💡Okay, so I have this problem where I need to find the maximum value of a function f(x). The function f(x) is defined as the minimum of three other functions: y₁ = 4x + 1, y₂ = x + 2, and y₃ = -2x + 4. Hmm, okay, so for any real number x, f(x) is the smallest value among these three functions. And I need to find the maximum value that this minimum can take. Interesting.Let me try to visualize this. Each of these functions is a straight line, right? So y₁ is a line with a slope of 4, y₂ has a slope of 1, and y₃ has a slope of -2. Since f(x) is the minimum of these three, it's like taking the lowest point on these three lines at each x-value. So, the graph of f(x) would be piecewise linear, switching from one line to another at the points where two lines intersect because that's where the minimum could switch.So, first, I think I need to find the intersection points of these lines because that's where the minimum function f(x) might change from one line to another. Let me find where y₁ intersects y₂, y₂ intersects y₃, and y₁ intersects y₃.Starting with y₁ and y₂: 4x + 1 = x + 2. Let me solve for x. Subtract x from both sides: 3x + 1 = 2. Subtract 1: 3x = 1. So x = 1/3. Then, plugging back into y₂, y = (1/3) + 2 = 7/3. So they intersect at (1/3, 7/3).Next, y₂ and y₃: x + 2 = -2x + 4. Let me solve for x. Add 2x to both sides: 3x + 2 = 4. Subtract 2: 3x = 2. So x = 2/3. Plugging back into y₂: y = (2/3) + 2 = 8/3. So they intersect at (2/3, 8/3).Lastly, y₁ and y₃: 4x + 1 = -2x + 4. Let me solve for x. Add 2x to both sides: 6x + 1 = 4. Subtract 1: 6x = 3. So x = 1/2. Plugging back into y₁: y = 4*(1/2) + 1 = 2 + 1 = 3. So they intersect at (1/2, 3).Wait, so now I have three intersection points: (1/3, 7/3), (2/3, 8/3), and (1/2, 3). Hmm, that seems a bit confusing because the intersection points are not in order on the x-axis. Let me list them in order of x:1. (1/3, 7/3) ≈ (0.333, 2.333)2. (1/2, 3) = (0.5, 3)3. (2/3, 8/3) ≈ (0.666, 2.666)Wait, so the intersection points are at x = 1/3, x = 1/2, and x = 2/3. So, between these x-values, the minimum function f(x) will switch from one line to another.But I need to figure out which line is the minimum in each interval. Let me test each interval:1. For x < 1/3: Let's pick x = 0. Then y₁ = 1, y₂ = 2, y₃ = 4. So the minimum is y₁ = 1.2. For 1/3 < x < 1/2: Let's pick x = 0.4. Then y₁ = 4*0.4 + 1 = 1.6 + 1 = 2.6, y₂ = 0.4 + 2 = 2.4, y₃ = -2*0.4 + 4 = -0.8 + 4 = 3.2. So the minimum is y₂ = 2.4.3. For 1/2 < x < 2/3: Let's pick x = 0.6. Then y₁ = 4*0.6 + 1 = 2.4 + 1 = 3.4, y₂ = 0.6 + 2 = 2.6, y₃ = -2*0.6 + 4 = -1.2 + 4 = 2.8. So the minimum is y₂ = 2.6.Wait, but at x = 1/2, which is 0.5, the intersection point is (0.5, 3). So at x = 0.5, y₁ = 3, y₂ = 2.5, y₃ = 3. So the minimum is y₂ = 2.5.Wait, so actually, between x = 1/3 and x = 2/3, the minimum is y₂ = x + 2. But at x = 1/2, y₁ and y₃ both equal 3, which is higher than y₂. So in the interval (1/3, 2/3), the minimum is y₂.But then, what about beyond x = 2/3? Let's pick x = 1. Then y₁ = 5, y₂ = 3, y₃ = 2. So the minimum is y₃ = 2.Wait, so to summarize:- For x ≤ 1/3, the minimum is y₁ = 4x + 1.- For 1/3 ≤ x ≤ 2/3, the minimum is y₂ = x + 2.- For x ≥ 2/3, the minimum is y₃ = -2x + 4.But hold on, at x = 1/2, which is between 1/3 and 2/3, y₁ and y₃ are both 3, but y₂ is 2.5, which is still the minimum. So yes, y₂ is the minimum in that interval.So, f(x) is a piecewise function:f(x) = 4x + 1, when x ≤ 1/3f(x) = x + 2, when 1/3 ≤ x ≤ 2/3f(x) = -2x + 4, when x ≥ 2/3Now, I need to find the maximum value of f(x). Since f(x) is the minimum of these three lines, its graph will consist of segments from each of these lines, switching at the intersection points.Looking at the pieces:1. For x ≤ 1/3, f(x) = 4x + 1. This is a line with a positive slope, so it's increasing as x increases.2. For 1/3 ≤ x ≤ 2/3, f(x) = x + 2. This is also a line with a positive slope, so it's increasing as x increases.3. For x ≥ 2/3, f(x) = -2x + 4. This is a line with a negative slope, so it's decreasing as x increases.Therefore, f(x) increases up to x = 2/3, and then starts decreasing. So the maximum value of f(x) should occur at x = 2/3, where the function switches from increasing to decreasing.Let me compute f(2/3):f(2/3) = (2/3) + 2 = 8/3 ≈ 2.666...Alternatively, checking with y₃ at x = 2/3: y₃ = -2*(2/3) + 4 = -4/3 + 12/3 = 8/3. So yes, consistent.Therefore, the maximum value of f(x) is 8/3.Wait, let me just double-check my reasoning. So, f(x) is increasing on (-∞, 2/3] and decreasing on [2/3, ∞). So the highest point is at x = 2/3, which is 8/3. That seems correct.Just to make sure, let me check the value at x = 1/3:f(1/3) = 4*(1/3) + 1 = 4/3 + 3/3 = 7/3 ≈ 2.333...And at x = 2/3, it's 8/3 ≈ 2.666..., which is higher. So yes, the function increases up to 2/3, then decreases. So 8/3 is indeed the maximum.I think that's solid. So the maximum value of f(x) is 8/3.**Final Answer**The maximum value of ( f(x) ) is boxed{dfrac{8}{3}}.

💡Okay, so I have this problem: Find all odd integers ( n geq 1 ) such that ( n ) divides ( 3^{n} + 1 ). Hmm, interesting. I need to figure out for which odd numbers ( n ), the expression ( 3^n + 1 ) is divisible by ( n ). Let me try to break this down step by step.First, let's understand what it means for ( n ) to divide ( 3^n + 1 ). If ( n ) divides ( 3^n + 1 ), then ( 3^n + 1 ) is a multiple of ( n ). In other words, ( 3^n equiv -1 pmod{n} ). That's a congruence relation, which might be useful.Since ( n ) is an odd integer greater than or equal to 1, let's consider small odd integers first to see if they satisfy the condition. Maybe starting with ( n = 1 ), ( n = 3 ), ( n = 5 ), and so on.For ( n = 1 ): ( 3^1 + 1 = 4 ). Does 1 divide 4? Yes, because any number is divisible by 1. So, ( n = 1 ) is a solution.For ( n = 3 ): ( 3^3 + 1 = 27 + 1 = 28 ). Does 3 divide 28? 28 divided by 3 is approximately 9.333, which is not an integer. So, 3 does not divide 28. Therefore, ( n = 3 ) is not a solution.For ( n = 5 ): ( 3^5 + 1 = 243 + 1 = 244 ). Does 5 divide 244? 244 divided by 5 is 48.8, which is not an integer. So, ( n = 5 ) is not a solution.For ( n = 7 ): ( 3^7 + 1 = 2187 + 1 = 2188 ). Does 7 divide 2188? Let's check: 7 times 312 is 2184, and 2188 minus 2184 is 4. So, 2188 divided by 7 is 312 with a remainder of 4. Therefore, 7 does not divide 2188. So, ( n = 7 ) is not a solution.Hmm, so far, only ( n = 1 ) works. Let's try ( n = 9 ): ( 3^9 + 1 = 19683 + 1 = 19684 ). Does 9 divide 19684? Let's see: 9 times 2187 is 19683, so 19684 minus 19683 is 1. Therefore, 19684 divided by 9 is 2187 with a remainder of 1. So, 9 does not divide 19684. Thus, ( n = 9 ) is not a solution.This pattern suggests that maybe ( n = 1 ) is the only solution. But I need to be sure. Perhaps I should look for a general approach rather than checking each number individually.Let me consider the congruence ( 3^n equiv -1 pmod{n} ). If I square both sides, I get ( 3^{2n} equiv 1 pmod{n} ). So, ( 3^{2n} equiv 1 pmod{n} ). This means that the order of 3 modulo ( n ) divides ( 2n ). Let's denote the order of 3 modulo ( n ) as ( d ). Then, ( d ) divides ( 2n ).But also, by Fermat's Little Theorem, if ( n ) is a prime, then ( 3^{n-1} equiv 1 pmod{n} ). So, the order ( d ) must divide ( n - 1 ) as well. Therefore, ( d ) divides both ( 2n ) and ( n - 1 ). Hence, ( d ) divides the greatest common divisor (gcd) of ( 2n ) and ( n - 1 ).Let's compute ( gcd(2n, n - 1) ). Since ( n ) is odd, ( 2n ) is even, and ( n - 1 ) is even as well because ( n ) is odd. So, ( gcd(2n, n - 1) ) is at least 2. Let me see: ( gcd(2n, n - 1) = gcd(n - 1, 2n mod (n - 1)) ). Since ( 2n mod (n - 1) = 2n - 2(n - 1) = 2 ). Therefore, ( gcd(2n, n - 1) = gcd(n - 1, 2) ). Since ( n ) is odd, ( n - 1 ) is even, so ( gcd(n - 1, 2) = 2 ).Thus, ( d ) divides 2. So, the order ( d ) is either 1 or 2. If ( d = 1 ), then ( 3 equiv 1 pmod{n} ), which implies ( n ) divides ( 3 - 1 = 2 ). But ( n ) is odd, so the only possibility is ( n = 1 ). If ( d = 2 ), then ( 3^2 equiv 1 pmod{n} ), which implies ( n ) divides ( 9 - 1 = 8 ). But ( n ) is odd, so the only possibility is ( n = 1 ) again because 8 is even, and its divisors are 1, 2, 4, 8. Only 1 is odd.Wait, but what if ( n ) is composite? The above reasoning assumes ( n ) is prime because I used Fermat's Little Theorem. But ( n ) could be composite. Let me think about that.If ( n ) is composite, say ( n = p_1^{k_1} p_2^{k_2} dots p_m^{k_m} ), where ( p_i ) are primes, then for ( n ) to divide ( 3^n + 1 ), each prime power ( p_i^{k_i} ) must divide ( 3^n + 1 ). So, the same reasoning as above applies to each prime ( p_i ). For each prime ( p_i ), the order of 3 modulo ( p_i ) must divide 2. Therefore, each ( p_i ) must satisfy that ( 3^2 equiv 1 pmod{p_i} ), so ( p_i ) divides 8. But the only odd prime dividing 8 is none, since 8 is ( 2^3 ). Therefore, there are no odd primes ( p_i ) such that ( p_i ) divides ( 3^2 - 1 = 8 ). Hence, ( n ) cannot have any prime factors, meaning ( n = 1 ).Therefore, the only odd integer ( n geq 1 ) such that ( n ) divides ( 3^n + 1 ) is ( n = 1 ).

💡First, I will factorize the expression ( xy^2 - 2xy ).I notice that both terms have a common factor of ( xy ). I will factor out ( xy ) from both terms:[xy^2 - 2xy = xy(y - 2)]Next, I will factorize the expression ( 6a(x + y) - 5b(x + y) ).I observe that both terms have a common factor of ( (x + y) ). I will factor out ( (x + y) ) from both terms:[6a(x + y) - 5b(x + y) = (x + y)(6a - 5b)]Therefore, the factorized forms of the expressions are:[xy(y - 2) quad text{and} quad (x + y)(6a - 5b)]

💡Okay, so I have this problem about a parabola. Let me try to understand it step by step. The problem says that the distance from the focus F of a parabola to its directrix is 4. Then, it mentions a point P on the parabola where the distance from P to the y-axis is 1. I need to find the distance from point P to the focus F, and the options are A) 2, B) 3, C) 4, D) 5.First, I remember that for a parabola, the distance from any point on the parabola to the focus is equal to the distance from that point to the directrix. This is the definition of a parabola. So, if I can find the distance from P to the directrix, that should be equal to the distance from P to F.The distance from the focus to the directrix is given as 4. I think this means the focal length, which is the distance from the vertex to the focus, is half of that. Wait, no, actually, the distance from the focus to the directrix is twice the focal length. Let me recall: for a parabola, the distance from the vertex to the focus is called 'p', and the distance from the vertex to the directrix is also 'p', so the total distance between the focus and the directrix is 2p. So, if the distance from focus to directrix is 4, then 2p = 4, which means p = 2. So, the focal length is 2.Now, I need to figure out the equation of the parabola. Since the distance from P to the y-axis is 1, that means the x-coordinate of P is either 1 or -1. So, P is at (1, y) or (-1, y). Let me assume the parabola opens upward because the focus is above the vertex and the directrix is below. So, the standard equation of a parabola that opens upward with vertex at (0,0) is x² = 4py. Since p = 2, the equation becomes x² = 8y.Now, if P is on the parabola and its x-coordinate is 1, then plugging into the equation: 1² = 8y, so y = 1/8. So, P is at (1, 1/8). Similarly, if P is at (-1, y), then y would also be 1/8 because x² is the same for both 1 and -1.Now, I need to find the distance from P to the focus F. The focus F is at (0, p) which is (0, 2). So, the coordinates of F are (0, 2). The coordinates of P are (1, 1/8).To find the distance between P(1, 1/8) and F(0, 2), I can use the distance formula: distance = sqrt[(x2 - x1)² + (y2 - y1)²]. Plugging in the values:Distance PF = sqrt[(0 - 1)² + (2 - 1/8)²] = sqrt[(-1)² + (15/8)²] = sqrt[1 + 225/64].Let me compute 225/64: 64 goes into 225 three times (192) with a remainder of 33, so 225/64 = 3 33/64. So, 1 + 225/64 = 64/64 + 225/64 = 289/64.So, sqrt[289/64] = 17/8. Wait, 17/8 is 2.125. Hmm, that's not one of the answer choices. Did I make a mistake?Wait, maybe I assumed the wrong orientation of the parabola. What if the parabola opens to the right or left instead of upward? Let me think. If the parabola opens to the right, the standard equation is y² = 4px. In this case, the distance from the focus to the directrix is 4, so 2p = 4, p = 2. So, the equation would be y² = 8x.Given that the distance from P to the y-axis is 1, the x-coordinate of P is 1. So, plugging into the equation: y² = 8*1 = 8, so y = sqrt(8) or -sqrt(8). So, P is at (1, 2√2) or (1, -2√2).The focus F of a right-opening parabola is at (p, 0) which is (2, 0). So, coordinates of F are (2, 0). Coordinates of P are (1, 2√2).Now, distance PF = sqrt[(2 - 1)² + (0 - 2√2)²] = sqrt[1 + ( -2√2)²] = sqrt[1 + 8] = sqrt[9] = 3.Ah, that's one of the answer choices, B) 3. So, maybe the parabola opens to the right instead of upward. But why did I initially think it opens upward? Because the focus was above the vertex, but if it opens to the right, the focus is to the right of the vertex.Wait, the problem didn't specify the orientation of the parabola. It just said the distance from the focus to the directrix is 4. So, it could be any orientation. But the distance from P to the y-axis is 1, so P is at x = 1 or x = -1. If the parabola opens to the right, then P is at (1, y), which makes sense. If it opens upward, P is at (x, 1/8). But in that case, the distance PF didn't match the answer choices.So, perhaps the correct orientation is to the right, giving PF = 3, which is option B.But wait, let me double-check. If the parabola opens to the right, then the directrix is a vertical line x = -p, which is x = -2. The distance from P(1, 2√2) to the directrix is |1 - (-2)| = 3. And the distance from P to F(2,0) is sqrt[(2-1)^2 + (0 - 2√2)^2] = sqrt[1 + 8] = 3. So, yes, that works.Alternatively, if the parabola opens upward, the distance from P(1, 1/8) to the directrix y = -2 is |1/8 - (-2)| = 1/8 + 2 = 17/8, which is 2.125, and the distance PF was also 17/8, which doesn't match the answer choices. So, the parabola must open to the right, giving PF = 3.Therefore, the answer is B) 3.

💡Okay, so I have this problem here: For any real number ( m ), the graph of the function ( f(x) = m cdot arcsin x - 1 ) always passes through a fixed point ( P ). I need to find the coordinates of point ( P ). Hmm, interesting. Let me think about how to approach this.First, I know that a fixed point means that no matter what value ( m ) takes, the point ( P ) will always lie on the graph of the function. So, regardless of ( m ), when I plug in the coordinates of ( P ) into the equation ( f(x) = m cdot arcsin x - 1 ), the equation should hold true.Let me denote the coordinates of point ( P ) as ( (a, b) ). That means when ( x = a ), ( f(a) = b ). So, substituting into the function, we get:[ b = m cdot arcsin(a) - 1 ]Now, since this equation must hold true for any real number ( m ), the term involving ( m ) must somehow cancel out or not affect the equation. Otherwise, changing ( m ) would change the value of ( b ), which contradicts the idea that ( P ) is a fixed point.Looking at the equation ( b = m cdot arcsin(a) - 1 ), I notice that ( m ) is multiplied by ( arcsin(a) ). For this product to not affect the equation regardless of ( m ), the coefficient of ( m ) must be zero. In other words, ( arcsin(a) ) must be zero. Because if ( arcsin(a) = 0 ), then the term ( m cdot arcsin(a) ) becomes zero, and the equation simplifies to ( b = -1 ).So, let's solve for ( a ) when ( arcsin(a) = 0 ). The arcsine function equals zero when its argument is zero. Therefore:[ arcsin(a) = 0 implies a = sin(0) = 0 ]So, ( a = 0 ). Now, substituting ( a = 0 ) back into the equation for ( b ):[ b = m cdot arcsin(0) - 1 = m cdot 0 - 1 = -1 ]Therefore, ( b = -1 ). So, the coordinates of point ( P ) are ( (0, -1) ).Let me double-check this result. If I plug ( x = 0 ) into the function ( f(x) ), regardless of the value of ( m ), I should get ( f(0) = m cdot arcsin(0) - 1 = 0 - 1 = -1 ). That makes sense because ( arcsin(0) ) is indeed 0, so the term involving ( m ) disappears, leaving ( -1 ) as the y-coordinate.Is there any other point that could satisfy this condition? Suppose there is another point ( (a, b) ) where ( a neq 0 ). Then, ( arcsin(a) ) would not be zero, and ( b = m cdot arcsin(a) - 1 ). But since ( m ) can be any real number, ( b ) would change as ( m ) changes, which contradicts the requirement that ( P ) is a fixed point. Therefore, the only possible fixed point is ( (0, -1) ).Another way to think about this is to consider the function ( f(x) = m cdot arcsin x - 1 ) as a linear function in terms of ( m ). For each ( x ), ( f(x) ) is a linear function of ( m ). The only way this function can pass through the same point ( P ) for all ( m ) is if the coefficient of ( m ) is zero at that point, which again leads us to ( arcsin(a) = 0 ) and thus ( a = 0 ).So, after considering different angles and verifying my reasoning, I'm confident that the fixed point ( P ) is indeed ( (0, -1) ).**Final Answer**The coordinates of point ( P ) are boxed{(0, -1)}.

💡Okay, so I have this problem about quadratic polynomials called "double-up polynomials." It means that the polynomial has two real roots, and one of them is twice the other. The polynomial is given by ( f(x) = x^2 + px + q ), where ( p ) and ( q ) are real numbers. There are three parts to this problem: (a), (b), and (c). I need to solve each part step by step.Starting with part (a): If a double-up polynomial ( f(x) ) has ( p = -15 ), determine the value of ( q ).Alright, so I know that for a quadratic polynomial ( x^2 + px + q ), the sum of the roots is ( -p ) and the product of the roots is ( q ). Since it's a double-up polynomial, one root is twice the other. Let me denote the roots as ( r ) and ( 2r ).So, the sum of the roots is ( r + 2r = 3r ), which should be equal to ( -p ). Given that ( p = -15 ), the sum of the roots is ( 3r = -(-15) = 15 ). Therefore, ( r = 15 / 3 = 5 ).Now, the product of the roots is ( r times 2r = 2r^2 ), which should be equal to ( q ). Substituting ( r = 5 ), we get ( q = 2 times 5^2 = 2 times 25 = 50 ).So, for part (a), ( q = 50 ).Moving on to part (b): If ( f(x) ) is a double-up polynomial with one of the roots equal to 4, determine all possible values of ( p + q ).Again, since it's a double-up polynomial, the roots are ( r ) and ( 2r ). One of the roots is 4, so there are two possibilities: either ( r = 4 ) or ( 2r = 4 ).Case 1: If ( r = 4 ), then the other root is ( 2r = 8 ). The sum of the roots is ( 4 + 8 = 12 ), so ( -p = 12 ) which means ( p = -12 ). The product of the roots is ( 4 times 8 = 32 ), so ( q = 32 ). Therefore, ( p + q = -12 + 32 = 20 ).Case 2: If ( 2r = 4 ), then ( r = 2 ). The other root is ( 2r = 4 ). The sum of the roots is ( 2 + 4 = 6 ), so ( -p = 6 ) which means ( p = -6 ). The product of the roots is ( 2 times 4 = 8 ), so ( q = 8 ). Therefore, ( p + q = -6 + 8 = 2 ).So, the possible values of ( p + q ) are 20 and 2.Now, part (c): Determine all double-up polynomials for which ( p + q = 9 ).Again, the roots are ( r ) and ( 2r ). The sum of the roots is ( 3r = -p ), so ( p = -3r ). The product of the roots is ( 2r^2 = q ). Therefore, ( p + q = -3r + 2r^2 ).We are given that ( p + q = 9 ), so:( 2r^2 - 3r = 9 )Let me write this as a quadratic equation:( 2r^2 - 3r - 9 = 0 )To solve for ( r ), I can use the quadratic formula:( r = frac{3 pm sqrt{(-3)^2 - 4 times 2 times (-9)}}{2 times 2} )Calculating the discriminant:( (-3)^2 = 9 )( 4 times 2 times (-9) = -72 )So, the discriminant is ( 9 - (-72) = 9 + 72 = 81 )Therefore,( r = frac{3 pm sqrt{81}}{4} = frac{3 pm 9}{4} )So, two solutions:1. ( r = frac{3 + 9}{4} = frac{12}{4} = 3 )2. ( r = frac{3 - 9}{4} = frac{-6}{4} = -frac{3}{2} )So, the roots can be either 3 and 6, or -3/2 and -3.For ( r = 3 ), the polynomial is ( (x - 3)(x - 6) = x^2 - 9x + 18 ). So, ( p = -9 ) and ( q = 18 ), which gives ( p + q = -9 + 18 = 9 ).For ( r = -frac{3}{2} ), the polynomial is ( (x + frac{3}{2})(x + 3) = x^2 + frac{9}{2}x + frac{9}{2} ). So, ( p = frac{9}{2} ) and ( q = frac{9}{2} ), which gives ( p + q = frac{9}{2} + frac{9}{2} = 9 ).Therefore, the double-up polynomials are ( x^2 - 9x + 18 ) and ( x^2 + frac{9}{2}x + frac{9}{2} ).I think that's all for each part. Let me just recap:(a) Found ( q = 50 ) by using the sum and product of roots.(b) Considered both cases where the given root is the smaller or the larger one, leading to two possible values for ( p + q ).(c) Set up the equation ( p + q = 9 ) in terms of ( r ), solved the quadratic, and found the corresponding polynomials.Everything seems to check out. I don't see any mistakes in my calculations or reasoning.**Final Answer**(a) (boxed{50})(b) (boxed{20}) and (boxed{2})(c) (boxed{x^2 - 9x + 18}) and (boxed{x^2 + frac{9}{2}x + frac{9}{2}})