💡Alright, so I have this problem about a quadratic function, and I need to solve two parts. Let me start with part (1).The function is given as f(x) = x² - 2a x + 5. I need to find the range of values for the real number a such that f(x) > 0 for any x in (0, +∞). Hmm, okay. So, this means that the quadratic function should always be positive when x is positive. First, I remember that for a quadratic function ax² + bx + c, the graph is a parabola. If the coefficient of x² is positive, the parabola opens upwards, and if it's negative, it opens downwards. In this case, the coefficient of x² is 1, which is positive, so the parabola opens upwards. Since the parabola opens upwards, the function will have a minimum point. For f(x) to be always positive, this minimum must be above the x-axis. So, the minimum value of f(x) should be greater than 0. To find the minimum value of f(x), I can use the vertex formula. The x-coordinate of the vertex is at -b/(2a). Wait, in this case, the quadratic is f(x) = x² - 2a x + 5, so a is 1, b is -2a. So, the x-coordinate of the vertex is -(-2a)/(2*1) = 2a/2 = a. So, the vertex is at x = a. Then, the minimum value of f(x) is f(a). Let me compute f(a):f(a) = a² - 2a*a + 5 = a² - 2a² + 5 = -a² + 5.So, the minimum value is -a² + 5. For f(x) to be always positive, this minimum must be greater than 0:-a² + 5 > 0=> -a² > -5=> a² < 5=> |a| < sqrt(5)=> -sqrt(5) < a < sqrt(5)Wait, but the question says "for any x in (0, +∞)". So, I need to make sure that the function is positive for all x > 0. The minimum occurs at x = a, but if a is negative, then the vertex is at a negative x, which is not in our interval of interest (0, +∞). So, in that case, the function is increasing on (0, +∞) because the vertex is to the left of 0. Therefore, if a is negative, the function is increasing on (0, +∞), so the minimum on (0, +∞) would be at x = 0. Let me compute f(0):f(0) = 0² - 2a*0 + 5 = 5.So, f(0) = 5, which is positive. Therefore, if a is negative, the function is increasing on (0, +∞) and starts at 5, so it will always be positive.But if a is positive, the vertex is at x = a, which is in (0, +∞). So, in that case, the minimum is at x = a, and we need f(a) > 0, which we already found is -a² + 5 > 0, so a² < 5, so |a| < sqrt(5). But since a is positive, that would mean 0 < a < sqrt(5).But wait, when a is negative, the function is increasing on (0, +∞), so f(x) > f(0) = 5 > 0, so that's always true. So, combining both cases, when a is negative, it's always positive on (0, +∞), and when a is positive, we need a < sqrt(5). So overall, the range of a is a < sqrt(5). But wait, when a is negative, it's allowed, so the range is all real numbers a such that a < sqrt(5). But wait, when a is negative, it's okay, but when a is positive, it's only okay if a < sqrt(5). So, the range is (-∞, sqrt(5)).Wait, but earlier I thought it was -sqrt(5) < a < sqrt(5). Hmm, maybe I made a mistake there. Let me think again.If a is negative, the vertex is at x = a < 0, so on (0, +∞), the function is increasing, starting from f(0) = 5, so it's always positive. So, for a negative, it's fine.If a is positive, the vertex is at x = a > 0, so the minimum is at x = a, and we need f(a) > 0, which is -a² + 5 > 0, so a² < 5, so a < sqrt(5). So, for a positive, a must be less than sqrt(5). So, combining both cases, a can be any real number less than sqrt(5). So, the range is a ∈ (-∞, sqrt(5)).Wait, but earlier I thought it was -sqrt(5) < a < sqrt(5). Maybe I confused something. Let me check.If a is greater than sqrt(5), then f(a) = -a² + 5 would be negative, so the function would dip below zero at x = a, which is in (0, +∞), so f(x) would not be always positive. So, a must be less than sqrt(5). But if a is negative, it's fine because the vertex is at x = a < 0, so on (0, +∞), the function is increasing from f(0) = 5, so it's always positive. Therefore, the correct range is a < sqrt(5). So, a ∈ (-∞, sqrt(5)).Wait, but in the initial calculation, I thought that a² < 5, so |a| < sqrt(5), but that's when the vertex is in (0, +∞). But if a is negative, the vertex is not in (0, +∞), so the condition is different. So, actually, the range is a < sqrt(5). Wait, let me test with a specific value. Let's say a = -1. Then f(x) = x² + 2x + 5. The vertex is at x = -1, which is not in (0, +∞). So, on (0, +∞), f(x) is increasing, starting from f(0) = 5, so it's always positive. So, a = -1 is acceptable.If a = 3, which is greater than sqrt(5) ≈ 2.236, then f(x) = x² - 6x + 5. The vertex is at x = 3, and f(3) = 9 - 18 + 5 = -4, which is negative. So, f(x) is negative at x = 3, which is in (0, +∞), so a = 3 is not acceptable.If a = 2, which is less than sqrt(5), then f(x) = x² - 4x + 5. The vertex is at x = 2, f(2) = 4 - 8 + 5 = 1 > 0, so it's acceptable.If a = sqrt(5), then f(a) = - (sqrt(5))² + 5 = -5 + 5 = 0. So, f(a) = 0, which is not greater than 0. So, a must be less than sqrt(5).Therefore, the range of a is all real numbers less than sqrt(5), so a ∈ (-∞, sqrt(5)).Okay, so that's part (1). Now, moving on to part (2).Part (2) says: If a > 1, and both the domain and range of the function f(x) are [1, a], find the value of the real number a.Hmm, so the function f(x) has domain [1, a] and range [1, a]. Also, a > 1.First, let me recall that f(x) is a quadratic function, which is a parabola. Since the coefficient of x² is positive, it opens upwards, so it has a minimum point.Given that the domain is [1, a], which is an interval, and the range is also [1, a], which is the same interval. So, the function maps [1, a] onto itself.Since the parabola opens upwards, the minimum value of f(x) on [1, a] will be at the vertex, and the maximum will be at one of the endpoints, either x = 1 or x = a.But the range is [1, a], so the minimum value of f(x) on [1, a] is 1, and the maximum is a.So, we need to have f(x) attains its minimum value 1 at some point in [1, a], and its maximum value a at either x = 1 or x = a.Also, since the function is continuous and the domain is [1, a], the range being [1, a] implies that f maps [1, a] onto [1, a] bijectively? Or at least surjectively.But since it's a quadratic, it's not injective over an interval that includes the vertex, unless the interval is on one side of the vertex.Wait, the vertex is at x = a, as we found earlier. Wait, no, wait. The vertex is at x = a, because f(x) = x² - 2a x + 5, so the vertex is at x = a.Wait, that's interesting. So, the vertex is at x = a, which is the right endpoint of the domain [1, a]. So, the function is decreasing on (-∞, a) and increasing on (a, +∞). But since our domain is [1, a], which is to the left of the vertex, the function is decreasing on [1, a].Wait, no, because the vertex is at x = a, so on the interval [1, a], the function is decreasing from x = 1 to x = a.Therefore, the maximum value of f(x) on [1, a] is at x = 1, and the minimum is at x = a.But the range is [1, a], so the minimum value is 1 and the maximum is a.So, f(a) = 1, and f(1) = a.Let me write that down:f(a) = a² - 2a*a + 5 = a² - 2a² + 5 = -a² + 5 = 1So, -a² + 5 = 1 => -a² = -4 => a² = 4 => a = ±2But since a > 1, we take a = 2.Also, f(1) = 1² - 2a*1 + 5 = 1 - 2a + 5 = 6 - 2aBut f(1) should be equal to a, so:6 - 2a = a => 6 = 3a => a = 2So, both conditions give a = 2.Let me verify:If a = 2, then f(x) = x² - 4x + 5.The domain is [1, 2]. Let's compute f(1) and f(2):f(1) = 1 - 4 + 5 = 2f(2) = 4 - 8 + 5 = 1So, f(1) = 2 and f(2) = 1. Since the function is decreasing on [1, 2], the range is [1, 2], which matches the given range [1, a] with a = 2.Therefore, a = 2 is the solution.So, summarizing:1. For part (1), the range of a is (-∞, sqrt(5)).2. For part (2), the value of a is 2.**Final Answer**(1) The range of values of ( a ) is (boxed{(-infty, sqrt{5})}).(2) The value of ( a ) is (boxed{2}).

💡Okay, so I have this problem here where I need to calculate the surface area of the interior of a box made from a rectangular sheet of cardboard. The sheet is 25 units by 35 units, and they're cutting out square corners that are each 7 units on a side. Then, they fold up the remaining material to form an open-top box. I need to find the surface area of the inside of this box.Alright, let me break this down step by step. First, I should visualize the original cardboard sheet. It's a rectangle, 25 units long and 35 units wide. If I imagine cutting out squares from each corner, each of those squares is 7 units by 7 units. So, from each corner, a small square is removed, and then the sides are folded up to form the walls of the box.Now, the question is about the surface area of the interior of the box. Hmm, so that would be the area of the inside surfaces of the box, right? Since it's an open-top box, there's no top surface, only the base and the four sides.Let me think about how the dimensions of the box are determined after cutting out the squares. When you remove a 7x7 square from each corner, the length and width of the base of the box are reduced by twice the size of the square because you're removing from both ends. So, the original length is 35 units, and after cutting out 7 units from both ends, the new length becomes 35 - 2*7 = 35 - 14 = 21 units. Similarly, the original width is 25 units, so the new width is 25 - 2*7 = 25 - 14 = 11 units.So, the base of the box is 21 units by 11 units. The height of the box is equal to the size of the square that was cut out, which is 7 units. That makes sense because when you fold up the sides, the height is determined by how much you've cut out from the corners.Now, to find the surface area of the interior, I need to calculate the areas of all the inside surfaces and add them up. The box has a base and four sides. The base is straightforward: it's just the area of the rectangle, which is length times width, so 21 * 11.For the sides, there are two pairs of identical sides. The longer sides will have a length equal to the length of the base (21 units) and a height of 7 units. The shorter sides will have a length equal to the width of the base (11 units) and the same height of 7 units. So, the area of each longer side is 21 * 7, and there are two of them, so that's 2 * (21 * 7). Similarly, the area of each shorter side is 11 * 7, and there are two of them, so that's 2 * (11 * 7).Let me write this out:- Base area: 21 * 11- Longer sides area: 2 * (21 * 7)- Shorter sides area: 2 * (11 * 7)Adding these up will give the total interior surface area.Wait a second, but the problem says "the surface area of the interior of this resulting box." Does that include the base? Because if it's the interior, I think it does include the base. So, yes, I should include all the surfaces that are inside the box, which are the base and the four sides.Let me calculate each part step by step.First, the base area:21 * 11 = 231 square units.Next, the longer sides:2 * (21 * 7) = 2 * 147 = 294 square units.Then, the shorter sides:2 * (11 * 7) = 2 * 77 = 154 square units.Now, adding them all together:231 (base) + 294 (longer sides) + 154 (shorter sides) = ?Let me add 231 and 294 first:231 + 294 = 525.Then, add 154 to that:525 + 154 = 679.So, the total interior surface area is 679 square units.Wait, but let me double-check my calculations to make sure I didn't make any mistakes.Base area: 21 * 11. 21 times 10 is 210, plus 21 is 231. That seems right.Longer sides: 21 * 7 is 147, times 2 is 294. Correct.Shorter sides: 11 * 7 is 77, times 2 is 154. Correct.Adding them up: 231 + 294 is 525, plus 154 is indeed 679.Hmm, looking at the answer choices, option C is 679 square units. So, that must be the correct answer.But just to be thorough, let me think about another approach to confirm. Maybe calculating the remaining area after cutting out the squares and see if it matches.The original area of the cardboard sheet is 25 * 35 = 875 square units.Each corner square is 7 * 7 = 49 square units, and there are four corners, so total area removed is 4 * 49 = 196 square units.Therefore, the remaining area is 875 - 196 = 679 square units.Oh, that's the same result as before! So, that confirms it. The surface area of the interior of the box is 679 square units.I think that's solid. Both methods lead to the same answer, so I'm confident that 679 is correct.

💡Okay, so I have this problem about an arithmetic sequence. Let me try to understand it step by step. First, the problem says that {a_n} is an arithmetic sequence with a common difference of -2. That means each term is 2 less than the previous term. So, if I denote the first term as a_1, then the second term a_2 is a_1 - 2, the third term a_3 is a_1 - 4, and so on. In general, the nth term can be written as a_n = a_1 + (n-1)(-2). Simplifying that, a_n = a_1 - 2(n - 1).Next, the problem states that a_7 is the geometric mean of a_3 and a_9. Hmm, geometric mean. I remember that the geometric mean of two numbers x and y is the square root of their product, so sqrt(x * y). Therefore, according to the problem, a_7 = sqrt(a_3 * a_9). But wait, since a_7 is a term in the arithmetic sequence, it's a real number, and the geometric mean is also a real number. So, squaring both sides should give me an equation without the square root. Let me write that down:a_7^2 = a_3 * a_9.Alright, so I can express a_3, a_7, and a_9 in terms of a_1. Let's do that.First, a_3 is the third term. Using the formula for the nth term:a_3 = a_1 + (3 - 1)(-2) = a_1 - 4.Similarly, a_7 is the seventh term:a_7 = a_1 + (7 - 1)(-2) = a_1 - 12.And a_9 is the ninth term:a_9 = a_1 + (9 - 1)(-2) = a_1 - 16.So, substituting these into the equation a_7^2 = a_3 * a_9:(a_1 - 12)^2 = (a_1 - 4)(a_1 - 16).Let me expand both sides.Left side: (a_1 - 12)^2 = a_1^2 - 24a_1 + 144.Right side: (a_1 - 4)(a_1 - 16) = a_1^2 - 16a_1 - 4a_1 + 64 = a_1^2 - 20a_1 + 64.So, setting left side equal to right side:a_1^2 - 24a_1 + 144 = a_1^2 - 20a_1 + 64.Hmm, let's subtract a_1^2 from both sides to simplify:-24a_1 + 144 = -20a_1 + 64.Now, let's bring all terms to one side. I'll add 24a_1 to both sides:144 = 4a_1 + 64.Then, subtract 64 from both sides:80 = 4a_1.Divide both sides by 4:a_1 = 20.Okay, so the first term is 20. Now, the problem asks for S_10, which is the sum of the first 10 terms of the arithmetic sequence.I remember the formula for the sum of the first n terms of an arithmetic sequence is S_n = n/2 * (2a_1 + (n - 1)d), where d is the common difference.Alternatively, it can also be written as S_n = n * (a_1 + a_n)/2.Either formula should work. Let me use the first one:S_10 = 10/2 * [2*20 + (10 - 1)*(-2)].Simplify step by step.First, 10/2 is 5.Then, inside the brackets: 2*20 is 40, and (10 - 1)*(-2) is 9*(-2) = -18.So, inside the brackets: 40 + (-18) = 22.Therefore, S_10 = 5 * 22 = 110.Wait, but let me double-check using the other formula to make sure I didn't make a mistake.Using S_n = n * (a_1 + a_n)/2.First, I need a_10. Using the nth term formula:a_10 = a_1 + (10 - 1)*(-2) = 20 + 9*(-2) = 20 - 18 = 2.So, a_10 is 2.Then, S_10 = 10*(20 + 2)/2 = 10*(22)/2 = 10*11 = 110.Same result. So, that seems consistent.But wait, the options are A: -110, B: -90, C:90, D:110.So, 110 is option D. Hmm, but let me think again. Did I make a mistake in the calculations?Wait, the common difference is -2, so each term is decreasing by 2. Starting from 20, the terms go 20, 18, 16, 14, 12, 10, 8, 6, 4, 2 for the first 10 terms. Let me add them up:20 + 18 = 3838 + 16 = 5454 + 14 = 6868 + 12 = 8080 + 10 = 9090 + 8 = 9898 + 6 = 104104 + 4 = 108108 + 2 = 110.Yes, adding them one by one gives 110. So, that's correct.But wait, in the initial problem statement, the options include negative numbers as well. Did I perhaps misinterpret something?Let me go back to the problem. It says that {a_n} is an arithmetic sequence with a common difference of -2, and a_7 is the geometric mean of a_3 and a_9. Then, S_n is the sum of the first n terms, and we need to find S_10.Wait, when I calculated a_1, I got 20, which is positive, and the terms are decreasing by 2 each time, so all terms are positive. Therefore, the sum should be positive, which is 110, which is option D.But wait, let me check my calculation for a_1 again because the options have negative numbers, so maybe I made a mistake there.We had the equation:(a_1 - 12)^2 = (a_1 - 4)(a_1 - 16)Expanding both sides:Left: a_1^2 -24a_1 +144Right: a_1^2 -20a_1 +64Subtracting right from left:(a_1^2 -24a_1 +144) - (a_1^2 -20a_1 +64) = 0Simplify:-24a_1 +144 - (-20a_1) -64 = 0Wait, that would be:-24a_1 +144 +20a_1 -64 = 0Combine like terms:(-24a_1 +20a_1) + (144 -64) = 0-4a_1 +80 =0-4a_1 = -80a_1 = 20.So, that seems correct. So, a_1 is 20.Wait, but let me think again. If a_7 is the geometric mean of a_3 and a_9, and the common difference is negative, does that affect anything?Wait, in the problem, it's given that a_7 is the geometric mean. So, a_7^2 = a_3 * a_9.But in my calculation, a_7 is 8, a_3 is 16, a_9 is 4.Wait, 8^2 is 64, and 16*4 is also 64. So, that's correct.But a_7 is 8, which is less than a_3 which is 16, which makes sense because the sequence is decreasing.So, all the terms are positive, so the sum is positive.Therefore, S_10 is 110, which is option D.Wait, but in the initial problem, the options are A: -110, B: -90, C:90, D:110.So, 110 is an option, so that's correct.Wait, but in the initial problem, the assistant had a different thought process, but arrived at the same answer. So, perhaps I'm overcomplicating.Alternatively, maybe the problem is designed to have negative sums, but in this case, the sum is positive.Alternatively, perhaps I made a mistake in the formula for the sum.Wait, let me check the formula again.The sum of the first n terms is S_n = n/2 [2a_1 + (n-1)d].So, for n=10, d=-2, a_1=20.So, S_10 = 10/2 [2*20 + 9*(-2)] = 5 [40 - 18] = 5*22=110.Yes, that's correct.Alternatively, using the other formula, S_n = n*(a_1 + a_n)/2.We found a_10=2, so S_10=10*(20+2)/2=10*22/2=10*11=110.Same result.So, I think the answer is D:110.But wait, let me check if a_7 is indeed the geometric mean.a_3=16, a_7=8, a_9=4.So, geometric mean of a_3 and a_9 is sqrt(16*4)=sqrt(64)=8, which is a_7. So, that checks out.Therefore, all steps are correct, and the sum S_10 is 110.So, the answer is D:110.

💡Okay, so I have this problem about a right-angled triangle where the point where the inscribed circle touches the hypotenuse divides it into two segments of lengths 5 and 12. I need to find the lengths of the legs of the triangle. Hmm, let me think about how to approach this.First, I remember that in a right-angled triangle, the inradius (radius of the inscribed circle) can be calculated using the formula ( r = frac{a + b - c}{2} ), where ( a ) and ( b ) are the legs, and ( c ) is the hypotenuse. But I'm not sure if that's directly applicable here. Maybe I need to use some other properties.I also recall that the point where the incircle touches a side divides that side into segments related to the triangle's inradius and the triangle's sides. Specifically, in any triangle, the lengths from the vertices to the points of tangency can be expressed in terms of the semiperimeter.Let me denote the triangle as ( ABC ) with the right angle at ( C ). The hypotenuse is ( AB ), and the point where the incircle touches ( AB ) divides it into segments ( AD = 5 ) and ( DB = 12 ). So, the hypotenuse ( AB = AD + DB = 5 + 12 = 17 ).Now, in a right-angled triangle, the inradius ( r ) can also be expressed as ( r = frac{a + b - c}{2} ), where ( a ) and ( b ) are the legs, and ( c ) is the hypotenuse. So, if I can find ( r ), I can relate it to the legs.But how do I find ( r )? I know that the segments ( AD ) and ( DB ) are related to the inradius and the other sides. In any triangle, the lengths from the vertices to the points of tangency are equal to ( s - ) the opposite side, where ( s ) is the semiperimeter.So, in triangle ( ABC ), the semiperimeter ( s = frac{a + b + c}{2} ). The lengths ( AD ) and ( DB ) should be equal to ( s - b ) and ( s - a ) respectively. Let me write that down:( AD = s - b = 5 )( DB = s - a = 12 )So, I have two equations:1. ( s - b = 5 )2. ( s - a = 12 )And I also know that ( c = 17 ).Since ( s = frac{a + b + c}{2} ), I can substitute ( c = 17 ) into that:( s = frac{a + b + 17}{2} )Now, let's substitute ( s ) into the first two equations.From equation 1:( frac{a + b + 17}{2} - b = 5 )Simplify:( frac{a + b + 17 - 2b}{2} = 5 )( frac{a - b + 17}{2} = 5 )Multiply both sides by 2:( a - b + 17 = 10 )So,( a - b = 10 - 17 )( a - b = -7 ) [Equation 3]From equation 2:( frac{a + b + 17}{2} - a = 12 )Simplify:( frac{a + b + 17 - 2a}{2} = 12 )( frac{-a + b + 17}{2} = 12 )Multiply both sides by 2:( -a + b + 17 = 24 )So,( -a + b = 24 - 17 )( -a + b = 7 ) [Equation 4]Now, I have two equations:3. ( a - b = -7 )4. ( -a + b = 7 )Hmm, these look similar. Let me add them together:( (a - b) + (-a + b) = -7 + 7 )Simplify:( 0 = 0 )Wait, that doesn't help. Maybe I need to approach this differently.Alternatively, let's solve equation 3 and equation 4 together.From equation 3: ( a = b - 7 )Substitute into equation 4:( -(b - 7) + b = 7 )Simplify:( -b + 7 + b = 7 )( 7 = 7 )Hmm, that's just an identity, which means the equations are dependent, and I need another equation to solve for ( a ) and ( b ). That other equation is the Pythagorean theorem, since it's a right-angled triangle.So, ( a^2 + b^2 = c^2 )We know ( c = 17 ), so:( a^2 + b^2 = 17^2 )( a^2 + b^2 = 289 )But from equation 3, ( a = b - 7 ). So, substitute ( a = b - 7 ) into the equation:( (b - 7)^2 + b^2 = 289 )Expand ( (b - 7)^2 ):( b^2 - 14b + 49 + b^2 = 289 )Combine like terms:( 2b^2 - 14b + 49 = 289 )Subtract 289 from both sides:( 2b^2 - 14b + 49 - 289 = 0 )( 2b^2 - 14b - 240 = 0 )Divide the entire equation by 2 to simplify:( b^2 - 7b - 120 = 0 )Now, solve this quadratic equation for ( b ).Using the quadratic formula:( b = frac{7 pm sqrt{(-7)^2 - 4(1)(-120)}}{2(1)} )Calculate discriminant:( D = 49 + 480 = 529 )So,( b = frac{7 pm sqrt{529}}{2} )( sqrt{529} = 23 )Thus,( b = frac{7 + 23}{2} = frac{30}{2} = 15 )or( b = frac{7 - 23}{2} = frac{-16}{2} = -8 )Since length can't be negative, ( b = 15 ).Then, from equation 3, ( a = b - 7 = 15 - 7 = 8 ).So, the legs are 8 and 15.Let me verify this. If the legs are 8 and 15, hypotenuse should be 17, which it is because ( 8^2 + 15^2 = 64 + 225 = 289 = 17^2 ). Good.Also, the inradius ( r = frac{a + b - c}{2} = frac{8 + 15 - 17}{2} = frac{6}{2} = 3 ).Then, the segments on the hypotenuse should be ( s - a ) and ( s - b ). Let's compute ( s ):( s = frac{a + b + c}{2} = frac{8 + 15 + 17}{2} = frac{40}{2} = 20 )So,( s - a = 20 - 8 = 12 )( s - b = 20 - 15 = 5 )Which matches the given segments of 5 and 12. Perfect, everything checks out.**Final Answer**The legs of the triangle are boxed{8} and boxed{15}.

💡Okay, so I have this problem here about a hyperbola S that shares one of its foci with the parabola y² = 12x. Also, one of the asymptotes of the hyperbola is given as y = -√2 x. I need to find the equation of the hyperbola S. Hmm, let me think about how to approach this step by step.First, I remember that for a parabola in the form y² = 4px, the focus is at (p, 0). Comparing this with the given equation y² = 12x, I can see that 4p = 12, so p = 3. That means the focus of the parabola is at (3, 0). Since the hyperbola shares one of its foci with this parabola, the hyperbola must also have a focus at (3, 0).Now, I need to recall the standard form of a hyperbola. Since the focus is on the x-axis at (3, 0), the hyperbola must open left and right, which means its standard form is (x²/a²) - (y²/b²) = 1. Here, 'a' is the distance from the center to the vertices, and 'b' is related to the conjugate axis. The foci of such a hyperbola are located at (±c, 0), where c² = a² + b².Given that one of the foci is at (3, 0), the center of the hyperbola must be at the origin (0, 0) because the focus is on the x-axis and the hyperbola is symmetric about both axes. So, the distance from the center to the focus is c = 3. Therefore, c² = 9 = a² + b². That gives me my first equation: a² + b² = 9.Next, the problem states that y = -√2 x is one of the asymptotes of the hyperbola. I remember that for a hyperbola of the form (x²/a²) - (y²/b²) = 1, the equations of the asymptotes are y = ±(b/a)x. Since one of the asymptotes is y = -√2 x, the slope here is -√2. The slopes of the asymptotes are ±(b/a), so this tells me that b/a = √2. Therefore, b = a√2.Now I have two equations:1. a² + b² = 92. b = a√2I can substitute the second equation into the first to solve for 'a'. Let's do that:From equation 2, b = a√2, so b² = (a√2)² = 2a².Substituting into equation 1:a² + 2a² = 93a² = 9a² = 3So, a² is 3. Then, from equation 2, b² = 2a² = 2*3 = 6.Therefore, the equation of the hyperbola is (x²/3) - (y²/6) = 1.Let me just double-check my steps to make sure I didn't make any mistakes. I found the focus of the parabola correctly, which is (3, 0). Then, I assumed the hyperbola is centered at the origin because the focus is on the x-axis, which makes sense. I used the standard form of the hyperbola and correctly identified the relationship between a, b, and c. The asymptote slope gave me the ratio of b/a, which I used to find the values of a² and b². Everything seems to add up.So, I think I'm confident that the equation of the hyperbola S is (x²/3) - (y²/6) = 1.**Final Answer**The equation of the hyperbola ( S ) is boxed{dfrac{x^2}{3} - dfrac{y^2}{6} = 1}.

💡Okay, so I need to find a function ( f(x) ) that meets three specific conditions. Let me try to understand each condition one by one and see how I can approach this.First, the domain of ( f(x) ) has to be all real numbers, ( mathbb{R} ). That means ( f(x) ) should be defined for every real number ( x ). So, I can't have any restrictions like ( x neq 0 ) or anything like that because that would limit the domain.Second, for any real number ( b ), the equation ( f(x) = b ) should have exactly one solution. This means that ( f(x) ) must be a one-to-one function, right? Because if it's one-to-one, every ( b ) will correspond to exactly one ( x ). So, ( f(x) ) needs to be injective.Third, for any ( a > 0 ) and any ( b in mathbb{R} ), the equation ( f(x) = ax + b ) should have at least two solutions. That means that when I set ( f(x) ) equal to a linear function with a positive slope, there should be at least two points where they intersect. So, ( f(x) ) must intersect any line with a positive slope at least twice.Hmm, this is interesting. So, I need a function that's defined everywhere, injective, and yet intersects any line with a positive slope at least twice. That seems a bit contradictory because if it's injective, it can't be periodic or anything like that, which usually causes multiple intersections.Let me think about some functions I know. Linear functions are injective, but they only intersect another linear function at one point, so that won't satisfy the third condition. Quadratic functions are not injective over the entire real line because they fail the horizontal line test. Cubic functions are injective if they are strictly increasing or decreasing, but again, they might not intersect a line with a positive slope twice.Wait, maybe a function that's injective but has some kind of "wobble" or curvature that allows it to intersect a line twice? But if it's injective, it can't have any local maxima or minima because that would mean it's not strictly increasing or decreasing, which would violate injectivity.This is confusing. Maybe I need to think of a function that's injective but still manages to cross a line with a positive slope twice. How is that possible?Let me consider the function ( f(x) = x + sin(x) ). Is this injective? Let's see. The derivative is ( f'(x) = 1 + cos(x) ). Since ( cos(x) ) ranges between -1 and 1, ( f'(x) ) ranges between 0 and 2. So, the function is not strictly increasing everywhere because the derivative can be zero. That means it's not injective over the entire real line. So, that won't work.What about ( f(x) = e^x )? It's injective because it's strictly increasing, and its domain is all real numbers. But does ( e^x = ax + b ) have at least two solutions for any ( a > 0 ) and ( b )? Let's see. For ( a > 0 ), the equation ( e^x = ax + b ) is a transcendental equation. Depending on the values of ( a ) and ( b ), it might have one or two solutions. But I'm not sure if it always has at least two solutions for any ( a > 0 ) and ( b ). Maybe not. For example, if ( a ) is very large, the line ( ax + b ) might only intersect ( e^x ) once.Hmm, maybe I need a different approach. What if I define a piecewise function? For example, a function that behaves differently on different intervals but still maintains injectivity.Let me think about a function that is strictly increasing but has different rates of increase in different regions. Maybe something like ( f(x) = x ) for ( x geq 0 ) and ( f(x) = x - 1 ) for ( x < 0 ). Wait, but that's not injective because it's just a shifted line on the negative side. It's still injective, but does it satisfy the third condition? Let's see. If I set ( f(x) = ax + b ), for ( x geq 0 ), it's ( x = ax + b ), which gives ( x = frac{b}{1 - a} ). For ( x < 0 ), it's ( x - 1 = ax + b ), which gives ( x = frac{b + 1}{1 - a} ). So, depending on the values of ( a ) and ( b ), we might get solutions in both regions or only one. But I'm not sure if it always gives at least two solutions. Maybe not.This is tricky. Maybe I need a function that is injective but has asymptotes or something that allows it to intersect a line twice. Wait, the function ( f(x) = frac{1}{x} ) is injective on its domain, which is ( mathbb{R} setminus {0} ), but it's not defined at ( x = 0 ), so it doesn't satisfy the first condition.But what if I define ( f(0) = 0 )? So, ( f(x) = frac{1}{x} ) for ( x neq 0 ) and ( f(0) = 0 ). Is this function injective? Let's see. For ( x neq 0 ), ( f(x) = frac{1}{x} ) is injective because it's strictly decreasing on both sides of zero. And at ( x = 0 ), it's defined as 0, which is unique. So, overall, it's injective.Does it satisfy the third condition? Let's see. If I set ( f(x) = ax + b ), then for ( x neq 0 ), it's ( frac{1}{x} = ax + b ), which can be rearranged to ( ax^2 + bx - 1 = 0 ). This is a quadratic equation, and since ( a > 0 ), the discriminant is ( b^2 + 4a ), which is always positive. So, there are two real solutions for ( x neq 0 ). Additionally, at ( x = 0 ), ( f(0) = 0 ), so if ( b = 0 ), ( x = 0 ) is also a solution. But wait, if ( b = 0 ), the equation ( f(x) = ax ) would have ( x = 0 ) and the two solutions from the quadratic, so actually three solutions? But the condition says at least two solutions, so that's okay.Wait, but if ( b neq 0 ), then ( x = 0 ) is not a solution because ( f(0) = 0 neq b ). So, in that case, we only have the two solutions from the quadratic. So, it seems to satisfy the third condition.But does it satisfy the first condition? The domain is all real numbers because we defined ( f(0) = 0 ). So, yes, the domain is ( mathbb{R} ).And for the second condition, for any ( b in mathbb{R} ), ( f(x) = b ) has exactly one solution. If ( b neq 0 ), then ( x = frac{1}{b} ). If ( b = 0 ), then ( x = 0 ). So, exactly one solution in each case.Therefore, this function seems to satisfy all three conditions.But wait, is ( f(x) = frac{1}{x} ) for ( x neq 0 ) and ( f(0) = 0 ) continuous? No, it's not continuous at ( x = 0 ). But the problem doesn't specify continuity, just the domain and the solution conditions. So, maybe that's acceptable.Alternatively, maybe I can think of a continuous function that satisfies all three conditions. But I'm not sure. The function ( f(x) = frac{1}{x} ) with ( f(0) = 0 ) seems to work, even though it's not continuous.So, maybe that's the example I need.But let me double-check. For ( f(x) = frac{1}{x} ) with ( f(0) = 0 ):1. Domain: All real numbers, yes.2. For any ( b ), ( f(x) = b ) has exactly one solution: If ( b neq 0 ), ( x = frac{1}{b} ). If ( b = 0 ), ( x = 0 ). So, exactly one solution.3. For any ( a > 0 ) and ( b ), ( f(x) = ax + b ) has at least two solutions: For ( x neq 0 ), it's a quadratic equation with two solutions. If ( b = 0 ), ( x = 0 ) is also a solution, but we only need at least two, so that's fine.Yes, this function seems to satisfy all the conditions.