💡Alright, so I have this problem here about a piecewise function and another function derived from it. Let me try to understand it step by step. First, the function f(x) is defined as:[ f(x) = begin{cases} x + 2, & text{if } x > a, x^2 + 5x + 2, & text{if } x leq a.end{cases} ]And then there's another function g(x) defined as:[ g(x) = f(x) - 2x. ]The question is asking for the range of the real number a such that g(x) has exactly three distinct zeros. The options are given as A: [-1,1), B: [0,2], C: [-2,2], D: [-1,2).Okay, so I need to find the values of a where g(x) has exactly three distinct zeros. Let me break this down.First, let's write out what g(x) is for both cases of f(x). When x > a, f(x) is x + 2, so:[ g(x) = (x + 2) - 2x = -x + 2. ]That simplifies to:[ g(x) = -x + 2 quad text{for } x > a. ]When x ≤ a, f(x) is x² + 5x + 2, so:[ g(x) = (x^2 + 5x + 2) - 2x = x^2 + 3x + 2. ]That simplifies to:[ g(x) = x^2 + 3x + 2 quad text{for } x leq a. ]So now, g(x) is a piecewise function:[ g(x) = begin{cases} -x + 2, & text{if } x > a, x^2 + 3x + 2, & text{if } x leq a.end{cases} ]Next, I need to find the zeros of g(x). That means solving g(x) = 0 for both cases.First, let's solve for x > a:[ -x + 2 = 0 ][ -x + 2 = 0 implies x = 2. ]So, one zero is at x = 2, but this is only valid if 2 > a. If 2 ≤ a, then this zero wouldn't exist in the domain x > a.Now, let's solve for x ≤ a:[ x^2 + 3x + 2 = 0 ]This is a quadratic equation, so let's factor it:[ (x + 1)(x + 2) = 0 ]So, the solutions are x = -1 and x = -2.Therefore, the zeros of g(x) are x = -2, x = -1, and x = 2. But we have to consider the conditions based on the value of a.Wait, hold on. The zero at x = 2 is only valid if 2 > a, right? Because g(x) = -x + 2 is only defined for x > a. So, if a is greater than or equal to 2, then x = 2 wouldn't be a zero because it wouldn't satisfy x > a. Similarly, the zeros at x = -1 and x = -2 are only valid if they are less than or equal to a.So, for g(x) to have exactly three distinct zeros, all three zeros x = -2, x = -1, and x = 2 must lie within their respective domains. That is:1. x = 2 must satisfy x > a, so 2 > a or a < 2.2. x = -1 must satisfy x ≤ a, so -1 ≤ a.3. x = -2 must satisfy x ≤ a, so -2 ≤ a.Therefore, combining these conditions:- a must be less than 2 (from the first condition),- a must be greater than or equal to -1 (from the second condition),- a must be greater than or equal to -2 (from the third condition).But wait, the third condition (-2 ≤ a) is less restrictive than the second condition (-1 ≤ a), because if a is greater than or equal to -1, it's automatically greater than or equal to -2. So, the more restrictive conditions are a < 2 and a ≥ -1.Therefore, the range of a is [-1, 2).Let me double-check. If a is in [-1, 2), then:- For a = -1: The zeros are x = -2, x = -1, and x = 2. But wait, when a = -1, x = -1 is the boundary point. So, does x = -1 count as a zero? Yes, because for x ≤ a, which is x ≤ -1, so x = -1 is included. Similarly, x = 2 is still a zero because 2 > a (-1). So, all three zeros are valid.If a is just below 2, say a = 1.9, then x = 2 is still a zero because 2 > 1.9. The zeros x = -2 and x = -1 are still less than or equal to a, so they are valid.If a is exactly 2, then x = 2 is not a zero because g(x) for x > 2 would be -x + 2, but x = 2 is not greater than 2, so it's not included. Instead, x = 2 would fall into the x ≤ a case, so g(2) = (2)^2 + 3*(2) + 2 = 4 + 6 + 2 = 12, which is not zero. So, at a = 2, we lose the zero at x = 2, so we only have two zeros: x = -2 and x = -1. Hence, a cannot be 2.Similarly, if a is less than -1, say a = -1.5, then x = -1 is greater than a, so x = -1 would not be a zero because it's not in the domain x ≤ a. Wait, no. If a = -1.5, then x = -1 is greater than a, so x = -1 would be in the x > a case, but in that case, g(x) = -x + 2. So, plugging x = -1 into g(x) = -x + 2, we get g(-1) = 1 + 2 = 3 ≠ 0. So, x = -1 is not a zero in this case. Therefore, if a < -1, then x = -1 is not a zero because it's in the x > a region, but it doesn't satisfy g(x) = 0 there. So, we only have two zeros: x = -2 and x = 2 (if a < 2). Wait, but if a is less than -1, say a = -2, then x = -2 is still a zero because x = -2 ≤ a = -2. But x = -1 is not a zero because it's in the x > a region, and plugging into g(x) = -x + 2 gives 3, not zero. So, in this case, we have two zeros: x = -2 and x = 2. Therefore, to have three zeros, a must be at least -1 so that x = -1 is included in the x ≤ a region and thus is a zero.Therefore, the range of a is from -1 (inclusive) to 2 (exclusive), which is [-1, 2). So, the correct answer should be D: [-1,2).

💡Okay, so I have this geometry problem here, and I need to figure out which statement is correct. Let me read it again.We have a line ( a ) in space and a plane ( alpha ) in space. Then, the question is asking which of the statements A, B, C, or D is correct. The options are:A: There definitely exists a plane ( beta ) passing through ( a ) such that ( beta parallel alpha ).B: There definitely exists a plane ( beta ) passing through ( a ) such that ( beta perp alpha ).C: There definitely does not exist a line ( b ) in plane ( alpha ) such that ( a perp b ).D: There definitely does not exist a line ( b ) in plane ( alpha ) such that ( a parallel b ).Hmm, okay. So, I need to analyze each statement one by one and see which one is necessarily true regardless of the position of line ( a ) and plane ( alpha ).Let me start with statement A: There definitely exists a plane ( beta ) passing through ( a ) such that ( beta parallel alpha ).So, for a plane ( beta ) to be parallel to plane ( alpha ), they must not intersect, and their normal vectors must be the same. Now, if line ( a ) is parallel to plane ( alpha ), then it's possible to construct such a plane ( beta ) that contains ( a ) and is parallel to ( alpha ). But what if line ( a ) is not parallel to ( alpha )? If ( a ) intersects ( alpha ), then any plane ( beta ) containing ( a ) would have to intersect ( alpha ) along the line of intersection, right? So, in that case, ( beta ) can't be parallel to ( alpha ). Therefore, statement A is not necessarily true because it depends on the position of ( a ) relative to ( alpha ). If ( a ) intersects ( alpha ), such a plane ( beta ) doesn't exist. So, A is incorrect.Moving on to statement B: There definitely exists a plane ( beta ) passing through ( a ) such that ( beta perp alpha ).Hmm, okay. So, regardless of the position of ( a ) and ( alpha ), can we always find such a plane ( beta ) that contains ( a ) and is perpendicular to ( alpha )?Let me think. If ( a ) is parallel to ( alpha ), then we can construct a plane ( beta ) containing ( a ) that is perpendicular to ( alpha ). Similarly, if ( a ) intersects ( alpha ), we can still construct such a plane. In fact, if ( a ) is not perpendicular to ( alpha ), we can rotate ( a ) around the point of intersection to form a plane ( beta ) that is perpendicular to ( alpha ). If ( a ) is already perpendicular to ( alpha ), then any plane containing ( a ) will automatically be perpendicular to ( alpha ). So, regardless of the initial position of ( a ), we can always find such a plane ( beta ). Therefore, statement B seems to be correct.But let me not jump to conclusions yet. Let me check the other options as well.Statement C: There definitely does not exist a line ( b ) in plane ( alpha ) such that ( a perp b ).So, this is saying that there is no line ( b ) in ( alpha ) that is perpendicular to ( a ). Is that necessarily true?Well, let's think. If line ( a ) is not perpendicular to plane ( alpha ), then its projection onto ( alpha ) is a line. If we take a line ( b ) in ( alpha ) that is perpendicular to this projection, then ( b ) would be perpendicular to ( a ). On the other hand, if ( a ) is perpendicular to ( alpha ), then every line in ( alpha ) is perpendicular to ( a ). So, in both cases, there does exist a line ( b ) in ( alpha ) such that ( a perp b ). Therefore, statement C is incorrect because such a line ( b ) does exist.Finally, statement D: There definitely does not exist a line ( b ) in plane ( alpha ) such that ( a parallel b ).So, this is saying that there is no line ( b ) in ( alpha ) that is parallel to ( a ). Is that necessarily true?Well, if line ( a ) is parallel to plane ( alpha ), then there exists a line ( b ) in ( alpha ) that is parallel to ( a ). For example, we can translate ( a ) into the plane ( alpha ) along the direction perpendicular to ( alpha ) to get such a line ( b ). However, if ( a ) is not parallel to ( alpha ), meaning it intersects ( alpha ), then any line ( b ) in ( alpha ) that is parallel to ( a ) would have to lie in both ( alpha ) and the plane containing ( a ) and ( b ), which would imply that ( a ) is parallel to ( alpha ). But since ( a ) intersects ( alpha ), such a line ( b ) cannot exist. Wait, so does that mean that if ( a ) is not parallel to ( alpha ), there is no line ( b ) in ( alpha ) parallel to ( a )? Hmm, that seems conflicting.Wait, actually, if ( a ) is not parallel to ( alpha ), it could still have a direction that is parallel to some line in ( alpha ). For instance, even if ( a ) intersects ( alpha ), the direction vector of ( a ) might still be parallel to some direction in ( alpha ). So, in that case, we can have a line ( b ) in ( alpha ) that is parallel to ( a ). Therefore, regardless of whether ( a ) is parallel to ( alpha ) or not, there can exist a line ( b ) in ( alpha ) that is parallel to ( a ). Hence, statement D is incorrect because such a line ( b ) can exist.Wait, hold on, I might be confusing something here. Let me think again.If ( a ) is parallel to ( alpha ), then yes, we can have a line ( b ) in ( alpha ) parallel to ( a ). But if ( a ) is not parallel to ( alpha ), meaning it intersects ( alpha ) at some point, can we still have a line ( b ) in ( alpha ) that is parallel to ( a )?I think yes, because the direction of ( a ) can still lie within the plane ( alpha ). For example, imagine ( a ) is a line that pierces through ( alpha ) at a point, but its direction vector is such that it's parallel to some line in ( alpha ). So, even if ( a ) intersects ( alpha ), as long as its direction vector is parallel to ( alpha ), there can be a line ( b ) in ( alpha ) that is parallel to ( a ). Therefore, statement D is incorrect because such a line ( b ) can exist.Wait, but if ( a ) is not parallel to ( alpha ), meaning it's not parallel and not intersecting? No, in space, two lines can be skew, but here ( a ) is a line in space, and ( alpha ) is a plane. So, if ( a ) is not parallel to ( alpha ), it must intersect ( alpha ) at a point. So, in that case, can there be a line ( b ) in ( alpha ) that is parallel to ( a )?Hmm, if ( a ) intersects ( alpha ), then ( a ) and ( b ) would both lie in some plane, but since ( b ) is in ( alpha ) and ( a ) is not, unless ( a ) is parallel to ( alpha ), ( a ) and ( b ) can't be parallel. Wait, no, that's not necessarily true. Let me think about the direction vectors.Suppose ( a ) has direction vector ( vec{v} ), and ( alpha ) has normal vector ( vec{n} ). If ( a ) is not parallel to ( alpha ), then ( vec{v} ) is not perpendicular to ( vec{n} ). However, if there exists a line ( b ) in ( alpha ) with direction vector ( vec{v} ), then ( vec{v} ) must lie within ( alpha ), meaning ( vec{v} ) is perpendicular to ( vec{n} ). But if ( a ) is not parallel to ( alpha ), ( vec{v} ) is not perpendicular to ( vec{n} ), so such a line ( b ) cannot exist. Therefore, if ( a ) is not parallel to ( alpha ), there does not exist a line ( b ) in ( alpha ) parallel to ( a ). But if ( a ) is parallel to ( alpha ), then such a line ( b ) does exist.So, statement D says: There definitely does not exist a line ( b ) in plane ( alpha ) such that ( a parallel b ).But this is not necessarily true because if ( a ) is parallel to ( alpha ), then such a line ( b ) does exist. Therefore, statement D is incorrect because it's not always true that such a line ( b ) does not exist.Wait, so going back, if ( a ) is parallel to ( alpha ), then such a line ( b ) exists, making statement D incorrect. If ( a ) is not parallel to ( alpha ), then such a line ( b ) does not exist. So, statement D is saying that "there definitely does not exist" such a line ( b ), which is only true when ( a ) is not parallel to ( alpha ). But since the statement is about "definitely does not exist regardless of the position of ( a )", it's incorrect because in some cases, such a line ( b ) does exist.Therefore, statement D is incorrect.So, summarizing:- Statement A: Incorrect, because if ( a ) intersects ( alpha ), such a plane ( beta ) does not exist.- Statement B: Correct, because regardless of the position of ( a ), we can always find a plane ( beta ) containing ( a ) that is perpendicular to ( alpha ).- Statement C: Incorrect, because such a line ( b ) does exist.- Statement D: Incorrect, because such a line ( b ) can exist if ( a ) is parallel to ( alpha ).Therefore, the correct answer is statement B.**Final Answer**boxed{B}

💡Alright, so I have this problem here where I need to find the number of non-empty sets ( M ) that satisfy certain conditions. The set ( M ) is a subset of ( {0, 1, 2, 3, 4, 5} ). The conditions are that for any element ( x ) in ( M ), neither ( x^2 ) nor ( sqrt{x} ) can be in ( M ). Hmm, okay, let's try to unpack this step by step.First off, ( M ) is a subset of ( {0, 1, 2, 3, 4, 5} ), and it's non-empty. So, ( M ) can have any number of elements from 1 to 6, but it has to follow those two rules about ( x^2 ) and ( sqrt{x} ).Let me think about what these conditions mean. For any ( x ) in ( M ), ( x^2 ) shouldn't be in ( M ). So, if I include a number in ( M ), I have to make sure that its square isn't also in ( M ). Similarly, the square root of ( x ) shouldn't be in ( M ). That means if I include a number, I can't include its square root either.Let me list out the elements and see how they relate in terms of squares and square roots. The set is ( {0, 1, 2, 3, 4, 5} ). Let's see:- ( 0 ): Its square is ( 0 ), and its square root is ( 0 ).- ( 1 ): Its square is ( 1 ), and its square root is ( 1 ).- ( 2 ): Its square is ( 4 ), and its square root is approximately ( 1.414 ), which isn't an integer.- ( 3 ): Its square is ( 9 ), which isn't in the set, and its square root is approximately ( 1.732 ), also not an integer.- ( 4 ): Its square is ( 16 ), not in the set, and its square root is ( 2 ).- ( 5 ): Its square is ( 25 ), not in the set, and its square root is approximately ( 2.236 ), not an integer.Okay, so from this, I can see that ( 0 ) and ( 1 ) are special because their squares and square roots are themselves. So, if I include ( 0 ) in ( M ), I can't include ( 0 ) again, which is kind of redundant because sets don't have duplicate elements. Similarly, including ( 1 ) would mean I can't include ( 1 ) again, which is also redundant.But wait, the problem says "for any ( x in M )", so if ( x ) is ( 0 ) or ( 1 ), their squares and square roots are themselves, which are already in ( M ). So, does that mean ( 0 ) and ( 1 ) can't be in ( M ) at all? Because if they are, then their squares and square roots would also have to be excluded, but since they are the same as themselves, it would mean they can't be in ( M ).Let me test this. Suppose ( M ) contains ( 0 ). Then, according to the condition, ( 0^2 = 0 ) shouldn't be in ( M ). But ( 0 ) is already in ( M ), so that violates the condition. Similarly, if ( M ) contains ( 1 ), then ( 1^2 = 1 ) shouldn't be in ( M ), but it is, which is a problem. Therefore, ( 0 ) and ( 1 ) cannot be elements of ( M ).So, that simplifies things a bit. ( M ) can only contain elements from ( {2, 3, 4, 5} ). Now, let's look at the remaining elements: ( 2, 3, 4, 5 ).Let's analyze the relationships between these elements:- ( 2 ): Its square is ( 4 ), which is in the set. Its square root is not an integer.- ( 3 ): Its square is ( 9 ), not in the set. Its square root is not an integer.- ( 4 ): Its square is ( 16 ), not in the set. Its square root is ( 2 ), which is in the set.- ( 5 ): Its square is ( 25 ), not in the set. Its square root is not an integer.So, the key relationships here are between ( 2 ) and ( 4 ). If I include ( 2 ) in ( M ), I can't include ( 4 ) because ( 2^2 = 4 ). Similarly, if I include ( 4 ) in ( M ), I can't include ( 2 ) because ( sqrt{4} = 2 ).What about the other elements? ( 3 ) and ( 5 ) don't have their squares or square roots in the set, so they don't impose any restrictions on each other or on ( 2 ) and ( 4 ). So, ( 3 ) and ( 5 ) can be included or excluded independently, as long as we follow the rules with ( 2 ) and ( 4 ).So, let's think about how to count the number of valid sets ( M ). Since ( M ) is a subset of ( {2, 3, 4, 5} ), and it's non-empty, we can consider all possible subsets and then exclude those that violate the conditions.But instead of listing all subsets, maybe we can approach this systematically. Let's consider the possible choices for ( 2 ) and ( 4 ):1. Neither ( 2 ) nor ( 4 ) is in ( M ).2. ( 2 ) is in ( M ), but ( 4 ) is not.3. ( 4 ) is in ( M ), but ( 2 ) is not.These are the only possibilities because we can't have both ( 2 ) and ( 4 ) in ( M ) together.For each of these cases, we can independently decide whether to include ( 3 ) and ( 5 ) or not, since they don't affect each other or the other elements.Let's break it down:**Case 1: Neither ( 2 ) nor ( 4 ) is in ( M ).**In this case, ( M ) can only contain ( 3 ) and/or ( 5 ). Since ( M ) is non-empty, we need to count all non-empty subsets of ( {3, 5} ).The number of subsets of ( {3, 5} ) is ( 2^2 = 4 ). Excluding the empty set, we have ( 3 ) subsets: ( {3} ), ( {5} ), and ( {3, 5} ).**Case 2: ( 2 ) is in ( M ), but ( 4 ) is not.**Here, ( M ) must include ( 2 ), and can include any combination of ( 3 ) and ( 5 ). Again, ( M ) is non-empty, but since ( 2 ) is already included, we just need to consider all subsets of ( {3, 5} ), including the empty set.The number of subsets of ( {3, 5} ) is ( 4 ). So, the possible sets are:- ( {2} )- ( {2, 3} )- ( {2, 5} )- ( {2, 3, 5} )That's ( 4 ) sets.**Case 3: ( 4 ) is in ( M ), but ( 2 ) is not.**This case is similar to Case 2, but with ( 4 ) instead of ( 2 ). So, ( M ) must include ( 4 ), and can include any combination of ( 3 ) and ( 5 ).Again, the number of subsets is ( 4 ):- ( {4} )- ( {4, 3} )- ( {4, 5} )- ( {4, 3, 5} )That's another ( 4 ) sets.Now, let's add up the number of sets from each case:- Case 1: ( 3 ) sets- Case 2: ( 4 ) sets- Case 3: ( 4 ) setsTotal: ( 3 + 4 + 4 = 11 ) sets.Wait, but let me double-check to make sure I haven't missed anything. So, in Case 1, we have subsets without ( 2 ) or ( 4 ), which are ( {3} ), ( {5} ), and ( {3, 5} ). That's 3 sets.In Case 2, we have subsets with ( 2 ) and any combination of ( 3 ) and ( 5 ). That's 4 sets.In Case 3, we have subsets with ( 4 ) and any combination of ( 3 ) and ( 5 ). That's another 4 sets.Adding them up gives 11 sets. Hmm, but let me think if there's any overlap or something I might have counted twice.Looking at the cases:- Case 1: No overlap with Cases 2 and 3 because they don't include ( 2 ) or ( 4 ).- Case 2: Includes ( 2 ), which is exclusive with Case 3.- Case 3: Includes ( 4 ), which is exclusive with Case 2.So, no overlaps between the cases. Therefore, the total is indeed 11.But just to be thorough, let me list all the possible sets to make sure:From Case 1:1. ( {3} )2. ( {5} )3. ( {3, 5} )From Case 2:4. ( {2} )5. ( {2, 3} )6. ( {2, 5} )7. ( {2, 3, 5} )From Case 3:8. ( {4} )9. ( {4, 3} )10. ( {4, 5} )11. ( {4, 3, 5} )Yep, that's 11 sets. Each of these sets satisfies the condition that for any ( x ) in ( M ), ( x^2 ) and ( sqrt{x} ) are not in ( M ).Wait a second, let me check one of these sets to make sure. Take ( {2, 3, 5} ). For ( x = 2 ), ( x^2 = 4 ) is not in the set, and ( sqrt{2} ) isn't an integer, so that's fine. For ( x = 3 ), ( x^2 = 9 ) isn't in the set, and ( sqrt{3} ) isn't an integer. For ( x = 5 ), same thing: ( x^2 = 25 ) isn't in the set, and ( sqrt{5} ) isn't an integer. So, that set is valid.Another one: ( {4, 3, 5} ). For ( x = 4 ), ( x^2 = 16 ) isn't in the set, and ( sqrt{4} = 2 ) isn't in the set. For ( x = 3 ) and ( x = 5 ), same as before. So, that's also valid.What about a single-element set like ( {3} )? For ( x = 3 ), ( x^2 = 9 ) isn't in the set, and ( sqrt{3} ) isn't an integer. So, that's fine too.And the empty set isn't considered because ( M ) has to be non-empty.So, after carefully considering each case and verifying some examples, I'm confident that the total number of valid sets ( M ) is 11.**Final Answer**boxed{A}

💡Alright, so I have this geometry problem here about planes and lines in space. It's asking which of the four statements are correct. Let me try to break it down step by step.First, let me restate the problem to make sure I understand it. We have a plane α, and three different lines l, m, and n. Then there are four statements, and I need to figure out which ones are correct.Okay, let's go through each statement one by one.**Statement ①:** If m is a subset of α, n is a subset of α, l is perpendicular to m, and l is perpendicular to n, then l is perpendicular to α.Hmm, so m and n are both lines lying on plane α. If line l is perpendicular to both m and n, does that mean l is perpendicular to the entire plane α?I remember that for a line to be perpendicular to a plane, it needs to be perpendicular to every line in that plane. But wait, that's a bit too strict. Actually, the correct criterion is that if a line is perpendicular to two non-parallel lines in the plane, then it's perpendicular to the plane. So, if m and n are intersecting lines in α, and l is perpendicular to both, then yes, l is perpendicular to α.But the problem doesn't specify that m and n intersect. They could be parallel lines in α. If m and n are parallel, then l being perpendicular to both doesn't necessarily mean it's perpendicular to the entire plane. It might just be perpendicular to the direction of those parallel lines.So, since m and n could be parallel, statement ① isn't necessarily true. Therefore, ① is incorrect.**Statement ②:** If m is a subset of α, n is a subset of α, then l is parallel to m.Wait, this seems a bit vague. If m and n are both in α, what does that say about l? The statement doesn't specify any relationship between l and the plane α or the lines m and n. It just says that if m and n are in α, then l is parallel to m.But l could be any line in space. It might intersect m, be skew to m, or maybe even lie on α. There's no information given about l's relationship to α or the lines m and n. So, I don't see how we can conclude that l is parallel to m.Therefore, statement ② is incorrect.**Statement ③:** If l is parallel to m, m is perpendicular to α, and n is perpendicular to α, then l is parallel to n.Okay, let's parse this. If l is parallel to m, and m is perpendicular to α, then l must also be perpendicular to α because parallel lines have the same relationship with planes. Similarly, n is also perpendicular to α.Now, if two lines are both perpendicular to the same plane, they must be parallel to each other. That's a standard result in geometry. So, since both l and n are perpendicular to α, they must be parallel.Therefore, statement ③ is correct.**Statement ④:** If l is perpendicular to m, and l is perpendicular to n, then n is parallel to m.Hmm, so l is perpendicular to both m and n. Does that mean m and n are parallel?I think not necessarily. If l is perpendicular to both m and n, m and n could be parallel, but they could also intersect or be skew lines. For example, in three-dimensional space, two lines can both be perpendicular to a third line without being parallel to each other. They could lie on different planes and intersect or not intersect at all.So, statement ④ is incorrect.**Summary:**- ①: Incorrect- ②: Incorrect- ③: Correct- ④: IncorrectTherefore, the only correct statement is ③.**Final Answer**The correct statement is boxed{③}.

💡Okay, so I have this problem where I need to find the equation of a circle that's symmetric to the given circle ((x + 2)^2 + y^2 = 2016) with respect to the line (x - y + 1 = 0). The options are A, B, C, and D, each with different equations. First, I remember that when you reflect a circle over a line, the radius stays the same, but the center gets reflected over that line. So, my main task is to find the reflection of the center of the original circle over the given line. The original circle is ((x + 2)^2 + y^2 = 2016), so its center is at ((-2, 0)) and the radius is (sqrt{2016}). I don't think the radius will affect the reflection, so I can focus on reflecting the center point.The line we're reflecting over is (x - y + 1 = 0). To find the reflection of a point over a line, I can use the formula for reflection over a line. I think the formula involves finding the perpendicular projection of the point onto the line and then moving the same distance on the other side.Let me recall the formula for reflecting a point ((x, y)) over the line (ax + by + c = 0). The reflected point ((x', y')) can be found using:[x' = x - frac{2a(ax + by + c)}{a^2 + b^2}][y' = y - frac{2b(ax + by + c)}{a^2 + b^2}]So, in this case, the line is (x - y + 1 = 0), which can be written as (1x - 1y + 1 = 0). Therefore, (a = 1), (b = -1), and (c = 1).The point to reflect is ((-2, 0)). Let me plug these into the formula.First, compute (ax + by + c):[ax + by + c = (1)(-2) + (-1)(0) + 1 = -2 + 0 + 1 = -1]Now, compute the denominators (a^2 + b^2):[a^2 + b^2 = (1)^2 + (-1)^2 = 1 + 1 = 2]Now, compute (x'):[x' = x - frac{2a(ax + by + c)}{a^2 + b^2} = -2 - frac{2(1)(-1)}{2} = -2 - frac{-2}{2} = -2 + 1 = -1]Similarly, compute (y'):[y' = y - frac{2b(ax + by + c)}{a^2 + b^2} = 0 - frac{2(-1)(-1)}{2} = 0 - frac{2}{2} = 0 - 1 = -1]So, the reflected point is ((-1, -1)). Therefore, the center of the symmetric circle is ((-1, -1)), and since the radius remains the same, the equation of the symmetric circle should be:[(x + 1)^2 + (y + 1)^2 = 2016]Looking at the options, that's option C. Wait, just to make sure I didn't make any calculation errors. Let me double-check the reflection formula.Another way to find the reflection is to find the midpoint between the original point and its reflection, which should lie on the line, and the line should be perpendicular to the segment connecting the original point and its reflection.So, let me denote the original point as (P(-2, 0)), the reflected point as (P'(m, n)), and the midpoint (M) between (P) and (P') should lie on the line (x - y + 1 = 0). Also, the line segment (PP') should be perpendicular to the given line.First, the midpoint (M) has coordinates:[Mleft( frac{-2 + m}{2}, frac{0 + n}{2} right)]Since (M) lies on the line (x - y + 1 = 0), substituting into the equation:[frac{-2 + m}{2} - frac{n}{2} + 1 = 0]Multiply both sides by 2:[(-2 + m) - n + 2 = 0][m - n = 0][m = n]So, the midpoint condition gives (m = n).Next, the line segment (PP') should be perpendicular to the given line (x - y + 1 = 0). The slope of the given line is 1 (since it's (y = x + 1)), so the slope of the perpendicular line should be -1.The slope of (PP') is:[frac{n - 0}{m - (-2)} = frac{n}{m + 2}]This slope should be equal to -1:[frac{n}{m + 2} = -1][n = - (m + 2)][n = -m - 2]But from the midpoint condition, we have (m = n). So substituting (n = m) into the above equation:[m = -m - 2][2m = -2][m = -1]Therefore, (n = m = -1). So, the reflected point is indeed ((-1, -1)), confirming my earlier result.Thus, the equation of the symmetric circle is ((x + 1)^2 + (y + 1)^2 = 2016), which is option C.I think that's solid. I used two different methods to confirm the reflection point, so I'm confident it's correct.**Final Answer**boxed{C}

💡Okay, so I have this probability problem here: there's a box with 5 white balls and 3 black balls. I need to find the probability that when I draw the balls one at a time, all the draws alternate in color, starting with a white ball. Hmm, okay. Let me try to figure this out step by step.First, I need to understand what it means for the draws to alternate in color starting with white. That would mean the sequence of draws goes white, black, white, black, and so on. Since there are 5 white balls and 3 black balls, the sequence would have to be WBWBWBWB. Wait, but hold on, there are 5 white and 3 black, so the sequence would actually be WBWBWBWB, right? Because after 3 black balls, we still have 2 white balls left. So the sequence would be WBWBWBWB, which is 8 balls in total. That makes sense because 5 white and 3 black add up to 8 balls.Now, I need to find the probability of this specific sequence happening when drawing the balls one by one. Probability is generally the number of successful outcomes divided by the total number of possible outcomes. So, I need to figure out how many successful outcomes there are and how many total possible outcomes there are.Let me think about the total number of possible outcomes first. Since there are 8 balls in total, and each draw is without replacement, the total number of ways to draw all 8 balls is 8 factorial, which is 8! That's 40320. But wait, actually, since the balls of the same color are indistinct, maybe I should think in terms of combinations instead of permutations. Hmm, but in probability, when dealing with sequences, order matters, so maybe I should stick with permutations.Wait, no. Let me clarify. If I consider each ball as unique, then the total number of possible sequences is indeed 8!. However, if the balls of the same color are identical, then the number of distinct sequences is the number of ways to arrange 5 white balls and 3 black balls, which is given by the combination formula: 8 choose 3, which is 56. Because choosing positions for the black balls among the 8 slots automatically determines the positions for the white balls.So, the total number of possible sequences is 56. That makes sense because we're only concerned with the color sequence, not the specific balls.Now, how many successful outcomes are there? A successful outcome is the specific sequence WBWBWBWB. But wait, is that the only successful sequence? Or are there multiple sequences that alternate starting with white?Let me think. If I start with white, the next must be black, then white, and so on. Given that there are 5 white and 3 black balls, the sequence would have to be WBWBWBWB. That's 8 balls in total. So, there's only one such sequence where the colors alternate starting with white.Wait, but hold on a second. Is that the case? Let me consider if there are other sequences that alternate starting with white. For example, could there be a sequence like WBWBWBWW? But that would mean two white balls in a row, which violates the alternating condition. Similarly, any deviation from WBWBWBWB would break the alternation. So, yes, only one specific sequence meets the criteria.Therefore, the number of successful outcomes is 1. So, the probability is 1 divided by the total number of possible sequences, which is 56. So, the probability is 1/56.But wait, let me double-check that. Maybe I'm oversimplifying. Let's think about it another way. Instead of considering all possible sequences, maybe I should calculate the probability step by step as I draw each ball.Starting with the first draw: the probability of drawing a white ball is 5/8. Then, the next draw should be black, so the probability is 3/7. Then, the next should be white again, so 4/6. Then black, 2/5. Then white, 3/4. Then black, 1/3. Then white, 2/2. Then, since we've already drawn all 3 black balls, the last two draws must be white, but wait, that's not possible because we need to alternate. Wait, hold on, I think I made a mistake here.Let me recount. If we start with white, then black, white, black, white, black, white, black, white. That's 5 white and 3 black, so the sequence is WBWBWBWB. So, the probability would be:(5/8) * (3/7) * (4/6) * (2/5) * (3/4) * (1/3) * (2/2) * (1/1)Wait, let me compute that step by step.First draw: 5/8 (white)Second draw: 3/7 (black)Third draw: 4/6 (white)Fourth draw: 2/5 (black)Fifth draw: 3/4 (white)Sixth draw: 1/3 (black)Seventh draw: 2/2 (white)Eighth draw: 1/1 (white)Wait, but hold on, after the seventh draw, we've already drawn all 3 black balls, so the eighth draw should be white, but we've already drawn 4 white balls, so there's only 1 white ball left. Wait, no, we started with 5 white balls. Let me recount the number of white balls drawn:First draw: 1 whiteThird draw: 2 whiteFifth draw: 3 whiteSeventh draw: 4 whiteEighth draw: 5 whiteWait, but in the seventh draw, we have 2 white balls left, so the probability is 2/2, which is 1. Then, the eighth draw, we have 1 white ball left, so the probability is 1/1, which is also 1. So, the eighth draw is certain.But wait, let me check the denominators:After first draw: 7 balls leftAfter second draw: 6 balls leftAfter third draw: 5 balls leftAfter fourth draw: 4 balls leftAfter fifth draw: 3 balls leftAfter sixth draw: 2 balls leftAfter seventh draw: 1 ball leftAfter eighth draw: 0 balls leftSo, the denominators are correct.Now, let's compute the product:(5/8) * (3/7) * (4/6) * (2/5) * (3/4) * (1/3) * (2/2) * (1/1)Simplify each fraction:5/8 remains3/7 remains4/6 simplifies to 2/32/5 remains3/4 remains1/3 remains2/2 is 11/1 is 1So, the product becomes:(5/8) * (3/7) * (2/3) * (2/5) * (3/4) * (1/3) * 1 * 1Now, let's multiply them step by step:First, 5/8 * 3/7 = (15)/(56)Then, 15/56 * 2/3 = (30)/(168) = 5/28Next, 5/28 * 2/5 = (10)/(140) = 1/14Then, 1/14 * 3/4 = 3/56Next, 3/56 * 1/3 = 1/56So, the final probability is 1/56.Wait, that's the same result as before. So, whether I think of it as the number of successful sequences over total sequences or calculate the step-by-step probability, I get 1/56.But let me make sure I didn't make a mistake in the step-by-step calculation. Let me go through it again:(5/8) * (3/7) = 15/5615/56 * (4/6) = 15/56 * 2/3 = (15*2)/(56*3) = 30/168 = 5/285/28 * (2/5) = (5*2)/(28*5) = 10/140 = 1/141/14 * (3/4) = 3/563/56 * (1/3) = 1/56Yes, that's correct. So, both methods give me the same result, 1/56.Alternatively, I could think of it as arranging the balls in a specific order. Since there are 8 balls, and we want a specific sequence where colors alternate starting with white, there's only one such sequence. The total number of possible sequences is the number of ways to arrange 5 white and 3 black balls, which is 8 choose 3 = 56. So, the probability is 1/56.Wait, but hold on, is that accurate? Because when arranging the balls, the number of distinct sequences is 8! / (5!3!) = 56. So, yes, that's correct. And only one of those sequences is WBWBWBWB. Therefore, the probability is 1/56.I think that makes sense. So, the probability that all draws alternate in color starting with a white ball is 1/56.