terms.Mutually exclusive events are those that cannot happen at the same time. In other words, their intersection is empty. So, if one event occurs, the other cannot. Contradictory events, I think, are events that are complements of each other. So, one must occur if the other doesn't, and vice versa. Therefore, contradictory events are not only mutually exclusive but also cover the entire sample space between them. So, if two events are mutually exclusive but not contradictory, they don't overlap, but their union doesn't cover all possible outcomes. Alright, let's look at each option one by one. The problem involves drawing two balls from a bag with 2 red and 2 black balls. So, the total number of balls is 4, and we draw two. Let me first figure out the sample space. The possible outcomes when drawing two balls would be combinations of red and black. The possible pairs are:1. Both red (RR)2. One red and one black (RB or BR)3. Both black (BB)But since we're dealing with combinations, the order doesn't matter. So, the sample space has C(4,2) = 6 possible outcomes. Let me list them out explicitly. The balls can be labeled as R1, R2, B1, B2. The possible pairs are:- R1 & R2- R1 & B1- R1 & B2- R2 & B1- R2 & B2- B1 & B2So, 6 outcomes. Now, let's categorize these into the events mentioned in each option.Option A: "At least one red ball" and "Both are black balls"Let's parse these. "At least one red ball" would include all outcomes where there's one or two red balls. So, that would be RR, RB (in any combination). Looking at the list:- RR: 1 outcome (R1 & R2)- RB: 4 outcomes (R1 & B1, R1 & B2, R2 & B1, R2 & B2)So, "At least one red" has 5 outcomes. "Both are black balls" is just BB, which is 1 outcome (B1 & B2). Now, are these two events mutually exclusive? Well, if you have at least one red ball, you can't have both black balls. Similarly, if you have both black balls, you can't have any red balls. So, yes, they are mutually exclusive because their intersection is empty. Now, are they contradictory? For that, their union should cover the entire sample space. Let's check. "At least one red" plus "both black" – does that cover all possibilities? The total outcomes are 6. "At least one red" has 5, "both black" has 1. 5 + 1 = 6. So their union is the entire sample space. That means they are contradictory events. But the question is asking for mutually exclusive but not contradictory. So, A is contradictory, hence not the answer.Wait, but let me double-check. If two events are mutually exclusive and their union is the entire sample space, they are complementary (contradictory). So, in this case, since "at least one red" and "both black" are mutually exclusive and their union is everything, they are contradictory. Therefore, A is out.Option B: "At least one black ball" and "Both are black balls"Similarly, "At least one black" would include outcomes with one or two black balls. That would be BB, BR. So:- BB: 1 outcome- BR: 4 outcomes (same as RB in the previous case)So, "At least one black" has 5 outcomes. "Both are black" is 1 outcome. Are these mutually exclusive? If you have both black, you certainly have at least one black, so their intersection is the "both black" event. Therefore, they are not mutually exclusive. So, B is not mutually exclusive, hence not the answer.Option C: "At least one black ball" and "At least one red ball"So, "At least one black" and "At least one red". Let's see. If you have at least one black and at least one red, that's the intersection of the two events. Wait, but are these events mutually exclusive? If they are mutually exclusive, they cannot occur together. Let's check. The event "at least one black" includes all outcomes except both red. "At least one red" includes all outcomes except both black. So, their intersection is outcomes with at least one black and at least one red, which is the two BR outcomes. Wait, no. Wait, actually, in our sample space, there's 4 outcomes with one red and one black. So, if we have "at least one black" and "at least one red", their intersection is the set of outcomes where there's one black and one red. Therefore, they are not mutually exclusive. So, if you draw one black and one red, both events occur. Therefore, these events are not mutually exclusive, so C is not the answer.Option D: "Exactly one black ball" and "Exactly two black balls"So, "Exactly one black" means one black and one red. "Exactly two black" means both black. Let's see. Are these mutually exclusive? If you have exactly one black, you can't have exactly two black, and vice versa. So, yes, they are mutually exclusive. Now, are they contradictory? For that, their union should cover all possibilities. Let's check. The union would be "exactly one black or exactly two black". What's left out? The case of exactly zero black balls, which is both red. So, their union does not cover the entire sample space, because there's also the possibility of both red. Therefore, they are mutually exclusive but not contradictory. So, D is the answer.Let me verify once more. The events in D are:- Exactly one black: which is 4 outcomes (all the BR combinations)- Exactly two black: 1 outcome (BB)So, total union is 5 outcomes. The remaining outcome is RR. Therefore, their union is not the entire sample space, so they are not contradictory. But they are mutually exclusive. Therefore, D is the correct answer.So, the answer should be D.Final Answerboxed{D}</think>To determine which pair of events are mutually exclusive but not contradictory, we need to analyze each option given the problem of drawing two balls from a bag containing 2 red and 2 black balls.Key Definitions:- Mutually Exclusive Events: Events that cannot occur simultaneously (their intersection is empty).- Contradictory Events: Events that are complements of each other (mutually exclusive and their union covers the entire sample space).Sample Space Analysis:The possible outcomes when drawing two balls are:1. Both red (RR)2. One red and one black (RB or BR)3. Both black (BB)The sample space has 6 outcomes: RR, R1B1, R1B2, R2B1, R2B2, BB.Option Analysis:- Option A: "At least one red ball" and "Both are black balls" - "At least one red" includes RR, RB, BR (5 outcomes). - "Both are black" includes BB (1 outcome). - These events are mutually exclusive but their union covers the entire sample space, making them contradictory. Not the answer.- Option B: "At least one black ball" and "Both are black balls" - "At least one black" includes BB, RB, BR (5 outcomes). - "Both are black" includes BB (1 outcome). - These events are not mutually exclusive (they share BB). Not the answer.- Option C: "At least one black ball" and "At least one red ball" - "At least one black" includes BB, RB, BR (5 outcomes). - "At least one red" includes RR, RB, BR (5 outcomes). - Their intersection is RB, BR (4 outcomes), so they are not mutually exclusive. Not the answer.- Option D: "Exactly one black ball" and "Exactly two black balls" - "Exactly one black" includes RB, BR (4 outcomes). - "Exactly two black" includes BB (1 outcome). - These events are mutually exclusive. Their union does not cover the entire sample space (excludes RR), so they are not contradictory. Correct answer.Final Answer:boxed{D},<think>Okay, so I need to figure out which of the options A, B, C, or D are mutually exclusive but not contradictory events when drawing two balls from a bag with 2 red and 2 black balls. Let me start by recalling what mutually exclusive events are. If two events are mutually exclusive, it means they cannot happen at the same time. In other words, their intersection is empty. But they are not contradictory, which I think means that their union doesn't necessarily cover the entire sample space. Wait, no, maybe contradictory means that one is the complement of the other? Hmm, actually, let me make sure. If events are contradictory, they are mutually exclusive and their union is the entire sample space. So, if the problem says "mutually exclusive but not contradictory," that means they can't occur together, but their union doesn't cover all possible outcomes. So, they are exclusive but not exhaustive.First, let me figure out the sample space for drawing two balls from the bag. There are 2 red (R) and 2 black (B) balls. The possible outcomes when drawing two balls are:1. First red, second red (RR)2. First red, second black (RB)3. First black, second red (BR)4. First black, second black (BB)But since the order might not matter here? Wait, when we draw two balls, the problem might consider combinations rather than permutations. Let me check. For example, "Both are black balls" would be the combination of two black balls, regardless of order. Similarly, "At least one red ball" would include all combinations where there's one or two red balls. So, maybe we need to consider the sample space as combinations.So the combinations are:- RR: 1 way (since there are two red balls)- RB: 1 way (one red and one black)- BB: 1 way (two black balls)Wait, but actually, the number of possible combinations when drawing two balls from four is C(4,2)=6. But since there are two red and two black, some combinations are duplicates. Let me list all possible unordered pairs:1. R1 and R22. R1 and B13. R1 and B24. R2 and B15. R2 and B26. B1 and B2So, in terms of color combinations:- RR: 1 (R1 and R2)- RB: 4 (R1-B1, R1-B2, R2-B1, R2-B2)- BB: 1 (B1 and B2)But wait, that's 6 outcomes. So the sample space has six outcomes. But when considering color, some are equivalent. But maybe in probability terms, each of these six outcomes is equally likely. So, each has a probability of 1/6.But let's confirm that. When you draw two balls without replacement, the number of possible outcomes is indeed C(4,2)=6. Each outcome is equally likely if the balls are identical except for color. But since the balls are distinct (they are labeled R1, R2, B1, B2), each pair is equally likely. So yes, each of the six pairs has probability 1/6.But when we talk about events like "At least one red ball," that would include all pairs that have at least one red ball. So, that would be RR (1 outcome) and RB (4 outcomes), totaling 5 outcomes. Therefore, the probability of "At least one red ball" is 5/6.Similarly, "Both are black balls" is just BB, which is 1 outcome, so 1/6.So, let's analyze each option:Option A: "At least one red ball" and "Both are black balls"Are these mutually exclusive? If one event is "at least one red," which includes RR and RB, and the other is "both black," which is BB. Since BB does not overlap with either RR or RB (since they have different colors), they can't occur at the same time. So they are mutually exclusive. Now, are they contradictory? That would require their union to be the entire sample space. The union of "at least one red" and "both black" would cover all outcomes except... Wait, "at least one red" includes all outcomes with at least one red, which is 5 outcomes, and "both black" is 1 outcome. So together, they cover all 6 outcomes. Wait, but that would make them exhaustive, hence contradictory? Wait, but I thought contradictory meant that they are complements? So if two events are mutually exclusive and their union is the entire sample space, they are complementary, hence contradictory. So in this case, since their union is the entire sample space, they are contradictory. But the question says "mutually exclusive but not contradictory." So Option A is mutually exclusive but contradictory, so it's not the answer.Option B: "At least one black ball" and "Both are black balls"Similarly, "At least one black ball" would include RB and BB. "Both are black balls" is BB. Are these mutually exclusive? Well, if an outcome is in both events, it has to be in BB. But "at least one black ball" includes BB, so their intersection is BB. Therefore, they are not mutually exclusive because there is an overlap (BB). So Option B is not mutually exclusive, hence not the answer.Option C: "At least one black ball" and "At least one red ball"Let's see. "At least one black" includes RB and BB. "At least one red" includes RB and RR. Their intersection is RB. So they share the RB outcomes. Therefore, they are not mutually exclusive. So Option C is not the answer.Option D: "Exactly one black ball" and "Exactly two black balls""Exactly one black ball" corresponds to RB (since in two draws, exactly one black and one red). "Exactly two black balls" is BB. These two events have no overlap, since you can't have exactly one and exactly two black balls in the same outcome. Therefore, they are mutually exclusive. Now, are they contradictory? Their union would be RB and BB. But the entire sample space is RR, RB, BB. So their union does not include RR. Therefore, their union is not the entire sample space, so they are mutually exclusive but not contradictory. Hence, Option D is the correct answer.Wait, let me verify once more. The sample space has three types of outcomes: RR, RB (which has four instances), and BB. But when considering the events:- "Exactly one black ball" is equivalent to "exactly one red ball," which is the same as one of each color. So that's four outcomes (RB1, RB2, RB3, RB4 if we label the balls, but in terms of color, it's just the combination of one red and one black).- "Exactly two black balls" is BB, one outcome.So the union of these two events is RB and BB. The remaining outcome is RR. Therefore, their union is missing RR, so they are not exhaustive. Thus, they are mutually exclusive (no overlap) but not contradictory (since they don't cover the entire sample space). Therefore, Option D is correct.Just to recap:A: Mutually exclusive but contradictory (since their union is everything) - not the answer.B: Not mutually exclusive (they share BB) - not the answer.C: Not mutually exclusive (they share RB) - not the answer.D: Mutually exclusive and not contradictory (since RR is left out) - correct.Yes, that makes sense.Final Answerboxed{D}</think>To determine which events are mutually exclusive but not contradictory, we first analyze the sample space of drawing two balls from a bag containing 2 red and 2 black balls. The possible outcomes are:1. RR (both red)2. RB (one red and one black)3. BB (both black)Each outcome is equally likely. We then evaluate each option:- Option A: "At least one red ball" (includes RR and RB) and "Both are black balls" (BB). These are mutually exclusive but their union covers the entire sample space, making them contradictory. Hence, not the answer.- Option B: "At least one black ball" (includes RB and BB) and "Both are black balls" (BB). These are not mutually exclusive as they share BB. Hence, not the answer.- Option C: "At least one black ball" (includes RB and BB) and "At least one red ball" (includes RR and RB). These are not mutually exclusive as they share RB. Hence, not the answer.- Option D: "Exactly one black ball" (RB) and "Exactly two black balls" (BB). These are mutually exclusive as they have no overlap. Their union (RB and BB) does not cover the entire sample space (leaves out RR), so they are not contradictory. Hence, this is the correct answer.boxed{D}]

🔑:Answer: For option A: The event "At least one red ball" and the event "Both are black balls" are contradictory events, therefore A is incorrect.For option B: The event "At least one black ball" and the event "Both are black balls" can occur simultaneously, such as one red ball and one black ball, therefore B is incorrect.For option C: The event "At least one black ball" and the event "At least one red ball" can occur simultaneously, such as one red ball and one black ball, therefore C is incorrect.For option D: The event "Exactly one black ball" and "Exactly two black balls" cannot occur at the same time, therefore these two events are mutually exclusive.Additionally, from drawing two balls from a bag containing 2 red balls and 2 black balls, all possible events are "Exactly one black ball", "Exactly two black balls", and "Exactly two red balls". Thus, these two events are not contradictory events.Therefore, the correct answer is boxed{text{D}}.Analysis: List each event's basic outcomes and combine the definitions of mutually exclusive and contradictory events to verify each option accordingly.

🧠:[<think>Okay, let's see. I need to figure out the relationship between propositions p and q. The question is asking whether p is a sufficient condition, necessary condition, both, or neither for q. Hmm. Let me break it down step by step.First, let's recall what the propositions are:- Proposition p: The dot product of vectors a and b is equal to negative the square of vector a. So mathematically, that's (overrightarrow{a} cdot overrightarrow{b} = -overrightarrow{a}^2).- Proposition q: Vector a is equal to negative vector b. So, (overrightarrow{a} = -overrightarrow{b}).The question is asking whether p is a sufficient condition, necessary condition, both, or neither for q. So, in logical terms, does p imply q (sufficient), does q imply p (necessary), both, or neither?Let me start by analyzing if p implies q. Suppose p is true; does that necessarily mean q is true?Given (overrightarrow{a} cdot overrightarrow{b} = -overrightarrow{a}^2).Let me write this out in terms of vector magnitudes and the angle between them. The dot product formula is (overrightarrow{a} cdot overrightarrow{b} = |overrightarrow{a}||overrightarrow{b}| cos theta), where θ is the angle between the two vectors.So according to p:( |overrightarrow{a}||overrightarrow{b}| cos theta = -|overrightarrow{a}|^2 ).Assuming vector a is not the zero vector (since if a is the zero vector, both sides are zero, but then vector b could be anything, which complicates things), we can divide both sides by |a|:( |overrightarrow{b}| cos theta = -|overrightarrow{a}| ).So, ( cos theta = -|overrightarrow{a}| / |overrightarrow{b}| ).Now, the cosine of an angle can range between -1 and 1. So, the right-hand side must satisfy:( -1 leq -|overrightarrow{a}| / |overrightarrow{b}| leq 1 ).Multiplying all parts by -1 (which reverses inequalities):( 1 geq |overrightarrow{a}| / |overrightarrow{b}| geq -1 ).But since magnitudes are non-negative, the lower bound here is actually redundant. The main condition is that ( |overrightarrow{a}| / |overrightarrow{b}| leq 1 ), so ( |overrightarrow{b}| geq |overrightarrow{a}| ).Therefore, for p to hold, the magnitude of b must be at least as large as the magnitude of a, and the angle between them must satisfy ( cos theta = -|overrightarrow{a}| / |overrightarrow{b}| ).But does this necessarily mean that ( overrightarrow{a} = -overrightarrow{b} )?Well, if ( overrightarrow{a} = -overrightarrow{b} ), then their magnitudes must be equal, right? Because the negative sign just flips the direction. So |a| = |b|. Then, in that case, the angle θ between them is 180 degrees, so cos θ = -1. Plugging into the equation for p:( |overrightarrow{a}||overrightarrow{b}| cos theta = |a||a| (-1) = -|a|² ), which matches p. So q implies p. Therefore, q is a sufficient condition for p, which would make p a necessary condition for q. Wait, maybe I need to be careful here.Wait, if q implies p, then p is a necessary condition for q. Because if q is true, then p must also be true. So, p is necessary for q. But is p sufficient for q? That is, does p being true imply that q is true?Suppose p is true. Then we have ( |overrightarrow{a}||overrightarrow{b}| cos theta = -|overrightarrow{a}|² ).If we take vector a and vector b such that |b| = |a|, then cos θ must be -1, which would mean θ = 180 degrees, so vectors a and b are in opposite directions. Hence, if |b| = |a| and θ = 180°, then (overrightarrow{a} = -overrightarrow{b}). So in that case, p implies q.But wait, but the condition p only requires that |b| >= |a| (as we saw earlier from ( |overrightarrow{b}| geq |overrightarrow{a}| )) and that ( cos theta = -|overrightarrow{a}| / |overrightarrow{b}| ). So even if |b| > |a|, as long as the angle is such that cos θ = -|a| / |b|, then p holds. But in that case, θ is not 180 degrees, unless |a| = |b|.For example, suppose |a| = 1, |b| = 2. Then cos θ = -1/2, which would mean θ = 120 degrees. So vectors a and b are not in opposite directions, just at 120 degrees. Then p would hold, since ( 1*2*(-1/2) = -1 ), which equals -|a|². But in this case, q is not true because vector a is not equal to -vector b. So in this case, p is true but q is false. Therefore, p does not necessarily imply q. Therefore, p is not a sufficient condition for q.But earlier, we saw that if q is true, then p must be true. So q implies p. Therefore, p is a necessary condition for q. But since p can be true without q being true, p is not sufficient. Therefore, p is a necessary but not sufficient condition for q. So the answer should be B: Necessary but not sufficient condition.Wait, but let me check again. Let me test with a specific example.Case 1: Let vector a = (1, 0), vector b = (-1, 0). Then, q is true because a = -b. Then, p: a · b = (1)(-1) + (0)(0) = -1. -a² = -(1² + 0²) = -1. So p holds. So q implies p.Case 2: Let vector a = (1, 0), vector b = ( -0.5, √3/2 ). Let's compute |b|: sqrt( (-0.5)^2 + (√3/2)^2 ) = sqrt(0.25 + 0.75) = sqrt(1) = 1. So |a| = 1, |b| = 1. Wait, but then if I compute a · b: (1)(-0.5) + 0*(√3/2) = -0.5. But -a² = -1. So in this case, a · b = -0.5 ≠ -1. So p is not satisfied. Wait, but in this case, θ is 120 degrees. Wait, if |a| = |b| = 1, θ = 120°, then a · b = |a||b|cos θ = 1*1*(-0.5) = -0.5. So p would require a · b = -1. So this doesn't satisfy p. So to satisfy p when |a| = |b|, θ must be 180°, right?Wait, maybe my previous example where |b| > |a| is better.Let me take vector a = (1, 0), vector b = (-2, 0). Then |a| = 1, |b| = 2. Then a · b = (1)(-2) + 0*0 = -2. -a² = -1. So here, a · b = -2 ≠ -1. Therefore, p is not satisfied. Wait, but according to our earlier analysis, if |b| = 2, |a| = 1, then cos θ = -1/2, so θ = 120 degrees, but vector b in this case is (-2, 0), which is colinear with a but with magnitude 2, so the angle θ is 180 degrees. Wait, no. Wait, if vector a is (1,0) and vector b is (-2,0), they are colinear but opposite in direction, so θ = 180 degrees. So cos θ = -1. Then a · b = |a||b|cos θ = 1*2*(-1) = -2. So p would require this to be -1. So p is not satisfied here.Wait, so perhaps my earlier reasoning was flawed. Let me think again.Wait, according to p: a · b = -a². Let's rearrange that.a · b + a · a = 0. Which is equivalent to a · (a + b) = 0.So the vector a + b is orthogonal to vector a.So, proposition p is equivalent to vector (a + b) being perpendicular to vector a.Now, proposition q is a = -b, which would mean a + b = 0 vector, which is trivially orthogonal to a, since the dot product of any vector with the zero vector is zero. So q implies p.But p can hold even if a + b is a non-zero vector orthogonal to a. For example, take a = (1, 0) and a + b = (0, 1), so b = (-1, 1). Then a · (a + b) = (1, 0) · (0, 1) = 0, so p is satisfied. But a is not equal to -b, since -b = (1, -1) ≠ a = (1,0). Therefore, p can hold without q being true.Therefore, p is a necessary condition for q (since q implies p) but p is not sufficient for q (since p can hold without q). Therefore, the answer should be B: Necessary but not sufficient condition.Wait, but let's confirm this with another example.Take a = (1, 0), and b = (-1, 1). Then a + b = (0,1). Then, a · (a + b) = 1*0 + 0*1 = 0, so p holds. But a ≠ -b. So here, p is true but q is false. Therefore, p is not sufficient.Conversely, if q is true, then a = -b, so a + b = 0, so a · (a + b) = a · 0 = 0, so p is true. Therefore, q implies p. Hence, p is necessary for q.Therefore, the correct answer is B.Final Answerboxed{B}</think>Given vectors (overrightarrow{a}) and (overrightarrow{b}), we need to determine the relationship between propositions (p) and (q):- Proposition (p): (overrightarrow{a} cdot overrightarrow{b} = -overrightarrow{a}^2)- Proposition (q): (overrightarrow{a} = -overrightarrow{b})To analyze if (p) implies (q) (sufficient condition) or if (q) implies (p) (necessary condition), we start by expressing the dot product in terms of magnitudes and the angle between the vectors. For (p) to hold:[overrightarrow{a} cdot overrightarrow{b} = |overrightarrow{a}||overrightarrow{b}| cos theta = -|overrightarrow{a}|^2]Assuming (overrightarrow{a} neq overrightarrow{0}), we divide both sides by (|overrightarrow{a}|):[|overrightarrow{b}| cos theta = -|overrightarrow{a}|]This implies (cos theta = -|overrightarrow{a}| / |overrightarrow{b}|). For (cos theta) to be valid, (|overrightarrow{b}| geq |overrightarrow{a}|).If (q) is true ((overrightarrow{a} = -overrightarrow{b})), then (|overrightarrow{a}| = |overrightarrow{b}|) and (theta = 180^circ), making (cos theta = -1). This satisfies (p), so (q) implies (p), making (p) a necessary condition for (q).However, (p) can hold without (q) being true. For example, if (overrightarrow{a} = (1, 0)) and (overrightarrow{b} = (-1, 1)), then (overrightarrow{a} cdot overrightarrow{b} = -1) and (-|overrightarrow{a}|^2 = -1), satisfying (p). But (overrightarrow{a} neq -overrightarrow{b}). Thus, (p) does not imply (q), so (p) is not sufficient for (q).Therefore, (p) is a necessary but not sufficient condition for (q).[boxed{B}],<think>Okay, so I need to figure out the relationship between proposition p and proposition q. Let me start by understanding what each proposition means.Proposition p is that the dot product of vectors a and b is equal to negative the square of vector a. In mathematical terms, that's:[overrightarrow{a} cdot overrightarrow{b} = -overrightarrow{a}^2]And proposition q is that vector a is equal to negative vector b:[overrightarrow{a} = -overrightarrow{b}]The question is asking whether p is a sufficient condition, necessary condition, both, or neither for q. So, I need to check if p implies q (sufficient), if q implies p (necessary), both, or neither.First, let's recall some vector properties. The dot product of two vectors a and b is given by:[overrightarrow{a} cdot overrightarrow{b} = |overrightarrow{a}||overrightarrow{b}| cos theta]where θ is the angle between them. The square of vector a, which is (overrightarrow{a}^2), is the dot product of a with itself, so that's:[overrightarrow{a} cdot overrightarrow{a} = |overrightarrow{a}|^2]So, proposition p can be rewritten as:[|overrightarrow{a}||overrightarrow{b}| cos theta = -|overrightarrow{a}|^2]Let me simplify this equation. If I divide both sides by (|overrightarrow{a}|) (assuming a is not the zero vector), we get:[|overrightarrow{b}| cos theta = -|overrightarrow{a}|]So,[cos theta = -frac{|overrightarrow{a}|}{|overrightarrow{b}|}]Hmm. That tells us something about the angle between a and b. Since cosine is between -1 and 1, the right-hand side must also lie in that interval. Therefore,[-1 leq -frac{|overrightarrow{a}|}{|overrightarrow{b}|} leq 1]Multiplying all parts by -1 (which reverses inequalities):[1 geq frac{|overrightarrow{a}|}{|overrightarrow{b}|} geq -1]But since magnitudes are non-negative, the lower bound is actually 0. So, (frac{|overrightarrow{a}|}{|overrightarrow{b}|}) must be between 0 and 1. Therefore,[0 leq frac{|overrightarrow{a}|}{|overrightarrow{b}|} leq 1 implies |overrightarrow{a}| leq |overrightarrow{b}|]So, for the equation to hold, the magnitude of a must be less than or equal to the magnitude of b.Now, let's consider proposition q: (overrightarrow{a} = -overrightarrow{b}). If this is true, then let's check if p holds. If a = -b, then their dot product is:[overrightarrow{a} cdot overrightarrow{b} = overrightarrow{a} cdot (-overrightarrow{a}) = - (overrightarrow{a} cdot overrightarrow{a}) = -|overrightarrow{a}|^2]Which is exactly proposition p. So, q implies p. Therefore, p is a necessary condition for q. Because if q is true, then p must also be true. So, p is necessary for q.But is p sufficient for q? In other words, does p being true necessarily mean that q is true? Let's check.Suppose that (overrightarrow{a} cdot overrightarrow{b} = -|overrightarrow{a}|^2). Does this force a to be equal to -b?Let me think of a counterexample. Let’s take vectors a and b such that a is not equal to -b, but their dot product still equals -|a|².Let’s choose specific vectors. Let’s let a be a non-zero vector. Let’s choose b such that it has a component in the direction of a, but maybe more components in other directions. Wait, but how to get the dot product equal to -|a|².Alternatively, consider a in 2D space. Let’s let a = (1, 0). Then |a|² = 1. Suppose b is such that a · b = -1. Let's find such a b.Let’s say b = (-1, y). Then, a · b = (1)(-1) + (0)(y) = -1. So, for any y, this holds. So, if b is (-1, y), then a · b = -1 = -|a|². Therefore, even if y ≠ 0, the dot product is -1, satisfying p. But in this case, a = (1, 0) and b = (-1, y) where y can be any real number. So, unless y is zero, a ≠ -b. For example, take y = 1: then b = (-1, 1), and -b = (1, -1) ≠ a. Therefore, even though p is satisfied, q is not. Therefore, p is not sufficient for q.Thus, p is a necessary but not sufficient condition for q. Therefore, the answer should be B.Wait, but let me verify again. If q implies p, then p is necessary for q. So, if q is true, p has to be true. But p can be true without q being true. As shown in the example above. So, yes, p is necessary but not sufficient. Therefore, option B.Final Answerboxed{B}</think>Given vectors (overrightarrow{a}) and (overrightarrow{b}), proposition (p) states that (overrightarrow{a} cdot overrightarrow{b} = -overrightarrow{a}^2), and proposition (q) states that (overrightarrow{a} = -overrightarrow{b}). We need to determine the relationship between (p) and (q).First, we note that if (q) is true (i.e., (overrightarrow{a} = -overrightarrow{b})), then the dot product (overrightarrow{a} cdot overrightarrow{b}) becomes:[overrightarrow{a} cdot (-overrightarrow{a}) = -(overrightarrow{a} cdot overrightarrow{a}) = -|overrightarrow{a}|^2]This confirms that (q) implies (p), making (p) a necessary condition for (q).Next, we check if (p) implies (q). Consider a counterexample where (overrightarrow{a} = (1, 0)) and (overrightarrow{b} = (-1, y)) for any (y neq 0). Here, the dot product is:[(1, 0) cdot (-1, y) = -1 + 0 = -1 = -|overrightarrow{a}|^2]This satisfies (p), but (overrightarrow{a} neq -overrightarrow{b}) since (-overrightarrow{b} = (1, -y) neq (1, 0)). Thus, (p) does not necessarily imply (q), showing that (p) is not a sufficient condition for (q).Therefore, (p) is a necessary but not sufficient condition for (q). The answer is (boxed{B}).]

🔑:1. We start with the given proposition p: overrightarrow{a} cdot overrightarrow{b} = -overrightarrow{a}^{2}.2. By the definition of the dot product, we can rewrite the equation as |overrightarrow{a}||overrightarrow{b}| cos = -overrightarrow{a}^{2}.3. This implies that overrightarrow{a} = overrightarrow{0} or |overrightarrow{b}| cos = -|overrightarrow{a}|.4. If overrightarrow{a} = -overrightarrow{b}, then overrightarrow{a} cdot overrightarrow{b} = -overrightarrow{a}^{2} holds true.5. However, if overrightarrow{a} cdot overrightarrow{b} = -overrightarrow{a}^{2}, it does not necessarily mean that overrightarrow{a} = -overrightarrow{b}.6. Therefore, proposition p is a necessary but not sufficient condition for proposition q.The answer is: boxed{B}

🧠:[<think>Okay, so I need to find the volume of a cone. The problem says that the total surface area is πS square units, and when you unfold the lateral surface (which I think is called the lateral surface area) into a plane, it forms a sector with a 60-degree angle. Hmm, let me try to visualize this. First, I remember that the volume of a cone is given by (1/3)πr²h, where r is the radius of the base and h is the height. So, to find the volume, I need to figure out both r and h. But how do I get those from the given information?The problem mentions the total surface area is πS. The total surface area of a cone is the sum of the base area and the lateral (side) surface area. The formula for total surface area is πr² + πrl, where l is the slant height. So, πr² + πrl = πS. If I divide both sides by π, that simplifies to r² + rl = S. So equation one is r² + rl = S.Now, the second part says that when the lateral surface is unfolded, it forms a sector with a 60-degree angle. When you unfold the lateral surface of a cone, you get a sector of a circle. The radius of this sector is the slant height of the cone, l, and the arc length of the sector corresponds to the circumference of the base of the cone. The circumference of the base of the cone is 2πr. When unfolded into a sector, the arc length of the sector is equal to that circumference. The sector has a radius of l and an angle of 60 degrees. But wait, sector arc length is given by (θ/360) * 2πL, where θ is the central angle and L is the radius of the sector. In this case, θ is 60 degrees, and L is the slant height l. So, the arc length of the sector is (60/360)*2πl = (1/6)*2πl = (πl)/3.But this arc length must equal the circumference of the base of the cone, which is 2πr. Therefore, (πl)/3 = 2πr. Let me solve this equation. Divide both sides by π: l/3 = 2r. Multiply both sides by 3: l = 6r. So, the slant height l is 6 times the radius r. That's a useful relationship.So now, from equation one: r² + rl = S. But l = 6r, so substituting that in: r² + r*(6r) = S. That simplifies to r² + 6r² = S, which is 7r² = S. Therefore, r² = S/7, so r = sqrt(S/7). But I need to be careful here; since radius can't be negative, we take the positive root. So r = √(S/7).Now, since l = 6r, l = 6√(S/7). But we also need the height h of the cone. In a right circular cone, the slant height l is related to the radius r and height h by the Pythagorean theorem: l² = r² + h². So, we can solve for h: h = sqrt(l² - r²).Plugging in the values of l and r:h = sqrt((6√(S/7))² - (√(S/7))²) = sqrt(36*(S/7) - (S/7)) = sqrt(35*(S/7)) = sqrt(5S). Wait, let's check that:First, calculate (6√(S/7))²: 6² is 36, (√(S/7))² is S/7. So, 36*(S/7) = 36S/7.Then, (√(S/7))² is S/7.So, l² - r² = 36S/7 - S/7 = 35S/7 = 5S. Therefore, h = sqrt(5S). Got it.So now, we have r = √(S/7) and h = √(5S). Now, plug these into the volume formula:Volume = (1/3)πr²h = (1/3)π*(S/7)*√(5S). Let me compute that step by step.First, r² is S/7. Then multiply by h, which is √(5S). So:(S/7) * √(5S) = (S/7) * (5S)^(1/2) = (S/7) * (5)^(1/2) * S^(1/2) = (5^(1/2)/7) * S^(3/2).Then multiply by (1/3)π:Volume = (1/3)π * (5^(1/2)/7) * S^(3/2) = (π √5 / 21) * S^(3/2).Alternatively, that can be written as (π √5 S^(3/2))/21.But perhaps we can rationalize or write it in a different form. Let's see.S^(3/2) is the same as S√S. So, Volume = (π √5 /21) * S√S = (π S√(5S))/21.But both forms are acceptable. However, to check if I did everything correctly, let me verify the steps again.Starting from the sector angle: 60 degrees. The arc length of the sector is (60/360)*2πl = (πl)/3. This equals the circumference of the base, 2πr. So πl/3 = 2πr => l = 6r. That seems right.Total surface area is πr² + πrl = πS. So, substitute l = 6r: πr² + πr*(6r) = π(r² + 6r²) = π7r² = πS => 7r² = S => r² = S/7. So, r = √(S/7). Correct.Then, l = 6r = 6√(S/7). Then, using Pythagoras: l² = r² + h² => h² = l² - r² = (36*(S/7)) - (S/7) = 35*(S/7) = 5S. So h = √(5S). Correct.Then, Volume = (1/3)πr²h = (1/3)π*(S/7)*√(5S). Yes. Multiply S/7 and √(5S):(S/7)*√(5S) = (S/7)*√5*√S = (S√S√5)/7 = (√5 S^(3/2))/7.Multiply by (1/3)π: Volume = (π √5 S^(3/2))/21. Which is the same as (π S √(5S))/21.So, that seems to be the answer. Let me check if there's a simpler way or if I made any mistakes.Wait, the problem states that the total surface area is πS. So, if S is a variable, maybe we need to write the answer in terms of S. But according to the problem, the answer should be in terms of S, so this expression is okay.Alternatively, perhaps I can write S^(3/2) as S√S, so the Volume is (π √5 S√S)/21 = (π S √(5S))/21. But both forms are equivalent.Alternatively, maybe factor out S:Volume = (π √5 /21) * S^(3/2). Depending on the preferred form, but both are correct.Alternatively, rationalizing the denominator? Not necessary here. Since the problem doesn't specify a particular form, either is acceptable. But perhaps the answer is expected in terms of S, so we can leave it as is.So, I think that's it. So the volume is (π √5 S^(3/2))/21. Which can also be written as (π S^(3/2) √5)/21. Either way, they are the same.Wait, just to check if all steps make sense. The unfolded lateral surface area is a sector with angle 60 degrees. The radius of the sector is the slant height l. The arc length of the sector is equal to the circumference of the base. So, (60/360)*2πl = 2πr. Therefore, l = 6r. Then total surface area πr² + πrl = πS. Then substituting l = 6r, leading to r = sqrt(S/7), and h = sqrt(5S). Then Volume = (1/3)πr²h = ... yes, all steps seem consistent.Therefore, the volume of the cone is (π √5 S^(3/2))/21.Final AnswerThe volume of the cone is boxed{dfrac{pi S sqrt{5S}}{21}}.</think>Given the total surface area of the cone is (pi S) square units, and the lateral surface forms a sector with a (60^circ) angle when unfolded, we need to determine the volume of the cone.1. Total Surface Area: The total surface area of the cone is given by (pi r^2 + pi r l = pi S). Dividing both sides by (pi), we get: [ r^2 + r l = S ]2. Unfolded Lateral Surface: The lateral surface forms a sector with a (60^circ) angle. The arc length of this sector is equal to the circumference of the base of the cone. The arc length of the sector is (frac{60}{360} times 2pi l = frac{pi l}{3}), which equals the circumference (2pi r). Solving for (l): [ frac{pi l}{3} = 2pi r implies l = 6r ]3. Substitute (l = 6r) into the Surface Area Equation: [ r^2 + r(6r) = S implies 7r^2 = S implies r^2 = frac{S}{7} implies r = sqrt{frac{S}{7}} ]4. Calculate Slant Height (l): [ l = 6r = 6sqrt{frac{S}{7}} ]5. Determine Height (h) using Pythagoras' Theorem: [ l^2 = r^2 + h^2 implies left(6sqrt{frac{S}{7}}right)^2 = left(sqrt{frac{S}{7}}right)^2 + h^2 implies frac{36S}{7} = frac{S}{7} + h^2 implies h^2 = 5S implies h = sqrt{5S} ]6. Calculate Volume: [ text{Volume} = frac{1}{3} pi r^2 h = frac{1}{3} pi left(frac{S}{7}right) sqrt{5S} = frac{pi S sqrt{5S}}{21} ]The volume of the cone is (boxed{dfrac{pi S sqrt{5S}}{21}}).,<think>Okay, so I have this problem here about a cone. The total surface area is given as πS square units, and when the lateral surface is unfolded into a plane, it forms a sector with a 60-degree angle. I need to find the volume of the cone. Hmm, let's break this down step by step.First, let me recall some formulas related to cones. The total surface area of a cone is the sum of the base area and the lateral (or curved) surface area. The formula for the total surface area is πr² + πrl, where r is the radius of the base and l is the slant height of the cone. So, according to the problem, πr² + πrl = πS. Maybe I can simplify that equation by dividing both sides by π, so it becomes r² + rl = S. That seems manageable.Next, the problem mentions that the lateral surface, when unfolded, forms a sector with a 60-degree angle. When you unfold the lateral surface of a cone, you get a sector of a circle. The radius of this sector is equal to the slant height l of the cone. The arc length of the sector corresponds to the circumference of the base of the cone. Wait, let me make sure. When you unfold a cone's lateral surface, you get a sector with radius equal to the slant height l. The arc length of this sector is equal to the circumference of the cone's base, which is 2πr. But the sector also has an angle θ (in this case, 60 degrees), and the arc length of a sector with angle θ (in radians) and radius l is θl. So, setting this equal to the base circumference: θl = 2πr. But θ is given in degrees here, 60 degrees. I need to convert that to radians because the formula for arc length uses radians. 60 degrees is π/3 radians. So, substituting into the equation: (π/3) * l = 2πr. Let's solve for one variable in terms of the other. Multiply both sides by 3: πl = 6πr, then divide both sides by π: l = 6r. So, the slant height l is 6 times the radius r. That's a useful relationship.Now, going back to the total surface area equation: r² + rl = S. Since l = 6r, substitute that in: r² + r*(6r) = S. Simplify that: r² + 6r² = S, so 7r² = S. Therefore, r² = S/7, and r = sqrt(S/7). Then, since l = 6r, l = 6*sqrt(S/7). But to find the volume of the cone, I need to know the height h. The volume of a cone is (1/3)πr²h. So, I need to find h in terms of S. From the Pythagorean theorem, in a right triangle formed by the radius r, height h, and slant height l, we have l² = r² + h². Therefore, h = sqrt(l² - r²). Let's substitute l and r with the expressions in terms of S.First, l = 6r, so l² = 36r². Then, h² = l² - r² = 36r² - r² = 35r². Therefore, h = sqrt(35r²) = r*sqrt(35). Since r = sqrt(S/7), substitute that in: h = sqrt(S/7) * sqrt(35) = sqrt((S/7)*35) = sqrt(5S). So, h = sqrt(5S).Now, plug r and h into the volume formula. The volume V is (1/3)πr²h. We know r² = S/7, and h = sqrt(5S). Therefore:V = (1/3)π*(S/7)*sqrt(5S) = (π S / 21) * sqrt(5S). Hmm, that simplifies to (π S * sqrt(5S)) / 21. Let's see if we can write that in a better form. sqrt(5S) is (5S)^(1/2), so multiplying by S gives (5^(1/2)) * S^(3/2). Therefore, V = (π * 5^(1/2) * S^(3/2)) / 21.Alternatively, we can factor out the S^(3/2): V = (π * sqrt(5) / 21) * S^(3/2). But maybe we can rationalize this or present it differently. Let me check the steps again to see if I made a mistake.Wait, let's verify the calculation for h. Let me recap:We had l = 6r.From Pythagoras, h = sqrt(l² - r²) = sqrt(36r² - r²) = sqrt(35r²) = r*sqrt(35). Then, since r = sqrt(S/7), h = sqrt(S/7)*sqrt(35). Let's compute that:sqrt(S/7) * sqrt(35) = sqrt( (S/7) * 35 ) = sqrt(5S). Yes, that's correct.So, h = sqrt(5S). Then, the volume is (1/3)πr²h = (1/3)π*(S/7)*sqrt(5S). Let's compute that step by step:Multiply (1/3) * (1/7) = 1/21. Then, S * sqrt(5S) = S * (5S)^(1/2) = 5^(1/2) * S^(3/2). So, Volume = π/(21) * 5^(1/2) * S^(3/2) = (π sqrt(5) / 21) * S^(3/2). Alternatively, writing S^(3/2) as S*sqrt(S), so Volume = (π sqrt(5) / 21) * S * sqrt(S) = (π sqrt(5) S^(3/2))/21. That seems as simplified as it gets.But let me check if there's a different way to express this. Maybe factor the sqrt(5) and sqrt(S) together. sqrt(5S) * S = S * sqrt(5S). Wait, but that's the same as before.Alternatively, express S^(3/2) as sqrt(S^3), but that might not be simpler. Alternatively, rationalize the denominator? Let's see:sqrt(5)/21 can be written as sqrt(5)/21, but I don't think that helps. Maybe factor out the constants:π * sqrt(5) / 21 is approximately π/(21) * 2.236, but exact form is better here. So, the exact volume is (π sqrt(5) S^(3/2))/21.But let me check if there's an error in the steps.Wait, the problem states that the total surface area is πS. So, when I divided by π, it became r² + rl = S. Then, with l = 6r, substituted to get 7r² = S, so r² = S/7. That seems correct.Then, l = 6r, so h = sqrt(35r²) = r sqrt(35). Then, since r = sqrt(S/7), substituting gives h = sqrt(S/7) * sqrt(35). Let's compute sqrt(S/7) * sqrt(35):sqrt(35) / sqrt(7) = sqrt(35/7) = sqrt(5). So, sqrt(S/7) * sqrt(35) = sqrt(S) * sqrt(5) = sqrt(5S). That's correct. So, h = sqrt(5S).So, Volume = (1/3)π*(S/7)*sqrt(5S). Let me compute this:(1/3) * (S/7) = S/(21). Then, multiply by sqrt(5S): S/(21) * sqrt(5S) = (S * sqrt(5S))/21. Since sqrt(5S) is sqrt(5)*sqrt(S), this becomes (S * sqrt(5) * sqrt(S))/21 = (sqrt(5) * S^(3/2))/21 * π. So, Volume = (π sqrt(5) S^(3/2))/21. Alternatively, S^(3/2) is S*sqrt(S), so we can write this as (π sqrt(5) S sqrt(S))/21. Either way, this seems to be the simplified form.But let me check if maybe the problem expects a rationalized form or a different expression. Let's see. Alternatively, since Volume is (1/3)πr²h, and we have expressions for r and h in terms of S, perhaps there's another way to compute it. But I think this is the correct answer.Wait, let me verify once again.Given:1. Total surface area: πr² + πrl = πS ⇒ r² + rl = S.2. When the lateral surface is unfolded, it's a sector with angle 60°, which is π/3 radians. The arc length of the sector is equal to the circumference of the base of the cone. So, arc length = 2πr. The length of the arc of the sector is also θ*l, where θ is in radians. So, (π/3)*l = 2πr ⇒ l = 6r. Correct.So, substituting l = 6r into the surface area equation: r² + 6r² = 7r² = S ⇒ r² = S/7 ⇒ r = sqrt(S/7). Then, slant height l = 6r = 6*sqrt(S/7). Then, height h = sqrt(l² - r²) = sqrt(36r² - r²) = sqrt(35r²) = r*sqrt(35). Substituting r: h = sqrt(S/7)*sqrt(35) = sqrt(S*35/7) = sqrt(5S). Correct.Thus, Volume = (1/3)πr²h = (1/3)π*(S/7)*sqrt(5S) = π/(21) * S * sqrt(5S) = π sqrt(5) S^(3/2)/21. Yes, that seems right.But maybe the answer can be written as (π S^(3/2) sqrt(5))/21, which is the same thing. So, unless there's a simplification I'm missing, this is the answer.Wait, maybe the problem expects the answer in terms of S without fractional exponents? Let's see. S^(3/2) is the same as S*sqrt(S). So, writing Volume = (π sqrt(5) S sqrt(S))/21. That's acceptable.Alternatively, to make it look neater, we can write it as (π S sqrt(5S))/21. But regardless, the answer is in terms of S raised to the 3/2 power multiplied by sqrt(5) and divided by 21.Alternatively, let's see if this can be written as (π √5 S^{3/2}) / 21, which is equivalent.Alternatively, if the problem expects a numerical coefficient multiplied by S^{3/2}, then that's the answer.Alternatively, perhaps rationalizing the denominator, but in this case, there's no denominator with a radical. The denominator is 21, and the numerator has sqrt(5). So, that's already rationalized.Alternatively, maybe the problem expects the answer in some other form. Let me check.Wait, perhaps I made a mistake in assuming the unfolded sector angle is 60 degrees. Wait, when you unfold the lateral surface, the angle of the sector is 60 degrees, which is θ. Then, the relationship between θ, l, and r is θ = (arc length)/l. But arc length is 2πr, so θ (in radians) = 2πr / l. But θ is given as 60 degrees, which is π/3 radians. Therefore, π/3 = 2πr / l. Solving for l: l = 2πr / (π/3) = 6r. Which is exactly what I did before. So, that's correct.Therefore, the steps are correct, and the answer is (π √5 S^{3/2}) / 21.Alternatively, maybe I can write this as (π S^{3/2} √5 ) / 21. Either way, the order of the terms doesn't matter. So, the answer is π times the square root of 5 times S to the power of 3/2, divided by 21.But let me check if the problem wants the answer in terms of S, perhaps using fractional exponents or roots. Since S is given as a parameter, expressing the Volume in terms of S is appropriate.Alternatively, maybe there's a miscalculation when combining the exponents. Let's see:Volume = (1/3)πr²h. We have r² = S/7, h = sqrt(5S). So, Volume = (1/3)π*(S/7)*sqrt(5S). Let's express sqrt(5S) as (5S)^(1/2). Then, Volume = (1/3)*(1/7)*π*S*(5S)^(1/2) = (1/21)π*S*(5)^(1/2)*S^(1/2) = (1/21)π*5^(1/2)*S^(3/2). Which is exactly the same as before.So, the answer is indeed (π √5 S^{3/2}) / 21.Therefore, the volume of the cone is (π √5 S^{3/2}) divided by 21. So, in boxed form, that would be:boxed{dfrac{pi sqrt{5} S^{3/2}}{21}}Final AnswerThe volume of the cone is boxed{dfrac{pi sqrt{5} S^{3/2}}{21}}.</think>The total surface area of the cone is given as (pi S) square units. The lateral surface of the cone, when unfolded, forms a sector with an angle of (60^circ). We need to determine the volume of the cone.1. Total Surface Area: [ pi r^2 + pi r l = pi S ] Dividing by (pi): [ r^2 + r l = S ]2. Unfolded Lateral Surface: The sector angle is (60^circ) (which is (frac{pi}{3}) radians). The arc length of the sector is equal to the circumference of the base of the cone: [ frac{pi}{3} l = 2 pi r ] Solving for (l): [ l = 6r ]3. Substitute (l = 6r) into the surface area equation: [ r^2 + r(6r) = S implies 7r^2 = S implies r^2 = frac{S}{7} implies r = sqrt{frac{S}{7}} ]4. Calculate the height (h): Using the Pythagorean theorem: [ l^2 = r^2 + h^2 implies (6r)^2 = r^2 + h^2 implies 36r^2 = r^2 + h^2 implies h^2 = 35r^2 implies h = rsqrt{35} ] Substituting (r = sqrt{frac{S}{7}}): [ h = sqrt{frac{S}{7}} cdot sqrt{35} = sqrt{5S} ]5. Volume of the cone: [ V = frac{1}{3} pi r^2 h ] Substituting (r^2 = frac{S}{7}) and (h = sqrt{5S}): [ V = frac{1}{3} pi left(frac{S}{7}right) sqrt{5S} = frac{pi S}{21} sqrt{5S} = frac{pi sqrt{5} S^{3/2}}{21} ]The volume of the cone is (boxed{dfrac{pi sqrt{5} S^{3/2}}{21}}).]

🔑:Solution.Let R be the radius of the base of the cone, l its slant height, and H its height. The lateral surface of the cone unfolds into a sector of a circle with radius l and a central angle of 60^{circ}, the length of the arc of which is 2 pi R. Then 2 pi R = frac{1}{6} cdot 2 pi l ; l = 6 R . H = sqrt{l^{2} - R^{2}} = R sqrt{35}. The total surface area of the cone pi S = pi R l + pi R^{2}. Therefore, S = R cdot 6 R + R^{2} = 7 R^{2}, R = sqrt{frac{S}{7}}, H = sqrt{frac{S}{7}} cdot sqrt{35} = sqrt{5 S}. The volume of the cone V = frac{1}{3} pi R^{2} H = frac{1}{3} pi frac{S}{7} sqrt{5 S} = frac{pi S sqrt{5 S}}{21}. Answer: frac{pi S sqrt{5 S}}{21}.

🧠:[<think>Alright, so I've got this problem here with three quadratic equations:1. ( a x^2 + b x + c = 0 )2. ( b x^2 + c x + a = 0 )3. ( c x^2 + a x + b = 0 )And it says they all have exactly one common real root. I need to find the value of ( frac{a^2}{bc} + frac{b^2}{ca} + frac{c^2}{ab} ). The options are 0, 1, 2, or 3. Hmm, okay, let's break this down step by step.First, if there's exactly one common real root, let's call that root ( alpha ). So, ( alpha ) satisfies all three equations. Let me write that out:1. ( a alpha^2 + b alpha + c = 0 ) …(1)2. ( b alpha^2 + c alpha + a = 0 ) …(2)3. ( c alpha^2 + a alpha + b = 0 ) …(3)So, all three equations equal zero when ( x = alpha ). Now, maybe I can set up some relationships between a, b, c using these equations. Let's see.If I subtract equation (2) from equation (1), maybe that will help. Let's try:Equation (1) - Equation (2):( (a alpha^2 + b alpha + c) - (b alpha^2 + c alpha + a) = 0 - 0 )Simplify:( (a - b)alpha^2 + (b - c)alpha + (c - a) = 0 )Similarly, maybe subtract equation (3) from equation (2):Equation (2) - Equation (3):( (b alpha^2 + c alpha + a) - (c alpha^2 + a alpha + b) = 0 - 0 )Simplify:( (b - c)alpha^2 + (c - a)alpha + (a - b) = 0 )And perhaps subtract equation (1) from equation (3):Equation (3) - Equation (1):( (c alpha^2 + a alpha + b) - (a alpha^2 + b alpha + c) = 0 - 0 )Simplify:( (c - a)alpha^2 + (a - b)alpha + (b - c) = 0 )Hmm, so now I have three new equations:1. ( (a - b)alpha^2 + (b - c)alpha + (c - a) = 0 ) …(4)2. ( (b - c)alpha^2 + (c - a)alpha + (a - b) = 0 ) …(5)3. ( (c - a)alpha^2 + (a - b)alpha + (b - c) = 0 ) …(6)I wonder if there's a pattern here. Each equation is a cyclic permutation of the previous one. That is, if you replace a with b, b with c, and c with a, equation (4) becomes equation (5), and so on.Since all these equations are equal to zero, maybe they are related in a way that can help us find a relationship between a, b, and c.Alternatively, maybe I can use the original equations (1), (2), and (3) to express a, b, c in terms of α. Let's see.From equation (1):( a alpha^2 = -b alpha - c )Similarly, from equation (2):( b alpha^2 = -c alpha - a )And from equation (3):( c alpha^2 = -a alpha - b )So, each coefficient can be expressed in terms of the others. Let me try substituting these into each other.From equation (1): ( a alpha^2 = -b alpha - c ). Let's express a here:( a = frac{ -b alpha - c }{ alpha^2 } )Similarly, from equation (2): ( b alpha^2 = -c alpha - a ). Substitute a from above into this equation.First, substitute a:( b alpha^2 = -c alpha - left( frac{ -b alpha - c }{ alpha^2 } right) )Simplify:( b alpha^2 = -c alpha + frac{ b alpha + c }{ alpha^2 } )Multiply both sides by ( alpha^2 ) to eliminate the denominator:( b alpha^4 = -c alpha^3 + b alpha + c )Bring all terms to one side:( b alpha^4 + c alpha^3 - b alpha - c = 0 )Factor terms:Group terms with b and terms with c:( b( alpha^4 - alpha ) + c( alpha^3 - 1 ) = 0 )Factor further:( b alpha ( alpha^3 - 1 ) + c ( alpha^3 - 1 ) = 0 )Notice that ( alpha^3 - 1 ) is a common factor:( ( alpha^3 - 1 )( b alpha + c ) = 0 )So, either ( alpha^3 - 1 = 0 ) or ( b alpha + c = 0 ).Case 1: ( alpha^3 - 1 = 0 implies alpha = 1 ). Since we are dealing with real roots, the other roots are complex, but α is real, so α=1 is the only real root here.Case 2: ( b alpha + c = 0 implies c = -b alpha ). Let's explore both cases.First, let's check Case 1: α = 1.If α=1, then substitute back into equations (1), (2), (3):Equation (1): a(1)^2 + b(1) + c = 0 => a + b + c = 0Equation (2): b(1)^2 + c(1) + a = 0 => b + c + a = 0, same as equation (1)Equation (3): c(1)^2 + a(1) + b = 0 => c + a + b = 0, same again.So, all three equations reduce to a + b + c = 0. Therefore, if α=1, then a + b + c = 0. Now, let's see if this leads us somewhere.We need to compute ( frac{a^2}{bc} + frac{b^2}{ca} + frac{c^2}{ab} ). Let's write this as:( frac{a^3 + b^3 + c^3}{abc} )Because:( frac{a^2}{bc} = frac{a^3}{abc} ), similarly for the others. So sum is ( frac{a^3 + b^3 + c^3}{abc} )Now, recall that if a + b + c = 0, then ( a^3 + b^3 + c^3 = 3abc ). This is a known identity. Let me verify that:The identity is: ( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) )So if a + b + c = 0, then the right side is zero, so ( a^3 + b^3 + c^3 = 3abc ). Therefore, ( frac{a^3 + b^3 + c^3}{abc} = frac{3abc}{abc} = 3 ). So the value would be 3, which is option D.But before concluding, let's check Case 2: ( b alpha + c = 0 implies c = -b alpha ). Let's see if this is possible.If c = -b α, then from equation (1): ( a alpha^2 + b alpha + c = 0 ), substitute c:( a alpha^2 + b alpha - b alpha = 0 implies a alpha^2 = 0 )Therefore, either a = 0 or α = 0. But if α = 0, then substituting into equation (1): 0 + 0 + c = 0 => c = 0. Then equation (2): 0 + 0 + a = 0 => a = 0. Similarly equation (3): 0 + 0 + b = 0 => b = 0. But if a = b = c = 0, then the equations are not quadratic. So this case is invalid because the equations must be quadratic (assuming a, b, c are coefficients of x² terms, so they can't be zero). Therefore, α ≠ 0, so a must be 0.But if a = 0, then from equation (1): 0 + b α + c = 0 => b α + c = 0, which is the same as c = -b α. Then equation (2): b α² + c α + a = 0. Substitute a = 0 and c = -b α:b α² + (-b α) α + 0 = 0 => b α² - b α² = 0 => 0 = 0. Okay, which is always true. Then equation (3): c α² + a α + b = 0. Substitute c = -b α and a = 0:(-b α) α² + 0 + b = -b α³ + b = 0 => b ( -α³ + 1 ) = 0So either b = 0 or α³ = 1. If b = 0, then from c = -b α, c = 0. Then equation (2): 0 + 0 + a = 0 => a = 0. Again, all coefficients zero, which is invalid. Therefore, must have α³ = 1. So α = 1 (since real root). Therefore, this reduces to Case 1: α = 1. Therefore, even in Case 2, we end up with α = 1 and a + b + c = 0. So regardless, the condition is a + b + c = 0.Therefore, in this case, ( frac{a^2}{bc} + frac{b^2}{ca} + frac{c^2}{ab} = 3 ), which is option D. But wait, the options given are (A)0, (B)1, (C)2, (D)3. So the answer should be D)3.But wait, let me double-check. Suppose a + b + c = 0, then the value is 3. But let's test with specific numbers to make sure.Let's pick a = 1, b = 1, c = -2. Then a + b + c = 0.Compute the expression:( frac{1^2}{1*(-2)} + frac{1^2}{(-2)*1} + frac{(-2)^2}{1*1} = frac{1}{-2} + frac{1}{-2} + frac{4}{1} = (-0.5) + (-0.5) + 4 = 3 ). Yep, that works.Alternatively, take a=2, b=3, c=-5. Then a + b + c=0.Compute:( frac{4}{3*(-5)} + frac{9}{(-5)*2} + frac{25}{2*3} = frac{4}{-15} + frac{9}{-10} + frac{25}{6} )Convert to decimals:≈ -0.2667 - 0.9 + 4.1667 ≈ (-1.1667) + 4.1667 ≈ 3. So yes, it works.Therefore, the value is 3, which is option D. But let me check if there are other possibilities where a + b + c ≠ 0 but the equations still have a common root. But according to our earlier deduction, when we subtracted the equations, we found that either α=1 (leading to a + b + c=0) or a relation that leads back to α=1. So maybe there's no other possibility. Hence, the answer should be 3, option D.Wait, but the problem says "exactly one common real root". If all three quadratics have exactly one common real root, that would mean that each quadratic can have other roots, but they share exactly one. However, since they are quadratics, each has at most two roots. If they all share exactly one real root, then each of them must have another root which is not shared by the others. But according to our analysis, if a + b + c = 0, then α=1 is a common root. Let's check if the quadratics can have another root.Take for example the first quadratic: ( a x^2 + b x + c = 0 ). If a + b + c =0, then 1 is a root. The other root is c/a (since product of roots is c/a). Similarly, the second quadratic: ( b x^2 + c x + a =0 ). Sum of roots is -c/b, product is a/b. Since 1 is a root, the other root is a/(b*1) = a/b. Similarly, third quadratic has other root b/c.So unless c/a = a/b = b/c, the other roots would be different. But c/a = a/b implies b = a² / c. Similarly, a/b = b/c implies c = b² / a. Then substituting into b = a² / c, we get b = a² / (b² / a) ) = a³ / b² => b³ = a³ => b = a (assuming real numbers). Then c = b² / a = a² / a = a. So all three coefficients are equal: a = b = c. But if a = b = c, then the original equations become a x² + a x + a = 0, which simplifies to x² + x + 1 = 0 (if a ≠ 0). But this quadratic has discriminant 1 - 4 = -3 < 0, so no real roots. Contradiction. Therefore, unless a = b = c = 0 (which isn't allowed), the other roots are different. Therefore, if a + b + c = 0, each quadratic has one real root at x=1 and another distinct root (since quadratics with real coefficients must have either two real roots or two complex roots). Wait, but if a + b + c =0, and the quadratic has x=1 as a root, then the quadratic can have another real root or another complex root. For example, take a =1, b=1, c=-2. Then quadratic 1: x² + x -2 =0. This factors as (x-1)(x+2)=0, roots 1 and -2. Quadratic 2: x² -2x +1=0. Wait, hold on, let's compute quadratic 2:If a=1, b=1, c=-2, then quadratic 2 is b x² + c x + a = 1 x² + (-2)x +1 = x² -2x +1 = (x-1)^2. So this quadratic has a repeated root at x=1. Similarly, quadratic 3: c x² + a x + b = (-2)x² +1 x +1 = -2x² +x +1. Let's compute its roots. The quadratic equation: roots at [ -1 ± sqrt(1 +8) ] / (-4) = [ -1 ± 3 ] / (-4). So (2)/(-4)= -0.5 and (-4)/(-4)=1. So quadratic 3 has roots at 1 and -0.5. Therefore, quadratic 1 has roots 1 and -2, quadratic 2 has a double root at 1, quadratic 3 has roots 1 and -0.5. So all three quadratics share the root 1, but quadratic 2 has it as a double root. But the problem states that they have "exactly one common real root". But in this case, quadratic 2 has only the root 1 (repeated), so does that count as exactly one common real root? Wait, the problem says "exactly one common real root". So if they all share root 1, and quadratic 2 has another root which is also 1 (but it's the same root), does that affect anything? Hmm. Wait, perhaps the problem means that there is exactly one real root that is common to all three quadratics, regardless of multiplicity. In that case, even if quadratic 2 has a double root at 1, the common root is still just 1. The other roots are different for each quadratic. So in that case, the answer is 3. So the example I took does satisfy the problem's condition. Therefore, the answer is D)3.But let me check another example where a + b + c ≠0. Wait, according to our previous reasoning, if there is a common root, then a + b + c must be zero. Because in Case 2, we ended up again with α=1. So in any case, a + b + c=0. Therefore, the only way for all three quadratics to have a common real root is that a + b + c=0, leading to the value 3. Therefore, the answer is D.But wait, just to be thorough, let's check if there's a case where a + b + c ≠0 but there's a common root. Suppose someone finds such a case. Let's assume that α is not 1. Then from the equations (4), (5), (6), which are cyclic. If α is not 1, perhaps we can find a non-trivial solution.But let's suppose α ≠1. Then from equation (4):( (a - b)alpha^2 + (b - c)alpha + (c - a) = 0 )Let me write this as:( (a - b)alpha^2 + (b - c)alpha + (c - a) = 0 )Let me rearrange terms:( (a - b)alpha^2 - (c - b)alpha - (a - c) = 0 )Not sure. Alternatively, maybe express equation (4), (5), (6) as a system. But since they are cyclic, perhaps if we assume that a ≠ b ≠ c, but given the symmetry, maybe a = b = c? Wait, if a = b = c, then all three quadratics become the same equation: a x² + a x + a =0, which is x² +x +1=0 if a ≠0. This equation has discriminant 1 -4 = -3, so no real roots. Therefore, invalid. Hence a, b, c cannot be all equal.Alternatively, maybe two of them are equal. Let's suppose a = b. Then, from equation (4):( (0)alpha^2 + (a - c)alpha + (c - a) = 0 implies (a - c)(α -1) =0 )So either a = c or α=1. If a = c, then since a = b, we have a = b = c. But as before, that leads to no real roots. So α must be 1. Therefore, even if a = b, we must have α=1. Then, from equation (1): a + a + c =0 => 2a + c=0 => c= -2a. Then, check equation (2): a x² + c x + a =0. Substitute c=-2a: a x² -2a x +a =0 => x² -2x +1=0 => (x-1)^2=0. So quadratic 2 has double root at 1. Quadratic 3: c x² +a x +b = (-2a)x² +a x +a = -2a x² +a x +a. Let's factor out a: a(-2x² +x +1)=0. The quadratic equation: -2x² +x +1=0. Roots at x = [ -1 ± sqrt(1 +8) ] / (-4) = [ -1 ±3 ] / (-4). So (2)/-4 = -0.5 and (-4)/-4=1. So roots at 1 and -0.5. So quadratic 3 has roots 1 and -0.5. Quadratic 1: a x² +a x +c = a x² +a x -2a =a(x² +x -2)=a(x+2)(x-1). So roots at 1 and -2. So all three quadratics have root 1, and others different. So even in this case, a + b + c = a + a -2a =0. So indeed, even if two variables are equal, the sum is zero. Therefore, the result still holds.Therefore, regardless of the case, if there's a common real root, then a + b + c=0, leading to the value of 3. Therefore, the answer is D)3.Final Answerboxed{D}</think>Given three quadratic equations in ( x ):[begin{array}{l}a x^{2} + b x + c = 0, b x^{2} + c x + a = 0, c x^{2} + a x + b = 0end{array}]they have exactly one common real root (alpha). We need to find the value of ( frac{a^{2}}{bc} + frac{b^{2}}{ca} + frac{c^{2}}{ab} ).1. Substitute (alpha) into each equation: - ( a alpha^2 + b alpha + c = 0 ) - ( b alpha^2 + c alpha + a = 0 ) - ( c alpha^2 + a alpha + b = 0 )2. Subtract pairs of equations to derive relationships: - Subtracting the second equation from the first: ( (a - b)alpha^2 + (b - c)alpha + (c - a) = 0 ) - Subtracting the third equation from the second: ( (b - c)alpha^2 + (c - a)alpha + (a - b) = 0 ) - Subtracting the first equation from the third: ( (c - a)alpha^2 + (a - b)alpha + (b - c) = 0 )3. These equations suggest cyclic permutations and lead to the conclusion that (alpha = 1) is a common root. Substituting (alpha = 1) into the original equations, we find: - ( a + b + c = 0 )4. Using the identity for ( a + b + c = 0 ), we know ( a^3 + b^3 + c^3 = 3abc ). Therefore: - ( frac{a^2}{bc} + frac{b^2}{ca} + frac{c^2}{ab} = frac{a^3 + b^3 + c^3}{abc} = frac{3abc}{abc} = 3 )Thus, the value is (boxed{D}).,<think>Okay, let's try to tackle this problem. So we have three quadratic equations:1. ( a x^{2} + b x + c = 0 )2. ( b x^{2} + c x + a = 0 )3. ( c x^{2} + a x + b = 0 )And they all have exactly one common real root. We need to find the value of ( frac{a^{2}}{b c} + frac{b^{2}}{c a} + frac{c^{2}}{a b} ). The options are 0, 1, 2, 3. Hmm, let's see.First, let's denote that common root as ( alpha ). So ( alpha ) satisfies all three equations. That means:1. ( a alpha^2 + b alpha + c = 0 ) ...(1)2. ( b alpha^2 + c alpha + a = 0 ) ...(2)3. ( c alpha^2 + a alpha + b = 0 ) ...(3)Our goal is to relate a, b, c using these equations. Maybe we can manipulate these equations to find some relationships between a, b, c. Let's try subtracting equations or combining them somehow.Let me start by subtracting the second equation from the first:Equation (1) - Equation (2):( a alpha^2 + b alpha + c - (b alpha^2 + c alpha + a) = 0 )Simplify:( (a - b) alpha^2 + (b - c) alpha + (c - a) = 0 )Hmm, maybe factor this? Let's see:Let me group terms:( (a - b)alpha^2 + (b - c)alpha + (c - a) )Not immediately obvious. Maybe factor out (a - b) or something. Alternatively, maybe if I write it as:( (a - b)alpha^2 + (b - c)alpha = (a - c) )But not sure. Alternatively, let's think about equations (2) - (3):Equation (2) - Equation (3):( b alpha^2 + c alpha + a - (c alpha^2 + a alpha + b) = 0 )Simplify:( (b - c)alpha^2 + (c - a)alpha + (a - b) = 0 )Hmm, this is similar in structure to the previous difference. Let me note that.Similarly, equations (3) - (1):( c alpha^2 + a alpha + b - (a alpha^2 + b alpha + c) = 0 )Simplify:( (c - a)alpha^2 + (a - b)alpha + (b - c) = 0 )So we have three equations from the differences:1. ( (a - b)alpha^2 + (b - c)alpha + (c - a) = 0 )2. ( (b - c)alpha^2 + (c - a)alpha + (a - b) = 0 )3. ( (c - a)alpha^2 + (a - b)alpha + (b - c) = 0 )Interesting, they seem cyclic permutations of each other. Let me denote these as Equations (4), (5), (6). Maybe we can find a relationship here.Alternatively, maybe if we consider adding all three original equations:Equation (1) + Equation (2) + Equation (3):( (a + b + c)alpha^2 + (b + c + a)alpha + (c + a + b) = 0 )Which simplifies to:( (a + b + c)(alpha^2 + alpha + 1) = 0 )So either ( a + b + c = 0 ) or ( alpha^2 + alpha + 1 = 0 ).But ( alpha ) is a real root. However, ( alpha^2 + alpha + 1 = 0 ) has discriminant ( 1 - 4 = -3 < 0 ), so no real roots. Therefore, we must have ( a + b + c = 0 ).Ah! So that's a key insight. Therefore, ( a + b + c = 0 ).So that's a crucial relation. Now, maybe we can use this to find the required expression.The expression we need is ( frac{a^{2}}{b c} + frac{b^{2}}{c a} + frac{c^{2}}{a b} ). Let me denote this as S.Note that S can be written as ( frac{a^3 + b^3 + c^3}{a b c} ).Because:( frac{a^2}{b c} = frac{a^3}{a b c} ), similarly for the others, so sum is ( frac{a^3 + b^3 + c^3}{a b c} ).So ( S = frac{a^3 + b^3 + c^3}{a b c} ).Now, recall that if ( a + b + c = 0 ), then ( a^3 + b^3 + c^3 = 3 a b c ). Is that correct?Yes, because the identity:( a^3 + b^3 + c^3 - 3 a b c = (a + b + c)(a^2 + b^2 + c^2 - a b - b c - c a) ).Therefore, if ( a + b + c = 0 ), then the left-hand side is 0, so ( a^3 + b^3 + c^3 = 3 a b c ).Therefore, substituting into S:( S = frac{3 a b c}{a b c} = 3 ). So S = 3, which is option D.But wait, let me check if there are any other possibilities or if I made a mistake here.Wait, the problem states that the three quadratics have exactly one common real root. So our deduction that ( a + b + c = 0 ) is based on adding the three equations and getting ( (a + b + c)(alpha^2 + alpha + 1) = 0 ), and since ( alpha ) is real, ( alpha^2 + alpha + 1 ne 0 ), so ( a + b + c = 0 ). Therefore, the conclusion is that ( a + b + c = 0 ).Then, using that, we have ( S = 3 ). So the answer should be D) 3.But let me check again. Suppose a + b + c = 0, then we can also have individual relationships.Alternatively, maybe I can verify with an example where a + b + c = 0 and check if the expression is indeed 3.Let me take a simple case where a = 1, b = 1, c = -2. So that a + b + c = 1 + 1 - 2 = 0.Then, compute S = (1^2)/(1*(-2)) + (1^2)/((-2)*1) + ((-2)^2)/(1*1) = 1/(-2) + 1/(-2) + 4/1 = (-0.5) + (-0.5) + 4 = 3. So indeed S = 3. So that's correct.But wait, in this case, do the quadratics have a common root?Let's check. Let's take the quadratics:1. ( x^2 + x - 2 = 0 ). Solving: x = [-1 ± sqrt(1 + 8)] / 2 = [-1 ± 3]/2. So roots are 1 and -2.2. ( x^2 - 2x + 1 = 0 ). This factors as (x - 1)^2 = 0. So double root at 1.3. ( -2x^2 + x + 1 = 0 ). Multiply by -1: 2x^2 - x -1 = 0. Solutions: [1 ± sqrt(1 + 8)] / 4 = [1 ± 3]/4. So roots are 1 and -0.5.So the common root is 1. But the first quadratic also has a root at -2, the second quadratic has only root 1 (double root), and the third quadratic has roots 1 and -0.5. So they all share the root 1, and the other roots are different. However, the problem states "exactly one common real root," which is satisfied here. So in this example, S = 3, which is option D. So seems correct.But let's verify another example. Let a = 2, b = -1, c = -1. So a + b + c = 0.Then, compute S = (4)/((-1)(-1)) + (1)/((-1)(2)) + (1)/(2*(-1)) = 4/1 + 1/(-2) + 1/(-2) = 4 - 0.5 -0.5 = 3. Again S = 3.Check the quadratics:1. 2x² - x -1 = 0. Roots: [1 ± sqrt(1 + 8)] / 4 = [1 ± 3]/4. So roots 1 and -0.5.2. -1x² -1x +2 = 0 → -x² -x +2 =0 → x² +x -2 =0. Roots: [-1 ± sqrt(1 + 8)] / 2 = [-1 ± 3]/2. Roots 1 and -2.3. -1x² +2x -1 =0 → -x² +2x -1 =0 → x² -2x +1=0. Root 1 (double).So again, common root 1. So this example also gives S=3, and satisfies the problem's conditions. So seems solid.Therefore, the answer is D) 3.But wait, the options given are (A)0, (B)1, (C)2, (D)3. So answer is D. But let me check again my reasoning.Wait, but the problem says "exactly one common real root". In the first example, the second equation had a double root at 1, so technically, does that count as exactly one common real root? Or does "exactly one" mean that each equation has only one real root, but they all share it?Wait, the problem states: "have exactly one common real root." So it's possible that each quadratic has two roots, but they all share exactly one real root. So even if one equation has a double root at the common root, but the other equations have another root, that's acceptable. So in the first example, quadratics 1 and 3 have two distinct roots each, quadratic 2 has a double root, but they all share the root 1. So that's exactly one common real root. So the example is valid.Therefore, given that the answer comes out as 3 in these examples, and through the algebraic manipulation (using the identity for a + b + c =0 leading to a^3 + b^3 + c^3 =3abc), the answer must be D)3.But just to make sure there's no alternative scenario where a + b + c ≠ 0. But we derived that from the equations by adding them, leading to (a + b + c)(α² + α +1)=0. Since α is real, and α² + α +1 cannot be zero (as discriminant is negative), it must be that a + b + c=0. So that's the only possibility. Hence, the conclusion is correct.Therefore, the answer is D)3.Final Answerboxed{D}</think>Given three quadratic equations in (x):[begin{array}{l}a x^{2} + b x + c = 0, b x^{2} + c x + a = 0, c x^{2} + a x + b = 0end{array}]they have exactly one common real root. Let this common root be (alpha). Then (alpha) satisfies all three equations:1. (a alpha^2 + b alpha + c = 0)2. (b alpha^2 + c alpha + a = 0)3. (c alpha^2 + a alpha + b = 0)Adding these three equations, we get:[(a + b + c)alpha^2 + (a + b + c)alpha + (a + b + c) = 0]This simplifies to:[(a + b + c)(alpha^2 + alpha + 1) = 0]Since (alpha) is a real root, (alpha^2 + alpha + 1 neq 0) (as its discriminant is negative). Therefore, we must have (a + b + c = 0).Using the identity for the sum of cubes, when (a + b + c = 0):[a^3 + b^3 + c^3 = 3abc]We need to find the value of (frac{a^2}{bc} + frac{b^2}{ca} + frac{c^2}{ab}). This can be rewritten as:[frac{a^3 + b^3 + c^3}{abc}]Substituting (a^3 + b^3 + c^3 = 3abc) into the expression, we get:[frac{3abc}{abc} = 3]Thus, the value is (boxed{D}).]

🔑:Let x_{0} be their common real root. Thenbegin{array}{l}a x_{0}^{2}+b x_{0}+c=0, b x_{0}^{2}+c x_{0}+a=0, c x_{0}^{2}+a x_{0}+b=0 .end{array}Adding and rearranging the above three equations, we get(a+b+c)left(x_{0}^{2}+x_{0}+1right)=0 text {. }Since x_{0}^{2}+x_{0}+1=left(x_{0}+frac{1}{2}right)^{2}+frac{3}{4}>0,it follows that a+b+c=0.begin{array}{l}text { Then } frac{a^{2}}{b c}+frac{b^{2}}{c a}+frac{c^{2}}{a b}=frac{a^{3}+b^{3}+c^{3}}{a b c} =frac{a^{3}+b^{3}-(a+b)^{3}}{a b c} =frac{-3 a b(a+b)}{a b c}=3 text {. } end{array}Therefore, the answer is D.

🧠:[<think>Okay, so I need to solve the inequality 1/x ≤ 2 and figure out which of the given options is correct. Let me take it step by step. Hmm, inequalities with fractions can be tricky because the denominator can be positive or negative, which affects the direction of the inequality when multiplying both sides. Also, x can't be zero because division by zero is undefined. So first, I should note that x ≠ 0.Starting with the inequality: 1/x ≤ 2.I need to consider two cases because the sign of x affects the inequality. Case 1: x is positive. Case 2: x is negative.Let's start with Case 1: x > 0.If x is positive, then multiplying both sides of the inequality by x doesn't change the direction of the inequality. So:1 ≤ 2xThen divide both sides by 2:1/2 ≤ xWhich is the same as x ≥ 1/2. So in this case, since x is positive, the solution here is x ≥ 1/2.Case 2: x < 0.Here, x is negative. If I multiply both sides of the inequality by x, which is negative, the inequality sign flips. So:1 ≥ 2xThen divide both sides by 2:1/2 ≥ xBut in this case, x is negative, so x < 0. So combining these two, x ≤ 1/2 and x < 0. Since x is already less than 0, the solution here is all x < 0.So combining both cases, the solution set is x < 0 or x ≥ 1/2.Looking at the answer choices, option B is (-∞,0) ∪ [1/2, ∞), which matches my solution. Let me double-check to make sure.Alternative approach: Maybe solve the inequality algebraically without splitting into cases.Starting with 1/x ≤ 2.Subtract 2 from both sides: 1/x - 2 ≤ 0.Combine the terms: (1 - 2x)/x ≤ 0.So the inequality becomes (1 - 2x)/x ≤ 0.To solve this, find the critical points where the numerator or denominator is zero: 1 - 2x = 0 when x = 1/2, and denominator x = 0. So critical points at x = 0 and x = 1/2.These divide the number line into intervals: (-∞, 0), (0, 1/2), and (1/2, ∞).Test each interval.1. Interval (-∞, 0): pick x = -1.(1 - 2*(-1))/(-1) = (1 + 2)/(-1) = 3/-1 = -3 ≤ 0. So this interval is part of the solution.2. Interval (0, 1/2): pick x = 1/4.(1 - 2*(1/4))/(1/4) = (1 - 1/2)/(1/4) = (1/2)/(1/4) = 2 > 0. Not part of the solution.3. Interval (1/2, ∞): pick x = 1.(1 - 2*1)/1 = (1 - 2)/1 = -1 ≤ 0. So this interval is part of the solution.Now check the critical points: x = 0 is excluded because division by zero is undefined. x = 1/2: plug in x = 1/2, we get (1 - 2*(1/2))/ (1/2) = (1 - 1)/(1/2) = 0/(1/2) = 0 ≤ 0. So x = 1/2 is included.Therefore, the solution is x < 0 or x ≥ 1/2, which is exactly option B. So the answer should be B. Let me make sure that none of the other options match. Option A is only [1/2, ∞), which misses the negative part. Option C is (-∞,1/2], which would include between 0 and 1/2 which isn't part of the solution. Option D is [2, ∞), which is even more restricted. So yes, B is the correct answer.Final Answerboxed{B}</think>To solve the inequality (frac{1}{x} leqslant 2), we need to consider the cases where (x) is positive and where (x) is negative because the direction of the inequality changes when multiplying by a negative number. Additionally, (x neq 0) because division by zero is undefined.Case 1: (x > 0)1. Multiply both sides by (x) (positive, so inequality direction remains): [ 1 leq 2x ]2. Divide by 2: [ frac{1}{2} leq x ] This gives (x geq frac{1}{2}).Case 2: (x < 0)1. Multiply both sides by (x) (negative, so inequality direction reverses): [ 1 geq 2x ]2. Divide by 2: [ frac{1}{2} geq x ] Since (x) is negative, this gives (x < 0).Combining both cases, the solution set is (x < 0) or (x geq frac{1}{2}).To verify, we can solve the inequality algebraically by rewriting it as:[frac{1 - 2x}{x} leq 0]The critical points are (x = 0) and (x = frac{1}{2}). Testing intervals around these points confirms the solution set is ((-infty, 0) cup left[frac{1}{2}, +inftyright)).Thus, the correct answer is (boxed{B}).,<think>Okay, so I need to solve the inequality 1/x ≤ 2 and choose the correct answer from the given options. Let me think step by step.First, I remember that when dealing with inequalities involving fractions, I have to be careful about the sign of the denominator because multiplying both sides by a negative number reverses the inequality sign. Also, x cannot be zero here because division by zero is undefined. So, x ≠ 0 is a critical point.The inequality is 1/x ≤ 2. Let me rewrite this to make it easier to solve. Maybe subtract 2 from both sides to get 1/x - 2 ≤ 0. Then, combine the terms into a single fraction: (1 - 2x)/x ≤ 0. That way, I can analyze where the expression (1 - 2x)/x is less than or equal to zero.To solve (1 - 2x)/x ≤ 0, I need to find the critical points where the numerator or denominator is zero. The numerator 1 - 2x is zero when x = 1/2. The denominator x is zero when x = 0. These critical points divide the real number line into intervals. The intervals are (-∞, 0), (0, 1/2], and [1/2, ∞). I need to test each interval to see the sign of the expression in that interval.Let me start with the interval (-∞, 0). Let's pick x = -1 as a test point. Plugging into (1 - 2x)/x: 1 - 2*(-1) = 1 + 2 = 3, and the denominator is -1. So, 3/-1 = -3, which is negative. Since the inequality is ≤ 0, this interval is part of the solution.Next, the interval (0, 1/2). Let's choose x = 1/4. Plugging into the expression: 1 - 2*(1/4) = 1 - 1/2 = 1/2. The denominator is 1/4. So, (1/2)/(1/4) = 2, which is positive. Since we want ≤ 0, this interval is not part of the solution.Then, the interval [1/2, ∞). Let's take x = 1. Plugging in: 1 - 2*1 = -1. Denominator is 1. So, -1/1 = -1, which is negative. Therefore, this interval is part of the solution. Also, at x = 1/2, the expression becomes (1 - 2*(1/2))/1/2 = (1 -1)/ (1/2) = 0/(1/2) = 0, which satisfies the inequality since it's ≤ 0. So, x = 1/2 is included.But wait, x = 0 is excluded because the original expression 1/x is undefined there. So, putting it all together, the solution set is (-∞, 0) union [1/2, ∞). Comparing with the options given:A: [1/2, +∞) – Missing the negative part, so not correct.B: (-∞,0) ∪ [1/2, +∞) – This matches our solution.C: (-∞,1/2] – Incorrect because between 0 and 1/2, the inequality isn't satisfied, and also includes positive numbers less than 1/2 which aren't part of the solution.D: [2, +∞) – Way off, not related.So, the correct answer should be B.Wait, but let me double-check. Maybe I made a mistake when combining the terms. Let me go back to the original inequality: 1/x ≤ 2.Another approach: Consider two cases, when x is positive or negative because multiplying both sides by x (to solve the inequality) will depend on the sign of x.Case 1: x > 0.If x is positive, multiplying both sides by x doesn't change the inequality direction: 1 ≤ 2x. Then, divide by 2: 1/2 ≤ x. So, in this case, x ≥ 1/2.Case 2: x < 0.If x is negative, multiplying both sides by x reverses the inequality: 1 ≥ 2x. Then, divide by 2: 1/2 ≥ x. But since x is negative, this is always true because 1/2 is greater than any negative number. So, for all x < 0, the inequality holds.So combining both cases: x < 0 or x ≥ 1/2. Which is exactly option B. So that confirms the solution.Therefore, the answer is B.Final Answerboxed{B}</think>To solve the inequality (frac{1}{x} leqslant 2), we start by rewriting it as (frac{1 - 2x}{x} leq 0). The critical points where the numerator or denominator is zero are (x = frac{1}{2}) and (x = 0). These points divide the real number line into intervals: ((-infty, 0)), ((0, frac{1}{2}]), and ([frac{1}{2}, infty)).1. Interval ((-infty, 0)): Choosing (x = -1) as a test point, the expression (frac{1 - 2x}{x}) becomes (frac{3}{-1} = -3), which is negative. This interval is part of the solution.2. Interval ((0, frac{1}{2})): Choosing (x = frac{1}{4}) as a test point, the expression becomes (frac{1/2}{1/4} = 2), which is positive. This interval is not part of the solution.3. Interval ([frac{1}{2}, infty)): Choosing (x = 1) as a test point, the expression becomes (frac{-1}{1} = -1), which is negative. This interval is part of the solution, including (x = frac{1}{2}) because the expression equals zero there.Additionally, considering the original inequality (frac{1}{x} leq 2) by cases:- For (x > 0), multiplying both sides by (x) gives (1 leq 2x), leading to (x geq frac{1}{2}).- For (x < 0), multiplying both sides by (x) (a negative number) reverses the inequality, giving (1 geq 2x), which is always true for negative (x).Thus, the solution set is ((-infty, 0) cup [frac{1}{2}, infty)), which corresponds to option B.(boxed{B})]

🔑:The given inequality can be rewritten in two cases, depending on the sign of x. We will need to consider these cases because the inequality includes a fraction which changes its direction based on whether x is positive or negative.Case 1: When x > 0, we have 2x geqslant 1. Dividing both sides by 2, we get x geqslant frac{1}{2}.Case 2: When x < 0, we do not need to consider the case where frac{1}{x} leq 2 since for negative values of x, the left side is already negative and hence always less than 2.Therefore, combining the results from both cases, the solution set for the inequality is all x such that x geqslant frac{1}{2} or x < 0. In set-builder notation, this can be expressed as {x | x geqslant frac{1}{2} text{ or } x < 0}.Hence, the correct option is boxed{B}.