🧠:[<think>Okay, so I have two parts to solve here. Let me start with the first one: solving the inequality ( f(x) + f(x - 1) leqslant 2 ) where ( f(x) = |x| ). Hmm, since ( f(x) ) is the absolute value function, this translates to ( |x| + |x - 1| leq 2 ). I need to find all real numbers x that satisfy this inequality.Absolute value inequalities can sometimes be tricky because they involve different cases depending on the value of x. Let me recall that ( |x| ) is equal to x when x is non-negative and -x when x is negative. Similarly, ( |x - 1| ) is x - 1 when x >= 1 and -(x - 1) when x < 1. So, maybe I should break this problem into different intervals based on the critical points where the expressions inside the absolute values change sign, which are x = 0 and x = 1.So, the critical points are at x = 0 and x = 1. Therefore, the number line is divided into three intervals:1. x < 02. 0 ≤ x < 13. x ≥ 1I need to consider each interval separately.Case 1: x < 0In this interval, |x| = -x because x is negative, and |x - 1| = -(x - 1) = -x + 1 because x - 1 is also negative (since x < 0 < 1). Therefore, the inequality becomes:(-x + (-x + 1) leq 2)Simplify:(-2x + 1 leq 2)Subtract 1 from both sides:(-2x leq 1)Divide by -2, remembering to reverse the inequality sign:(x geq -frac{1}{2})But wait, this case is for x < 0. So the solution in this interval is the intersection of x < 0 and x ≥ -1/2. Therefore, ( -frac{1}{2} leq x < 0 ).Case 2: 0 ≤ x < 1Here, |x| = x because x is non-negative, and |x - 1| = -(x - 1) = -x + 1 because x - 1 is still negative (since x < 1). So the inequality becomes:(x + (-x + 1) leq 2)Simplify:(0x + 1 leq 2)Which simplifies to:(1 leq 2)This is always true. So in this interval, all x between 0 and 1 satisfy the inequality. Therefore, 0 ≤ x < 1 is part of the solution.Case 3: x ≥ 1In this interval, both |x| = x and |x - 1| = x - 1 because both x and x - 1 are non-negative. So the inequality becomes:(x + (x - 1) leq 2)Simplify:(2x - 1 leq 2)Add 1 to both sides:(2x leq 3)Divide by 2:(x leq frac{3}{2})But this case is for x ≥ 1, so the solution here is the intersection of x ≥ 1 and x ≤ 3/2. Therefore, (1 leq x leq frac{3}{2}).Combining all cases:From Case 1: ( -frac{1}{2} leq x < 0 )From Case 2: ( 0 leq x < 1 )From Case 3: (1 leq x leq frac{3}{2} )Combining these, the entire solution set is ( -frac{1}{2} leq x leq frac{3}{2} ).So the solution to part (1) is ( x in left[ -frac{1}{2}, frac{3}{2} right] ). Let me double-check.For x between -1/2 and 0: The sum |x| + |x - 1| = -x + (-x +1) = -2x +1. At x = -1/2, this is -2*(-1/2) +1 = 1 +1 =2. At x approaching 0 from the left, it's approaching 0 +1 =1. So between -1/2 and 0, the sum goes from 2 down to 1, which is ≤2. Correct.Between 0 and1: The sum is x + (1 -x) =1, which is always 1 ≤2. Correct.Between 1 and 3/2: The sum is x + (x -1) =2x -1. At x=1, that's 1, and at x=3/2, that's 3 -1=2. So the sum increases from1 to2, which is all ≤2. Correct.So yes, the solution is from -1/2 to 3/2. So part (1) is done.Now part (2): For ( a + b =1 ) with ( a, b >0 ), and for all real numbers x, the inequality ( f(x - m) - |x + 2| leq frac{1}{a} + frac{1}{b} ) always holds. Find the range of real number m.First, let's parse the problem. We have ( f(x) = |x| ), so ( f(x - m) = |x - m| ). Therefore, the inequality is:( |x - m| - |x + 2| leq frac{1}{a} + frac{1}{b} )Given that ( a + b =1 ) and ( a, b >0 ), and this inequality must hold for all real x. We need to find the range of m such that this is true.First, note that ( frac{1}{a} + frac{1}{b} ) can be expressed in terms of a and b. Since ( a + b =1 ), we can write:( frac{1}{a} + frac{1}{b} = frac{a + b}{ab} = frac{1}{ab} )So the inequality becomes:( |x - m| - |x + 2| leq frac{1}{ab} )But since ( a + b =1 ), we can also note that by AM-GM inequality, ( ab leq left( frac{a + b}{2} right)^2 = frac{1}{4} ), so ( frac{1}{ab} geq 4 ). Equality holds when ( a = b = frac{1}{2} ).But perhaps that's a detour. Wait, the problem states that for given a and b with ( a + b =1 ), and for all x, the inequality must hold. So the right-hand side ( frac{1}{a} + frac{1}{b} ) is a constant depending on a and b. But the problem says "for a + b =1 (a, b >0) and ∀x∈ℝ, the inequality... always holds". So perhaps m must be chosen such that, regardless of a and b (as long as a + b =1), the inequality holds? Wait, maybe not.Wait, the problem states: "For a + b =1 (a, b >0) and ∀x∈ℝ, the inequality ... always holds. Find the range of real number m."Wait, maybe it's for given a and b with a + b =1, and the inequality holds for all x. Then we need to find m such that, for all a, b >0 with a + b=1, the inequality is true for all x. Or maybe it's for each a,b with a + b=1, the inequality holds for all x, and find m. Hmm, the wording is a bit ambiguous.Wait, the problem says: "For a + b =1 (a, b >0) and ∀x∈ℝ, the inequality ... always holds. Find the range of real number m."So "for a + b=1 and for all x in R, the inequality holds". So the inequality must hold for all x and for all a, b >0 with a + b=1. Therefore, m must be such that regardless of a and b (as long as they are positive and sum to 1), the inequality is true for all x. Therefore, m has to satisfy that the left-hand side ( |x - m| - |x + 2| ) is less than or equal to ( frac{1}{a} + frac{1}{b} ) for all x and for all a, b >0 with a + b=1.But since a and b can vary (as long as a + b=1), the right-hand side ( frac{1}{a} + frac{1}{b} ) can take different values. However, the inequality needs to hold for all such a and b. Therefore, to satisfy the inequality for all a and b, the left-hand side must be less than or equal to the minimal possible value of ( frac{1}{a} + frac{1}{b} ).Wait, but if the inequality has to hold for all a and b, then ( |x - m| - |x +2| leq frac{1}{a} + frac{1}{b} ) for all x and for all a, b with a + b=1. That means that the left-hand side, which is independent of a and b, must be less than or equal to the minimal value of ( frac{1}{a} + frac{1}{b} ) over all a, b >0 with a + b=1.Wait, because if the left-hand side is ≤ the minimal value of the right-hand side, then it will automatically be ≤ any larger value. But since the inequality must hold for all a and b, the left-hand side must be ≤ the infimum of ( frac{1}{a} + frac{1}{b} ). But the infimum of ( frac{1}{a} + frac{1}{b} ) given a + b=1 is 4, achieved when a = b = 1/2.Wait, let me confirm:Given a + b =1, then ( frac{1}{a} + frac{1}{b} = frac{a + b}{ab} = frac{1}{ab} ). To minimize ( frac{1}{ab} ), we need to maximize ab. By AM-GM, ab ≤ ( (a + b)/2 )² = (1/2)² = 1/4. So maximum ab is 1/4, hence minimum ( frac{1}{ab} =4 ).Therefore, the minimal value of ( frac{1}{a} + frac{1}{b} ) is 4. Therefore, if the inequality ( |x - m| - |x +2| leq 4 ) holds for all x, then since for any a, b, ( frac{1}{a} + frac{1}{b} geq4 ), the original inequality would hold. Conversely, if the left-hand side exceeds 4 for some x, then choosing a = b =1/2 would make the right-hand side 4, and the inequality would fail. Therefore, the inequality holds for all a, b if and only if ( |x - m| - |x +2| leq4 ) for all x. Therefore, the problem reduces to finding m such that for all x, ( |x - m| - |x +2| leq4 ). But we also need to ensure that this inequality holds for all x. Therefore, m must be chosen such that the maximum value of ( |x - m| - |x +2| ) over all x is ≤4. However, since ( |x - m| - |x +2| ) is a function of x, we can analyze its maximum value.But actually, we need that for all x, ( |x - m| - |x +2| leq4 ). Therefore, the maximum of ( |x - m| - |x +2| ) over x must be ≤4.Alternatively, we can think of this as the difference between two absolute values. The expression ( |x - m| - |x + 2| ) can be bounded using the triangle inequality. Recall that ( |x - m| leq |x + 2| + |m + 2| ), by triangle inequality ( |x - m| = |(x + 2) - (m + 2)| leq |x + 2| + |m + 2| ). Therefore,( |x - m| - |x + 2| leq |m + 2| )Similarly, ( |x - m| - |x + 2| geq -|m - (-2)| = -|m + 2| ). Wait, but maybe a different approach.Alternatively, the expression ( |x - m| - |x + 2| ) can be considered as a function of x. Let's denote ( h(x) = |x - m| - |x + 2| ). We need to find the maximum value of h(x) over all x, and set that maximum to be ≤4.To find the maximum of h(x), we can analyze the function h(x). Let's consider different cases based on the critical points where the expressions inside the absolute values change sign, which are x = m and x = -2.So, the critical points are x = m and x = -2. Therefore, depending on the relation between m and -2, the intervals will change. Let's suppose m >= -2 and m < -2.But maybe another approach: the function h(x) can be rewritten depending on the position of x relative to m and -2. Let's consider different intervals.Case 1: x ≥ max(m, -2)In this case, both |x - m| = x - m and |x + 2| = x + 2. So h(x) = (x - m) - (x + 2) = -m -2.Case 2: x ≤ min(m, -2)Here, |x - m| = m - x and |x + 2| = -x -2. So h(x) = (m - x) - (-x -2) = m + 2.Case 3: Between -2 and m (assuming m > -2)If -2 < m, then the interval -2 ≤ x ≤ m. Here, |x - m| = m - x and |x +2| = x +2. So h(x) = (m - x) - (x +2) = m -2 -2x.Case 4: Between m and -2 (if m < -2)If m < -2, then the interval m ≤ x ≤ -2. Here, |x - m| = x - m and |x +2| = -x -2. So h(x) = (x - m) - (-x -2) = 2x - m +2.So depending on whether m is greater than or less than -2, the function h(x) has different expressions in the middle interval.Let me first assume m >= -2.Case when m ≥ -2:Then the critical points are x = -2 and x = m. So intervals are:1. x ≤ -2: h(x) = m +22. -2 ≤ x ≤ m: h(x) = m -2 -2x3. x ≥ m: h(x) = -m -2Now, let's find the maximum of h(x) in each interval.1. For x ≤ -2, h(x) = m +2. Since m >= -2, this is a constant value. The maximum here is m +2.2. For -2 ≤ x ≤ m: h(x) = m -2 -2x. This is a linear function in x with slope -2, decreasing as x increases. Therefore, the maximum in this interval occurs at x = -2: h(-2) = m -2 -2*(-2) = m -2 +4 = m +2. The minimum in this interval occurs at x = m: h(m) = m -2 -2m = -m -2.3. For x >= m: h(x) = -m -2. This is a constant, which is less than or equal to h(x) in the first interval since -m -2 <= m +2 only if 2m +4 >=0, which is m >= -2, which is our assumption here.Therefore, the maximum value of h(x) is m +2. Therefore, to have h(x) <=4 for all x, we need m +2 <=4 => m <=2.But we assumed m >= -2. So combining both, m must be in [-2, 2].Case when m < -2:Now the critical points are x = m and x = -2. So intervals:1. x ≤ m: h(x) = m +22. m ≤ x ≤ -2: h(x) = 2x - m +23. x >= -2: h(x) = -m -2Again, let's find the maximum in each interval.1. For x <=m: h(x) = m +2. Since m < -2, this is m +2 < 0.2. For m <=x <=-2: h(x) =2x -m +2. This is linear in x with slope 2, increasing. So maximum at x=-2: h(-2)= 2*(-2) -m +2= -4 -m +2= -m -2. The minimum at x=m: h(m)=2m -m +2=m +2.3. For x >= -2: h(x) = -m -2. Since m < -2, -m is positive, so -m -2 >0.Therefore, the maximum value of h(x) is max(m +2, -m -2). Since m < -2, m +2 is less than 0, and -m -2 is positive. So the maximum is -m -2. Therefore, to have h(x) <=4, we need -m -2 <=4 => -m <=6 => m >= -6.But since we assumed m < -2, combining these gives m in [-6, -2).Combining both cases:If m >= -2, then m <=2.If m < -2, then m >=-6.Therefore, the range of m is [-6, 2].But let's verify this.Wait, when m is between -6 and 2, inclusive, then the maximum of h(x) is either m +2 (when m >=-2, maximum m +2 <=4 implies m <=2) or -m -2 (when m < -2, maximum -m -2 <=4 implies m >= -6). Thus, combining both, m ∈ [-6, 2].But we need to check if this is correct.Let me take m =2. Then h(x) = |x -2| - |x +2|. Let's check when x is large positive: |x -2| - |x +2| = (x -2) - (x +2) = -4, which is ≤4. When x is large negative: |x -2| - |x +2| = (-x +2) - (-x -2) =4, which is exactly 4. So the maximum is 4, which is equal to the required 4. So m=2 is acceptable.Similarly, m=-6. Then h(x) = |x +6| - |x +2|. Let's check x=-2: | -2 +6 | - | -2 +2 | =4 -0=4. x=-6: | -6 +6 | - | -6 +2 | =0 -4= -4. For x approaching infinity: |x +6| - |x +2| approaches 0. For x approaching -infinity: |x +6| - |x +2| approaches ( -x -6 ) - ( -x -2 )= -4. The maximum is 4 at x=-2, so m=-6 is acceptable.If m is between -6 and2, then the maximum is either m +2 (for m >=-2) or -m -2 (for m <=-2). For example, take m=0. Then h(x) = |x| - |x +2|. The maximum of this function is at x=-2, where it's |-2| -0=2. Which is less than 4. Wait, but according to our previous analysis, for m >=-2, the maximum is m +2. But when m=0, m +2=2, which is indeed the maximum. So since 2<=4, this is okay. But wait, in the problem statement, the RHS is 4. So if the maximum of h(x) is 4 when m=2 or m=-6, but for other m in between, the maximum is less than 4, then those m would satisfy the inequality h(x) <=4, which is required. Therefore, the allowable m is from -6 to2 inclusive.Hence, the range of m is [-6, 2].Therefore, the answer to part (2) is m ∈ [-6, 2].But let me check another value. Take m=-3 (which is between -6 and -2). Then h(x)=|x +3| - |x +2|. Let's find the maximum of this.When x >=-2: |x +3| - |x +2| = (x +3) - (x +2)=1.When x <=-3: |x +3| - |x +2|= (-x -3) - (-x -2)= -1.When -3 <=x <=-2: |x +3| - |x +2|=(x +3) - (-x -2)=2x +5. This is a linear function from x=-3: 2*(-3)+5=-1 to x=-2:2*(-2)+5=1.Therefore, the maximum is 1, which is less than4. Wait, but according to our previous analysis, the maximum should be -m -2=3 -2=1, which is correct. So h(x) <=1 in this case, which is ≤4. Therefore, m=-3 is acceptable. Wait, but according to our conclusion m must be >=-6. So m=-3 is in the range [-6, 2], which is correct. Similarly, take m=-6. Then h(x)=|x +6| - |x +2|. As checked earlier, maximum is4, which is equal to the required bound. Similarly, m=2 gives maximum4. So all m between -6 and2, inclusive, will have h(x) <=4. Therefore, the answer is m ∈ [-6,2].Thus, part (2)'s answer is m ∈ [-6,2].Final Answer(1) The solution is boxed{left[ -dfrac{1}{2}, dfrac{3}{2} right]}.(2) The range of ( m ) is boxed{[-6, 2]}.</think>(1) To solve the inequality ( f(x) + f(x - 1) leqslant 2 ) where ( f(x) = |x| ), we translate this to ( |x| + |x - 1| leq 2 ). We consider different intervals based on the critical points ( x = 0 ) and ( x = 1 ):- Case 1: ( x < 0 ) [ -x + (-x + 1) leq 2 implies -2x + 1 leq 2 implies x geq -frac{1}{2} ] Solution: ( -frac{1}{2} leq x < 0 )- Case 2: ( 0 leq x < 1 ) [ x + (-x + 1) leq 2 implies 1 leq 2 ] Solution: ( 0 leq x < 1 )- Case 3: ( x geq 1 ) [ x + (x - 1) leq 2 implies 2x - 1 leq 2 implies x leq frac{3}{2} ] Solution: ( 1 leq x leq frac{3}{2} )Combining all cases, the solution is ( left[ -frac{1}{2}, frac{3}{2} right] ).(2) For the inequality ( |x - m| - |x + 2| leq frac{1}{a} + frac{1}{b} ) with ( a + b = 1 ) and ( a, b > 0 ), we need to find the range of ( m ). The right-hand side simplifies to ( frac{1}{ab} geq 4 ) by AM-GM inequality. We need ( |x - m| - |x + 2| leq 4 ) for all ( x ).Analyzing the function ( h(x) = |x - m| - |x + 2| ):- Case 1: ( m geq -2 ) - Maximum value is ( m + 2 ). To satisfy ( m + 2 leq 4 ), we get ( m leq 2 ).- Case 2: ( m < -2 ) - Maximum value is ( -m - 2 ). To satisfy ( -m - 2 leq 4 ), we get ( m geq -6 ).Combining both cases, the range of ( m ) is ( [-6, 2] ).Final answers:(1) boxed{left[ -dfrac{1}{2}, dfrac{3}{2} right]}(2) boxed{[-6, 2]},<think>Okay, let's tackle these two problems step by step. Starting with part (1): Solve the inequality ( f(x) + f(x - 1) leqslant 2 ) where ( f(x) = |x| ). So, substituting the function, the inequality becomes ( |x| + |x - 1| leqslant 2 ). Hmm, absolute value inequalities can sometimes be tricky because they involve different cases depending on the value of ( x ). Let me remember that the expression ( |x| + |x - 1| ) is the sum of two absolute values. This sum represents the distance from ( x ) to 0 plus the distance from ( x ) to 1 on the number line. I need to find all ( x ) where this total distance is less than or equal to 2.But maybe breaking it into cases based on the critical points where the expressions inside the absolute values change sign, which are at ( x = 0 ) and ( x = 1 ). So, we'll consider three intervals: ( x < 0 ), ( 0 leqslant x leqslant 1 ), and ( x > 1 ).Case 1: ( x < 0 ) In this case, ( |x| = -x ) (since ( x ) is negative) and ( |x - 1| = -(x - 1) = -x + 1 ) (since ( x - 1 ) is also negative when ( x < 0 )). So, substituting into the inequality:[- x + (-x + 1) leqslant 2 -2x + 1 leqslant 2 -2x leqslant 1 x geqslant -frac{1}{2}]But remember we are in the case where ( x < 0 ), so combining these, ( -frac{1}{2} leqslant x < 0 ).Case 2: ( 0 leqslant x leqslant 1 ) Here, ( |x| = x ) (since ( x ) is non-negative) and ( |x - 1| = -(x - 1) = -x + 1 ) (since ( x - 1 leqslant 0 )). So:[x + (-x + 1) leqslant 2 0x + 1 leqslant 2 1 leqslant 2]This is always true. Therefore, all ( x ) in ( [0, 1] ) satisfy the inequality.Case 3: ( x > 1 ) Now, both ( |x| = x ) and ( |x - 1| = x - 1 ). So:[x + (x - 1) leqslant 2 2x - 1 leqslant 2 2x leqslant 3 x leqslant frac{3}{2}]Since we're in the case ( x > 1 ), combining gives ( 1 < x leqslant frac{3}{2} ).Putting all three cases together:- From Case 1: ( -frac{1}{2} leqslant x < 0 )- From Case 2: ( 0 leqslant x leqslant 1 )- From Case 3: ( 1 < x leqslant frac{3}{2} )Combining these intervals, the solution is ( -frac{1}{2} leqslant x leqslant frac{3}{2} ). Let me double-check by testing a value in each interval.Testing ( x = -0.5 ):( | -0.5 | + | -0.5 - 1 | = 0.5 + 1.5 = 2 ), which is equal to 2. Good.Testing ( x = 0 ):( 0 + |0 - 1| = 1 ), which is less than 2. Okay.Testing ( x = 1 ):( 1 + 0 = 1 ), also less than 2.Testing ( x = 1.5 ):( 1.5 + 0.5 = 2 ), which is equal to 2. Good.What about outside the interval? Let's try ( x = -1 ):( 1 + 2 = 3 > 2 ). Not okay. And ( x = 2 ):( 2 + 1 = 3 > 2 ). Correct. So the solution seems to be from -0.5 to 1.5. So part (1)'s answer is ( -frac{1}{2} leqslant x leqslant frac{3}{2} ).Moving on to part (2): Given ( a + b = 1 ) with ( a, b > 0 ), and for all real ( x ), the inequality ( f(x - m) - |x + 2| leqslant frac{1}{a} + frac{1}{b} ) holds. Find the range of real number ( m ).First, let's parse the problem. We have ( f(x - m) = |x - m| ), so the inequality becomes:[|x - m| - |x + 2| leqslant frac{1}{a} + frac{1}{b}]And this must hold for all ( x in mathbb{R} ). Also, ( a + b = 1 ), with ( a, b > 0 ). We need to find the possible values of ( m ) such that the inequality is always true, regardless of ( x ).Hmm. Since the inequality has to hold for all ( x ), the left-hand side (LHS) must be less than or equal to the right-hand side (RHS) for every ( x ). Therefore, the maximum value of the LHS over all ( x ) must be less than or equal to the RHS. So, if we can find the maximum of ( |x - m| - |x + 2| ), then set that maximum to be ( leqslant frac{1}{a} + frac{1}{b} ), and then find ( m ) accordingly.But we also have the condition ( a + b = 1 ). The RHS ( frac{1}{a} + frac{1}{b} ) can be rewritten using the given that ( a + b = 1 ). Let's see:( frac{1}{a} + frac{1}{b} = frac{a + b}{ab} = frac{1}{ab} ). So the RHS is ( frac{1}{ab} ). Since ( a + b = 1 ), by AM-GM inequality, ( ab leqslant left( frac{a + b}{2} right)^2 = frac{1}{4} ). Therefore, ( frac{1}{ab} geqslant 4 ), with equality when ( a = b = frac{1}{2} ). So the minimum possible value of ( frac{1}{ab} ) is 4, but since ( a ) and ( b ) can be any positive numbers adding to 1, ( frac{1}{ab} ) can be made arbitrarily large by making either ( a ) or ( b ) approach 0. However, in the problem statement, it's given that the inequality holds for all ( x in mathbb{R} ). So the RHS ( frac{1}{a} + frac{1}{b} ) can be as large as needed, but we need the inequality to hold for all ( a, b > 0 ) with ( a + b = 1 ). Wait, no—actually, re-reading the problem:"For ( a + b = 1 ) ( ( a, b > 0 ) ) and ( forall x in mathbb{R} ), the inequality ( f(x - m) - |x + 2| leqslant frac{1}{a} + frac{1}{b} ) always holds. Find the range of real number ( m )."Wait, the wording is a bit ambiguous. Is the inequality supposed to hold for all ( x in mathbb{R} ) given some fixed ( a, b > 0 ) with ( a + b = 1 ), or is it that the inequality must hold for all ( x in mathbb{R} ) and for all ( a, b > 0 ) with ( a + b = 1 )? The problem says: "For ( a + b = 1 ) ( ( a, b > 0 ) ) and ( forall x in mathbb{R} ), the inequality... always holds." So I think it's for given ( a, b ), with ( a + b = 1 ), the inequality must hold for all ( x ). Then, find the range of ( m ) such that for all such ( a, b ), the inequality holds. Wait, no. Wait, actually, the problem says: "For ( a + b = 1 ) ( ( a, b > 0 ) ) and ( forall x in mathbb{R} ), the inequality... always holds." So the parameters are ( a, b ) with ( a + b = 1 ), and for all ( x in mathbb{R} ), the inequality is true. Then, we need to find the range of ( m ). So perhaps ( a ) and ( b ) are fixed, and ( m ) must be chosen such that for that ( a, b ), the inequality holds for all ( x ). But the wording is a bit unclear. Wait, maybe it's that ( a ) and ( b ) are given as positive real numbers with ( a + b = 1 ), and the inequality must hold for all real numbers ( x ). Then, find the range of ( m ). But the problem says "Find the range of real number ( m )" given that for all ( x in mathbb{R} ), the inequality holds. So maybe the problem is: Given that ( a + b = 1 ) with ( a, b > 0 ), and for all ( x in mathbb{R} ), the inequality ( |x - m| - |x + 2| leq frac{1}{a} + frac{1}{b} ). Find the range of ( m ).But the question is a bit ambiguous on whether ( a ) and ( b ) are fixed or variable. If ( a ) and ( b ) are fixed, then ( frac{1}{a} + frac{1}{b} ) is a constant, and we need to find ( m ) such that ( |x - m| - |x + 2| leq C ) for all ( x ), where ( C = frac{1}{a} + frac{1}{b} ). However, if ( a ) and ( b ) are variables satisfying ( a + b = 1 ), then ( frac{1}{a} + frac{1}{b} ) can vary, and the inequality must hold for all such ( a, b ). But in that case, since ( frac{1}{a} + frac{1}{b} ) can be made arbitrarily large (by making ( a ) approach 0 and ( b ) approach 1, or vice versa), unless there's a minimum value. Wait, but the inequality is supposed to hold for all ( a, b > 0 ) with ( a + b = 1 ). Therefore, the inequality must be true for any such ( a ) and ( b ), but the LHS ( |x - m| - |x + 2| ) doesn't involve ( a ) or ( b ). So, for the inequality ( |x - m| - |x + 2| leq frac{1}{a} + frac{1}{b} ) to hold for all ( x in mathbb{R} ) and for all ( a, b > 0 ) with ( a + b = 1 ), then the LHS must be less than or equal to the minimal possible value of ( frac{1}{a} + frac{1}{b} ), which is 4, as we saw earlier. Because if it's less than or equal to 4, then since all other ( frac{1}{a} + frac{1}{b} geq 4 ), the inequality would hold. Alternatively, if the problem is that for some ( a, b ), the inequality must hold for all ( x ), then ( m ) must satisfy that ( |x - m| - |x + 2| leq frac{1}{a} + frac{1}{b} ) for all ( x ), but since ( a ) and ( b ) can vary, perhaps we need ( |x - m| - |x + 2| leq 4 ) for all ( x ), since 4 is the minimal RHS. Because if we require the inequality to hold for all ( a, b ), then the RHS can be as small as 4 (when ( a = b = 1/2 )), hence the LHS must be ≤4. Alternatively, if the problem is that for given ( a, b ), the inequality holds for all ( x ), then the answer would depend on ( a, b ), but since the problem asks for the range of ( m ), regardless of ( a, b ), perhaps it's the former interpretation. This is a bit confusing. Let me try to clarify.The problem states: "For ( a + b = 1 ) ( ( a, b > 0 ) ) and ( forall x in mathbb{R} ), the inequality ( f(x - m) - |x + 2| leq frac{1}{a} + frac{1}{b} ) always holds. Find the range of real number ( m )."The way it's phrased is that both ( a, b ) (satisfying ( a + b = 1 )) and ( x ) are universally quantified. That is, "For all ( a, b > 0 ) with ( a + b = 1 ), and for all ( x in mathbb{R} ), the inequality holds." Therefore, the inequality must hold for every ( x ) and every such ( a, b ). Therefore, the LHS ( |x - m| - |x + 2| ) must be less than or equal to ( frac{1}{a} + frac{1}{b} ) for all ( x in mathbb{R} ) and all ( a, b > 0 ) with ( a + b = 1 ).But since ( frac{1}{a} + frac{1}{b} ) can be made arbitrarily large (by choosing ( a ) close to 0 and ( b ) close to 1, or vice versa), the only constraint comes from the minimal value of ( frac{1}{a} + frac{1}{b} ), which is 4. Because if ( |x - m| - |x + 2| leq 4 ) for all ( x ), then since for any other ( a, b ), ( frac{1}{a} + frac{1}{b} geq 4 ), the inequality would automatically hold. Therefore, the problem reduces to finding ( m ) such that ( |x - m| - |x + 2| leq 4 ) for all ( x in mathbb{R} ). But actually, since the inequality must hold for all ( a, b ), including the case when ( a = b = 1/2 ), leading to ( frac{1}{a} + frac{1}{b} = 4 ), so the LHS must be ≤4 for all ( x ). Therefore, the maximum value of ( |x - m| - |x + 2| ) over all ( x ) must be ≤4. But actually, we need the LHS to be ≤ the RHS for all ( x ). However, the RHS varies depending on ( a, b ). Wait, no, in the inequality, the RHS is ( frac{1}{a} + frac{1}{b} ), which is fixed once ( a, b ) are fixed. But the problem states that the inequality must hold for all ( a, b ) with ( a + b =1 ), and for all ( x ). Therefore, for each ( x ), and each ( a, b ), the inequality must hold. Therefore, the LHS (which depends on ( x ) and ( m )) must be ≤ the RHS (which depends on ( a, b )) for all ( x ), ( a ), ( b ).But how can this be possible? Because for a fixed ( x ), the LHS is a fixed number, but the RHS can be made as large as we like by choosing ( a ) approaching 0 or 1. Therefore, if we require that for each ( x ), ( |x - m| - |x + 2| leq frac{1}{a} + frac{1}{b} ) holds for all ( a, b >0 ) with ( a + b =1 ), then since ( frac{1}{a} + frac{1}{b} ) can be made larger than any fixed number, the only real constraint is that the LHS must be ≤ the minimal possible RHS, which is 4. Because if the LHS is ≤4, then for all ( a, b ), since ( frac{1}{a} + frac{1}{b} geq4 ), the inequality will hold. If the LHS were greater than 4 for some ( x ), then choosing ( a, b ) such that ( frac{1}{a} + frac{1}{b} ) is just slightly above 4 (i.e., ( a, b ) close to 1/2) would make the inequality fail. Therefore, the necessary and sufficient condition is that the maximum of ( |x - m| - |x + 2| ) over all ( x ) is ≤4. But actually, wait, no. Wait, for the inequality to hold for all ( a, b ), including those that make ( frac{1}{a} + frac{1}{b} ) equal to 4. So, the maximum value of the LHS over ( x ) must be ≤4. Because if the maximum LHS is ≤4, then for all ( x ), the LHS is ≤4, and since ( frac{1}{a} + frac{1}{b} geq4 ), the inequality holds. Conversely, if the maximum LHS exceeds 4, then for ( a, b =1/2 ), the RHS is exactly 4, so the inequality would fail for that ( x ).Therefore, the key is to compute the maximum of ( |x - m| - |x + 2| ) over all ( x in mathbb{R} ), and set this maximum to be ≤4. Then, find the values of ( m ) for which this holds.So, first, let's find the maximum of ( |x - m| - |x + 2| ). To do this, we can analyze the function ( g(x) = |x - m| - |x + 2| ). The maximum of this function will depend on ( m ).Let me recall that expressions of the form ( |x - a| - |x - b| ) have their maximum and minimum values determined by the positions of ( a ) and ( b ). Specifically, the maximum value is ( |a - b| ) and the minimum is ( -|a - b| ). Wait, is that correct?Wait, more carefully, consider ( h(x) = |x - a| - |x - b| ). Let's analyze this function.Case 1: ( x geq max(a, b) ). Then, ( |x - a| = x - a ), ( |x - b| = x - b ). So, ( h(x) = (x - a) - (x - b) = b - a ).Case 2: ( x leq min(a, b) ). Then, ( |x - a| = a - x ), ( |x - b| = b - x ). So, ( h(x) = (a - x) - (b - x) = a - b ).Case 3: If ( a < b ), then between ( a ) and ( b ), when ( a leq x leq b ), ( |x - a| = x - a ), ( |x - b| = b - x ). Then, ( h(x) = (x - a) - (b - x) = 2x - a - b ), which is a linear function increasing from ( - (b - a) ) to ( b - a ).Similarly, if ( b < a ), between ( b ) and ( a ), it's linear decreasing from ( a - b ) to ( - (a - b) ).Therefore, the maximum value of ( h(x) = |x - a| - |x - b| ) is ( |a - b| ) if ( a neq b ), otherwise it's 0. Wait, no. Wait, if ( a < b ), then the maximum is when ( x geq b ), giving ( b - a ), and the minimum is when ( x leq a ), giving ( a - b ). But in the middle, between ( a ) and ( b ), it varies from ( a - b ) to ( b - a ). Wait, but that can't be. Wait, in Case 3 when ( a < b ), between ( a ) and ( b ), the function ( h(x) = 2x - a - b ), which when ( x = a ), is ( 2a - a - b = a - b ), and when ( x = b ), is ( 2b - a - b = b - a ). So, as ( x ) increases from ( a ) to ( b ), ( h(x) ) increases from ( a - b ) to ( b - a ). Therefore, overall, the maximum value of ( h(x) ) is ( |a - b| ), occurring when ( x geq max(a, b) ), and the minimum is ( -|a - b| ), occurring when ( x leq min(a, b) ).But in our problem, ( g(x) = |x - m| - |x + 2| ). So this is ( |x - m| - |x - (-2)| ). So comparing to ( h(x) ), it's ( h(x) = |x - a| - |x - b| ) with ( a = m ), ( b = -2 ). Therefore, the maximum value of ( g(x) ) is ( |m - (-2)| = |m + 2| ), and the minimum value is ( -|m + 2| ).Therefore, the maximum of ( g(x) = |x - m| - |x + 2| ) over all ( x ) is ( |m + 2| ). Therefore, to ensure that ( |x - m| - |x + 2| leqslant frac{1}{a} + frac{1}{b} ) for all ( x in mathbb{R} ) and all ( a, b > 0 ) with ( a + b = 1 ), we need that the maximum value ( |m + 2| leqslant frac{1}{a} + frac{1}{b} ) for all such ( a, b ).But as we established earlier, ( frac{1}{a} + frac{1}{b} geq 4 ), with equality when ( a = b = 1/2 ). Therefore, to have ( |m + 2| leqslant frac{1}{a} + frac{1}{b} ) for all ( a, b ), we must have ( |m + 2| leqslant 4 ). Because if ( |m + 2| leq 4 ), then since all ( frac{1}{a} + frac{1}{b} geq4 ), the inequality holds. Conversely, if ( |m + 2| >4 ), then when ( a = b = 1/2 ), the RHS is 4, so the inequality ( |m + 2| leq4 ) must hold. Therefore, the necessary and sufficient condition is ( |m + 2| leq4 ), which simplifies to ( -4 leq m + 2 leq4 ), so ( -6 leq m leq2 ).Wait, let me verify this conclusion. If the maximum value of ( |x - m| - |x + 2| ) is ( |m + 2| ), then requiring that this maximum is ≤ the minimal possible RHS, which is 4. Therefore, ( |m + 2| leq4 ), which leads to ( -4 leq m + 2 leq4 implies -6 leq m leq2 ). But let me double-check this maximum value. Let's take specific examples. Suppose ( m = 0 ). Then ( |x| - |x + 2| ). The maximum of this function would be ( |0 + 2| = 2 ). So according to our formula, ( |0 + 2| = 2 ), which is correct. If ( m = -6 ), then ( |x - (-6)| - |x + 2| = |x + 6| - |x + 2| ). The maximum value is ( | -6 + 2 | = 4 ). Similarly, if ( m = 2 ), then ( |x - 2| - |x + 2| ). The maximum value is ( |2 + 2| = 4 ). So indeed, when ( m = -6 ) or ( m =2 ), the maximum is 4, and for ( m ) between them, the maximum is less than 4.Therefore, the range of ( m ) is ( -6 leq m leq2 ).But let me make sure there are no mistakes here. Let's pick a value of ( m ) outside this range, say ( m =3 ). Then the maximum value should be ( |3 + 2| =5 ). So for ( x geq3 ), ( |x -3| - |x + 2| = (x -3) - (x +2)= -5 ). Wait, that's negative. Wait, but according to our earlier analysis, the maximum is when ( x geq max(m, -2) ). Wait, in this case, ( m =3 ), so max(3, -2)=3. So when ( x geq3 ), the function becomes ( (x -3) - (x + 2) = -5 ). Wait, that can't be. Wait, but according to the earlier analysis, the maximum is |m + 2|. Wait, m =3, so |3 +2|=5. But how?Wait, maybe I made a mistake in the earlier analysis. Let me re-examine.The function ( g(x) = |x - m| - |x + 2| ). To find its maximum.Let me consider different cases based on the relative positions of ( m ) and -2.Case 1: ( x geq max(m, -2) ). Then, ( |x - m| = x - m ), ( |x + 2| = x + 2 ). Therefore, ( g(x) = (x - m) - (x +2) = -m -2 ).Case 2: ( x leq min(m, -2) ). Then, ( |x - m| = m - x ), ( |x + 2| = -x -2 ). Therefore, ( g(x) = (m - x) - (-x -2) = m +2 ).Case 3: If ( m > -2 ), then between -2 and m, ( x ) is in [-2, m]. Here, ( |x - m| = m - x ), ( |x +2| = x +2 ). So, ( g(x) = (m - x) - (x +2) = m -2x -2 ). This is a linear function decreasing from when ( x = -2 ): ( m -2*(-2) -2 = m +4 -2 = m +2 ), to when ( x = m ): ( m -2m -2 = -m -2 ).Similarly, if ( m < -2 ), between m and -2, ( x in [m, -2] ). Then, ( |x - m| = x - m ), ( |x +2| = -x -2 ). So, ( g(x) = (x - m) - (-x -2) = 2x - m +2 ), which is linear increasing from ( 2m - m +2 = m +2 ) when ( x = m ), to ( 2*(-2) -m +2 = -4 -m +2 = -m -2 ) when ( x = -2 ).Therefore, in all cases:- When ( x leq min(m, -2) ), ( g(x) = m +2 ).- When ( x geq max(m, -2) ), ( g(x) = -m -2 ).- In between, the function is linear, connecting these two values.Therefore, the maximum value of ( g(x) ) is ( m +2 ) if ( m +2 geq -m -2 ), i.e., if ( m geq -2 ), otherwise it's ( -m -2 ). Wait, no. Wait, the maximum is either the value when ( x leq min(m, -2) ), which is ( m +2 ), or the value when ( x geq max(m, -2) ), which is ( -m -2 ), depending on which is larger.So comparing ( m +2 ) and ( -m -2 ):If ( m +2 geq -m -2 ), then the maximum is ( m +2 ). Solving:( 2m +4 geq0 implies m geq -2 ).If ( m +2 < -m -2 ), then the maximum is ( -m -2 ). This occurs when ( m < -2 ).Therefore, the maximum of ( g(x) = |x - m| - |x +2| ) is:- ( m +2 ) if ( m geq -2 ),- ( -m -2 ) if ( m < -2 ).But in either case, the maximum is ( |m +2| ). Because:If ( m geq -2 ), then ( |m +2| = m +2 ).If ( m < -2 ), then ( |m +2| = - (m +2 ) = -m -2 ).Therefore, the maximum value of ( g(x) ) is indeed ( |m +2| ). Therefore, my initial conclusion was correct: to ensure that ( |x - m| - |x +2| leq frac{1}{a} + frac{1}{b} ) for all ( x in mathbb{R} ) and all ( a, b >0 ) with ( a +b =1 ), we need ( |m +2| leq frac{1}{a} + frac{1}{b} ). Since the minimal value of ( frac{1}{a} + frac{1}{b} ) is 4, we must have ( |m +2| leq4 ), leading to ( -6 leq m leq2 ).Therefore, the range of ( m ) is from -6 to 2, inclusive.Let me verify with an example. Take ( m = -6 ). Then the maximum of ( |x - (-6)| - |x +2| = |x +6| - |x +2| ). The maximum occurs when ( x leq min(-6, -2) = -6 ), so when ( x leq -6 ), ( |x +6| - |x +2| = (-x -6) - (-x -2) = -4 ). Wait, that's the minimum. Wait, but according to our previous analysis, the maximum when ( m = -6 leq -2 ), the maximum should be ( -m -2 = 6 -2 =4 ). Let's check at ( x = -6 ):( |-6 +6| - |-6 +2| = 0 -4 = -4 ). Wait, but according to the previous cases, when ( m < -2 ), the maximum is achieved when ( x leq m ), which is ( m +2 ). But for ( m = -6 ), that would be ( -6 +2 = -4 ). Wait, but that's contradictory. Wait, perhaps I messed up the signs.Wait, when ( m < -2 ), the maximum is ( -m -2 ). For ( m = -6 ), that's ( -(-6) -2 =6 -2=4 ). How is that achieved? Let's compute ( g(x) = |x - (-6)| - |x +2| = |x +6| - |x +2| ).Let me check different intervals:1. ( x geq -2 ): Here, both |x +6| and |x +2| are x +6 and x +2. So, ( g(x) = (x +6) - (x +2) =4 ).2. ( -6 leq x leq -2 ): |x +6| = x +6, |x +2| = -x -2. So, ( g(x) = (x +6) - (-x -2) =2x +8 ). At ( x = -2 ), this is 2*(-2) +8=4. At ( x = -6 ), 2*(-6) +8= -12 +8= -4.3. ( x leq -6 ): |x +6| = -x -6, |x +2|= -x -2. So, ( g(x) = (-x -6) - (-x -2)= -4 ).Therefore, the maximum of ( g(x) ) is 4, achieved when ( x geq -2 ). So for ( m = -6 ), the maximum value is indeed 4, which matches our previous conclusion. So even though ( m = -6 ) is less than -2, the maximum is achieved when ( x geq -2 ), giving 4. Similarly, if ( m =2 ), then ( |x -2| - |x +2| ). Let's check the maximum:For ( m =2 geq -2 ), the maximum is ( m +2 =4 ). Let's verify:When ( x leq min(2, -2) = -2 ), ( |x -2| - |x +2| = ( -x +2 ) - (-x -2 ) =4.When ( x geq2 ), ( |x -2| - |x +2| =0 - (x +2 ) = -x -2 ), which decreases to negative infinity. In between, from -2 to2, ( |x -2| - |x +2| = (2 -x ) - (x +2 )= -2x ), which varies from 4 to -4. So maximum at x <= -2 is 4. So yes, the maximum is 4. Hence, when ( m =2 ), the maximum is 4.Thus, our conclusion that the range of ( m ) is ( -6 leq m leq2 ) is correct.Therefore, summarizing:1. The solution to the inequality ( |x| + |x -1| leq2 ) is ( -frac{1}{2} leq x leq frac{3}{2} ).2. The range of ( m ) such that ( |x - m| - |x +2| leq frac{1}{a} + frac{1}{b} ) holds for all ( x in mathbb{R} ) and all ( a, b >0 ) with ( a +b =1 ) is ( -6 leq m leq2 ).Final Answer(1) boxed{left[ -dfrac{1}{2}, dfrac{3}{2} right]}(2) boxed{[-6, 2]}</think>(1) To solve the inequality ( |x| + |x - 1| leqslant 2 ), we consider three cases based on the critical points ( x = 0 ) and ( x = 1 ):- Case 1: ( x < 0 ) The inequality becomes ( -2x + 1 leqslant 2 ), leading to ( -frac{1}{2} leqslant x < 0 ).- Case 2: ( 0 leqslant x leqslant 1 ) The inequality simplifies to ( 1 leqslant 2 ), which is always true, so ( 0 leqslant x leqslant 1 ).- Case 3: ( x > 1 ) The inequality becomes ( 2x - 1 leqslant 2 ), leading to ( 1 < x leqslant frac{3}{2} ).Combining these intervals, the solution is ( -frac{1}{2} leqslant x leqslant frac{3}{2} ).(2) Given ( a + b = 1 ) with ( a, b > 0 ), we need the inequality ( |x - m| - |x + 2| leqslant frac{1}{a} + frac{1}{b} ) to hold for all ( x in mathbb{R} ). The right-hand side ( frac{1}{a} + frac{1}{b} ) is minimized when ( a = b = frac{1}{2} ), giving 4. Thus, the maximum value of ( |x - m| - |x + 2| ) must be less than or equal to 4. The function ( |x - m| - |x + 2| ) reaches its maximum value ( |m + 2| ). To ensure this maximum is at most 4, we solve ( |m + 2| leqslant 4 ), leading to ( -6 leqslant m leqslant 2 ).Final answers:(1) boxed{left[ -dfrac{1}{2}, dfrac{3}{2} right]}(2) boxed{[-6, 2]}]

🔑: Solution:# Part (1)Given the function f(x) = |x|, we need to solve the inequality f(x) + f(x-1) leqslant 2.First, let's express f(x) + f(x-1) in terms of x:- For x 1, we have f(x) = x and f(x-1) = x-1, so f(x) + f(x-1) = x + (x-1) = 2x - 1.Now, let's solve the inequality in each case:- When x 1, we have 2x - 1 leqslant 2, which simplifies to x leqslant frac{3}{2}. Thus, the solution in this interval is 1 0, we need to find the range of m such that for all x in mathbb{R}, the inequality f(x-m) - |x+2| leqslant frac{1}{a} + frac{1}{b} always holds.First, let's simplify frac{1}{a} + frac{1}{b} using a + b = 1:frac{1}{a} + frac{1}{b} = (a + b)left(frac{1}{a} + frac{1}{b}right) = 2 + frac{b}{a} + frac{a}{b} geqslant 2 + 2sqrt{frac{b}{a} cdot frac{a}{b}} = 4The equality holds when frac{b}{a} = frac{a}{b}, which implies a = b = frac{1}{2}. Thus, the minimum value of frac{1}{a} + frac{1}{b} is 4.Now, considering the inequality |x-m| - |x+2| leqslant 4, we observe that:|x-m| - |x+2| leqslant |(x-m) - (x+2)| = |m + 2|The equality holds when x = m. Therefore, we have |m + 2| leqslant 4, which simplifies to -6 leqslant m leqslant 2.Hence, the range of m is boxed{m in [-6, 2]}.

🧠:[<think>Okay, so I have this trigonometry problem here: Given that cos(π/4 - x) = -3/5, I need to find the value of sin 2x, and then choose the correct answer from the options A to D. Hmm, let's think about how to approach this step by step. First, I remember that there are trigonometric identities that relate angles in different forms. The problem gives me cosine of an angle that's a difference of π/4 and x, and I need to find sin 2x. Maybe I can use some angle addition or subtraction formulas here. Let me recall the cosine of a difference identity. The formula is:cos(A - B) = cos A cos B + sin A sin BSo, applying this to cos(π/4 - x), that would be:cos(π/4)cos x + sin(π/4)sin xI know that cos(π/4) and sin(π/4) are both √2/2. So substituting those in:(√2/2)cos x + (√2/2)sin x = -3/5Hmm, maybe I can factor out the √2/2:√2/2 (cos x + sin x) = -3/5Then, multiplying both sides by 2/√2 to isolate (cos x + sin x):cos x + sin x = (-3/5)*(2/√2) = (-6)/(5√2)I can rationalize the denominator here:(-6)/(5√2) * (√2/√2) = (-6√2)/10 = (-3√2)/5So, cos x + sin x = -3√2/5Okay, so now I have that cos x + sin x equals this value. How can I relate this to sin 2x? Let me recall that sin 2x is 2 sin x cos x. So if I can find sin x cos x, then multiplying by 2 will give me sin 2x.Alternatively, I might consider squaring both sides of the equation cos x + sin x = -3√2/5. Let me try that.(cos x + sin x)^2 = (-3√2/5)^2Expanding the left side:cos^2 x + 2 sin x cos x + sin^2 x = (9*2)/25Simplify the left side. Since cos^2 x + sin^2 x = 1:1 + 2 sin x cos x = 18/25Subtract 1 from both sides:2 sin x cos x = 18/25 - 1 = 18/25 - 25/25 = -7/25But 2 sin x cos x is exactly sin 2x. Therefore, sin 2x = -7/25. Wait, but looking at the answer choices, option D is -7/25. But wait, let me check my steps again because I might have made a mistake.Wait, hold on. Let me verify each step carefully. Starting with the given:cos(π/4 - x) = -3/5Using the identity, we get:cos π/4 cos x + sin π/4 sin x = -3/5Since cos π/4 = sin π/4 = √2/2, this becomes:√2/2 cos x + √2/2 sin x = -3/5Factor out √2/2:√2/2 (cos x + sin x) = -3/5Multiply both sides by 2/√2:cos x + sin x = (-3/5) * (2/√2) = (-6)/(5√2)Rationalizing the denominator:Multiply numerator and denominator by √2:(-6√2)/(5*2) = (-6√2)/10 = (-3√2)/5So cos x + sin x = -3√2/5. Then, squaring both sides:(cos x + sin x)^2 = (-3√2/5)^2Left side:cos^2 x + 2 sin x cos x + sin^2 x = 1 + 2 sin x cos xRight side:(9 * 2)/25 = 18/25So:1 + 2 sin x cos x = 18/25Subtract 1:2 sin x cos x = 18/25 - 25/25 = -7/25Therefore, sin 2x = 2 sin x cos x = -7/25. So that's option D. But wait, the options are A: 7/25, B: -24/25, C:24/25, D: -7/25. So D is -7/25, which would be the answer.But let me think again. The answer given by this method is D. But I want to make sure that there are no errors in the steps. Let me verify again.Alternatively, perhaps there's another method to check. For example, we can find x first, then compute sin 2x. Let's try that.Given cos(π/4 - x) = -3/5. Let θ = π/4 - x. So cos θ = -3/5, which means θ is an angle in either the second or third quadrant since cosine is negative there. But since π/4 - x is θ, depending on x, θ could be in different quadrants. Let's find θ.First, arccos(-3/5) gives us the reference angle. The reference angle would be arccos(3/5). Let's compute that. Let's say α = arccos(3/5), so θ is π - α or π + α. But θ = π/4 - x. Wait, maybe this approach is more complicated.Alternatively, since we have θ = π/4 - x, then x = π/4 - θ. Therefore, 2x = π/2 - 2θ. Then sin 2x = sin(π/2 - 2θ) = cos 2θ. So sin 2x = cos 2θ. But θ is such that cos θ = -3/5, so cos 2θ = 2 cos^2 θ - 1 = 2*(9/25) -1 = 18/25 -25/25 = -7/25. Therefore, sin 2x = cos 2θ = -7/25. So same answer, D. Therefore, this method also gives D. So that's confirming the result. Therefore, the answer is D. So why does the original problem have options including -24/25? Maybe someone might make a mistake in the process. Let me check if maybe I messed up a sign somewhere.Wait, in the first method, when I squared both sides, is there a possibility that squaring introduced an extraneous solution? But in this case, we were given that cos(π/4 - x) is negative, which is -3/5, so π/4 - x is in the second or third quadrant. However, when we computed sin 2x, we arrived at -7/25. Let me see. Alternatively, maybe I should check with actual numbers.Suppose cos(π/4 - x) = -3/5. Let's let θ = π/4 - x. Then θ is an angle whose cosine is -3/5. So θ is in the second or third quadrant. The reference angle is arccos(3/5). Let me compute arccos(3/5). Let's say that angle is approximately 53.13 degrees. Therefore, θ is either π - 53.13 degrees ≈ 126.87 degrees, or π + 53.13 ≈ 233.13 degrees. But θ = π/4 - x, which is 45 degrees - x (if we measure in degrees). So solving for x, x = 45 degrees - θ. Therefore, if θ is 126.87 degrees, then x = 45 - 126.87 = -81.87 degrees. Alternatively, if θ is 233.13 degrees, then x = 45 - 233.13 = -188.13 degrees. Both possibilities. Let's take the first case: θ ≈ 126.87 degrees, so x ≈ -81.87 degrees. Then 2x ≈ -163.74 degrees. The sine of that angle is sin(-163.74) = sin(180 - 163.74) = sin(16.26) but with a negative sign because it's in the third quadrant. Wait, sin(-163.74) is the same as sin(180 - 163.74) = sin(16.26) but adjusted for the quadrant. Wait, -163.74 degrees is equivalent to 360 - 163.74 = 196.26 degrees, which is in the third quadrant. The sine of 196.26 degrees is negative. The reference angle is 196.26 - 180 = 16.26 degrees. So sin(196.26) = -sin(16.26). Let's compute sin(16.26) ≈ 0.28, so sin(196.26) ≈ -0.28. But according to our previous answer, sin 2x ≈ -7/25 = -0.28. So that matches. Alternatively, if θ is 233.13 degrees, then x = 45 - 233.13 = -188.13 degrees. Then 2x = -376.26 degrees, which is equivalent to -376.26 + 360 = -16.26 degrees, which is the same as 360 - 16.26 = 343.74 degrees. The sine of that is sin(343.74) = -sin(16.26) ≈ -0.28, which is the same as before. So in either case, sin 2x ≈ -0.28, which is -7/25. So this numerical example confirms that the answer is indeed D. Therefore, despite the answer choices including options like -24/25, which might correspond to a different error in calculation, the correct answer is D, -7/25.Wait, but let me just think once more. Suppose someone thought that sin 2x is 2 sin x cos x, so maybe tried to find sin x and cos x individually. But in that case, you would need more information. Alternatively, perhaps using another identity. Let me try that approach as well, just to check.Given cos(π/4 - x) = -3/5. Let me use the identity for cosine of difference again:cos(π/4 - x) = cos π/4 cos x + sin π/4 sin x = √2/2 (cos x + sin x) = -3/5Therefore, as before, cos x + sin x = -3√2/5Alternatively, perhaps we can write this as sin x + cos x = -3√2/5. Let me consider sin x + cos x = k, then (sin x + cos x)^2 = k^2 = sin^2 x + 2 sin x cos x + cos^2 x = 1 + sin 2x. Wait, 2 sin x cos x is sin 2x. Therefore, 1 + sin 2x = (sin x + cos x)^2So substituting in, sin 2x = (sin x + cos x)^2 - 1 = (-3√2/5)^2 - 1 = (9*2)/25 - 1 = 18/25 - 25/25 = -7/25. So same result. So regardless of the approach, the answer is D. Therefore, I think the answer is definitely D. Maybe some people might miscalculate the squaring step or make a mistake in the identity, leading them to other options, but based on multiple methods, D is correct.Final Answerboxed{D}</think>Given (cosleft( frac{pi}{4} - x right) = -frac{3}{5}), we need to find the value of (sin 2x).First, we use the cosine of a difference identity:[cosleft( frac{pi}{4} - x right) = cos frac{pi}{4} cos x + sin frac{pi}{4} sin x]Since (cos frac{pi}{4} = sin frac{pi}{4} = frac{sqrt{2}}{2}), we substitute these values:[frac{sqrt{2}}{2} cos x + frac{sqrt{2}}{2} sin x = -frac{3}{5}]Factoring out (frac{sqrt{2}}{2}):[frac{sqrt{2}}{2} (cos x + sin x) = -frac{3}{5}]Multiplying both sides by (frac{2}{sqrt{2}}):[cos x + sin x = -frac{3}{5} cdot frac{2}{sqrt{2}} = -frac{6}{5sqrt{2}} = -frac{3sqrt{2}}{5}]Next, we square both sides:[(cos x + sin x)^2 = left( -frac{3sqrt{2}}{5} right)^2]Expanding the left side and simplifying:[cos^2 x + 2 sin x cos x + sin^2 x = frac{18}{25}]Using the Pythagorean identity (cos^2 x + sin^2 x = 1):[1 + 2 sin x cos x = frac{18}{25}]Solving for (2 sin x cos x):[2 sin x cos x = frac{18}{25} - 1 = frac{18}{25} - frac{25}{25} = -frac{7}{25}]Thus, (sin 2x = 2 sin x cos x = -frac{7}{25}).The final answer is (boxed{D}).,<think>Okay, so I have this trigonometry problem here: Given that cos(π/4 - x) = -3/5, I need to find the value of sin 2x, and then choose the correct option from A to D. Let's see how to approach this. Hmm, I remember some trigonometric identities that might help here. First, maybe I can use the cosine of a difference identity. The formula for cos(A - B) is cos A cos B + sin A sin B. So if I let A be π/4 and B be x, then cos(π/4 - x) = cos π/4 cos x + sin π/4 sin x. Given that cos π/4 and sin π/4 are both √2/2, this simplifies to (√2/2)(cos x + sin x). So the equation becomes (√2/2)(cos x + sin x) = -3/5. Let me write that down step by step.Starting with the given equation:cos(π/4 - x) = -3/5Expanding using the cosine difference identity:cos π/4 cos x + sin π/4 sin x = -3/5Since cos π/4 = sin π/4 = √2/2:(√2/2) cos x + (√2/2) sin x = -3/5Factor out √2/2:(√2/2)(cos x + sin x) = -3/5Then, multiply both sides by 2/√2 to isolate (cos x + sin x):cos x + sin x = (-3/5) * (2/√2) = (-6)/(5√2)Hmm, maybe rationalize the denominator here:Multiply numerator and denominator by √2:(-6√2)/(5*2) = (-6√2)/10 = (-3√2)/5So, cos x + sin x = -3√2/5Okay, so I have cos x + sin x expressed in terms of a single value. Now, I need to find sin 2x. The double angle formula for sine is sin 2x = 2 sin x cos x. So if I can find sin x cos x, I can multiply by 2 to get sin 2x.Alternatively, maybe I can square both sides of the equation cos x + sin x = -3√2/5 to find sin x cos x.Let me try that. Squaring both sides:(cos x + sin x)^2 = (-3√2/5)^2Left side expands to cos²x + 2 sin x cos x + sin²x. Right side is (9*2)/25 = 18/25.So:cos²x + sin²x + 2 sin x cos x = 18/25But cos²x + sin²x = 1, so substitute that in:1 + 2 sin x cos x = 18/25Subtract 1 from both sides:2 sin x cos x = 18/25 - 1 = 18/25 - 25/25 = -7/25Therefore, 2 sin x cos x = -7/25. But 2 sin x cos x is sin 2x. Therefore, sin 2x = -7/25. Wait, but looking at the options, D is -7/25, so is that the answer? But hold on, let me check my steps again because the answer might be positive or negative. Let me go through the steps again carefully.Given cos(π/4 - x) = -3/5. Then, expanding using the identity:cos π/4 cos x + sin π/4 sin x = -3/5Which becomes (√2/2)(cos x + sin x) = -3/5. Then, multiplying both sides by 2/√2:cos x + sin x = (-3/5)*(2/√2) = -6/(5√2). Then rationalizing:-6/(5√2) * √2/√2 = -6√2/10 = -3√2/5. So that seems correct.Then, squaring both sides: (cos x + sin x)^2 = (-3√2/5)^2Left side: cos²x + 2 sin x cos x + sin²x = 1 + 2 sin x cos x.Right side: (9*2)/25 = 18/25. So:1 + 2 sin x cos x = 18/25Therefore, 2 sin x cos x = 18/25 - 1 = -7/25. So, sin 2x = -7/25.Looking back at the options, D is -7/25, so that seems to be the answer. However, looking at the answer choices, option A is 7/25, B is -24/25, C is 24/25, D is -7/25. So the answer should be D?Wait, but I recall that in some cases, depending on the quadrant, the value might be positive or negative, but here we squared the equation, so we might have introduced an extraneous solution. Wait, but in this problem, we squared both sides of an equation where the left side was cos x + sin x = -3√2/5. So the left side is equal to a negative number. Therefore, cos x + sin x is negative, which might tell us something about the angle x.But when we squared it, we might have lost the sign information. However, the result we obtained was 2 sin x cos x = -7/25, so sin 2x = -7/25. That would be answer D.But wait, let me check again. Alternatively, perhaps the answer is A: 7/25, but with a positive sign. Hmm. Let me verify if there's a mistake in my calculation.Wait, if cos(π/4 - x) is -3/5, then π/4 - x is an angle whose cosine is -3/5. So that angle must be in either the second or third quadrant since cosine is negative there. Then, π/4 - x is in second or third quadrant. Let's think about possible values.Suppose π/4 - x is in the second quadrant. Then, the sine of that angle would be positive. If it's in the third quadrant, sine would be negative. But since we need to find sin 2x, perhaps we can use another identity.Alternatively, maybe use the identity for sin 2x in terms of cos(π/4 - x). Let me think. Let me denote θ = π/4 - x. Then, cos θ = -3/5, and we can find sin θ. Then, since θ = π/4 - x, x = π/4 - θ. Then, 2x = π/2 - 2θ. Therefore, sin 2x = sin(π/2 - 2θ) = cos 2θ. So sin 2x = cos 2θ. Since cos θ = -3/5, then cos 2θ = 2 cos²θ - 1 = 2*(9/25) - 1 = 18/25 - 1 = -7/25. Therefore, sin 2x = -7/25. Which is answer D. So that seems to confirm the previous result. Therefore, the answer is D. Hmm, but the options are given as A:7/25, B:-24/25, C:24/25, D:-7/25. Therefore, D is correct.Wait, but let me check this again. If θ = π/4 - x, then x = π/4 - θ. Therefore, 2x = π/2 - 2θ. Then, sin 2x = sin(π/2 - 2θ) = cos 2θ. Because sin(π/2 - α) = cos α. Therefore, sin 2x = cos 2θ. Since θ is such that cos θ = -3/5. So cos 2θ = 2 cos²θ - 1 = 2*(9/25) - 1 = 18/25 - 25/25 = -7/25. Therefore, sin 2x = -7/25. So that's the same result as before. Therefore, the answer is D.But wait, the answer given here is -7/25. Let me cross-check using another method. Let's use the first method again. The initial equation was cos(π/4 - x) = -3/5. Expanding, we got cos x + sin x = -3√2/5. Then, squaring gives 1 + sin 2x = 18/25, leading to sin 2x = -7/25. So that's correct.Alternatively, maybe I can use another identity. Let's think. Let's denote θ = x. Then, the given equation is cos(π/4 - θ) = -3/5. Then, sin 2θ is what we need to find. Alternatively, maybe express π/4 - θ as an angle, and then relate to θ.Alternatively, let's find sin(π/4 - x). Since we know cos(π/4 - x) = -3/5, we can find sin(π/4 - x) using the identity sin²α + cos²α = 1. Therefore, sin(π/4 - x) = sqrt(1 - cos²(π/4 - x)) = sqrt(1 - 9/25) = sqrt(16/25) = 4/5 or -4/5. But the sign depends on the quadrant of π/4 - x.Since cos(π/4 - x) is -3/5, which is negative, π/4 - x must be in either the second or third quadrant. If π/4 - x is in the second quadrant, then sin(π/4 - x) is positive. If it's in the third quadrant, sin(π/4 - x) is negative. However, π/4 is 45 degrees, so π/4 - x. The original angle π/4 - x has cosine -3/5, so it's in the second or third quadrant. But π/4 - x could be in the second quadrant (if π/4 - x is between π/2 and π) or third quadrant (if π/4 - x is between π and 3π/2). But since π/4 is 45 degrees, subtracting x (which could be any angle depending on x), but without knowing the range of x, perhaps we can't determine the exact quadrant. However, in this problem, maybe we don't need to know the quadrant because we can express sin 2x in terms of the given value. Let's see.Alternatively, using the sine addition formula. Let me think. Since we need sin 2x, perhaps we can express sin 2x in terms of π/4 - x. Let me see.We know that π/4 - x = angle θ, so x = π/4 - θ. Therefore, 2x = π/2 - 2θ. Then, sin 2x = sin(π/2 - 2θ) = cos 2θ, which we already found to be -7/25. Therefore, regardless of the quadrant, we can express sin 2x as cos 2θ where θ is such that cos θ = -3/5. Then, cos 2θ is 2 cos²θ - 1 = -7/25. Therefore, sin 2x = -7/25.Therefore, the answer is D.Wait, but let me check once more. Let me compute sin 2x using another approach. Let's compute sin x and cos x. From the original equation, we have cos x + sin x = -3√2/5. Let me call this Equation (1). Also, we know that sin^2x + cos^2x = 1. Let me call this Equation (2). Let me solve Equations (1) and (2) for sin x and cos x.Let me denote S = sin x and C = cos x. Then:C + S = -3√2/5 ...(1)C² + S² = 1 ...(2)We can square Equation (1):(C + S)^2 = ( -3√2/5 )² => C² + 2CS + S² = 18/25But from Equation (2), C² + S² = 1, so substitute into the above:1 + 2CS = 18/25 => 2CS = 18/25 - 1 = -7/25 => CS = -7/50But sin 2x = 2 sin x cos x = 2CS = -7/25. So that's the same result. Therefore, sin 2x = -7/25.Therefore, the answer is D. But wait, in the options given, D is -7/25, so that's correct. But let me check again why the answer isn't, say, A or B. Maybe there's a miscalculation.Wait, let me check squaring (-3√2/5):(-3√2/5)^2 = (9*2)/25 = 18/25. Correct. Then 1 + 2 sin x cos x = 18/25, so 2 sin x cos x = -7/25. So that's correct. Therefore, sin 2x = -7/25. So D is the answer.Alternatively, maybe there's an error in the problem statement. Let me check the problem again. It says cos(π/4 - x) = -3/5, and asks for sin 2x. The steps seem correct, so unless there's a miscalculation I can't see. Let me verify once again.Using the identity: sin 2x = 2 sin x cos x. If we can find sin x cos x, then we can find sin 2x. From Equation (1): cos x + sin x = -3√2/5. From Equation (2): sin²x + cos²x = 1. Then, (cos x + sin x)^2 = cos²x + 2 sin x cos x + sin²x = 1 + 2 sin x cos x = 18/25. Therefore, 2 sin x cos x = 18/25 - 1 = -7/25. So sin 2x = -7/25. Therefore, D.Alternatively, let's suppose that I made an error in the sign when rationalizing. Wait, let me check that step again. Original equation:(√2/2)(cos x + sin x) = -3/5Multiply both sides by 2/√2:cos x + sin x = (-3/5)*(2/√2) = (-6)/(5√2)Rationalizing:Multiply numerator and denominator by √2:(-6√2)/(5*2) = (-6√2)/10 = (-3√2)/5. That's correct. So the sign is negative, so cos x + sin x is negative. Squaring it gives 18/25, which is positive. Then, 1 + 2 sin x cos x = 18/25. Then, 2 sin x cos x = 18/25 - 1 = -7/25, so sin 2x = -7/25. So that's correct.Alternatively, let me check the other method where I set θ = π/4 - x, so x = π/4 - θ. Then 2x = π/2 - 2θ, so sin 2x = cos 2θ. Since cos θ = -3/5, then cos 2θ = 2 cos²θ -1 = 2*(9/25) -1 = 18/25 -1 = -7/25. So that's the same result. Therefore, the answer is D.Therefore, the answer should be D.Final Answerboxed{D}</think>Given (cosleft( frac{pi}{4} - x right) = -frac{3}{5}), we need to find the value of (sin 2x).First, we use the cosine difference identity:[cosleft( frac{pi}{4} - x right) = cos frac{pi}{4} cos x + sin frac{pi}{4} sin x]Since (cos frac{pi}{4} = sin frac{pi}{4} = frac{sqrt{2}}{2}), we substitute these values:[frac{sqrt{2}}{2} cos x + frac{sqrt{2}}{2} sin x = -frac{3}{5}]Factor out (frac{sqrt{2}}{2}):[frac{sqrt{2}}{2} (cos x + sin x) = -frac{3}{5}]Multiply both sides by (frac{2}{sqrt{2}}):[cos x + sin x = -frac{3}{5} cdot frac{2}{sqrt{2}} = -frac{6}{5sqrt{2}} = -frac{3sqrt{2}}{5}]Next, we square both sides of the equation (cos x + sin x = -frac{3sqrt{2}}{5}):[(cos x + sin x)^2 = left( -frac{3sqrt{2}}{5} right)^2][cos^2 x + 2 sin x cos x + sin^2 x = frac{18}{25}]Using the Pythagorean identity (cos^2 x + sin^2 x = 1):[1 + 2 sin x cos x = frac{18}{25}]Subtract 1 from both sides:[2 sin x cos x = frac{18}{25} - 1 = frac{18}{25} - frac{25}{25} = -frac{7}{25}]Thus, (sin 2x = 2 sin x cos x = -frac{7}{25}).Therefore, the value of (sin 2x) is (boxed{D}).]

🔑:Solution: Since cosleft(frac {pi}{4}-xright) = frac {sqrt {2}}{2}(sin x+cos x) = - frac {3}{5},Squaring both sides gives: frac {1}{2}(sin x+cos x)^2 = frac {1}{2}(1+sin 2x) = frac {9}{25},Thus, sin 2x = - frac {7}{25}.Therefore, the correct answer is boxed{text{D}}.This problem involves using the cosine formula for the sum and difference of two angles and the special values of trigonometric functions to simplify and find the value of sin x + cos x. Squaring both sides then allows us to find the value of sin 2x. This question tests the formula for the sine of double angles, the cosine formula for the sum and difference of angles, and the basic relationships between trigonometric functions of the same angle. Mastering these formulas is key to solving this problem.

🧠:[<think>Alright, let's tackle this problem step by step. We have four propositions about lines and planes in space, and we need to determine which ones are false. The options given are A through D, corresponding to propositions 1 through 4. Let me think through each one carefully.Starting with proposition (1): "For any two skew lines in space, there exist infinitely many planes that form equal angles with these two lines." Hmm, skew lines are lines that neither intersect nor are parallel. So, they're not in the same plane. The question is about planes that form equal angles with both lines. I remember that when dealing with two lines, the angle between a line and a plane is the complement of the angle between the line and the normal to the plane. So, if we want a plane to form equal angles with both skew lines, we need to find a plane such that the angle between each line and the plane is the same. Alternatively, maybe there's a way to think about bisectors. In two dimensions, two intersecting lines have angle bisectors. In three dimensions, for skew lines, maybe there are analogous bisecting planes? Wait, I think there's a result that says there are infinitely many planes that are angle bisectors between two skew lines. Since the lines are skew, they don't intersect, but we can imagine moving a plane such that it maintains equal angles with both lines. Because there are infinitely many positions where such a plane can exist, maybe rotating around some axis or something. So proposition (1) might be true. But I need to verify.Alternatively, consider that for each point on one of the lines, there might be a plane passing through that point and forming equal angles. Since there are infinitely many points, perhaps there are infinitely many such planes. However, I need to be careful here. The angle between a line and a plane depends on the orientation of the plane relative to the line, not just its position. So maybe even though there are infinitely many points, the orientation could be restricted? Hmm.Wait, here's another approach. If we fix two skew lines, we can parametrize the problem. Let’s take two skew lines, say, line L1 and line L2. Since they are skew, there's a unique common perpendicular line connecting them. Let's denote the distance between them along this common perpendicular as d, and the angle between their direction vectors as θ.Now, a plane that forms equal angles with both lines would need to be positioned such that its normal vector makes equal angles with both lines. Alternatively, the dihedral angle between the plane and each line is equal. But I'm getting confused here. Maybe another way: for a plane to form equal angles with two lines, the angle between the plane's normal and each line should be equal. Because the angle between a line and a plane is 90 degrees minus the angle between the line and the normal. So if the angles between the normals and each line are equal, then the angles between the plane and each line would also be equal.Therefore, the set of all planes whose normals make equal angles with both lines. Since the lines are skew, they don't lie in the same plane, but their direction vectors are fixed. The set of normals that make equal angles with both direction vectors would form a double cone or something. Wait, the set of vectors that make equal angles with two given vectors is a cone. So the normals lie on such a cone, which is a quadratic surface. Therefore, there are infinitely many normals (directions) that satisfy this, which means there are infinitely many planes. So proposition (1) is true. So (1) is true, so A is not the answer.Moving on to proposition (2): "For any 3 pairwise skew lines in space, there exist infinitely many lines that intersect all three lines." Hmm, three skew lines. So pairwise skew, meaning no two intersect and none are parallel. The question is about lines that intersect all three. I recall something called a regulus or maybe the theory of lines in three dimensions. In three-dimensional space, given three skew lines, there can be zero, one, or infinitely many lines that intersect all three. Wait, but the proposition says "for any 3 pairwise skew lines," there are infinitely many lines intersecting all three. But is that the case?Wait, there's a theorem called the "Regulus Theorem" which states that three skew lines either lie on a ruled surface called a regulus, in which case there are infinitely many lines intersecting all three, or they don't, in which case there are at most two lines intersecting all three. So the answer depends on the configuration of the three skew lines. Therefore, the proposition says "for any 3 pairwise skew lines," which would not necessarily lie on a regulus. Therefore, in some cases, there might be only two or one or even none. Wait, but the theorem says at most two if they are not on a regulus. Wait, no. Let me check.Wait, the exact statement: If three skew lines are such that they are not all generators of a regulus, then there are either two or no lines intersecting all three. If they lie on a regulus, then there are infinitely many. Therefore, the proposition (2) claims that for any three skew lines, there are infinitely many lines intersecting all three. But this is only true if the three lines lie on a regulus. But since the problem says "for any 3 pairwise skew lines," which includes those not lying on a regulus, then the proposition is false. Therefore, proposition (2) is false, so B is a candidate.Wait, but maybe I'm misremembering. Let me think again. In three-dimensional space, given three skew lines, when can a line intersect all three? If you have three skew lines, you can use the following method: take two of the lines, say L1 and L2. There are infinitely many lines that intersect both L1 and L2 (these are called transversals). For each transversal, check if it intersects the third line L3. Depending on the configuration, how many such transversals intersect L3. But since L1, L2, L3 are all skew, the transversals between L1 and L2 form a regulus. If L3 is part of the complementary regulus, then each transversal of L1 and L2 will intersect L3. Wait, but if three skew lines lie on a hyperboloid of one sheet, which has two reguli, then each line from one regulus intersects all lines of the other regulus. So in that case, there are infinitely many lines intersecting all three. But if three skew lines do not lie on such a surface, then there might be only a finite number. Therefore, if the three lines lie on a hyperboloid of one sheet (i.e., a regulus), then there are infinitely many transversals. Otherwise, only two. Therefore, the answer depends on the configuration. The proposition says "for any 3 pairwise skew lines," which would include those not on a hyperboloid. Therefore, the proposition is false. Hence, proposition (2) is false. So B is correct.But wait, actually, the exact theorem is: Given three skew lines, there are either 0, 1, 2, or infinitely many lines intersecting all three. If they lie on a hyperboloid, then infinitely many. Otherwise, two or none. Therefore, the statement (2) says "there exist infinitely many lines that intersect all three," but this is only true if the three lines lie on a hyperboloid. Since the problem states "for any 3 pairwise skew lines," which may or may not lie on a hyperboloid, so the existence of infinitely many lines is not guaranteed. Therefore, proposition (2) is false. So B is correct.Moving to proposition (3): "For any 4 pairwise skew lines in space, there does not exist a line that intersects all four lines." So, the claim is that no line can intersect four pairwise skew lines. Is this true?I know that in three-dimensional space, given four skew lines, it's generally impossible for a single line to intersect all four. However, there might be special configurations where such a line exists. But is it always impossible?Wait, let's think. If we have four lines on a hyperboloid of one sheet, which contains two reguli. Each line in one regulus intersects all lines in the other regulus. So, if we have four lines, two from each regulus, then a line in one regulus will intersect the two lines in the other regulus, but not the ones in its own regulus. Therefore, a line can intersect at most two lines from the opposite regulus. So, if you have four lines, two in each regulus, then no line can intersect all four. But what if the four lines are arranged in some other way?Alternatively, suppose we have four lines, not on a hyperboloid. Is there a possibility for a line to intersect all four? It seems highly unlikely. In 3D space, a line is determined by two parameters. To intersect four skew lines would require satisfying four conditions, which would overdetermine the system, making it impossible in general. However, maybe in some special cases?Wait, but the proposition says "for any 4 pairwise skew lines," there does not exist such a line. So even if there exists some configuration where such a line exists, the proposition would be false. So we need to check if there exists a set of four pairwise skew lines with a common transversal.I think in some cases, you can have four skew lines with a common transversal. For example, take a line L, and four skew lines each intersecting L. But wait, if they all intersect L, then they can't be pairwise skew. Because if two lines both intersect L, then they might be coplanar with L, hence intersecting each other or being parallel. Wait, no. If two lines both intersect L at different points and are not coplanar, they can still be skew. Wait, suppose L is a vertical line. Take line M1 intersecting L at point A, and line M2 intersecting L at point B, with M1 and M2 not coplanar. Then M1 and M2 are skew. Similarly, lines M3 and M4 intersecting L at different points and not coplanar with each other. Then all four lines M1, M2, M3, M4 are pairwise skew, and all intersect L. Therefore, line L intersects all four. Therefore, such a configuration exists. Hence, proposition (3) is false, because there exists at least one set of four pairwise skew lines with a common transversal. Therefore, proposition (3) is false, so C is a candidate.But wait, the problem says "for any 4 pairwise skew lines," there does not exist a line intersecting all four. The existence of even a single example where such a line exists would make the proposition false. Since I just thought of such an example, proposition (3) is indeed false. Therefore, C is correct.But wait, let me confirm. Suppose we have four lines, each intersecting a common line L, but all pairwise skew. Let's construct them. Let L be the z-axis. Then, take four lines, each in different planes containing L but not overlapping. For example, line M1: x=1, z=0; line M2: x=2, z=1; line M3: y=1, z=2; line M4: y=2, z=3. Wait, no, those lines would not be skew. Wait, actually, if each of these lines intersects L (the z-axis) at different points, but are not coplanar with each other. Let me parametrize them properly.Take L as the z-axis. Then, for line M1, let it pass through (0,0,0) with direction vector (1,0,1). For line M2, pass through (0,0,1) with direction vector (1,0,2). For line M3, pass through (0,0,2) with direction vector (0,1,1). For line M4, pass through (0,0,3) with direction vector (0,1,2). Now, check if these lines are pairwise skew. Wait, M1 and M2: both have x and z components. Let's see if they intersect. Parametrize M1: (t, 0, t). M2: (s, 0, 1 + 2s). To find intersection, set t = s, 0=0, t = 1 + 2s. Then t = 1 + 2t => -t = 1 => t = -1. Then s = -1. Check if points are the same: M1 at t=-1 is (-1,0,-1). M2 at s=-1 is (-1,0,1 + 2*(-1)) = (-1,0,-1). So they intersect at (-1,0,-1). Therefore, M1 and M2 intersect, which violates the pairwise skew condition. Oops, so that's a problem.Therefore, constructing four lines intersecting L without being pairwise intersecting is tricky. Let me think differently. Take four lines, each intersecting L at a different point, but with different direction vectors such that no two are coplanar.Let’s take L as the z-axis. For line M1, take a line through (0,0,0) with direction vector (1,1,0). For line M2, a line through (0,0,1) with direction vector (1,-1,0). For line M3, a line through (0,0,2) with direction vector (1,0,1). For line M4, a line through (0,0,3) with direction vector (-1,0,1). Now, check pairwise.Check M1 and M2: direction vectors (1,1,0) and (1,-1,0). The lines pass through (0,0,0) and (0,0,1). Are they skew? Let's parametrize M1: (t, t, 0). M2: (s, -s, 1). To check intersection: t = s, t = -s, 0 = 1. Impossible. So no intersection. Similarly, check if they're parallel. Direction vectors are different. So M1 and M2 are skew.Check M1 and M3: M1 is (t, t, 0), direction (1,1,0). M3 is through (0,0,2) with direction (1,0,1). Parametrize M3 as (s, 0, 2 + s). Check if they intersect. Equations: t = s, t = 0, 0 = 2 + s. From t = 0, s = 0, but then 0 = 2 + 0 = 2, which is false. So no intersection. Not parallel. So skew.Similarly, check M1 and M4: M1 is (t, t, 0), M4 is (-s, 0, 3 + s). Equations: t = -s, t = 0, 0 = 3 + s. From t=0, s= -0, but 0 = 3 + 0 = 3, which is false. So no intersection. Skew.M2 and M3: M2 is (s, -s, 1), M3 is (t, 0, 2 + t). Equations: s = t, -s = 0, 1 = 2 + t. From -s=0, s=0, then t=0, 1=2+0=2: nope. Skew.M2 and M4: M2 is (s, -s,1), M4 is (-t, 0,3 + t). Equations: s = -t, -s = 0, 1 = 3 + t. From -s=0, s=0, then t=0, 1=3+0=3: nope. Skew.M3 and M4: M3 is (s,0,2 + s), M4 is (-t,0,3 + t). Equations: s = -t, 0=0, 2 + s = 3 + t. Substitute s = -t into third equation: 2 - t = 3 + t => 2 -3 = 2t => -1 = 2t => t = -0.5. Then s = 0.5. Check coordinates: M3 at s=0.5: (0.5, 0, 2.5). M4 at t=0.5: (-0.5, 0, 3 -0.5) = (-0.5, 0, 2.5). Different points. So no intersection. Not parallel. Skew.Therefore, all four lines M1, M2, M3, M4 are pairwise skew, and all intersect the z-axis L. Therefore, line L intersects all four. Thus, in this configuration, there exists a line intersecting all four pairwise skew lines. Therefore, proposition (3) is false. So C is correct.Now, proposition (4): "If 3 planes intersect pairwise, producing 3 distinct intersection lines, then these 3 lines are concurrent." So, three planes, each pair intersecting in a line, and the three lines are distinct. The claim is that these three lines meet at a single point (are concurrent). Is this true?Let me visualize. Three planes intersecting pairwise in three distinct lines. The question is whether these three lines must meet at a common point.In three-dimensional space, three planes can intersect in different ways. If each pair of planes intersects in a line, and the three lines are distinct, then the three lines might form a triangular prism or meet at a common line or point.Wait, if three planes intersect pairwise in three distinct lines, those lines must either be concurrent or form a triangular prism (i.e., be mutually skew). Wait, but planes in space can't have all three pairwise intersections as skew lines. Wait, if two planes intersect along a line, then the third plane can intersect each of the first two planes along different lines. If the three lines are to be concurrent, there must be a common point where all three planes meet.Alternatively, think of the three planes as having a common point. Suppose three planes intersect at a single point, then their pairwise intersections would be three lines passing through that point. Hence, concurrent. But if the three planes intersect in three lines that are not concurrent, then they form a triangular prism, but in 3D space, can three planes intersect pairwise in three non-concurrent lines?Wait, let's consider the standard case. Take three coordinate planes: xy, yz, xz. They intersect along the coordinate axes, which are concurrent at the origin. So in that case, concurrent.But suppose we take three planes: x=0, y=0, and x+y=1. Then, the intersections are x=0 and y=0: z-axis; x=0 and x+y=1: line x=0, y=1, z arbitrary; y=0 and x+y=1: line y=0, x=1, z arbitrary. These three lines are the z-axis, the line (0,1,z), and the line (1,0,z). These three lines are all parallel to the z-axis but offset, so they are mutually skew? Wait, no. The z-axis is (0,0,z). The line (0,1,z) is parallel to the z-axis and offset in y. The line (1,0,z) is parallel to the z-axis and offset in x. So all three are parallel to the z-axis but not coplanar, hence pairwise skew. Wait, but planes x=0, y=0, x+y=1. Each pair of planes intersect in a line. However, the three lines are not concurrent. They are all parallel to the z-axis but lie on different positions. Hence, in this case, the three lines are not concurrent. Therefore, proposition (4) is false. Therefore, D is correct.But wait, let's check if the example is valid. The three planes: x=0 (yz-plane), y=0 (xz-plane), and x+y=1. Their pairwise intersections:x=0 and y=0: z-axis, which is (0,0,z).x=0 and x+y=1: substituting x=0 into x+y=1 gives y=1, so the line is x=0, y=1, z arbitrary.y=0 and x+y=1: substituting y=0 into x+y=1 gives x=1, so the line is x=1, y=0, z arbitrary.So the three lines are:1. (0, 0, z)2. (0, 1, z)3. (1, 0, z)These three lines are indeed all parallel to the z-axis but shifted, so none of them intersect each other. Thus, they are not concurrent. Hence, proposition (4) is false. Therefore, D is also a candidate.But wait, the problem statement says: "If 3 planes intersect pairwise, producing 3 distinct intersection lines, then these 3 lines are concurrent." So in my example, three planes intersect pairwise to produce three distinct lines which are not concurrent. Therefore, proposition (4) is false. Hence, D is correct.But wait, in the problem statement, the answer choices are A, B, C, D, each corresponding to propositions 1,2,3,4. The question is asking "Among these, the false propositions are ( )." So the answer could be multiple. But looking back at the original problem, the options are A through D, each corresponding to one proposition. Wait, no. Wait, actually, the problem is presented as "Among these, the false propositions are ( ). A. 1 B. 2 C. 3 D. 4." So the options are A to D, each being a single proposition. Wait, but the way the problem is phrased is ambiguous. Wait, the original question in Chinese might have different formatting, but based on the translation given, it says "the false propositions are ( ). A. 1 B. 2 C. 3 D. 4." So it's possible that multiple options can be chosen, but the way it's presented here is A to D, each being a single proposition. So maybe the answer is multiple of them. But the original problem might have a different setup. Wait, in the original problem statement, the user wrote:"Among these, the false propositions are ( ). A. 1 B. 2 C. 3 D. 4"Typically, in Chinese exams, this format is usually multiple choice, where you can have multiple correct answers, and the options are labeled A, B, C, D, each corresponding to a different proposition. So if propositions 2,3,4 are false, then the answer would be B,C,D. But since the options are given as A.1, B.2, C.3, D.4, the user is likely expecting multiple selections. However, in the translation provided, the instruction says "put your final answer within boxed{}," which usually expects a single boxed answer. Hmm, conflicting information.Wait, maybe in the original problem, it's a multiple-choice question where the options are combinations, like (A) 1, (B) 2, (C) 3, (D) 4, (E) 1 and 2, etc., but the user has translated it as options A to D corresponding to each proposition. However, given the translation provided, we have to work with what's given.Assuming that the answer is multiple and we have to select all that are false, but the way to represent it in the box is with multiple letters. But in the Chinese exam system, often answers are given as a combination, like boxed{BCD} if B, C, D are false. However, the user instruction says "put your final answer within boxed{}", which in previous problems in this style often takes a single letter. Wait, confusion arises. Wait, let me check the original problem statement again:"Among these, the false propositions are ( ). A. 1 B. 2 C. 3 D. 4"Given that the options are labeled A to D, each referring to a single proposition, it's possible that the question expects multiple answers, but in the Chinese context, sometimes they use multiple-choice where multiple letters can be correct, and you have to select all that apply, and the answer is boxed as, for example, boxed{BCD}. However, considering the user's previous problems, sometimes even when multiple answers are correct, the system might expect a single boxed answer. But this is conflicting.Alternatively, maybe the original problem is a multiple-select question, but in the translation, it's presented as options A-D, each being one proposition. In Chinese exams, sometimes they use letters to denote each proposition as options, and you select all that apply. For example, the answer might be B, C, D, so you write boxed{BCD}. However, since the user hasn't specified the exact format, but given the options as A-D, each being a single proposition, perhaps in the original test, it's a multiple-select question where you can choose more than one, and the answer is a combination. Given that, and based on the analysis:Propositions:1. True2. False3. False4. FalseTherefore, the false propositions are 2,3,4, which correspond to options B, C, D. So the answer should be boxed{BCD}. However, if the system only allows one boxed answer, maybe the intended answer is different. Wait, but the user's initial problem statement says:"You will be given a problem. Please reason step by step, and put your final answer within boxed{}:1. There are 4 propositions... Among these, the false propositions are ( ). A. 1 B. 2 C. 3 D. 4"So the user is following the format where the answer is among A-D, each corresponding to one of the propositions. However, in Chinese exams, often for such questions, each option is a combination, but here, it's possible that only one of them is false. But according to my analysis, propositions 2,3,4 are false. However, in the initial analysis, I might have messed up.Wait, let me recap:1. (1) is true. Infinitely many planes forming equal angles with two skew lines.2. (2) is false. Not every three skew lines have infinitely many common transversals.3. (3) is false. There exists configurations of four skew lines with a common transversal.4. (4) is false. Three planes can intersect pairwise in three non-concurrent lines.Therefore, three propositions are false: 2,3,4. So if the options are A.1, B.2, C.3, D.4, then the answer should be B, C, D. But if the question expects only one answer, maybe there's a mistake in my analysis.Wait, let me check proposition (3) again. The proposition says "For any 4 pairwise skew lines in space, there does not exist a line that intersects all four lines." If there exists even one example where such a line exists, then the proposition is false. I constructed four pairwise skew lines all intersecting a common line, so that common line intersects all four. Hence, proposition (3) is indeed false.Similarly, for proposition (4), I constructed three planes whose pairwise intersections are three non-concurrent lines, so (4) is false.Therefore, three propositions are false: 2,3,4. If the options are separate, then answer B, C, D. But the problem may have a different intended answer. Wait, maybe my analysis is incorrect.Wait, let's check another source. For proposition (4): three planes intersecting pairwise in three lines, are those lines necessarily concurrent?According to geometric principles, three planes intersecting in three lines, those three lines may or may not be concurrent. If the three planes intersect at a common point, then the three lines pass through that point. But if not, they can form a triangular prism arrangement where the lines are parallel or something. Wait, but in three-dimensional space, three planes can intersect in three lines that form a triangular prism, which are parallel? No, in a triangular prism, the three lines are the edges, which are parallel. Wait, no. A triangular prism has two triangular bases and three rectangular faces. The intersection lines of the three rectangular faces are parallel. But if three planes are arranged like the three rectangular faces, their lines of intersection are parallel. However, planes in a prism are not intersecting pairwise in those lines. Wait, maybe not. Let me think.Alternatively, three planes can intersect such that each pair intersects in a separate line, and all three lines are parallel. For example, take three planes: z=0, z=1, and x+y=0. Wait, no, z=0 and z=1 are parallel and don't intersect. So that's not good. Alternatively, take three planes: x=0, y=0, and x+y+z=0. The intersections would be:x=0 and y=0: z-axis.x=0 and x+y+z=0: line x=0, y + z = 0.y=0 and x+y+z=0: line y=0, x + z = 0.These three lines: z-axis, x=0, y=-z; and y=0, x=-z. These three lines all pass through the origin. Wait, z-axis is x=0,y=0,z arbitrary. Wait, no, z-axis is x=0, y=0. Wait, the line x=0, y + z = 0 is x=0, y = -z. The line y=0, x + z = 0 is y=0, x = -z. These two lines intersect at the origin (z=0 gives x=0,y=0). So all three lines pass through the origin. Therefore, concurrent.Hmm, maybe my previous example was incorrect. Let me check again.Previously, I had planes x=0, y=0, x+y=1. Their intersections are:x=0 and y=0: z-axis.x=0 and x+y=1: line x=0, y=1, z arbitrary.y=0 and x+y=1: line y=0, x=1, z arbitrary.But these three lines are the z-axis (0,0,z), (0,1,z), and (1,0,z). These lines are not concurrent because they don't all pass through a common point. For example, the point (0,0,0) is on the z-axis but not on the other two lines. The point (0,1,0) is on the second line but not on the others. Similarly, (1,0,0) is on the third line but not the others. So these three lines are mutually skew? Wait, no. Wait, in three-dimensional space, two lines that are not parallel and do not intersect are skew. The lines (0,0,z) and (0,1,z) are both vertical lines in the x=0 plane, but offset along y. So they are parallel? No, they have direction vector (0,0,1), same direction. So they are parallel. Wait, the lines (0,0,z) and (0,1,z) are parallel, since their direction vectors are the same. Similarly, (1,0,z) has direction vector (0,0,1), so all three lines are parallel. Wait, in that case, three parallel lines. But the z-axis is (0,0,z), direction (0,0,1). The line x=0, y=1, z arbitrary is (0,1,z), direction (0,0,1). Similarly, (1,0,z) direction (0,0,1). So all three lines are parallel. Therefore, the three lines are parallel, hence do not intersect, so are not concurrent. Therefore, in this case, three planes intersect pairwise in three parallel lines, which are not concurrent. Hence, proposition (4) is false. Therefore, my initial analysis holds. So (4) is false.Therefore, propositions 2,3,4 are false. So if the options are A.1, B.2, C.3, D.4, the answer is B, C, D. However, depending on the problem's design, maybe only one is supposed to be selected, but given the analysis, three are false. But the user might have intended a different answer.Wait, let me cross-verify. Maybe proposition (3) is actually true? Let me think again.Proposition (3): For any 4 pairwise skew lines in space, there does not exist a line that intersects all four lines.But I constructed an example with four pairwise skew lines that are all intersected by a common line. Therefore, proposition (3) is false. So unless my construction is flawed.Wait, in my construction, I tried to create four lines that all intersect a common line L, but are pairwise skew. However, when I first tried, two of them ended up intersecting, but then I adjusted the directions and they became pairwise skew. So yes, such a configuration is possible, meaning (3) is false.Similarly, for proposition (4), the three planes can intersect in three parallel lines, which are not concurrent, so (4) is false.So indeed, propositions 2, 3, 4 are false, and only proposition 1 is true.But looking back at the initial problem statement, the options are labeled A to D, each corresponding to a single proposition. If the question is presented as "the false propositions are ( )" with options A-D, each being one proposition, then in the Chinese exam context, this might be a multiple-select question, where you can choose multiple options, and the answer is the combination of B, C, D. However, in many cases, without specific instructions, it's ambiguous. However, given that the user asks to put the final answer within boxed{}, and in previous examples, single answers are boxed, but sometimes multiple letters are boxed. For example, if the answer is B, C, D, then boxed{BCD}. However, if the system requires only one, maybe there's an error.But according to the strict analysis, propositions 2,3,4 are false. Therefore, the answer should be B, C, D. But given the options are separate, maybe the intended answer is C and D, but no, the analysis shows 2,3,4 are false. Hmm.Alternatively, perhaps my analysis is wrong. Let me check another angle.For proposition (4): three planes intersecting pairwise in three lines. If the three planes do not have a common point of intersection, then their lines of intersection must be parallel or form a triangular prism.Wait, if three planes intersect pairwise in three lines, those three lines must either be concurrent or parallel. Is that the case?Wait, no. If three planes are such that each pair intersects in a line, and the three lines are not concurrent, they can either be parallel or form a triangular prism (which would require them to be parallel), but in three-dimensional space, three planes can intersect in three lines that are not concurrent and not parallel.Wait, for example, take the planes:1. z = 02. z = y3. z = xThen, the intersections:1 and 2: line z=0, z=y → y=0, z=0 (x-axis)1 and 3: line z=0, z=x → x=0, z=0 (y-axis)2 and 3: line z=y, z=x → x=y, z=x. This is the line x=y=z.So the three lines are the x-axis, y-axis, and the line x=y=z. These three lines are not all concurrent. The x-axis and y-axis meet at the origin, but the line x=y=z passes through the origin but is not coplanar with the x and y axes. Wait, actually, all three lines pass through the origin. So they are concurrent. So this example doesn't work.Wait, maybe take three planes:1. x=02. y=03. x + y + z = 1Intersections:1 and 2: z-axis (x=0, y=0)1 and 3: x=0, y + z =12 and 3: y=0, x + z =1These three lines:1. z-axis: (0,0,z)2. x=0, y =1 - z3. y=0, x=1 - zThese three lines:- The z-axis (0,0,z)- Line (0,1 - z, z)- Line (1 - z, 0, z)Do these three lines intersect at a common point?Suppose there is a common point. Then, for some z1, z2, z3:From line 1: (0,0,z1)From line 2: (0,1 - z2, z2)From line 3: (1 - z3, 0, z3)For these to be the same point:0 = 0 = 1 - z3 → 1 - z3 = 0 → z3 =10 =1 - z2 =0 → 1 - z2=0 → z2=1z1 = z2 = z3=1So the point would be (0,0,1). Check line 2 at z2=1: (0,0,1). Line 3 at z3=1: (0,0,1). So all three lines intersect at (0,0,1). Therefore, in this case, the lines are concurrent. Hmm, so my previous example didn't work.Wait, maybe another example. Let me take three planes:1. z=02. z=13. x + y =0But planes 1 and 2 are parallel, so their intersection is empty. Not allowed. Need planes that intersect pairwise.Take three planes:1. x=02. y=03. z=0These intersect at the origin, so the three coordinate axes, which are concurrent.Alternatively, take planes:1. x=02. y=03. x + y =1As before, intersections are z-axis, x=0 y=1, y=0 x=1. These three lines are parallel to the z-axis but offset. Wait, but in this case, these three lines are parallel to each other (all have direction vector (0,0,1)), and thus are parallel lines. So in this case, the three lines are parallel, hence not concurrent. But earlier, when I tried this example, the lines actually were parallel, which is another case. So three planes can intersect pairwise in three parallel lines, which are not concurrent. Therefore, proposition (4) is false.Yes, that example works. The three lines are parallel but not coinciding, hence no common point. Therefore, proposition (4) is indeed false.Thus, consolidating:1. True2. False3. False4. FalseHence, the false propositions are 2,3,4, which correspond to options B, C, D. Therefore, the answer is boxed{BCD}.However, considering the user might expect a single letter answer, but given the problem statement structure, I think the correct answer should be B, C, D. Therefore, boxed as boxed{BCD}.Final Answerboxed{BCD}</think>1. Proposition (1): For any two skew lines in space, there exist infinitely many planes that form equal angles with these two lines. - Analysis: Skew lines have infinitely many planes that form equal angles with both lines. The set of normals to these planes forms a double cone, leading to infinitely many solutions. Therefore, proposition (1) is true.2. Proposition (2): For any 3 pairwise skew lines in space, there exist infinitely many lines that intersect all three lines. - Analysis: According to the Regulus Theorem, three skew lines must lie on a hyperboloid of one sheet to have infinitely many common transversals. Since the proposition states "for any 3 pairwise skew lines," which includes configurations not on a hyperboloid, this is false. Therefore, proposition (2) is false.3. Proposition (3): For any 4 pairwise skew lines in space, there does not exist a line that intersects all four lines. - Analysis: A configuration of four skew lines intersecting a common line can be constructed. For example, four lines intersecting the z-axis at different points with different direction vectors can be pairwise skew. Thus, proposition (3) is false.4. Proposition (4): If 3 planes intersect pairwise, producing 3 distinct intersection lines, then these 3 lines are concurrent. - Analysis: Three planes can intersect pairwise in three parallel or non-concurrent lines. An example with planes x=0, y=0, and x+y=1 results in three non-concurrent lines. Therefore, proposition (4) is false.Conclusion: Propositions (2), (3), and (4) are false. The corresponding options are B, C, and D.boxed{BCD},<think>Okay, let's try to figure out which of these four propositions are false. I need to check each one step by step. Starting with Proposition (1): "For any two skew lines in space, there exist infinitely many planes that form equal angles with these two lines." Hmm. Skew lines are lines that don't intersect and aren't parallel. So, they're sort of... diagonal in 3D space. Now, a plane forming equal angles with two lines... I think that means the angle between each line and the plane is the same. Wait, but how do you measure the angle between a line and a plane? Isn't it defined as the complement of the angle between the line and the normal to the plane? So, if the angle between the line and the normal is θ, then the angle between the line and the plane would be 90° - θ. So, forming equal angles with both lines... Maybe it's similar to bisecting the angle between two lines, but in 3D? But since the lines are skew, they don't lie on the same plane. Hmm. I recall that for two skew lines, there's exactly one common perpendicular line between them. But how does that relate to planes forming equal angles? Alternatively, if we consider the set of all planes that contain the angle bisector between the two skew lines. Wait, but in 3D, angle bisectors aren't straightforward. Maybe there's a way to construct such planes. Let me think. If we fix a point on each line, maybe there's a plane that bisects the angle between the two lines at those points. But since the lines are skew, moving along the lines, the angle could vary? Or maybe not. Alternatively, maybe for each point on one line, you can find a corresponding point on the other line and construct a plane that forms equal angles. Since there are infinitely many points, there might be infinitely many such planes. I'm not sure. I need to verify this. Wait, I think there's a theorem related to this. For two skew lines, the locus of all lines that form equal angles with both skew lines is a regulus, which is a set of lines forming a ruled surface. And since a regulus is part of a hyperboloid, which is a quadric surface, it has infinitely many lines. Therefore, there might be infinitely many planes that contain these lines and form equal angles with the original skew lines. So, maybe proposition (1) is true? Hmm. But I'm not entirely certain. Let's hold off on that and check the other propositions.Proposition (2): "For any 3 pairwise skew lines in space, there exist infinitely many lines that intersect all three lines." Okay, three skew lines. In 3D space, if three lines are pairwise skew, under what conditions can a line intersect all three? I remember something called the regulus or maybe the hyperboloid of one sheet. If three skew lines lie on a hyperboloid of one sheet, then there are infinitely many lines (the generators of the hyperboloid) that intersect all three. But is it always the case that three pairwise skew lines lie on such a hyperboloid? Wait, no. Three skew lines don't necessarily lie on a hyperboloid. There's a theorem that says three pairwise skew lines lie on a unique hyperboloid of one sheet if and only if there exists a transversal line intersecting all three. Wait, but actually, if three lines are such that no two are parallel and not all three lie on a hyperboloid, then there might be only finitely many transversals. Alternatively, I think the number of lines intersecting three skew lines depends on their configuration. In general, for three arbitrary skew lines, there can be 0, 1, 2, or infinitely many lines intersecting all three. For example, if the three lines are generators of a hyperboloid, then there are infinitely many. But if they are not, then maybe only a finite number. But the proposition claims "for any 3 pairwise skew lines", there are "infinitely many" lines intersecting all three. That seems too strong. So, maybe proposition (2) is false. But I need to be careful here. Wait, in classical geometry, there is a result called the "threeline theorem" or something similar. Let me recall. If three lines are pairwise skew and not all on a hyperboloid, then there are either two or one transversal lines. But if they lie on a hyperboloid, then there are infinitely many. Therefore, unless the three lines lie on a hyperboloid, there are only finitely many transversals. Therefore, since the proposition says "for any 3 pairwise skew lines", it includes those not lying on a hyperboloid. Hence, for such lines, there might be only finitely many transversals. Therefore, the proposition (2) is false. So, (2) is false. Hmm. But wait, maybe I'm mixing things up. Let me check. In 3D projective geometry, three skew lines can have 0, 1, 2, or infinitely many common transversals. So, it's not necessarily always infinitely many. Therefore, proposition (2) is false because it's stated for "any" three skew lines. So, if there exists even one set of three skew lines with only finitely many transversals, then (2) is false. Hence, (2) is false. So, B is one of the answers. Proposition (3): "For any 4 pairwise skew lines in space, there does not exist a line that intersects all four lines." Hmm. So, the question is, can there be a line that intersects four pairwise skew lines? I know that in 3D space, four skew lines generally do not have a common transversal. However, there might be specific configurations where a line can intersect all four. For example, if the four lines lie on a hyperboloid of one sheet, which has two families of generators. Each line from one family intersects all lines of the other family. So, if the four lines are from one family, then each line from the other family intersects all four. But wait, in that case, the four lines are part of a regulus, so there exists infinitely many lines intersecting all four. Therefore, proposition (3) is false in that case. But the proposition states "for any 4 pairwise skew lines", so if there exists even one set of four skew lines that has a common transversal, then the proposition is false. But wait, the proposition says "there does not exist a line that intersects all four lines". So, if there exists a single quadruple of skew lines for which such a line exists, then the proposition is false. But in general, if you take four lines on a hyperboloid, as I mentioned, there are lines intersecting all four. Therefore, such a quadruple exists. Hence, proposition (3) is false. Therefore, (3) is false. So, C is another answer. Wait, but the options are A, B, C, D. The question says "the false propositions are ( )", so possibly multiple answers. Wait, the original problem says "the false propositions are ( ). Among these, the false propositions are ( ). A. 1 B. 2 C. 3 D. 4". Wait, the options are individual letters, but the answer might be a combination. Wait, the original problem is in Chinese, but translated here as "Among these, the false propositions are ( ). A. 1 B. 2 C. 3 D. 4". So, the options are given as A to D, each corresponding to a single proposition. But the question is "the false propositions are ( )", so possibly multiple can be selected. But in the original problem statement, maybe it's a multiple choice question where the answer is a combination. But here, the user hasn't specified whether the answer is multiple or single. Wait, in Chinese exams, sometimes they use letters with multiple possible answers, like (B, C) or something. But the way the problem is presented here is: the options are A, B, C, D, each corresponding to one proposition, and the question is "the false propositions are ( )". So, likely, it's a multiple choice question where the answer is a combination. But the user instructions say "put your final answer within boxed{}". So, in Chinese exams, for such questions, the answer is written as the letters corresponding to the false propositions. So, if propositions 2 and 3 are false, the answer would be BC, boxed as boxed{BC}. But let's check the original problem again. Wait, the original problem says: "Among these, the false propositions are ( ). A. 1 B. 2 C. 3 D. 4". So, each option is a single proposition. So, maybe the answer is multiple options. But in the Chinese exam context, sometimes they have multiple correct answers, and you have to select all that apply. But the way the problem is presented here, the user might expect the answer to be multiple letters, like BC, but the way the options are given is each is a single letter. Hmmm. Wait, perhaps the question is designed such that only one of them is false, but according to our analysis, both 2 and 3 might be false. Hmm. Wait, maybe I made a mistake. Let's re-examine.Starting again. Proposition (1): For any two skew lines, infinitely many planes form equal angles with both. Earlier, I thought maybe it's true because of some regulus or hyperboloid. But let's think differently. For two skew lines, can we find such planes?Suppose we have two skew lines. Let's fix a coordinate system where one line is along the x-axis, and the other is some skew line. Let me parameterize them. Let’s say line 1 is x-axis: points (t, 0, 0), and line 2 is, say, (0, t, 1) + t(0, 1, 0). Wait, that might not be skew. Wait, to make them skew, line 2 could be (1, t, 0) + (0, 0, t). Hmm, maybe better to use standard example: line 1 is x-axis, line 2 is (1, 0, 0) + t(0, 1, 1). So they don't intersect and aren't parallel.Now, we need to find planes that form equal angles with both lines. The angle between a line and a plane is the complement of the angle between the line and the normal to the plane. So, equal angles would mean that the angle between each line and the normal to the plane is equal, or complementary. Wait, the angle between the line and the plane is defined as the smallest angle between the line and any line in the plane, which is equal to 90° minus the angle between the line and the normal. So, if we want equal angles between the plane and each line, that would mean that 90° - θ1 = 90° - θ2, where θ1 and θ2 are the angles between the normals and the lines. So, θ1 = θ2. Therefore, the plane's normal makes equal angles with both lines. So, the problem reduces to finding planes whose normals make equal angles with both skew lines. How many such planes are there? For two lines in space, the set of all lines (normals) that make equal angles with both lines would form a certain locus. Alternatively, think of the angle bisectors. For two lines in space, the set of lines that are angle bisectors. But in 3D, the angle bisectors between two skew lines form a surface. Therefore, there are infinitely many such lines, each corresponding to a plane (since each line is a normal to some plane). Hence, there are infinitely many planes whose normals make equal angles with both skew lines, hence forming equal angles with the lines. Therefore, proposition (1) is true. So, (1) is true. Hence, A is not an answer.Proposition (2): For any three skew lines, infinitely many lines intersect all three. As discussed earlier, this is not necessarily true. If three skew lines lie on a hyperboloid, then there are infinitely many transversals. But if they don't, there might be only a finite number. Therefore, since the proposition says "for any", it's false. So, (2) is false. Therefore, B is an answer.Proposition (3): For any four skew lines, there does not exist a line intersecting all four. But as per the hyperboloid example, if four lines lie on a hyperboloid, then each generator of the other regulus intersects all four. So, such lines do exist. Hence, proposition (3) is false because it claims that for any four skew lines, such a line doesn't exist, which is not true. Therefore, (3) is false. So, C is an answer.Proposition (4): If three planes intersect pairwise, producing three distinct intersection lines, then these three lines are concurrent. Hmm. Let's visualize. Three planes intersecting pairwise. Each pair of planes intersects in a line. So, three planes can intersect in three different lines. Are these lines necessarily concurrent? That is, do all three lines meet at a single point?Let me take three planes. For example, the xy-plane, xz-plane, and yz-plane. Their pairwise intersections are the x-axis, y-axis, and z-axis, which are concurrent at the origin. So, in this case, they are concurrent. But what if I take three different planes? Suppose plane 1: z = 0, plane 2: x = 0, plane 3: x + y = 1. Then, the intersection lines are:Plane 1 & 2: z=0, x=0 (the y-axis)Plane 1 & 3: z=0, x + y = 1 (a line in the xy-plane)Plane 2 & 3: x=0, x + y = 1 => y = 1 (a line along x=0, y=1, z arbitrary)These three lines: y-axis, the line x + y = 1 in z=0, and the line x=0, y=1. Are they concurrent? The y-axis is (0, t, 0). The line x + y =1 in z=0 is (t, 1 - t, 0). The line x=0, y=1 is (0, 1, t). The common point would need to satisfy all three. The y-axis and x + y =1 intersect at (0,1,0). The line x=0, y=1 also passes through (0,1,0). So, all three lines meet at (0,1,0). So, they are concurrent. Wait, but maybe another example. Let me try three planes where the lines are not concurrent. Let’s consider plane 1: z = 0, plane 2: z = 1, plane 3: x + y = 0. Then, the intersection lines:Plane1 & Plane2: They are parallel and don't intersect. Wait, but planes z=0 and z=1 are parallel, so they don't intersect. Therefore, they don't form a line. So, the problem states "three planes intersect pairwise, producing 3 distinct intersection lines". So, if planes 1 and 2 are parallel, they don't intersect. So, that example doesn't satisfy the condition. Another example. Let's take three planes that pairwise intersect in lines, but the lines are not concurrent. For instance, take three planes:Plane 1: y = 0Plane 2: z = 0Plane 3: x = 0Their lines of intersection are:Plane1 & Plane2: x-axis (y=0, z=0)Plane1 & Plane3: z-axis (y=0, x=0)Plane2 & Plane3: y-axis (z=0, x=0)These three axes meet at the origin, so concurrent.Another try. Let's take three planes:Plane1: z = 0Plane2: y = 1Plane3: x + y = 0Intersection lines:Plane1 & Plane2: z=0, y=1 (a horizontal line)Plane1 & Plane3: z=0, x + y = 0 (a line in the xy-plane)Plane2 & Plane3: y=1, x + y =0 ⇒ x = -1 (a vertical line at x=-1, y=1, z arbitrary)Are these three lines concurrent? The first line is (t, 1, 0). The second line is (t, -t, 0). The third line is (-1, 1, s). Do they all intersect at a common point?The first and second lines: (t, 1, 0) and (t, -t, 0). Setting equal: t = t, 1 = -t, 0 = 0 ⇒ t = -1. So, the point (-1, 1, 0) is on both lines. The third line is (-1, 1, s), which passes through (-1,1,0) when s=0. So, all three lines meet at (-1,1,0). Therefore, they are concurrent. Hmm. Maybe it's challenging to find three planes that pairwise intersect in three non-concurrent lines. Wait, maybe if the three planes are such that their lines of intersection are all parallel to each other? But if three planes are such that each pair intersects in a line, and all three lines are parallel, then they would be concurrent at infinity, but in projective geometry, maybe. But in standard Euclidean geometry, they are not concurrent. Wait, but can three planes pairwise intersect in three parallel but distinct lines?Let me see. Let's take three planes:Plane1: z = 0Plane2: z = yPlane3: z = 2yThen, the intersections:Plane1 & Plane2: z=0, z=y ⇒ y=0, so the x-axis.Plane1 & Plane3: z=0, z=2y ⇒ y=0, same as above. Wait, this is not three distinct lines. The problem states "producing 3 distinct intersection lines". So, this example is invalid.Another try. Let's take Plane1: z = 0, Plane2: z = y, Plane3: z = x.Then:Plane1 & Plane2: z=0, y=0 (the x-axis)Plane1 & Plane3: z=0, x=0 (the y-axis)Plane2 & Plane3: Solve z = y and z = x ⇒ y = x. So, the line x=y, z=x (or z=y). This is a line in space. These three lines: x-axis, y-axis, and the line x=y=z. These three lines are not parallel, and they intersect at the origin. So, again concurrent.Wait, is it possible to have three planes intersecting pairwise in three distinct lines that are not concurrent? Suppose Plane1: x = 0Plane2: y = 0Plane3: x + y + z = 1Their intersections:Plane1 & Plane2: z-axis (x=0, y=0)Plane1 & Plane3: x=0, y + z =1 ⇒ line (0, t, 1 - t)Plane2 & Plane3: y=0, x + z =1 ⇒ line (t, 0, 1 - t)These three lines: z-axis, (0, t, 1 - t), and (t, 0, 1 - t). Do these lines concur? The z-axis is (0,0,t). The line (0, t, 1 - t) passes through (0,0,1) when t=0. The line (t,0,1 - t) passes through (0,0,1) when t=0. The z-axis passes through (0,0,1) when t=1. Wait, no. The z-axis is (0,0,t). So, when t=1, it's (0,0,1). The line (0, t, 1 - t) at t=0 is (0,0,1), and the line (t,0,1 - t) at t=0 is (0,0,1). Therefore, all three lines pass through (0,0,1). So, they are concurrent.Hmm, maybe it's impossible? Wait, maybe all three lines must be concurrent? Wait, think in terms of linear algebra. Three planes are given by equations:a1x + b1y + c1z = d1a2x + b2y + c2z = d2a3x + b3y + c3z = d3If each pair of planes intersects in a line, then the three planes do not have a common intersection (unless all three planes intersect along a common line). But the problem states "three planes intersect pairwise, producing three distinct intersection lines". So, each pair intersects in a different line. Now, the question is: are these three lines necessarily concurrent? That is, is there a common point lying on all three lines?Suppose that the three planes intersect pairwise along three different lines. Let’s denote the lines as L12 (intersection of plane1 and plane2), L13 (intersection of plane1 and plane3), and L23 (intersection of plane2 and plane3). If there is a point P that lies on all three lines, then P is common to all three planes. Therefore, P is the solution to all three plane equations. However, if the three planes do not have a common point (i.e., the system of equations is inconsistent), then the three lines L12, L13, L23 do not concur. But wait, if three planes pairwise intersect in three lines, can the system of three equations be inconsistent? Let me check.Suppose we have three planes:Plane1: z = 0Plane2: z = 1Plane3: x + y + z = 0Then, Plane1 & Plane2 are parallel, so they don't intersect. So, this doesn't satisfy the problem's condition of producing three distinct intersection lines. So, invalid example.Another example. Let’s take:Plane1: z = 0Plane2: y = 1Plane3: x = 1Then, intersections:Plane1 & Plane2: z=0, y=1 (line x can be anything)Plane1 & Plane3: z=0, x=1Plane2 & Plane3: y=1, x=1 (line z can be anything)These three lines are:1. (t, 1, 0)2. (1, t, 0)3. (1, 1, t)These three lines all pass through the point (1,1,0). So, they are concurrent.Wait, but what if we take three planes that form a triangular prism? Like, the three faces of a prism. But in that case, the planes would be parallel lines, but in space, the intersections would be parallel lines, which don't intersect. But such planes would have parallel lines of intersection, which are not concurrent. But in Euclidean geometry, parallel lines don't meet, so the three lines would not be concurrent. But the problem states that the three planes produce three distinct intersection lines. If the planes are arranged like a prism, their pairwise intersections would be three parallel lines, which are distinct. But in that case, the three lines are parallel and non-concurrent. However, in reality, three planes forming a prism would have their lines of intersection being parallel, hence not concurrent. Wait, let me try to formalize this. Suppose we have three planes:Plane1: z = 0Plane2: z = yPlane3: z = -yBut then, Plane1 & Plane2: z=0, y=0 (x-axis)Plane1 & Plane3: z=0, y=0 (x-axis again). So, not distinct lines. Invalid.Alternatively, let's take:Plane1: z = 0Plane2: z = 1Plane3: x + y = 0But as before, Plane1 and Plane2 are parallel, no intersection. Not allowed.Wait, perhaps three planes whose normals are not coplanar. Suppose:Plane1: z = 0Plane2: x = 0Plane3: y = 0These intersect at the origin. So, concurrent.Another example. Plane1: z = 1Plane2: x + z = 0Plane3: y + z = 0Intersections:Plane1 & Plane2: z=1, x = -1 (line at x=-1, z=1, y can be anything)Plane1 & Plane3: z=1, y = -1 (line at y=-1, z=1, x can be anything)Plane2 & Plane3: Solve x + z = 0 and y + z = 0 ⇒ x = -z, y = -z ⇒ line parametrized as ( -t, -t, t )These three lines:1. (-1, t, 1)2. (t, -1, 1)3. (-t, -t, t )Do these lines intersect at a common point?Check if there's a point (x,y,z) common to all three.From line 1: x=-1, z=1, y is arbitrary.From line 2: y=-1, z=1, x is arbitrary.So, the intersection of lines 1 and 2 requires x=-1, y=-1, z=1. Does this point lie on line 3? Plug into line 3's parametric equations: x=-t, y=-t, z=t. So, -t = -1 ⇒ t=1, then y=-t = -1, z=t=1. So, yes, point (-1, -1, 1) is on line 3 when t=1. Therefore, all three lines meet at (-1, -1, 1). Therefore, they are concurrent.This is frustrating. It seems hard to find three planes with three distinct, non-concurrent intersection lines. Maybe it's impossible? If so, then proposition (4) is true.Wait, but I need to recall. In three-dimensional space, three planes can intersect in three lines that form a triangle, i.e., are not concurrent and not parallel. But how?Wait, in order to form a triangular arrangement, the three lines would need to form the edges of a triangle. But in 3D space, a triangle is a planar figure. However, three lines forming a triangle would have to lie on the same plane, but if each line is the intersection of two planes, then all three planes would have to intersect along lines lying on that same plane. But if the three planes are not all intersecting along a common line, then their lines of intersection can't lie on a single plane. Wait, this is getting confusing. Let me think differently.Suppose three planes intersect pairwise in three lines, and these three lines are not concurrent. Then, the three lines form a "triangle" in space, but not lying on a single plane. However, in 3D space, three lines can form a spatial triangle, but such a configuration requires that each pair of lines is skew, which is not the case here. Because the three lines are intersections of planes, each pair of lines lies on a plane (the original two planes). Therefore, each pair of lines must either intersect or be parallel. Wait, if three planes intersect pairwise in three lines, then each pair of lines (from different intersections) must either intersect or be parallel. Because each pair of lines lies on a common plane. For example, line L12 (intersection of Plane1 and Plane2) and line L13 (intersection of Plane1 and Plane3) both lie on Plane1. Therefore, they must either intersect or be parallel. Similarly, L12 and L23 lie on Plane2, so they must intersect or be parallel. Therefore, if the three lines are not concurrent, then at least two of them must be parallel. But if two lines are parallel and lie on the same plane, they don't intersect. However, in our case, lines L12 and L13 lie on Plane1. If they are parallel, then they don't intersect, meaning that Plane1 contains two parallel lines. Similarly, if all three lines are parallel, then each pair is parallel, and they don't intersect. However, in this case, the three lines would be parallel, hence not concurrent. But the problem states that the three intersection lines are distinct. So, if three planes intersect pairwise in three distinct, parallel lines, then those lines are not concurrent. Hence, the proposition (4) would be false. But can three planes intersect pairwise in three distinct parallel lines? Yes, for example, take three planes:Plane1: z = 0Plane2: z = 1Plane3: z = 2But then, the intersections between Plane1 & Plane2: empty, since they are parallel. So, invalid.Alternatively, take three planes such that each pair intersects in a parallel line. For instance:Plane1: z = 0Plane2: z = yPlane3: z = 2yThen, Plane1 & Plane2: line where z=0 and z=y ⇒ y=0, z=0 (x-axis)Plane1 & Plane3: z=0 and z=2y ⇒ y=0, z=0 (x-axis again). Not distinct.Alternatively, take three planes arranged like the pages of a book, but each page is shifted. Let me think.Take Plane1: z = 0Plane2: z = xPlane3: z = x + 1Then, Plane1 & Plane2: z=0, x=0 (y-axis)Plane1 & Plane3: z=0, x=-1 (line x=-1, z=0, y arbitrary)Plane2 & Plane3: z=x and z=x +1 ⇒ x = x +1 ⇒ impossible, no intersection. So, invalid.Another approach. Let’s consider three planes, each pair of which intersect in a line, and all three lines are parallel. For example:Plane1: y = 0Plane2: y = 1Plane3: y = 2Then, intersections are empty between the planes. Not valid.Alternatively, take three planes with equations:Plane1: y = 0Plane2: y = xPlane3: y = 2xThen, intersections:Plane1 & Plane2: y=0, y=x ⇒ x=0, y=0 (z-axis)Plane1 & Plane3: y=0, y=2x ⇒ x=0, y=0 (z-axis again)Plane2 & Plane3: y=x, y=2x ⇒ x=0, y=0 (z-axis again). Not distinct lines.This is difficult. Maybe it's not possible to have three distinct parallel lines as pairwise intersections of three planes. Because if two planes intersect in a line L, then a third plane can only intersect each of the first two planes in lines that are either intersecting L or parallel to L. But if the third plane's intersection lines are parallel to L, then they must also be parallel to each other. But to have three distinct parallel lines, all three planes would need to be "stacked" in some way, but planes are infinite, so parallel lines would never meet, but the planes themselves would intersect elsewhere? Wait, perhaps three planes can be set such that each pair intersects in a distinct parallel line. For example:Let’s take three planes:Plane1: z = 0Plane2: z = 1Plane3: z = 2But these are all parallel, so they don't intersect. So, not valid.Alternatively, take three planes:Plane1: z = 0Plane2: x = 0Plane3: x + z = 0Intersections:Plane1 & Plane2: z=0, x=0 (y-axis)Plane1 & Plane3: z=0, x=0 (y-axis again)Plane2 & Plane3: x=0, x + z =0 ⇒ z=0, x=0 (y-axis again). Not distinct.Hmm, this is tricky. Let's try another approach. Suppose we have three planes:Plane1: z = 0Plane2: z = m y + cPlane3: z = n y + dChoose m, c, n, d such that the intersections are three distinct parallel lines.Intersection of Plane1 & Plane2: z=0, z=my + c ⇒ my + c =0 ⇒ y = -c/m. So, the line is y = -c/m, z=0, x arbitrary.Intersection of Plane1 & Plane3: y = -d/n, z=0, x arbitrary.Intersection of Plane2 & Plane3: Solve my + c = n y + d ⇒ (m - n)y = d - c ⇒ y = (d - c)/(m - n), z= m y + c. This is a line with fixed y and z depending on y, x arbitrary.To have all three lines parallel, we need them to be along the x-direction (since x is arbitrary). So, the lines from Plane1 & Plane2 and Plane1 & Plane3 are along x-axis at different y positions. The third line, from Plane2 & Plane3, is a vertical line (fixed y and z?), no. Wait, if the line from Plane2 & Plane3 is x arbitrary, y = (d - c)/(m - n), z = m y + c. So, this is a line parallel to the x-axis at y = (d - c)/(m - n), z = m*(d - c)/(m - n) + c. So, if we set this line to also be parallel to the x-axis, then all three lines are parallel. So, if we choose m, n, c, d such that the three lines are parallel (all along x-axis direction), but at different y and z positions. For example:Let’s set Plane2: z = yPlane3: z = y + 1Then:Plane1 & Plane2: z=0, z=y ⇒ y=0, so the line x arbitrary, y=0, z=0 (x-axis)Plane1 & Plane3: z=0, z=y +1 ⇒ y = -1, so the line x arbitrary, y=-1, z=0Plane2 & Plane3: Solve z = y and z = y +1 ⇒ y = y +1 ⇒ impossible, no solution. So, Plane2 and Plane3 are parallel? Wait, no. Wait, the equations z = y and z = y +1 are not parallel planes. Wait, actually, they are parallel? Because their normals are both (0, -1, 1), right? Wait, plane2: z - y =0, normal vector (0, -1, 1). Plane3: z - y =1, normal vector (0, -1, 1). So, yes, parallel planes. Therefore, they don't intersect, so invalid.So, to have three planes where each pair intersects in a line, and all three lines are parallel, the third plane would have to intersect the first two planes in lines parallel to their original intersection. But unless the third plane is arranged in a way that its intersection lines with the first two are parallel. Let’s try:Plane1: z = 0Plane2: y = 0Plane3: y = 1Then, Plane1 & Plane2: z=0, y=0 (x-axis)Plane1 & Plane3: z=0, y=1 (a parallel line to x-axis)Plane2 & Plane3: no intersection, since planes y=0 and y=1 are parallel. So, invalid.Another example. Plane1: z =0Plane2: y =0Plane3: z =1Then, intersections:Plane1 & Plane2: x-axisPlane1 & Plane3: no intersection, since parallel.Invalid.Alternatively, Plane1: z =0Plane2: z = yPlane3: z = y +1Then, intersections:Plane1 & Plane2: z=0, y=0 (x-axis)Plane1 & Plane3: z=0, y= -1 (parallel line)Plane2 & Plane3: no intersection, since parallel. So, invalid.Hmm. It seems challenging to create three planes that intersect pairwise in three distinct, non-concurrent lines. Therefore, perhaps proposition (4) is true. Wait, but if three planes intersect pairwise in three lines, then those three lines must either be concurrent or all parallel. Because each pair of lines lies on a common plane (from the original three planes). If two lines lie on a plane, they can either intersect or be parallel. So, for three lines L12, L13, L23, if L12 and L13 lie on Plane1, they must intersect or be parallel. Similarly, L12 and L23 lie on Plane2, so they must intersect or be parallel. If all three lines are to be non-concurrent and non-parallel, that would require that, for example, L12 and L13 intersect at point P, L12 and L23 intersect at point Q, but P ≠ Q, and L13 and L23 either intersect at another point or are parallel. But this seems impossible. Wait, suppose L12 and L13 intersect at P (on Plane1), and L12 and L23 intersect at Q (on Plane2). If P ≠ Q, then L13 and L23 are two lines not on the same plane, unless they intersect at some point. But if L13 is on Plane1 and Plane3, and L23 is on Plane2 and Plane3, then their intersection would be on Plane3. If they intersect, then that point would also be on L12, because it's on both Plane1 and Plane2. Therefore, leading to a common intersection point. Therefore, if three planes intersect pairwise in three lines, then any two of those lines intersect at a point common to all three planes. Hence, the three lines must concur at that point. Therefore, proposition (4) is true. Hence, (4) is true. Therefore, D is not an answer.So, in summary:Proposition (1): TrueProposition (2): FalseProposition (3): FalseProposition (4): TrueTherefore, the false propositions are (2) and (3), which correspond to options B and C. But the original problem lists options A, B, C, D as individual propositions. So, the answer should be B and C. However, in the problem statement, the options are given as A. 1, B. 2, C. 3, D. 4. So, if the user expects multiple answers, the answer would be BC. However, in some testing formats, you might have to select all that apply. Since the user's instructions say to put the final answer within boxed{}, and typically for multiple choice in Chinese exams, the answer would be written as the letters corresponding to the incorrect propositions. In this case, B and C. So, boxed BC.But let me confirm once again:Proposition (3): "For any 4 pairwise skew lines in space, there does not exist a line that intersects all four lines." Is this false? Yes, because if four lines lie on a hyperboloid, there are infinitely many lines intersecting all four. Therefore, proposition (3) is false. Hence, (3) is false. Therefore, the answer should be options B and C. So, boxed{BC}Final Answerboxed{BC}</think>To determine which propositions are false, we analyze each one step by step:1. Proposition (1): For any two skew lines in space, there exist infinitely many planes that form equal angles with these two lines. - Skew lines do not intersect and are not parallel. The angle between a line and a plane is the complement of the angle between the line and the normal to the plane. For two skew lines, the set of normals forming equal angles with both lines forms a regulus, leading to infinitely many planes. This proposition is true.2. Proposition (2): For any 3 pairwise skew lines in space, there exist infinitely many lines that intersect all three lines. - Three pairwise skew lines do not necessarily lie on a hyperboloid of one sheet. If they do, there are infinitely many transversals. However, if they do not, there might be only finitely many transversals. Since the proposition claims "for any," it is false.3. Proposition (3): For any 4 pairwise skew lines in space, there does not exist a line that intersects all four lines. - Four skew lines can lie on a hyperboloid of one sheet, where each generator of the other regulus intersects all four. Thus, such lines exist, making this proposition false.4. Proposition (4): If 3 planes intersect pairwise, producing 3 distinct intersection lines, then these 3 lines are concurrent. - Three planes intersecting pairwise in three lines must have those lines concurrent. This is because any two lines from different intersections must meet at a common point. This proposition is true.The false propositions are (2) and (3), corresponding to options B and C.boxed{BC}]

🔑:1. B(1) Through a point in space, it is evident that two parallel lines to the given two lines can be constructed, thus the described plane can be formed, correct.(2) Let the three pairwise skew lines be a, b, c. Among all planes passing through a, there is only one that is parallel to b and only one that is parallel to c. Therefore, there are infinitely many planes passing through a and intersecting b, c, with intersection points B, C. Line BC is either parallel or intersects with a, but there is at most one plane such that BC parallel a. Thus, there are infinitely many planes where BC intersects with a, meaning there are infinitely many lines BC that intersect with a, b, c, correct.(3) In a plane, construct 4 concurrent lines, with the common point being K. Draw a perpendicular line l through K to the plane. Take 4 different points on l, and through each point, draw a parallel line to one of the 4 lines, ensuring that parallel lines to all 4 lines are drawn. The 4 lines thus constructed serve as a counterexample, incorrect.(4) These 3 intersection lines may be mutually parallel, incorrect.

🧠:[<think>Okay, so I need to find the equation of the circumcircle of triangle ABC with vertices at A(8,0), B(0,6), and O(0,0). Then, I have to find the equation of a tangent line from point P(-1,5) to this circle. Let's take this step by step.Starting with part (1): the circumcircle of triangle ABC. The circumcircle is the circle that passes through all three vertices of the triangle. To find its equation, I need to determine the center (circumcenter) and the radius.First, let me recall that the circumcenter is the intersection point of the perpendicular bisectors of the sides of the triangle. So, maybe I can find the perpendicular bisectors of two sides and then find their intersection, which will be the circumcenter. Then, the radius will be the distance from this center to any of the three vertices.Let me label the points again for clarity: A(8,0), B(0,6), and O(0,0). So, the triangle is formed by these three points. Let's find the perpendicular bisectors of two sides. Let's take sides OA and OB first. Wait, but OA is from O(0,0) to A(8,0), which is a horizontal line. The perpendicular bisector of OA would be a vertical line passing through the midpoint of OA. Similarly, OB is from O(0,0) to B(0,6), which is a vertical line, so its perpendicular bisector would be a horizontal line passing through the midpoint of OB.But wait, maybe taking sides OA and AB would be better? Let me see. Maybe sides OA and OB are easier because they are along the axes. Let me try that.First, find the midpoint of OA. OA is from (0,0) to (8,0). The midpoint is ((8+0)/2, (0+0)/2) = (4,0). The slope of OA is (0-0)/(8-0) = 0, so it's a horizontal line. Therefore, the perpendicular bisector is vertical, passing through (4,0). So, the equation is x = 4.Next, find the midpoint of OB. OB is from (0,0) to (0,6). Midpoint is (0,3). The slope of OB is (6-0)/(0-0), wait, that's undefined. So, OB is vertical, so the perpendicular bisector is horizontal, passing through (0,3). So, the equation is y = 3.Wait, but if we take the perpendicular bisectors of OA and OB, their intersection is (4,3). So, is (4,3) the circumcenter? Let me check if this point is equidistant from all three vertices.Distance from (4,3) to O(0,0): sqrt((4-0)^2 + (3-0)^2) = sqrt(16 + 9) = sqrt(25) = 5.Distance from (4,3) to A(8,0): sqrt((8-4)^2 + (0-3)^2) = sqrt(16 + 9) = sqrt(25) = 5.Distance from (4,3) to B(0,6): sqrt((0-4)^2 + (6-3)^2) = sqrt(16 + 9) = sqrt(25) = 5.Wow, so all distances are 5. That seems correct. So, the circumradius is 5, and the circumcenter is at (4,3). Therefore, the equation of the circumcircle is (x - 4)^2 + (y - 3)^2 = 25. That should be part (1)'s answer.Wait, but let me verify once more. Maybe there's a mistake here. Let me check with another method. Sometimes, using perpendicular bisectors can be tricky if the triangle is not right-angled. Wait, is this triangle right-angled? Let's check.Compute the vectors OA, OB, and AB. OA is (8,0), OB is (0,6), and AB is (-8,6). The dot product of OA and OB is (8)(0) + (0)(6) = 0. So, OA and OB are perpendicular. So, triangle OAB is right-angled at O. Therefore, the circumradius should be half the hypotenuse. The hypotenuse is AB. Let's compute the length of AB. From A(8,0) to B(0,6): sqrt((0-8)^2 + (6-0)^2) = sqrt(64 + 36) = sqrt(100) = 10. Therefore, the circumradius should be 10/2 = 5, which matches our previous result. For a right-angled triangle, the circumcenter is at the midpoint of the hypotenuse. Wait, so if the hypotenuse is AB, then the midpoint of AB should be the circumcenter. Let's compute the midpoint of AB: ((8+0)/2, (0+6)/2) = (4,3). Which is the same point we found earlier. Perfect. So, indeed, the circumcircle is centered at (4,3) with radius 5. So, part (1) is settled.Now, moving on to part (2): Draw a tangent line l from point P(-1,5) to circle C, and find the equation of the tangent line l.To find the equations of the tangent lines from an external point to a circle, there are a few methods. One common approach is to use the condition that the distance from the point to the center of the circle is equal to the square root of the sum of the squares of the radius and the distance from the point to the line. Wait, maybe another way is better.Alternatively, we can parametrize the tangent lines as lines passing through P(-1,5) and tangent to the circle. The general equation of a line passing through P can be written as y - 5 = m(x + 1), where m is the slope. Then, the condition for this line to be tangent to the circle (x - 4)^2 + (y - 3)^2 = 25 is that the distance from the center (4,3) to the line is equal to the radius, which is 5.So, let's use this method.First, the equation of the tangent line: y - 5 = m(x + 1). Let's write this in standard form: mx - y + (m + 5) = 0. Wait, let me check:Starting from y - 5 = m(x + 1), expanding gives y = mx + m + 5. Rearranged to standard form: mx - y + (m + 5) = 0. So, coefficients are A = m, B = -1, C = m + 5.The distance from center (4,3) to this line is |A*4 + B*3 + C| / sqrt(A^2 + B^2) = |4m - 3 + m + 5| / sqrt(m^2 + 1) = |5m + 2| / sqrt(m^2 + 1). This distance must equal the radius 5.Therefore, |5m + 2| / sqrt(m^2 + 1) = 5.Squaring both sides to remove the absolute value and the square root:(5m + 2)^2 = 25(m^2 + 1)Expand left side: 25m² + 20m + 4 = 25m² + 25Subtract 25m² +25 from both sides: 20m + 4 -25 = 0 => 20m -21 =0 => 20m =21 => m=21/20.Wait, but this gives only one value of m, but from a point outside the circle, we usually expect two tangent lines. Hmm. Did I make a mistake here?Wait, let's verify the algebra step again.Left side after squaring: (5m + 2)^2 = 25m² + 20m +4Right side: 25(m² +1)=25m² +25Subtracting right side from left side: (25m² +20m +4) - (25m² +25) = 20m -21 =0Thus, 20m =21 => m=21/20. Hmm, so only one solution. That suggests that maybe the point P is such that there is only one tangent line, but that's impossible unless the point is on the circle. Wait, but let me check if point P(-1,5) is on the circle.Plug into the circle equation: (x -4)^2 + (y -3)^2 = (-1 -4)^2 + (5 -3)^2 = (-5)^2 + (2)^2 =25 +4=29. Which is greater than 25, so P is outside the circle. Therefore, there should be two tangent lines. So, perhaps there was a mistake in the setup.Wait, maybe I made a mistake in converting the equation to standard form. Let's check the equation again.Starting from y -5 = m(x +1). Let's rearrange it to standard form Ax + By + C =0.Subtract y and add 5: m(x +1) - y +5 =0 => mx + m - y +5 =0 => mx - y + (m +5)=0. So, coefficients are A=m, B=-1, C=m +5. Then the distance formula is |A*4 + B*3 + C| / sqrt(A² + B²) = |4m -3 + m +5| / sqrt(m² +1) = |5m +2| / sqrt(m² +1). Set equal to 5.So equation |5m +2| =5 sqrt(m² +1). Squared: 25m² +20m +4=25m² +25 => 20m +4=25 =>20m=21 => m=21/20. Hmm, only one solution. But we should have two. Wait, maybe when squaring, we sometimes lose solutions, but in this case, the absolute value was handled by squaring, so maybe there's another case where 5m +2 is negative? Wait, but squaring takes care of absolute value. Wait, perhaps the line is vertical, which would have an undefined slope. Let me check.Wait, in the parametrization, I assumed the line has a slope m, but vertical lines (infinite slope) are not included. So, maybe one of the tangent lines is vertical. Let's check if there's a vertical tangent line passing through P(-1,5). A vertical line has equation x = -1. Let's see if this line is tangent to the circle.Substitute x = -1 into the circle equation: (-1 -4)^2 + (y -3)^2 =25 => 25 + (y -3)^2=25 => (y -3)^2=0 => y=3. So, the vertical line x=-1 touches the circle at (-1,3). But point P is (-1,5). The line x=-1 passes through P(-1,5) and touches the circle at (-1,3). So, this is indeed a tangent line from P to the circle. But in our previous calculation, we didn't get this solution because vertical lines have undefined slope. So, we have two tangent lines: one with slope 21/20 and the vertical line x = -1.Therefore, we need to consider both possibilities: lines with slope m and vertical lines. Since in our initial approach we considered lines with slope m, we missed the vertical tangent. So, the equations of the tangent lines are x = -1 and the line with slope 21/20. Let's confirm this.First, check the vertical line x=-1: as above, it touches the circle at (-1,3) and passes through P(-1,5), so that's valid.Now, the other tangent line with slope 21/20: let's write its equation. Using point-slope form: y -5 = (21/20)(x +1). To convert this to standard form:Multiply both sides by 20: 20(y -5) =21(x +1)20y -100 =21x +21Rearranged: -21x +20y -121 =0, or 21x -20y +121 =0. Wait, sign choice. Let me check:Starting from y -5 = (21/20)(x +1)Bring all terms to left: (21/20)x - y + (21/20 +5) =0Multiply through by 20 to eliminate fractions: 21x -20y +21 +100 =0 =>21x -20y +121=0So the equation is 21x -20y +121=0. Let's verify if this line is tangent to the circle.Compute the distance from center (4,3) to the line 21x -20y +121=0.Distance is |21*4 -20*3 +121| / sqrt(21² + (-20)²) = |84 -60 +121| / sqrt(441 +400)= |145| / sqrt(841)=145/29=5. Which is equal to the radius. So, yes, this line is indeed tangent to the circle.Therefore, the two tangent lines are x = -1 and 21x -20y +121=0. But the problem says "the equation of the tangent line l". Wait, maybe they want both equations? The problem says "Draw a tangent line l from point P(-1,5) to circle C, and find the equation of the tangent line l." Hmm, the wording is a bit ambiguous. Since there are two tangent lines, perhaps we need to provide both. But maybe in the problem's original language (Chinese?), it might specify if it's one or both. But given the translation, it just says "a tangent line l", but since there are two, perhaps both are required.But let me check again. Maybe there's a miscalculation. Wait, when I solved for m, I only found one solution, but there should be two. But since we considered vertical line separately, we have two. So, the answer should include both equations.However, let's check if there's another non-vertical tangent line. Wait, we have the vertical line x=-1 and the line with slope 21/20. Let's check another way: perhaps parametrize the tangent points.Alternatively, another method to find the tangent lines is to find the points of tangency on the circle, then write the equations of the lines connecting P to those points.Let me try this approach. Let the circle be (x-4)^2 + (y-3)^2=25. Let point P(-1,5). The tangent lines from P to the circle will touch the circle at points T1 and T2. The line PT1 is tangent to the circle at T1, so PT1 is perpendicular to the radius OT1 (where O is the center (4,3)). So, the vector from O to T1 is perpendicular to the vector from T1 to P.Let me denote T1 as (x,y). Then, the vector OT1 is (x-4, y-3), and the vector PT1 is (x - (-1), y -5) = (x+1, y-5). These two vectors must be perpendicular, so their dot product is zero:(x -4)(x +1) + (y -3)(y -5) =0Also, since T1 lies on the circle, we have (x -4)^2 + (y -3)^2 =25.So, we have two equations:1. (x -4)(x +1) + (y -3)(y -5) =02. (x -4)^2 + (y -3)^2 =25Let me expand equation 1:(x² -4x +x -4) + (y² -3y -5y +15) =0 => x² -3x -4 + y² -8y +15=0 => x² + y² -3x -8y +11=0Equation 2 is:x² -8x +16 + y² -6y +9=25 =>x² + y² -8x -6y +25=25 =>x² + y² -8x -6y=0Subtract equation 2 from equation 1:(x² + y² -3x -8y +11) - (x² + y² -8x -6y)=0 => (-3x -8y +11) - (-8x -6y)=0 => (5x -2y +11)=0So, 5x -2y +11=0. Therefore, the points of tangency lie on the line 5x -2y +11=0 and on the circle (x-4)^2 + (y-3)^2=25.So, solve the system:5x -2y +11=0(x-4)^2 + (y-3)^2=25Let me solve 5x -2y +11=0 for y:5x +11=2y => y=(5x +11)/2Substitute into circle equation:(x-4)^2 + [(5x +11)/2 -3]^2 =25Simplify the y-term:(5x +11)/2 -3 = (5x +11 -6)/2 = (5x +5)/2 =5(x +1)/2Thus, the equation becomes:(x -4)^2 + [5(x +1)/2]^2 =25Compute each term:(x -4)^2 =x² -8x +16[5(x +1)/2]^2 =25(x +1)^2 /4 =25(x² +2x +1)/4So, combining:x² -8x +16 +25(x² +2x +1)/4 =25Multiply through by 4 to eliminate denominator:4x² -32x +64 +25x² +50x +25 =100Combine like terms:(4x² +25x²) + (-32x +50x) + (64 +25) =10029x² +18x +89 =10029x² +18x -11 =0Now, solve this quadratic equation for x:x = [-18 ± sqrt(18² -4*29*(-11))]/(2*29)Compute discriminant:D=324 +4*29*11=324 +1276=1600sqrt(D)=40Thus, x=(-18 ±40)/58First solution: (-18 +40)/58=22/58=11/29Second solution: (-18 -40)/58=-58/58=-1Thus, x=11/29 and x=-1Now, find corresponding y:For x=11/29:y=(5*(11/29) +11)/2 = (55/29 + 319/29)/2 = (374/29)/2=374/(29*2)=187/29≈6.448So, point T1 is (11/29,187/29)For x=-1:y=(5*(-1) +11)/2=( -5 +11)/2=6/2=3So, point T2 is (-1,3)Therefore, the two points of tangency are (11/29,187/29) and (-1,3). Therefore, the equations of the tangent lines are the lines connecting P(-1,5) to these points.First, for T2(-1,3): the line from P(-1,5) to T2(-1,3) is vertical line x=-1, which we already found.Second, for T1(11/29,187/29): let's find the equation of the line through P(-1,5) and T1(11/29,187/29).Compute the slope m:m=(187/29 -5)/(11/29 -(-1))=(187/29 -145/29)/(11/29 +29/29)=(42/29)/(40/29)=42/40=21/20.Thus, the slope is 21/20, so the equation is y -5=(21/20)(x +1), which is the same as before. So, this confirms the previous result.Therefore, the two tangent lines are x=-1 and y=(21/20)x + (21/20)+5, which simplifies to y=(21/20)x + (21/20 +100/20)= y=(21/20)x +121/20. Multiplying by 20: 21x -20y +121=0, same as before.So, the two tangent lines are x = -1 and 21x -20y +121 =0.But the problem says "the equation of the tangent line l". Maybe in the problem's context, there are two tangent lines, so we need to present both. But maybe check if both are required.Alternatively, maybe there's an error in assuming both are valid. Let me verify if both lines actually pass through P(-1,5).For x=-1: yes, x=-1 passes through (-1,5).For 21x -20y +121=0: plug in x=-1, y=5: 21*(-1) -20*5 +121= -21 -100 +121=0. Yes, it passes through P.Therefore, both lines are valid tangent lines from P to the circle.Therefore, the answer to part (2) should include both equations: x = -1 and 21x -20y +121 =0. But let me check the problem statement again.The problem says: "Draw a tangent line l from point P(-1,5) to circle C, and find the equation of the tangent line l." The wording uses "a tangent line l", but since there are two, maybe both are required. However, in Chinese problems, sometimes only one might be expected, but since mathematically there are two, we should provide both.Alternatively, maybe there is a miscalculation here, but given that both methods yield two tangent lines, and verification shows they are correct, I think both are required.But let me check if x=-1 is indeed a tangent. As computed before, substituting x=-1 into the circle equation gives y=3, so the point of tangency is (-1,3), and the line x=-1 passes through P(-1,5), so yes, it's a valid tangent line.Therefore, the two equations are:1. x = -12. 21x -20y +121 =0Alternatively, the second equation can be written as y = (21/20)x + 121/20, but the standard form is probably preferred.Therefore, the final answer for part (2) should present both equations. But let me check the problem statement once more: it says "the equation of the tangent line l". The use of "the" might be confusing, but given that there are two, maybe they expect both. If the original problem is in Chinese, sometimes even if it says "the equation", they might expect both. Alternatively, maybe I made a mistake in thinking there are two, but according to the calculation, there are two.Alternatively, perhaps the vertical line x=-1 is not considered a "tangent line" in some contexts, but in reality, it is. So, to be thorough, both should be included.Alternatively, maybe the problem expects the non-vertical tangent line, but since x=-1 is a valid tangent, we have to include both.Therefore, the equations are x = -1 and 21x -20y +121 =0.But let me write them in a boxed format as per instructions.For part (1): boxed{(x - 4)^2 + (y - 3)^2 = 25}For part (2): Since there are two tangent lines, we need to box both. Depending on the problem's expectation, but since the user hasn't specified the number of answers, but mathematically there are two, so:The equations are boxed{x = -1} and boxed{21x - 20y + 121 = 0}But maybe in some contexts, they write multiple equations in one box, but I think separate boxes are better.Alternatively, check if 21x -20y +121=0 can be simplified. Let's see, 21 and 20 are coprime, and 121 is 11². Doesn't seem to simplify further. So, the equations are correct.Final Answer(1) The equation of the circumcircle is boxed{(x - 4)^2 + (y - 3)^2 = 25}.(2) The equations of the tangent lines are boxed{x = -1} and boxed{21x - 20y + 121 = 0}.</think>Given the vertices of (triangle ABC) are (A(8,0)), (B(0,6)), and (O(0,0)):1. Finding the equation of the circumcircle (C) of (triangle ABC): - The triangle is right-angled at (O) because the vectors (OA) and (OB) are perpendicular. - The hypotenuse (AB) has a length of 10, so the circumradius is (5). - The circumcenter is the midpoint of the hypotenuse (AB), which is ((4,3)). - The equation of the circumcircle is ((x - 4)^2 + (y - 3)^2 = 25).2. Finding the equation of the tangent line (l) from point (P(-1,5)) to circle (C): - The general equation of a line passing through (P(-1,5)) is (y - 5 = m(x + 1)). - Using the condition that the distance from the center ((4,3)) to this line equals the radius (5), we solve for (m): [ frac{|5m + 2|}{sqrt{m^2 + 1}} = 5 ] Solving this gives (m = frac{21}{20}). - The equation of the tangent line with this slope is (21x - 20y + 121 = 0). - Additionally, the vertical line (x = -1) is a tangent line passing through (P(-1,5)).Final Answers:1. The equation of the circumcircle is (boxed{(x - 4)^2 + (y - 3)^2 = 25}).2. The equations of the tangent lines are (boxed{x = -1}) and (boxed{21x - 20y + 121 = 0}).,<think>Alright, so I have this problem about triangle ABC with vertices at A(8,0), B(0,6), and O(0,0). There are two parts: first, finding the equation of the circumcircle of triangle ABC, and second, drawing a tangent line from point P(-1,5) to this circle and finding its equation. Let me tackle each part step by step.Starting with part (1): Finding the equation of the circumcircle of triangle ABC. Hmm, the circumcircle is the circle that passes through all three vertices of the triangle. To find its equation, I need to determine the center (which is the circumcenter) and the radius.First, let me recall that the circumcenter of a triangle is the intersection point of the perpendicular bisectors of the sides. Since the triangle is defined by three points, I can find the perpendicular bisectors of two sides and find their intersection, which will be the circumcenter. Then, the radius is the distance from this center to any of the three vertices.Let me label the points again: A(8,0), B(0,6), and O(0,0). So, triangle ABO. Let's consider the sides OA, OB, and AB. Wait, OA is from O(0,0) to A(8,0), so that's along the x-axis. OB is from O(0,0) to B(0,6), along the y-axis. AB is from A(8,0) to B(0,6). Interesting, so OA and OB are the legs of a right triangle, and AB is the hypotenuse. Wait, is triangle ABO a right triangle?Let me check: OA has length 8, OB has length 6, and AB has length sqrt((8-0)^2 + (0-6)^2) = sqrt(64 + 36) = sqrt(100) = 10. So, OA^2 + OB^2 = 8^2 + 6^2 = 64 + 36 = 100 = AB^2. Therefore, by the Pythagorean theorem, triangle ABO is a right triangle with the right angle at O(0,0). Ah, that simplifies things! For a right triangle, the circumcenter is at the midpoint of the hypotenuse. Because in a right triangle, the hypotenuse is the diameter of the circumcircle. Therefore, the circumradius is half the length of the hypotenuse. So, since AB is the hypotenuse, which is 10 units long, the radius is 5 units. The midpoint of AB will be the circumcenter.So, let's compute the midpoint of AB. Point A is (8,0), point B is (0,6). The midpoint M has coordinates ((8+0)/2, (0+6)/2) = (4, 3). Therefore, the circumcenter is at (4,3), and the radius is 5. Therefore, the equation of the circumcircle is (x - 4)^2 + (y - 3)^2 = 5^2, which simplifies to (x - 4)^2 + (y - 3)^2 = 25.Wait, let me verify that. If the midpoint is (4,3), then the distance from (4,3) to O(0,0) should be 5. Let's check: sqrt((4-0)^2 + (3-0)^2) = sqrt(16 + 9) = sqrt(25) = 5. Correct. Similarly, distance from (4,3) to A(8,0): sqrt((8-4)^2 + (0-3)^2) = sqrt(16 + 9) = 5. Same with distance to B(0,6): sqrt((0-4)^2 + (6-3)^2) = sqrt(16 + 9) = 5. Perfect, so all three points lie on the circle centered at (4,3) with radius 5. So part (1) seems done. The equation is (x - 4)^2 + (y - 3)^2 = 25.Moving on to part (2): Drawing a tangent line l from point P(-1,5) to circle C, and finding the equation of the tangent line l. So, point P is outside the circle, and there are two tangent lines from P to the circle. The problem says "draw a tangent line l" but asks to "find the equation of the tangent line l". Wait, maybe in Chinese, it's clear whether it's one or both? But the original problem says "the tangent line l", maybe implying a specific one? Hmm. But maybe in the answer, both tangent lines are expected? Wait, let me check the original problem again.The problem says: "Draw a tangent line l from point P(-1,5) to circle C, and find the equation of the tangent line l." The translation might be a bit ambiguous, but in Chinese problems, sometimes even if there are two tangent lines, they might ask for both. However, since the problem uses "the tangent line l", maybe only one is required? Wait, perhaps the problem is expecting both equations? Hmm.Alternatively, maybe due to translation, "a tangent line" is acceptable, but in Chinese math problems, usually, both tangent lines are found. Let me check. But first, let's recall the method to find the equations of tangent lines from an external point to a circle.Given a circle with center (h,k) and radius r, and an external point P(x1,y1), the tangent lines from P to the circle can be found using the following approach:1. The equation of any line through P is y - y1 = m(x - x1), where m is the slope.2. The condition for this line to be tangent to the circle is that the distance from the center (h,k) to the line is equal to the radius r.Alternatively, we can use the system of equations: solve the line equation and the circle equation, set the discriminant of the resulting quadratic to zero for tangency.Alternatively, another method is using the fact that the tangent line is perpendicular to the radius at the point of contact. So, the vectors from the center to the point of contact and from P to the point of contact are perpendicular.But perhaps the first method is more straightforward here. Let me proceed.Given point P(-1,5), circle C: (x - 4)^2 + (y - 3)^2 = 25. So, center at (4,3), radius 5. Let me check if P is outside the circle. The distance from P to the center: sqrt((4 - (-1))^2 + (3 - 5)^2) = sqrt(5^2 + (-2)^2) = sqrt(25 + 4) = sqrt(29), which is approximately 5.385, which is greater than the radius 5. So, P is indeed outside the circle, so there are two tangent lines.To find the equations of these tangent lines, let's use the method of setting the distance from the center to the line equal to the radius.Let’s denote the equation of the tangent line as y = mx + c. But since it passes through P(-1,5), we can write it as y - 5 = m(x + 1), so y = mx + m + 5. So, the equation is y = mx + (m + 5). Let me write this as mx - y + (m + 5) = 0.The distance from the center (4,3) to this line must be equal to the radius 5. The formula for the distance from a point (x0,y0) to the line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a^2 + b^2).So, applying this to our line mx - y + (m + 5) = 0:Distance = |m*4 - 1*3 + (m + 5)| / sqrt(m^2 + (-1)^2) = |4m - 3 + m + 5| / sqrt(m^2 + 1) = |5m + 2| / sqrt(m^2 + 1)Set this equal to 5:|5m + 2| / sqrt(m^2 + 1) = 5Multiply both sides by sqrt(m^2 + 1):|5m + 2| = 5 sqrt(m^2 + 1)Square both sides to eliminate the absolute value and the square root:(5m + 2)^2 = 25(m^2 + 1)Expand left side:25m^2 + 20m + 4 = 25m^2 + 25Subtract 25m^2 from both sides:20m + 4 = 25Subtract 4:20m = 21Divide by 20:m = 21/20Wait, but squaring both sides can sometimes introduce extraneous solutions, so we need to check. However, since we know there are two tangent lines, but here we got only one solution. Hmm, that's odd. Maybe I made a mistake in the setup.Wait, let's see. The equation |5m + 2| = 5 sqrt(m^2 + 1). When we square both sides, we get (5m + 2)^2 = 25(m^2 + 1). Let's expand that again:25m² + 20m + 4 = 25m² + 25Subtract 25m²:20m + 4 = 2520m = 21m = 21/20So, only one solution? But there should be two tangent lines. Hmm. Maybe the other solution comes from the absolute value. Wait, the absolute value equation |5m + 2| = 5 sqrt(m² + 1) would have two cases:Case 1: 5m + 2 = 5 sqrt(m² + 1)Case 2: -(5m + 2) = 5 sqrt(m² + 1) => 5m + 2 = -5 sqrt(m² + 1)But sqrt(m² +1 ) is non-negative, so the right-hand side of case 2 is non-positive. But 5 sqrt(m² +1 ) is positive, so -5 sqrt(m² +1 ) is negative. However, 5m +2 in case 2 would have to equal a negative number. Let's check if that's possible.So solving both cases:Case 1: 5m + 2 = 5 sqrt(m² +1 )Case 2: 5m + 2 = -5 sqrt(m² +1 )So we need to solve both equations.But when we square both sides, we might get both solutions.Wait, let's proceed again. Let's consider both cases.Case 1: 5m + 2 = 5 sqrt(m² +1 )Case 2: 5m + 2 = -5 sqrt(m² +1 )Then, square both sides in both cases.For Case 1:(5m + 2)^2 = 25(m² +1 )Which leads to 25m² +20m +4 =25m² +25Which simplifies to 20m +4 =25 =>20m=21 =>m=21/20=1.05For Case 2:(5m +2)^2 =25(m² +1 )Same as case 1, leading to 25m² +20m +4=25m² +25Again, 20m +4=25 =>20m=21 =>m=21/20.Wait, that's the same equation. So squaring both cases leads to the same equation. Therefore, we only get one solution here. But that contradicts the fact that there are two tangent lines. Therefore, my approach must be flawed.Wait, perhaps parametrizing the line as y = mx + c passing through P(-1,5) is better. Wait, but we did use that. Wait, perhaps there's another way. Let me think.Alternatively, the tangent lines from point P(-1,5) to the circle C can be found using the geometric approach. The tangent points on the circle satisfy the condition that the vector from the center to the tangent point is perpendicular to the vector from P to the tangent point.Let me denote the center of the circle as O'(4,3). Let the tangent point be T(x,y). Then, vector O'T = (x - 4, y - 3), and vector PT = (x +1, y -5). These two vectors must be perpendicular, so their dot product is zero:(x - 4)(x +1) + (y -3)(y -5) =0Also, since T(x,y) lies on the circle C: (x -4)^2 + (y -3)^2 =25So, we have two equations:1. (x -4)(x +1) + (y -3)(y -5) =02. (x -4)^2 + (y -3)^2 =25Let me expand equation 1:(x -4)(x +1) = x^2 + x -4x -4 = x^2 -3x -4(y -3)(y -5) = y^2 -5y -3y +15 = y^2 -8y +15So, equation 1 becomes:x^2 -3x -4 + y^2 -8y +15 =0Simplify:x^2 + y^2 -3x -8y +11 =0But equation 2 is:(x -4)^2 + (y -3)^2 =25Expanding equation 2:x^2 -8x +16 + y^2 -6y +9 =25Simplify:x^2 + y^2 -8x -6y +25 =25So, x^2 + y^2 -8x -6y =0Now, subtract equation 1 from equation 2:(x^2 + y^2 -8x -6y) - (x^2 + y^2 -3x -8y +11) =0 -0Simplify:(-8x -6y) - (-3x -8y +11) =0-8x -6y +3x +8y -11 =0(-5x +2y -11) =0Thus, -5x +2y -11 =0 => 5x -2y +11 =0So, 5x -2y +11 =0 is the equation obtained by combining the two equations. So, the tangent points lie on both the circle and the line 5x -2y +11 =0. Therefore, to find the tangent points, we can solve the system:1. (x -4)^2 + (y -3)^2 =252. 5x -2y +11 =0Let me solve equation 2 for y:5x +11 =2y => y = (5x +11)/2Substitute this into equation 1:(x -4)^2 + ((5x +11)/2 -3)^2 =25Simplify the second term:((5x +11)/2 -6/2) = (5x +5)/2So:(x -4)^2 + [(5x +5)/2]^2 =25Compute each term:(x -4)^2 = x² -8x +16[(5x +5)/2]^2 = (25x² +50x +25)/4So, equation becomes:x² -8x +16 + (25x² +50x +25)/4 =25Multiply all terms by 4 to eliminate the denominator:4x² -32x +64 +25x² +50x +25 =100Combine like terms:(4x² +25x²) + (-32x +50x) + (64 +25) =10029x² +18x +89 =100Subtract 100:29x² +18x -11 =0Now, solve this quadratic equation for x:Quadratic in form: 29x² +18x -11 =0Use quadratic formula: x = [-b ± sqrt(b² -4ac)]/(2a)Here, a=29, b=18, c=-11Discriminant D=18² -4*29*(-11)=324 + 1276=1600sqrt(D)=sqrt(1600)=40Thus, x = [-18 ±40]/(2*29)First solution: x = (-18 +40)/58 =22/58=11/29Second solution: x=(-18 -40)/58= -58/58= -1Thus, x=11/29 and x=-1.Now, substitute back into y=(5x +11)/2:For x=11/29:y=(5*(11/29) +11)/2 = (55/29 + 319/29)/2 = (374/29)/2= 374/58=187/29≈6.448For x=-1:y=(5*(-1) +11)/2=(-5 +11)/2=6/2=3So, the tangent points are T1(11/29, 187/29) and T2(-1,3).Wait, let me check T2(-1,3). Is this point on the circle?Plug into (x -4)^2 + (y -3)^2: (-1 -4)^2 + (3 -3)^2 = (-5)^2 +0=25. Yes, so it's on the circle. Also, check if PT2 is tangent.The line PT2 connects P(-1,5) to T2(-1,3). That's a vertical line x=-1. Wait, but x=-1 passes through (-1,3) and (-1,5). Wait, but this line is vertical, and the circle center is at (4,3). The distance from center (4,3) to the line x=-1 is |4 - (-1)|=5, which is equal to the radius. So, x=-1 is indeed a tangent line. But wait, the point T2 is (-1,3). The line x=-1 touches the circle at (-1,3). So that's one tangent line.For T1(11/29, 187/29), let me check if the line PT1 is tangent. Let's compute the slope from P(-1,5) to T1(11/29, 187/29):Slope m=(187/29 -5)/(11/29 - (-1))=(187/29 -145/29)/(11/29 +29/29)=(42/29)/(40/29)=42/40=21/20.So, the slope is 21/20. Therefore, the equation is y -5 = (21/20)(x +1). Let me write that as y = (21/20)x + 21/20 +5 => y = (21/20)x + 21/20 +100/20= (21x +121)/20. So, the equation is 21x -20y +121=0.Wait, let me check the distance from center (4,3) to this line:Line:21x -20y +121=0Distance: |21*4 -20*3 +121| / sqrt(21² + (-20)^2) = |84 -60 +121| / sqrt(441 +400)= |145| / sqrt(841)=145/29=5. Which matches the radius. So that's correct.Therefore, the two tangent lines are x=-1 and 21x -20y +121=0. But wait, when x=-1, the line is vertical, so it's not captured by the slope-intercept form we initially tried. That's probably why the earlier method using slope gave only one solution. Because when the tangent line is vertical, the slope is undefined, so it can't be expressed as y=mx +c. Therefore, the problem with the first method was assuming a non-vertical line with a finite slope. Hence, we missed the vertical tangent line x=-1.Therefore, to find all tangent lines from P(-1,5), we need to consider both possibilities: vertical lines and non-vertical lines. Since vertical lines can't be expressed in the form y=mx +c, we need to check separately.Alternatively, another approach is to parametrize all lines through P(-1,5). However, in this case, we found both tangent lines using the geometric approach by finding the points of tangency.Therefore, the two tangent lines are:1. x = -1 (vertical line)2. 21x -20y +121=0, which can be written as y = (21/20)x + 121/20.But let me verify the second equation. Let's take the point T1(11/29, 187/29):Check if it lies on the line 21x -20y +121=0:21*(11/29) -20*(187/29) +121 = (231 -3740)/29 +121 = (-3509)/29 +121.Wait, 231 - 3740 = -3509. Then, -3509/29 +121. Since 3509 ÷29: 29*121=3509, right? Because 29*120=3480, plus 29=3509. Therefore, -3509/29= -121. Therefore, -121 +121=0. Correct. So, the equation holds.Therefore, both tangent lines are x = -1 and 21x -20y +121=0.However, the problem says "the tangent line l". Since there are two tangent lines, maybe the answer expects both? But in Chinese problems, sometimes both are required. The original problem says "draw a tangent line l from point P(-1,5) to circle C, and find the equation of the tangent line l". The wording is a bit ambiguous, but since there are two, perhaps both should be given. But maybe in the problem statement in Chinese, it's specified whether one or both. Since in the translation, it's stated as "the tangent line l", but in Chinese, depending on the measure word, it might be more explicit. Given that there are two solutions, I think the answer expects both equations.Therefore, the equations are x = -1 and 21x -20y +121=0. Let me check if these are correct.For x = -1: As calculated earlier, distance from center (4,3) is 5, equal to radius. And passes through P(-1,5) and T(-1,3). So, correct.For the other line: 21x -20y +121=0. Let's check point P(-1,5): 21*(-1) -20*5 +121 = -21 -100 +121=0. Correct, so P lies on this line.Distance from center (4,3) to the line is |21*4 -20*3 +121| / sqrt(21² +20²)= |84 -60 +121| /sqrt(841)= |145| /29=5, which is correct.Therefore, both lines are valid tangent lines.Thus, the answer to part (2) should include both equations: x = -1 and 21x -20y +121=0.But let me present them in standard form. The vertical line is x = -1. The other line can be written as 21x -20y +121=0, or simplified if possible. Let's check if 21 and 121 have common factors. 21=3*7, 121=11². No common factors, so it's already simplified.Alternatively, we can write the second equation in slope-intercept form: y = (21/20)x + 121/20. But perhaps the problem expects standard form. The original circle equation is in standard form, so maybe the tangent lines should be presented similarly. Therefore, the equations are x = -1 and 21x -20y +121 =0.But let me check once more if I didn't make any calculation mistakes.Yes, using the point T1(11/29, 187/29):Plug into the line 21x -20y +121=0:21*(11/29) -20*(187/29) +121 = (231 - 3740)/29 +121 = (-3509)/29 +121 = -121 +121=0. Correct.Similarly, T2(-1,3):Plug into x = -1, which is true. Also, check the line from P(-1,5) to T2(-1,3): vertical line x=-1.Therefore, both equations are correct.Hence, part (2) has two tangent lines: x = -1 and 21x -20y +121=0.But the problem says "the tangent line l". Maybe in the original problem, there's a figure or specific instruction, but given the translation, it's safer to provide both equations. So I think the answer requires both equations.Therefore, summarizing:1. The equation of the circumcircle is (x -4)^2 + (y -3)^2 =25.2. The equations of the tangent lines from P(-1,5) are x = -1 and 21x -20y +121=0.But let me check once again for any miscalculations. For part (1), since it's a right triangle, circumradius is half the hypotenuse, which is 5, center at midpoint of hypotenuse, which is (4,3). So correct.For part (2), the vertical line x=-1 is indeed a tangent line, and the other line is 21x -20y +121=0. The slope 21/20 seems reasonable. Let me check the second tangent line with another point. For example, let's take x=0 in 21x -20y +121=0:21*0 -20y +121=0 => y=121/20=6.05. The point (0,6.05) is on the line. The distance from center (4,3) to this point is sqrt((4)^2 + (6.05-3)^2)=sqrt(16 +9.3025)=sqrt(25.3025)=5.03, which is approximately 5, as expected due to rounding. But exact value would be 5, as computed before.Therefore, I think all steps are correct. Therefore, the final answers are as above.Final Answer(1) The equation of the circumcircle is boxed{(x - 4)^2 + (y - 3)^2 = 25}.(2) The equations of the tangent lines are boxed{x = -1} and boxed{21x - 20y + 121 = 0}.</think>Given the vertices of (triangle ABC) are (A(8,0)), (B(0,6)), and (O(0,0)):1. Finding the equation of the circumcircle of (triangle ABC): - (triangle ABC) is a right triangle with the right angle at (O(0,0)). - The hypotenuse (AB) has a length of 10, so the circumradius is half of this, which is 5. - The circumcenter is the midpoint of the hypotenuse (AB), which is ((4, 3)). - Therefore, the equation of the circumcircle is: [ boxed{(x - 4)^2 + (y - 3)^2 = 25} ]2. Finding the equations of the tangent lines from point (P(-1,5)) to the circumcircle: - The distance from (P(-1,5)) to the center ((4,3)) is (sqrt{29}), which is greater than the radius 5, confirming (P) is outside the circle. - Using the geometric approach, the points of tangency are found by solving the system of equations derived from the circle and the condition that vectors from the center to the tangent points are perpendicular to vectors from (P) to the tangent points. - The tangent points are found to be ((11/29, 187/29)) and ((-1, 3)). - The equations of the tangent lines are: - Vertical line (x = -1) - Line with slope (21/20) passing through (P(-1,5)): [ 21x - 20y + 121 = 0 ] - Therefore, the equations of the tangent lines are: [ boxed{x = -1} quad text{and} quad boxed{21x - 20y + 121 = 0} ]]

🔑:Solution: (1) Since the vertices of triangle ABC are A(8,0), B(0,6), and O(0,0), the diameter of the circumcircle C of triangle ABC is |AB| = sqrt{8^2 + 6^2} = 10, and the center of the circle is at (4,3), thus, the equation of the circumcircle C of triangle ABC is: (x-4)^2 + (y-3)^2 = 25; (2) When the slope of line l does not exist, x=-1, the distance from the center (4,3) to the line x=-1 is 4 - (-1) = 5, hence the line x=-1 is a tangent line to the circle; When the slope of line l exists, let it be k, then the equation of line l passing through point P(-1,5) is y-5=k(x+1), which simplifies to kx-y+k+5=0, according to the problem, the distance d from the center (4,3) to line l is d= frac{|4k-3+k+5|}{sqrt{1+k^2}} = frac{|5k+2|}{sqrt{1+k^2}} = 5, solving this gives: k= frac{21}{20}, thus, the equation of line l is: 21x-20y+121=0, In summary, the equation of the tangent line l from point P(-1,5) to circle C is: x=-1 or 21x-20y+121=0.Therefore, the final answers are:(1) The equation of the circumcircle C of triangle ABC is boxed{(x-4)^2 + (y-3)^2 = 25}.(2) The equation of the tangent line l from point P(-1,5) to circle C is boxed{x=-1 text{ or } 21x-20y+121=0}.

🧠:[<think>Okay, let's try to figure out this problem. So, we have this sequence {2n - 1}, which I know is just the sequence of odd numbers: 1, 3, 5, 7, 9, 11, etc. The problem says to divide this sequence into groups where the nth group contains n numbers. The example given is (1), (3, 5), (7, 9, 11), and so on. We need to find the third number in the 100th group. Hmm, okay.First, let me make sure I understand the grouping. The first group has 1 number: (1). The second group has 2 numbers: (3, 5). The third group has 3 numbers: (7, 9, 11). So each subsequent group has one more number than the previous group. That makes sense. So the 100th group will have 100 numbers, and we need the third one in that group.To find the third number in the 100th group, I need to figure out two things: where the 100th group starts, and then the third number from there. So, maybe first, I need to determine how many numbers are there before the 100th group starts. That is, the total number of numbers in the first 99 groups. Since the nth group has n numbers, the total number of numbers before the 100th group is the sum from 1 to 99. That's a triangular number. The formula for the sum of the first n natural numbers is n(n + 1)/2. So for n = 99, it would be 99*100/2 = 4950. So, there are 4950 numbers before the 100th group. Therefore, the 100th group starts at the 4951st number in the sequence.But wait, the sequence here is the sequence of odd numbers. Each term is given by 2n - 1, where n is the position in the sequence. So, the 1st term is 1, the 2nd term is 3, the 3rd term is 5, and so on. So, the kth term in the sequence is 2k - 1. Therefore, the first number of the 100th group is the 4951st term in the sequence, which would be 2*4951 - 1. Let me compute that: 2*4951 = 9902, minus 1 is 9901. So, the first number in the 100th group is 9901.But wait, let's verify this with the example. The first group starts at term 1: 2*1 -1 =1. The second group starts after the first term, which is term 2. Let's check: the first number in the second group is 3. Term 2 is 2*2 -1 = 3. Correct. The third group starts after the first two groups, which have 1 + 2 = 3 terms. So the third group starts at term 4. The first number in the third group is 2*4 -1 =7, which matches the example. So this seems right.So, applying the same logic, the 100th group starts at term 4951, which is 2*4951 -1 =9901. Then, since the 100th group has 100 numbers, the numbers in that group will be 9901, 9903, 9905, ..., each subsequent number being 2 more than the previous. So, the third number would be 9901 + 2*2 = 9905. Wait, but let's check with the example again. In the third group, the first number is 7, then 9, then 11. So 7 + 2*(3-1) = 7 +4=11. Wait, actually, the third number is the first number plus 2*(n-1) where n is the position in the group. So, the third number in the group would be 9901 + 2*2 = 9905. Hmm, but in the example, the first number in the third group is 7, then 7 + 2 =9, then 9 +2=11. So, the third number is 11, which is 7 + 2*(3-1) =7 +4=11. So, yes, that works.Therefore, the third number in the 100th group would be 9901 + 4 =9905. So, is 9905 the answer? Wait, but let me verify again. Maybe I need to compute step by step.Alternatively, perhaps there's a formula to find the starting number of the nth group. Let's think. The starting number of the nth group is equal to the last number of the previous group plus 2. So, if we can find the last number of the 99th group, then the starting number of the 100th group is that plus 2. Alternatively, since the starting number of the nth group is 2*(sum_{k=1}^{n-1} k) +1. Wait, let's see.Each group is made up of consecutive odd numbers. The number of terms before the nth group is (n-1)n/2, so the starting term is the ( (n-1)n/2 +1 )th term in the sequence of odd numbers. Since each term is 2k -1, where k is the term number, so the starting number is 2*( (n-1)n/2 +1 ) -1 = (n-1)n +2 -1 = (n-1)n +1. Wait, let me check that.Wait, the kth term in the sequence is 2k -1. So, if the starting term of the nth group is the T +1 term, where T is the total number of terms before, then that term is 2*(T +1) -1. The total terms before the nth group is sum_{k=1}^{n-1} k = (n-1)n/2. Therefore, starting number is 2*( (n-1)n/2 +1 ) -1 = (n-1)n +2 -1 = (n-1)n +1. So, starting number is (n-1)n +1. Let's test this with the examples.For n=1: starting number is (0)(1)+1=1. Correct. For n=2: (1)(2)+1=3. Correct. For n=3: (2)(3)+1=7. Correct. So, this formula works. Therefore, for the 100th group, starting number is (99)(100)+1=9900 +1=9901. So that's confirmed. Then, the third number in the 100th group is 9901 + 2*(3-1) =9901 +4=9905. So, the answer should be 9905. But let's check again with the third group.Third group starts at 7. Third number is 11. Using the formula: 7 + 4=11. Correct. So, this seems to hold. Therefore, same logic applies. So, 9901 +4=9905. So, is that the answer?Wait, but let me make sure there isn't a miscalculation here. Let's check the steps again.First, total numbers before the 100th group: sum from 1 to 99. That's (99*100)/2=4950. Therefore, the starting term is the 4951st term in the sequence. The 4951st odd number. The first odd number is 1, which is 2*1 -1. The nth odd number is 2n -1. Therefore, the 4951st odd number is 2*4951 -1=9902 -1=9901. Correct. So the first number in the 100th group is 9901. Then, each subsequent number increases by 2. So the numbers are 9901, 9903, 9905, ..., up to the 100th term. So, the third number is 9905. Therefore, the answer is 9905.But wait, let me think once more. Is there a possibility that the third number is 9905? Let's list the first few groups:Group 1: 1 (1 number)Group 2: 3,5 (2 numbers)Group3:7,9,11 (3 numbers)Group4:13,15,17,19 (4 numbers)Group5:21,23,25,27,29 (5 numbers)So, if we look at Group3, the numbers are 7,9,11. The third number is 11. Following the formula: starting number is (3-1)*3 +1=6+1=7. Then third number is 7 + 2*(3-1)=7+4=11. Correct. For Group4, starting number is (4-1)*4 +1=12+1=13. The fourth number in Group4 would be 13 +2*(4-1)=13+6=19. Which matches.Therefore, the formula seems to hold. So, applying the same to Group100: starting number is (100-1)*100 +1=99*100+1=9901. Then, the third number is 9901 + 2*(3-1)=9901 +4=9905.Alternatively, if we consider the numbers in the 100th group: starting at 9901, each subsequent number is +2. So the numbers are 9901, 9903, 9905, 9907, ..., up to 9901 + 2*99=9901 +198=10099. So the 100th number in the group is 10099, which is 2*5050 -1=10100 -1=10099. Wait, 2*5050 -1. Wait, the 5050th term in the sequence would be 2*5050 -1=10100 -1=10099, which is indeed the last term of the 100th group. So that's consistent.Therefore, the third term in the 100th group is 9901 + 4=9905. So, I think that's the answer.But just to be absolutely thorough, let's check another group. Let's check Group4. Starting number is (4-1)*4 +1=13. Then the third number in Group4 is 13 + 2*(3-1)=13 +4=17. Which is correct as per the listing above. So, yes, 17 is the third number in Group4. Similarly, the fifth number in Group5 would be starting number (5-1)*5 +1=21, then fifth number is 21 + 2*(5-1)=21 +8=29. Which is correct. So, the formula works.Therefore, the answer is 9905. So, the third number in the 100th group is 9905.Final AnswerThe third number in the 100th group is boxed{9905}.</think>To solve the problem of finding the third number in the 100th group of the sequence {2n-1} divided into groups where the n-th group contains n numbers, we follow these steps:1. Determine the total number of terms before the 100th group: - The sum of the first 99 natural numbers is given by the formula for the sum of an arithmetic series: [ sum_{k=1}^{99} k = frac{99 times 100}{2} = 4950 ] - This means there are 4950 terms before the 100th group.2. Find the starting term of the 100th group: - The 100th group starts at the 4951st term in the sequence. The k-th term in the sequence of odd numbers is given by 2k - 1. - Therefore, the starting term of the 100th group is: [ 2 times 4951 - 1 = 9901 ]3. Calculate the third number in the 100th group: - The numbers in the 100th group are consecutive odd numbers starting from 9901. Each subsequent number increases by 2. - The third number in the 100th group is: [ 9901 + 2 times (3 - 1) = 9901 + 4 = 9905 ]Thus, the third number in the 100th group is (boxed{9905}).,<think>Okay, so I need to figure out the third number in the 100th group of this sequence where each group has n numbers, and the sequence is the odd numbers {2n-1}. Let me try to break this down step by step.First, let me make sure I understand the problem. The sequence is the odd numbers: 1, 3, 5, 7, 9, 11, 13, 15, 17,... and these are divided into groups where the first group has 1 number, the second group has 2 numbers, the third group has 3 numbers, and so on. So the groups look like (1), (3, 5), (7, 9, 11), etc. The question is asking for the third number in the 100th group. That would be the 100th group, which has 100 numbers, and within that group, the third number. So, first, I need to determine where the 100th group starts. Once I find the starting number of the 100th group, I can just count two more numbers ahead (since the third number is two positions after the first) to get the answer. Alternatively, maybe there's a formula to find the starting number of each group. Let's think about how the groups are structured.Each group n contains n numbers. So the total number of numbers before the 100th group is the sum of the sizes of the first 99 groups. Since each group k has k numbers, this total is 1 + 2 + 3 + ... + 99. I remember that the sum of the first m integers is given by m(m + 1)/2. So for m = 99, the sum is 99*100/2 = 4950. That means there are 4950 numbers before the 100th group. Therefore, the first number of the 100th group is the 4951st number in the sequence of odd numbers.But wait, the sequence of odd numbers is 2n - 1. So the nth term of this sequence is 2n - 1. Therefore, the 4951st term would be 2*4951 - 1. Let me compute that. But hold on, before I do that, let me confirm whether the first group starts at n=1. The first group is (1), which is 2*1 - 1 = 1, correct. The second group is (3, 5), which are 2*2 -1 = 3 and 2*3 -1 =5, so the next two numbers. Similarly, the third group is (7,9,11), which are 2*4 -1=7, 2*5 -1=9, and 2*6 -1=11. Wait, so maybe each group starts at a certain position. Let me check.Wait, the first group has 1 number: term 1: 1. The second group has 2 numbers: terms 2 and 3: 3, 5. The third group has 3 numbers: terms 4,5,6: 7,9,11. The fourth group would then have 4 numbers: terms 7,8,9,10:13,15,17,19. So yes, the starting term of the k-th group is equal to 1 + sum_{i=1}^{k-1} i. So for the 100th group, the starting term is 1 + sum_{i=1}^{99} i = 1 + 99*100/2 = 1 + 4950 = 4951. Therefore, the starting term is the 4951st term in the odd number sequence.Therefore, the starting number of the 100th group is 2*4951 - 1. Let me compute that:2*4951 = 9902, so 9902 -1 = 9901. Therefore, the first number in the 100th group is 9901. Then, since the 100th group has 100 numbers, each subsequent number is 2 more than the previous. So the numbers in the 100th group are 9901, 9903, 9905, ..., up to the 100th number.Since we need the third number in the 100th group, that would be 9901 + 2*2 = 9901 + 4 = 9905. Wait, but let me verify that. If the first number is 9901, then the second is 9903, the third is 9905. So yes, the third number is 9905. Is this correct?Wait, let's check with a smaller group. For example, the third group: starting number is term 4. Term 4 is 2*4 -1 =7. Then the third number in the third group is 7 + 2*2 =11. Which matches (7,9,11). So that works. Similarly, the second group starts at term 2, which is 3, and the second number is 3 + 2*1=5. Correct. So the formula seems to hold.Therefore, generalizing, the starting number of the k-th group is 2*(sum_{i=1}^{k-1} i +1) -1 = 2*( (k-1)*k/2 +1 ) -1 = (k-1)*k +2 -1 = k(k-1) +1. So starting number is k(k-1)+1. Let me check with k=1: 1*0 +1=1. Correct. k=2: 2*1 +1=3. Correct. k=3:3*2 +1=7. Correct. So yes, that formula works. Therefore, for the 100th group, starting number is 100*99 +1=9900 +1=9901. Then the third number is 9901 + 2*2=9905. So the answer is 9905. Therefore, the third number in the 100th group is 9905. Wait, but let me double-check. Let's compute the starting number again. The number of terms before the 100th group is 99*100/2=4950. So the 4951st term is the first term of the 100th group. The nth term of the odd sequence is 2n -1. Therefore, 2*4951 -1=9902 -1=9901. Correct. Then, the third term in the group is 9901 + 2*(3-1)=9901 +4=9905. Correct. So that seems right.Alternatively, maybe another approach. Since each group is consecutive odd numbers, each group starts where the previous left off. The first group ends at 1, the second ends at 5, the third at 11, fourth at 19, etc. So the last number of the k-th group is the sum of the first (sum_{i=1}^k i) odd numbers. But wait, the sum of the first m odd numbers is m^2. So sum of first m odd numbers is m². Therefore, the last number in the k-th group is (sum_{i=1}^{sum_{j=1}^k j} 1) squared? Wait, maybe this is complicating.Alternatively, the last number in the k-th group would be the (sum_{i=1}^k i)-th odd number, which is 2*(sum_{i=1}^k i) -1. So for example, the first group (k=1): sum=1, last number=2*1 -1=1. The second group (k=2): sum=3, last number=2*3 -1=5. The third group (k=3): sum=6, last number=2*6 -1=11. Correct. So for k=100, the last number in the 100th group would be 2*(sum_{i=1}^{100} i) -1. Sum from 1 to 100 is 5050. Therefore, last number is 2*5050 -1=10100 -1=10099. Therefore, the 100th group ends at 10099, which is the 5050th odd number. Then, since the 100th group has 100 numbers, the starting number is 10099 - 2*(100 -1)=10099 -198=9901. Which matches our previous result. So the first number is 9901, and the third number is 9905. So this cross-verification works.Therefore, I feel confident that the third number in the 100th group is 9905.Final Answerboxed{9905}</think>To determine the third number in the 100th group of the sequence {2n-1} divided into groups where the n-th group contains n numbers, we follow these steps:1. Calculate the total number of terms before the 100th group: The sum of the first 99 natural numbers is given by: [ sum_{i=1}^{99} i = frac{99 times 100}{2} = 4950 ] This means there are 4950 numbers before the 100th group.2. Determine the starting position of the 100th group: The first number in the 100th group is the 4951st term in the sequence of odd numbers. The n-th term of the sequence {2n-1} is: [ 2 times 4951 - 1 = 9902 - 1 = 9901 ]3. Find the third number in the 100th group: The sequence of numbers in the 100th group starts at 9901 and each subsequent number increases by 2. The third number in this group is: [ 9901 + 2 times (3 - 1) = 9901 + 4 = 9905 ]Thus, the third number in the 100th group is (boxed{9905}).]

🔑:For n geq 2, the total number of odd numbers in the first n-1 groups is 1+2+ldots+(n-1) = frac{(n-1)n}{2}. The last odd number of these is 2 times frac{(n-1)n}{2} - 1 = n^2 - n - 1. Therefore, the last odd number in the 100th group is 9809, thus, the third number in the 100th group is boxed{9905}. For n geq 2, the total number of odd numbers in the first n-1 groups is 1+2+ldots+(n-1) = frac{(n-1)n}{2}. The last odd number of these is 2 times frac{(n-1)n}{2} - 1 = n^2 - n - 1. By finding out that the last odd number in the 100th group is 9809, we can conclude. This problem tests the formula for the general term and the sum of the first n terms of an arithmetic sequence, as well as reasoning and computational skills, and is considered a medium-level question.

🧠:[<think>Okay, so I need to find the values of a and b such that when the polynomial x³ - 2x² + a x + b is divided by (x - 1)(x - 2), the remainder is 2x + 1. The options are given as A, B, C, D with different pairs for a and b. Hmm, let's see how to approach this.First, I remember that when a polynomial P(x) is divided by a divisor D(x), the remainder R(x) will have a degree less than D(x). Here, the divisor is (x - 1)(x - 2), which is a quadratic polynomial, so its degree is 2. Therefore, the remainder should be a linear polynomial, which is given here as 2x + 1. That makes sense.Another thing I recall is the Remainder Theorem. But since the divisor is a quadratic, maybe I should use the Polynomial Remainder Theorem, which states that P(x) = Q(x) * D(x) + R(x), where Q(x) is the quotient and R(x) is the remainder. In this case, since the remainder is 2x + 1, we can write:x³ - 2x² + a x + b = (x - 1)(x - 2) * Q(x) + 2x + 1.But I don't know Q(x), so maybe plugging in the roots of the divisor polynomial will help. The roots of (x - 1)(x - 2) are x = 1 and x = 2. If I substitute these values into the polynomial P(x), the term with Q(x) will become zero, so P(1) should equal the remainder evaluated at x = 1, and similarly P(2) should equal the remainder evaluated at x = 2.Let me check that. So, when x = 1:P(1) = (1)³ - 2*(1)² + a*(1) + b = 1 - 2 + a + b = (1 - 2) + a + b = -1 + a + b.But the remainder when x = 1 is 2*1 + 1 = 3. So, P(1) = 3. Therefore:-1 + a + b = 3 ⇒ a + b = 4. (Equation 1)Similarly, when x = 2:P(2) = (2)³ - 2*(2)² + a*(2) + b = 8 - 8 + 2a + b = 0 + 2a + b = 2a + b.The remainder when x = 2 is 2*2 + 1 = 5. So, P(2) = 5. Therefore:2a + b = 5. (Equation 2)Now, we have a system of equations:1. a + b = 42. 2a + b = 5Subtract equation 1 from equation 2:(2a + b) - (a + b) = 5 - 4Simplify left side: 2a + b - a - b = a = 1So, a = 1. Then substitute back into equation 1:1 + b = 4 ⇒ b = 3.So, a = 1 and b = 3, which is option A.Wait, but let me verify this. Let me plug a = 1 and b = 3 back into the polynomial and check if dividing by (x - 1)(x - 2) gives remainder 2x + 1.First, write out the polynomial: x³ - 2x² + x + 3.Let me perform polynomial long division or use synthetic division. Alternatively, maybe use substitution again. Since we already checked x = 1 and x = 2, and both satisfy P(1) = 3 and P(2) = 5, which matches the remainder 2x + 1 at those points. But wait, the remainder when divided by a quadratic is a linear polynomial, which is determined uniquely by its values at two distinct points. Since the remainder R(x) = 2x + 1 agrees with P(1) and P(2), then the remainder must indeed be 2x + 1. Therefore, the answer should be A.But let me cross-check using another method. Maybe expanding the polynomial division expression.Suppose x³ - 2x² + a x + b = (x - 1)(x - 2)Q(x) + 2x + 1.Let me denote Q(x) as a polynomial. Since the left-hand side is a cubic, and (x - 1)(x - 2) is quadratic, Q(x) must be linear. Let's suppose Q(x) = c x + d.Then, expanding the right-hand side:(x - 1)(x - 2)(c x + d) + 2x + 1.First, multiply (x - 1)(x - 2) = x² - 3x + 2.Multiply this by (c x + d):(x² - 3x + 2)(c x + d) = c x³ + d x² - 3c x² - 3d x + 2c x + 2d.Combine like terms:c x³ + (d - 3c) x² + (-3d + 2c) x + 2d.Then add 2x + 1:c x³ + (d - 3c) x² + (-3d + 2c + 2) x + (2d + 1).Now, set this equal to the original polynomial x³ - 2x² + a x + b.So, equate coefficients:1. Coefficient of x³: c = 1.2. Coefficient of x²: d - 3c = -2.3. Coefficient of x: -3d + 2c + 2 = a.4. Constant term: 2d + 1 = b.From the first equation, c = 1.Substitute c = 1 into the second equation:d - 3(1) = -2 ⇒ d - 3 = -2 ⇒ d = 1.Now, substitute c = 1 and d = 1 into the third equation:-3(1) + 2(1) + 2 = -3 + 2 + 2 = (-3 + 2) + 2 = (-1) + 2 = 1. So, a = 1.Fourth equation: 2(1) + 1 = 2 + 1 = 3. So, b = 3.Thus, a = 1, b = 3, which is option A. So that matches my previous answer. Therefore, I can be confident that the answer is A.Wait, but the options given are:A: a=1, b=3B: a=-1, b=3C: a=1, b=-3D: a=-1, b=-3So, yes, A is correct. But just to make sure, let me check if there's another possible way this could go wrong.Suppose I made a mistake in setting up the equations. Let me check the first substitution again.When x = 1, P(1) = 1 - 2 + a + b = -1 + a + b. The remainder is 2(1) +1 = 3. So, -1 + a + b = 3 ⇒ a + b = 4. Correct.When x = 2, P(2) = 8 - 8 + 2a + b = 0 + 2a + b. Remainder is 5. So, 2a + b = 5. Correct.Solving:a + b = 42a + b = 5Subtract the first from the second: a = 1. Then, b = 3. Correct.Another way: if we have a + b = 4 and 2a + b = 5, subtract the first equation from the second: (2a + b) - (a + b) = 5 - 4 ⇒ a = 1. So, correct.Therefore, the answer must be A. But let me look back at the problem statement to ensure I didn't misread anything."Given that the remainder of x³ - 2x² + a x + b divided by (x -1)(x -2) is 2x +1, the values of a and b are (　　)"Yes, that's exactly what I solved. The polynomial is divided by (x -1)(x -2), remainder 2x +1. So, values a=1, b=3. So option A. So, I think that's correct. Therefore, the answer is A.Final Answerboxed{A}</think>Given that the remainder of (x^3 - 2x^2 + ax + b) divided by ((x-1)(x-2)) is (2x + 1), we need to find the values of (a) and (b).First, we use the Remainder Theorem. When the polynomial (P(x) = x^3 - 2x^2 + ax + b) is divided by ((x-1)(x-2)), the remainder is (2x + 1). Therefore, substituting the roots (x = 1) and (x = 2) into (P(x)) should give the same results as substituting them into the remainder (2x + 1).1. For (x = 1): [ P(1) = 1^3 - 2 cdot 1^2 + a cdot 1 + b = 1 - 2 + a + b = -1 + a + b ] The remainder when (x = 1) is: [ 2 cdot 1 + 1 = 3 ] Therefore: [ -1 + a + b = 3 implies a + b = 4 quad text{(Equation 1)} ]2. For (x = 2): [ P(2) = 2^3 - 2 cdot 2^2 + a cdot 2 + b = 8 - 8 + 2a + b = 2a + b ] The remainder when (x = 2) is: [ 2 cdot 2 + 1 = 5 ] Therefore: [ 2a + b = 5 quad text{(Equation 2)} ]Solving the system of equations:1. (a + b = 4)2. (2a + b = 5)Subtract Equation 1 from Equation 2:[(2a + b) - (a + b) = 5 - 4 implies a = 1]Substitute (a = 1) back into Equation 1:[1 + b = 4 implies b = 3]Thus, the values of (a) and (b) are (a = 1) and (b = 3), which corresponds to option A.[boxed{A}],<think>Okay, so I need to find the values of a and b such that when the polynomial x³ - 2x² + a x + b is divided by (x - 1)(x - 2), the remainder is 2x + 1. Hmm, let me recall how polynomial division and remainders work. First, when a polynomial P(x) is divided by a divisor D(x), the remainder R(x) has a degree less than D(x). In this case, the divisor is (x - 1)(x - 2), which is a quadratic polynomial, so its degree is 2. Therefore, the remainder should be a linear polynomial, which it is here: 2x + 1. That makes sense.Now, one way to approach this problem is by using the Remainder Factor Theorem. Since the divisor is (x - 1)(x - 2), we know that when P(x) is divided by this, the remainder can be expressed as R(x) = 2x + 1. But another way to think about it is that substituting the roots of the divisor into P(x) should give the same result as substituting them into the remainder. So, the roots of (x - 1)(x - 2) are x = 1 and x = 2. Therefore, plugging x = 1 into P(x) should give R(1), and plugging x = 2 into P(x) should give R(2). Let me write that down.For x = 1:P(1) = (1)³ - 2(1)² + a(1) + b = 1 - 2 + a + b = (1 - 2) + a + b = -1 + a + b.And the remainder R(1) = 2(1) + 1 = 3.So, equation 1: -1 + a + b = 3.For x = 2:P(2) = (2)³ - 2(2)² + a(2) + b = 8 - 8 + 2a + b = 0 + 2a + b.And the remainder R(2) = 2(2) + 1 = 5.So, equation 2: 2a + b = 5.Now, I have a system of two equations:1. -1 + a + b = 32. 2a + b = 5Let me simplify equation 1. Adding 1 to both sides gives:a + b = 4.So now, the system is:1. a + b = 42. 2a + b = 5To solve this, I can subtract equation 1 from equation 2. Subtracting equation 1 from equation 2:(2a + b) - (a + b) = 5 - 4Which simplifies to:a = 1.Now, substitute a = 1 into equation 1:1 + b = 4Subtract 1 from both sides:b = 3.So, a = 1 and b = 3. Let me check if these values satisfy both equations.Plugging into equation 1: 1 + 3 = 4, which is correct.Plugging into equation 2: 2(1) + 3 = 2 + 3 = 5, which is also correct.But wait, let me cross-verify with the original polynomial. Maybe I made a mistake in setting up the equations.Original polynomial: x³ - 2x² + a x + b. When divided by (x - 1)(x - 2), the remainder is 2x + 1.Alternatively, another way to approach this is by expressing the polynomial as:P(x) = (x - 1)(x - 2)Q(x) + 2x + 1Where Q(x) is the quotient polynomial. Then, since (x - 1)(x - 2) is a quadratic polynomial, Q(x) must be linear (since P(x) is cubic). Let’s say Q(x) = c x + d. But maybe this is complicating things. Alternatively, since we already found a = 1 and b = 3, let me plug these back into the original polynomial and see if when divided by (x - 1)(x - 2), the remainder is indeed 2x + 1.So, P(x) becomes x³ - 2x² + x + 3.Let me perform the division or check the remainder by substituting x = 1 and x = 2.For x = 1: P(1) = 1 - 2 + 1 + 3 = (1 - 2) + (1 + 3) = (-1) + 4 = 3. Which is equal to 2(1) + 1 = 3. Good.For x = 2: P(2) = 8 - 8 + 2 + 3 = (8 - 8) + (2 + 3) = 0 + 5 = 5. Which is equal to 2(2) + 1 = 5. Also good.Therefore, the values a = 1 and b = 3 satisfy both conditions. Looking at the options, that's option A.But wait, let me check the options again. A: a=1, b=3. B: a=-1, b=3. C: a=1, b=-3. D: a=-1, b=-3.So, the correct answer is A. But hold on, let me double-check my equations because sometimes when setting up equations with remainders, there's a chance of error.Wait, when we have a divisor (x - 1)(x - 2), the remainder when divided by a quadratic is a linear polynomial, which they gave as 2x + 1. Then, by the Remainder Theorem, P(1) = R(1) and P(2) = R(2). So, P(1) should equal 3, which with a=1 and b=3, it does. P(2)=5, which with a=1 and b=3, it does. So, seems correct.But wait, why is the answer not C or B? Let me check if perhaps I miscalculated P(1) and P(2). Let me recalculate P(1):Original polynomial: x³ - 2x² + a x + b.At x = 1: 1 - 2 + a + b. 1 - 2 is -1. Then, -1 + a + b. Correct.At x = 2: 8 - 8 + 2a + b. 8 - 8 is 0. Then, 0 + 2a + b. Correct.So, equations are:-1 + a + b = 3 --> a + b = 42a + b = 5Subtracting first equation from second: a = 1. Then, b = 3. So, that's correct.Therefore, answer is A. But let me check if perhaps the question was stated differently. Wait, the problem says "the remainder is 2x + 1". So, when divided by (x - 1)(x - 2), the remainder is 2x + 1. So, the setup is correct.Alternatively, perhaps the answer given in the options is different. Let me see. The options are:A: a=1, b=3B: a=-1, b=3C: a=1, b=-3D: a=-1, b=-3So, A is a=1, b=3, which is what we found. So, the answer should be A. Wait, but sometimes there might be a trick here. Let me confirm by another method.Another approach: express the polynomial as (x - 1)(x - 2)Q(x) + 2x + 1. Since the original polynomial is cubic, Q(x) must be linear. Let’s suppose Q(x) = px + q. Then, expanding:(x - 1)(x - 2)(px + q) + 2x + 1.First, expand (x - 1)(x - 2):(x² - 3x + 2)(px + q) = px³ + qx² - 3px² - 3qx + 2px + 2q.Combine like terms:px³ + (q - 3p)x² + (-3q + 2p)x + 2q.Then, add 2x + 1:px³ + (q - 3p)x² + (-3q + 2p + 2)x + (2q + 1).Now, this should be equal to the original polynomial: x³ - 2x² + a x + b.Therefore, equate coefficients:For x³: p = 1.For x²: q - 3p = -2. Since p = 1, then q - 3(1) = -2 => q = -2 + 3 => q = 1.For x term: -3q + 2p + 2 = a.Plugging q = 1 and p = 1:-3(1) + 2(1) + 2 = -3 + 2 + 2 = 1. Therefore, a = 1.For constant term: 2q + 1 = b.Plugging q = 1: 2(1) + 1 = 3. Therefore, b = 3.Thus, a = 1 and b = 3, which is option A. So, this confirms the previous result. Therefore, the answer is indeed A.But let me check once more. Maybe I made a mistake in expanding the terms. Let's redo the expansion step.Starting over:Q(x) = px + q.Multiply (x - 1)(x - 2)(px + q):First, multiply (x - 1)(x - 2) = x² - 3x + 2.Then multiply by (px + q):= x²(px + q) - 3x(px + q) + 2(px + q)= px³ + qx² - 3px² - 3qx + 2px + 2q.Combine terms:px³ + (q - 3p)x² + (-3q + 2p)x + 2q.Adding the remainder 2x + 1:px³ + (q - 3p)x² + (-3q + 2p + 2)x + (2q + 1).Set equal to x³ - 2x² + a x + b.Therefore:px³ = 1x³ ⇒ p = 1.(q - 3p)x² = -2x² ⇒ q - 3(1) = -2 ⇒ q = 1.(-3q + 2p + 2)x = a x ⇒ -3(1) + 2(1) + 2 = -3 + 2 + 2 = 1 ⇒ a = 1.(2q + 1) = b ⇒ 2(1) + 1 = 3 ⇒ b = 3.Yep, same result. So, definitely, the correct answer is A: a=1, b=3. So, why is there an option B, C, D? Maybe to test if someone miscalculates.Wait, let me check if plugging a=-1 and b=3 (option B) into P(1) and P(2):If a=-1 and b=3,P(1) = 1 - 2 + (-1) + 3 = 1 -2 -1 +3 = 1. But the remainder at x=1 should be 3. So, 1 ≠ 3. Hence, B is invalid.Similarly, for a=1 and b=-3 (option C),P(1) = 1 - 2 +1 + (-3) = (1 -2) + (1 -3) = -1 -2 = -3. Which is not 3. So invalid.For D: a=-1, b=-3,P(1) = 1 -2 -1 + (-3) = (1-2) + (-1-3) = -1 -4 = -5. Not 3. So, definitely, only A works.Therefore, the answer is A.Final Answerboxed{A}</think>Given the polynomial ( P(x) = x^3 - 2x^2 + ax + b ) and the remainder when divided by ((x-1)(x-2)) is (2x + 1), we need to find the values of (a) and (b).First, we use the Remainder Factor Theorem. The roots of the divisor ((x-1)(x-2)) are (x = 1) and (x = 2). Substituting these roots into (P(x)) should give the same results as substituting them into the remainder (2x + 1).For (x = 1):[P(1) = 1^3 - 2 cdot 1^2 + a cdot 1 + b = 1 - 2 + a + b = -1 + a + b]The remainder at (x = 1) is:[2 cdot 1 + 1 = 3]Thus, we have the equation:[-1 + a + b = 3 implies a + b = 4 quad text{(Equation 1)}]For (x = 2):[P(2) = 2^3 - 2 cdot 2^2 + a cdot 2 + b = 8 - 8 + 2a + b = 2a + b]The remainder at (x = 2) is:[2 cdot 2 + 1 = 5]Thus, we have the equation:[2a + b = 5 quad text{(Equation 2)}]We solve the system of equations:1. (a + b = 4)2. (2a + b = 5)Subtracting Equation 1 from Equation 2:[(2a + b) - (a + b) = 5 - 4 implies a = 1]Substituting (a = 1) into Equation 1:[1 + b = 4 implies b = 3]Thus, the values of (a) and (b) are (a = 1) and (b = 3), which corresponds to option A.To verify, we express the polynomial as ((x-1)(x-2)Q(x) + 2x + 1) and equate coefficients. This confirms the values of (a) and (b) are correct.[boxed{A}]]

🔑:Since the remainder of x^3 - 2x^2 + ax + b divided by (x-1)(x-2) is 2x + 1,it follows that (x^3 - 2x^2 + ax + b) - (2x + 1) has (x-1)(x-2) as a factor.When x=1, (x^3 - 2x^2 + ax + b) - (2x + 1) = (1 - 2 + a + b) - (2 + 1) = a + b - 4 = 0 ①When x=2, (x^3 - 2x^2 + ax + b) - (2x + 1) = (8 - 8 + 2a + b) - (4 + 1) = 2a + b - 5 = 0 ②Subtracting ① from ②, we get a - 1 = 0,Therefore, a = 1.Substituting a = 1 into ①, we get b = 3.Hence, the correct choice is boxed{text{A}}.