🧠:<think>First, identify the given complex numbers:- ( z_1 = -2 - i )- ( z_2 = i )Next, substitute these values into the expression ( z_1 - 2z_2 ):[(-2 - i) - 2(i)]Distribute the (-2) in the second term:[-2 - i - 2i]Combine the like terms (the imaginary parts):[-2 - 3i]The final result is (-2 - 3i), which corresponds to option D.</think>

🧠:<think>Alright, so I have this problem about a quadratic function and some related questions. Let me try to break it down step by step.First, the problem says that the graph of the quadratic function ( y = f(x) ) intersects the line ( y = -6 ) at only one point. That probably means that the quadratic touches the line ( y = -6 ) at exactly one point, which suggests that it's tangent to the line. So, this implies that the quadratic equation ( f(x) = -6 ) has exactly one solution. That means the discriminant of ( f(x) + 6 = 0 ) must be zero.Next, it says that ( f(0) = -2 ). So, when ( x = 0 ), the value of the function is -2. That gives me a specific point on the graph.Also, it mentions that the function ( f(x - 2) ) is an even function. An even function satisfies ( f(-x) = f(x) ). So, if ( f(x - 2) ) is even, that means ( f(-x - 2) = f(x - 2) ). This suggests that the graph of ( f(x) ) is symmetric about the line ( x = -2 ). So, the vertex of the quadratic must be at ( x = -2 ).Given that, I can write the quadratic function in vertex form. The general vertex form is ( f(x) = a(x - h)^2 + k ), where ( (h, k) ) is the vertex. Since the vertex is at ( x = -2 ), ( h = -2 ). So, ( f(x) = a(x + 2)^2 + k ).Now, since the graph intersects ( y = -6 ) at only one point, that means when ( f(x) = -6 ), the equation has exactly one solution. So, substituting ( f(x) = -6 ) into the vertex form:( a(x + 2)^2 + k = -6 )Which simplifies to:( a(x + 2)^2 = -6 - k )For this to have exactly one solution, the right-hand side must be zero because the left-hand side is a square term. So, ( -6 - k = 0 ) which gives ( k = -6 ).So, now the function is ( f(x) = a(x + 2)^2 - 6 ).We also know that ( f(0) = -2 ). Let's substitute ( x = 0 ) into the function:( f(0) = a(0 + 2)^2 - 6 = 4a - 6 = -2 )Solving for ( a ):( 4a - 6 = -2 )( 4a = 4 )( a = 1 )So, the quadratic function is ( f(x) = (x + 2)^2 - 6 ). Let me expand that to standard form:( f(x) = x^2 + 4x + 4 - 6 = x^2 + 4x - 2 )Okay, so that's part (1) done. The quadratic function is ( f(x) = x^2 + 4x - 2 ).Moving on to part (2): We have ( g(x) = frac{f(x)}{x} ). So, substituting ( f(x) ):( g(x) = frac{x^2 + 4x - 2}{x} = x + 4 - frac{2}{x} )The problem states that for any ( x in [1, 2] ) and ( t in [-4, 4] ), ( g(x) geq -m^2 + tm ) always holds. We need to find the range of the real number ( m ).First, let's analyze ( g(x) ) on the interval ( [1, 2] ). Since ( g(x) = x + 4 - frac{2}{x} ), let's find its minimum value on this interval because the inequality must hold for all ( x ) in [1,2].To find the minimum, we can take the derivative of ( g(x) ):( g'(x) = 1 + frac{2}{x^2} )Since ( g'(x) ) is always positive for ( x > 0 ), ( g(x) ) is increasing on [1,2]. Therefore, the minimum value occurs at ( x = 1 ):( g(1) = 1 + 4 - 2 = 3 )So, the minimum value of ( g(x) ) on [1,2] is 3. Therefore, the inequality ( g(x) geq -m^2 + tm ) simplifies to:( 3 geq -m^2 + tm )Rearranging:( m^2 - tm + 3 geq 0 )This inequality must hold for all ( t in [-4, 4] ). So, we need to ensure that for all ( t ) in that interval, the quadratic in ( m ) is non-negative.Let's consider the quadratic in ( t ):( -m^2 + tm + 3 leq 0 )Wait, actually, the inequality is ( m^2 - tm + 3 geq 0 ). So, treating this as a quadratic in ( t ):( -t m + (m^2 + 3) geq 0 )But perhaps it's better to think of it as a quadratic in ( m ):( m^2 - t m + 3 geq 0 )For this quadratic in ( m ) to be non-negative for all ( t in [-4, 4] ), the discriminant must be non-positive because if the discriminant is positive, the quadratic would take negative values between its roots. So, the discriminant ( D ) must satisfy ( D leq 0 ).The discriminant of ( m^2 - t m + 3 ) is:( D = t^2 - 12 )For the quadratic to be always non-negative, ( D leq 0 ):( t^2 - 12 leq 0 )( t^2 leq 12 )( |t| leq 2sqrt{3} approx 3.464 )But ( t ) is given to be in [-4,4]. So, for ( t ) in [-4,4], the discriminant condition ( t^2 leq 12 ) must hold. However, since ( 4^2 = 16 > 12 ), the discriminant will be positive for ( |t| > 2sqrt{3} ). That means for ( t ) such that ( |t| > 2sqrt{3} ), the quadratic ( m^2 - t m + 3 ) will have real roots, and thus, it will take negative values between those roots. Therefore, to ensure ( m^2 - t m + 3 geq 0 ) for all ( t in [-4,4] ), we need to restrict ( m ) such that even when the discriminant is positive, the quadratic doesn't dip below zero in the interval.Alternatively, perhaps another approach is to consider the inequality ( m^2 - t m + 3 geq 0 ) for all ( t in [-4,4] ). This can be viewed as a linear inequality in ( t ):( -m t + (m^2 + 3) geq 0 )For this to hold for all ( t in [-4,4] ), the expression must be non-negative at both endpoints ( t = -4 ) and ( t = 4 ).So, let's plug in ( t = -4 ):( -m(-4) + (m^2 + 3) geq 0 )( 4m + m^2 + 3 geq 0 )Similarly, plug in ( t = 4 ):( -m(4) + (m^2 + 3) geq 0 )( -4m + m^2 + 3 geq 0 )So, we have two inequalities:1. ( m^2 + 4m + 3 geq 0 )2. ( m^2 - 4m + 3 geq 0 )Let's solve each inequality.First inequality: ( m^2 + 4m + 3 geq 0 )Factorizing:( (m + 1)(m + 3) geq 0 )So, the solutions are ( m leq -3 ) or ( m geq -1 ).Second inequality: ( m^2 - 4m + 3 geq 0 )Factorizing:( (m - 1)(m - 3) geq 0 )So, the solutions are ( m leq 1 ) or ( m geq 3 ).To satisfy both inequalities, we need the intersection of the solutions:From the first inequality: ( m leq -3 ) or ( m geq -1 )From the second inequality: ( m leq 1 ) or ( m geq 3 )The intersection is:( m leq -3 ) or ( m geq 3 )Because:- For ( m leq -3 ), both inequalities are satisfied.- For ( m geq 3 ), both inequalities are satisfied.- Between ( -3 ) and ( -1 ), the first inequality is not satisfied.- Between ( -1 ) and ( 1 ), the second inequality is not satisfied.- Between ( 1 ) and ( 3 ), the second inequality is not satisfied.Therefore, the range of ( m ) is ( (-infty, -3] cup [3, +infty) ).Okay, that seems reasonable. So, part (2) is done.Now, part (3): The function ( y = g(|x| + 3) + k cdot frac{2}{|x| + 3} - 11 ) has exactly three zeros. We need to find the value of ( k ) and the zeros of the function.First, let's express ( g(|x| + 3) ). From part (1), ( g(x) = x + 4 - frac{2}{x} ). So,( g(|x| + 3) = (|x| + 3) + 4 - frac{2}{|x| + 3} = |x| + 7 - frac{2}{|x| + 3} )So, the function becomes:( y = |x| + 7 - frac{2}{|x| + 3} + frac{2k}{|x| + 3} - 11 )Simplify:Combine the constant terms: ( 7 - 11 = -4 )Combine the terms with ( frac{1}{|x| + 3} ): ( -frac{2}{|x| + 3} + frac{2k}{|x| + 3} = frac{2(k - 1)}{|x| + 3} )So, the function is:( y = |x| - 4 + frac{2(k - 1)}{|x| + 3} )Let me denote ( n = |x| + 3 ). Since ( |x| geq 0 ), ( n geq 3 ).Express ( y ) in terms of ( n ):First, ( |x| = n - 3 ). So,( y = (n - 3) - 4 + frac{2(k - 1)}{n} )Simplify:( y = n - 7 + frac{2(k - 1)}{n} )So, the equation ( y = 0 ) becomes:( n - 7 + frac{2(k - 1)}{n} = 0 )Multiply both sides by ( n ):( n^2 - 7n + 2(k - 1) = 0 )So, we have the quadratic equation:( n^2 - 7n + 2k - 2 = 0 )We need this equation to have exactly three zeros for ( x ). Since ( n = |x| + 3 geq 3 ), each positive solution for ( n ) will correspond to two solutions for ( x ) (positive and negative), except when ( n = 3 ), which corresponds to ( x = 0 ).Therefore, to have exactly three zeros, the quadratic equation in ( n ) must have one solution at ( n = 3 ) and another solution ( n > 3 ). Because if ( n = 3 ) is a solution, it gives one zero at ( x = 0 ), and the other solution ( n > 3 ) will give two zeros ( x = pm (n - 3) ). So, total three zeros.Alternatively, if the quadratic has a repeated root at ( n = 3 ), then it would give only one zero at ( x = 0 ), which is not enough. So, we need one root at ( n = 3 ) and another distinct root ( n > 3 ).So, let's substitute ( n = 3 ) into the quadratic equation:( 3^2 - 7 cdot 3 + 2k - 2 = 0 )Simplify:( 9 - 21 + 2k - 2 = 0 )( -14 + 2k = 0 )( 2k = 14 )( k = 7 )So, ( k = 7 ).Now, let's verify that with ( k = 7 ), the quadratic equation has exactly one root at ( n = 3 ) and another root greater than 3.Substitute ( k = 7 ):( n^2 - 7n + 14 - 2 = n^2 - 7n + 12 = 0 )Factorizing:( (n - 3)(n - 4) = 0 )So, the roots are ( n = 3 ) and ( n = 4 ).Therefore, ( n = 3 ) corresponds to ( x = 0 ), and ( n = 4 ) corresponds to ( |x| = 1 ), so ( x = pm 1 ).Thus, the zeros of the function are ( x = 0 ), ( x = 1 ), and ( x = -1 ).So, part (3) is done.Final Answer(1) The quadratic function is (boxed{f(x) = x^2 + 4x - 2}).(2) The range of ( m ) is (boxed{(-infty, -3] cup [3, +infty)}).(3) The value of ( k ) is (boxed{7}) and the zeros of the function are (boxed{0}), (boxed{1}), and (boxed{-1}).</think>

🧠:<think>Alright, so I have this problem about a triangle with integer side lengths and an integer area. One of the sides is 30, and the perimeter is 72. I need to find the shortest side, and the options are from 12 to 16. Hmm, okay, let me think about how to approach this.First, I know that in a triangle, the sum of any two sides must be greater than the third side. That's the triangle inequality theorem. So, if one side is 30, the other two sides must add up to 72 - 30 = 42. Let me denote the other two sides as 'b' and 'c', so b + c = 42. Also, since we're looking for the shortest side, and one side is already 30, the shortest side must be either 'b' or 'c'.Now, the area is given to be an integer. That makes me think of Heron's formula, which relates the sides of a triangle to its area. Heron's formula states that the area of a triangle with sides a, b, and c is sqrt[s(s - a)(s - b)(s - c)], where 's' is the semi-perimeter. In this case, the semi-perimeter would be 72 / 2 = 36.So, plugging in the values, the area A would be sqrt[36(36 - 30)(36 - b)(36 - c)]. Simplifying that, it becomes sqrt[36 * 6 * (36 - b) * (36 - c)]. But since b + c = 42, I can express c as 42 - b. So, substituting that in, the area becomes sqrt[36 * 6 * (36 - b) * (b - 6)].Wait, let me make sure I did that correctly. If c = 42 - b, then 36 - c = 36 - (42 - b) = b - 6. So, yeah, that seems right. So, the area is sqrt[36 * 6 * (36 - b) * (b - 6)].Simplifying further, 36 * 6 is 216, so the area is sqrt[216 * (36 - b) * (b - 6)]. Now, since the area has to be an integer, the expression inside the square root must be a perfect square. So, 216 * (36 - b) * (b - 6) must be a perfect square.Hmm, okay, so I need to find integer values of 'b' such that 216 * (36 - b) * (b - 6) is a perfect square. Also, since b and c are sides of the triangle, they must satisfy the triangle inequality. So, b must be greater than 6 (since c = 42 - b must be positive), and less than 36 (since 36 - b must be positive). But actually, since b + c = 42, and one side is 30, the other sides must be such that the sum of any two sides is greater than the third.So, let's think about the possible values of 'b'. Since we're looking for the shortest side, and the options are from 12 to 16, let's test these values and see which one makes the area an integer.Starting with option A) 12. If b = 12, then c = 42 - 12 = 30. Wait, so the sides would be 30, 12, and 30. That's an isoceles triangle. Let's check the area. Plugging into Heron's formula: sqrt[36 * (36 - 30) * (36 - 12) * (36 - 30)] = sqrt[36 * 6 * 24 * 6]. Calculating that: 36 * 6 = 216, 24 * 6 = 144, so 216 * 144 = 31104. The square root of 31104 is 176.4, which is not an integer. So, 12 doesn't work.Next, option B) 13. If b = 13, then c = 42 - 13 = 29. So, sides are 30, 13, 29. Checking the area: sqrt[36 * 6 * (36 - 13) * (13 - 6)] = sqrt[36 * 6 * 23 * 7]. Calculating that: 36 * 6 = 216, 23 * 7 = 161, so 216 * 161 = 34,776. The square root of 34,776 is approximately 186.4, which is not an integer. So, 13 doesn't work either.Option C) 14. If b = 14, then c = 42 - 14 = 28. So, sides are 30, 14, 28. Checking the area: sqrt[36 * 6 * (36 - 14) * (14 - 6)] = sqrt[36 * 6 * 22 * 8]. Calculating that: 36 * 6 = 216, 22 * 8 = 176, so 216 * 176 = 37,824. The square root of 37,824 is 194.4, which is not an integer. So, 14 doesn't work.Option D) 15. If b = 15, then c = 42 - 15 = 27. So, sides are 30, 15, 27. Checking the area: sqrt[36 * 6 * (36 - 15) * (15 - 6)] = sqrt[36 * 6 * 21 * 9]. Calculating that: 36 * 6 = 216, 21 * 9 = 189, so 216 * 189 = 40,824. The square root of 40,824 is 202, which is an integer. So, 15 works!Just to be thorough, let's check option E) 16. If b = 16, then c = 42 - 16 = 26. So, sides are 30, 16, 26. Checking the area: sqrt[36 * 6 * (36 - 16) * (16 - 6)] = sqrt[36 * 6 * 20 * 10]. Calculating that: 36 * 6 = 216, 20 * 10 = 200, so 216 * 200 = 43,200. The square root of 43,200 is approximately 207.8, which is not an integer. So, 16 doesn't work.Therefore, the shortest side is 15, which is option D.</think>

🧠:<think>Okay, so I'm trying to figure out which of the given options allows me to determine that triangle ABC is congruent to triangle DEF using the SAS (Side-Angle-Side) Postulate. I remember that the SAS Postulate states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent.Let's go through each option one by one.Option A: AB = DE, ∠A = ∠D, BC = EFAlright, so here we have AB equal to DE, angle A equal to angle D, and BC equal to EF. Let me visualize triangle ABC and triangle DEF. In triangle ABC, AB is one side, BC is another side, and angle A is the angle at vertex A. In triangle DEF, DE is one side, EF is another side, and angle D is the angle at vertex D.Now, for SAS, we need two sides and the included angle. Here, AB corresponds to DE, and BC corresponds to EF. But the angle given is angle A, which is not the included angle between AB and BC in triangle ABC. The included angle between AB and BC would be angle B, not angle A. Similarly, in triangle DEF, the included angle between DE and EF would be angle E, not angle D. So, since the angle given isn't the included angle between the two sides, this doesn't satisfy the SAS Postulate. So, option A is out.Option B: AB = EF, ∠A = ∠D, AC = DFOkay, now we have AB equal to EF, angle A equal to angle D, and AC equal to DF. Let me see. In triangle ABC, AB is one side, AC is another side, and angle A is the angle between them. In triangle DEF, EF is one side, DF is another side, and angle D is the angle between them.Wait, so AB corresponds to EF, AC corresponds to DF, and angle A corresponds to angle D. But in triangle ABC, angle A is between sides AB and AC, and in triangle DEF, angle D is between sides DE and DF. However, we don't have DE equal to anything here. We have AB equal to EF and AC equal to DF, but the sides adjacent to angle A and angle D are not matching up correctly. Specifically, angle A is between AB and AC, but in triangle DEF, angle D is between DE and DF, and DE isn't given to be equal to any side in triangle ABC. So, this doesn't seem to satisfy SAS either because the sides adjacent to the given angles aren't both equal. So, option B is also out.Option C: AB = BC, ∠B = ∠E, DE = EFHmm, this one says AB equals BC, angle B equals angle E, and DE equals EF. Let me parse this. In triangle ABC, AB equals BC, so it's an isosceles triangle with AB = BC. Angle B is the angle at vertex B, which is between sides AB and BC. In triangle DEF, DE equals EF, so it's also an isosceles triangle with DE = EF. Angle E is the angle at vertex E, between sides DE and EF.So, angle B equals angle E, and the sides adjacent to these angles are AB = BC and DE = EF. But wait, in triangle ABC, AB and BC are the sides adjacent to angle B, and in triangle DEF, DE and EF are the sides adjacent to angle E. So, if AB = BC and DE = EF, and angle B = angle E, does that satisfy SAS?Wait, but in triangle ABC, AB and BC are equal, and in triangle DEF, DE and EF are equal. So, if AB = BC and DE = EF, and angle B = angle E, then by SAS, triangle ABC should be congruent to triangle DEF. But hold on, is that correct?Wait, no. Because in triangle ABC, AB and BC are sides, but in triangle DEF, DE and EF are sides. So, if AB = BC and DE = EF, and angle B = angle E, then yes, that would satisfy SAS because we have two sides and the included angle equal in both triangles. So, maybe option C is a candidate.But let me double-check. In triangle ABC, sides AB and BC are equal, and the included angle is angle B. In triangle DEF, sides DE and EF are equal, and the included angle is angle E. So, if AB = DE and BC = EF, and angle B = angle E, then yes, SAS would apply. But wait, in option C, it's AB = BC and DE = EF. So, AB is equal to BC in triangle ABC, and DE is equal to EF in triangle DEF. But we don't know if AB = DE or BC = EF. It just says AB = BC and DE = EF. So, unless AB = DE and BC = EF, we can't directly apply SAS. So, maybe option C isn't sufficient because we don't know if the corresponding sides are equal, only that the sides adjacent to the given angles are equal within each triangle. So, perhaps option C doesn't satisfy SAS because we don't have the corresponding sides equal between the two triangles. So, maybe option C is out as well.Option D: BC = EF, ∠C = ∠F, AC = DFAlright, let's look at option D. Here, BC equals EF, angle C equals angle F, and AC equals DF. So, in triangle ABC, BC is one side, AC is another side, and angle C is the angle between them. In triangle DEF, EF is one side, DF is another side, and angle F is the angle between them.So, BC corresponds to EF, AC corresponds to DF, and angle C corresponds to angle F. Now, in triangle ABC, angle C is between sides BC and AC, and in triangle DEF, angle F is between sides EF and DF. So, if BC = EF, AC = DF, and angle C = angle F, then by SAS, triangle ABC is congruent to triangle DEF. That seems to fit perfectly because we have two sides and the included angle equal in both triangles.Wait, but let me make sure. So, in triangle ABC, sides BC and AC meet at angle C, and in triangle DEF, sides EF and DF meet at angle F. If BC = EF, AC = DF, and angle C = angle F, then yes, SAS applies. So, option D satisfies the SAS Postulate.So, after going through all the options, it seems that option D is the correct one because it provides two sides and the included angle equal in both triangles, which is exactly what the SAS Postulate requires.</think>