answer：Okay, so I have this problem where I need to find the value of ( m ) such that the function ( y = x^2 + 2mx + m ) is always greater than ( frac{3}{16} ) for all values of ( x ). Hmm, let me think about how to approach this.First, I remember that quadratic functions have a parabola shape when graphed. Since the coefficient of ( x^2 ) is positive (it's 1), the parabola opens upwards. That means the function has a minimum point, and the rest of the graph goes upwards from there. So, if I can find the minimum value of this function and ensure that it's greater than ( frac{3}{16} ), then the entire function will be greater than ( frac{3}{16} ) for all ( x ).To find the minimum value of the quadratic function ( y = x^2 + 2mx + m ), I think I need to complete the square. Completing the square helps in converting the quadratic into vertex form, which makes it easy to identify the vertex (the minimum point in this case).Let me try completing the square for ( y = x^2 + 2mx + m ).Starting with:[ y = x^2 + 2mx + m ]I can rewrite this as:[ y = (x^2 + 2mx) + m ]To complete the square for ( x^2 + 2mx ), I take half of the coefficient of ( x ), which is ( m ), and square it, getting ( m^2 ). So, I add and subtract ( m^2 ) inside the equation to maintain equality.[ y = (x^2 + 2mx + m^2 - m^2) + m ][ y = (x + m)^2 - m^2 + m ]So, the equation becomes:[ y = (x + m)^2 + (m - m^2) ]Now, this is in vertex form ( y = a(x - h)^2 + k ), where the vertex is at ( (-m, m - m^2) ). Since the parabola opens upwards, the minimum value of ( y ) is ( m - m^2 ).We need this minimum value to be greater than ( frac{3}{16} ). So, setting up the inequality:[ m - m^2 > frac{3}{16} ]Let me rearrange this inequality to make it easier to solve. I'll bring all terms to one side:[ -m^2 + m - frac{3}{16} > 0 ]Multiplying both sides by -1 (and remembering to reverse the inequality sign):[ m^2 - m + frac{3}{16} < 0 ]Now, I have a quadratic inequality ( m^2 - m + frac{3}{16} < 0 ). To solve this, I need to find the roots of the quadratic equation ( m^2 - m + frac{3}{16} = 0 ).Using the quadratic formula:[ m = frac{-b pm sqrt{b^2 - 4ac}}{2a} ]where ( a = 1 ), ( b = -1 ), and ( c = frac{3}{16} ).Plugging in the values:[ m = frac{-(-1) pm sqrt{(-1)^2 - 4 cdot 1 cdot frac{3}{16}}}{2 cdot 1} ][ m = frac{1 pm sqrt{1 - frac{12}{16}}}{2} ][ m = frac{1 pm sqrt{frac{4}{16}}}{2} ][ m = frac{1 pm frac{2}{4}}{2} ][ m = frac{1 pm frac{1}{2}}{2} ]So, the two solutions are:[ m = frac{1 + frac{1}{2}}{2} = frac{frac{3}{2}}{2} = frac{3}{4} ]and[ m = frac{1 - frac{1}{2}}{2} = frac{frac{1}{2}}{2} = frac{1}{4} ]These roots divide the number line into three intervals:1. ( m < frac{1}{4} )2. ( frac{1}{4} < m < frac{3}{4} )3. ( m > frac{3}{4} )Since the quadratic ( m^2 - m + frac{3}{16} ) opens upwards (because the coefficient of ( m^2 ) is positive), the quadratic expression will be less than zero between its roots. Therefore, the solution to the inequality ( m^2 - m + frac{3}{16} < 0 ) is:[ frac{1}{4} < m < frac{3}{4} ]So, for ( m ) in this interval, the minimum value of the function ( y = x^2 + 2mx + m ) is greater than ( frac{3}{16} ), which means the entire function is greater than ( frac{3}{16} ) for all ( x ).I should double-check my steps to make sure I didn't make any mistakes. Completing the square seems correct, and solving the quadratic inequality also looks right. The roots were calculated accurately, and the interval where the quadratic is negative is between the roots because the parabola opens upwards. So, I think my conclusion is correct.**Final Answer**The value of ( m ) must satisfy ( boxed{frac{1}{4} < m < frac{3}{4}} ).

answer：Alright, so I've got this problem here about a function f(x) whose domain is all positive real numbers, ℝ₊. The problem says that f(x + y) equals f(x) plus f(y), and it also tells me that f(8) is 3. I need to find the value of f(2). Hmm, okay, let's break this down step by step.First, the function f(x) is defined for all positive real numbers, which means x and y in the equation f(x + y) = f(x) + f(y) must also be positive. That makes sense because if x and y were negative, their sum could potentially be zero or negative, which isn't in the domain. So, I don't have to worry about negative numbers here.Now, the equation f(x + y) = f(x) + f(y) looks familiar. I think this is called Cauchy's functional equation. From what I remember, functions that satisfy this equation are linear functions of the form f(x) = kx, where k is a constant. But wait, is that always true? I think it's true if we assume some regularity conditions like continuity, which might be implied here since the domain is ℝ₊ and we're dealing with real numbers.So, if f(x) is linear, then f(x) = kx. Let's test this assumption. If f(x) = kx, then f(x + y) = k(x + y) = kx + ky = f(x) + f(y), which satisfies the given condition. Great, so it seems like f(x) is indeed a linear function.Now, the problem gives me f(8) = 3. If f(x) = kx, then f(8) = 8k = 3. To find k, I can solve for it: k = 3/8. So, the function must be f(x) = (3/8)x.But wait, the question is asking for f(2), not f(8). So, plugging x = 2 into f(x) = (3/8)x, I get f(2) = (3/8)*2 = 6/8 = 3/4. So, f(2) should be 3/4.Hold on, let me make sure I didn't make any mistakes here. I assumed that f(x) is linear because it satisfies Cauchy's equation, and with the given domain, that seems reasonable. But is there another way to approach this without assuming linearity?Let me think. If I don't assume f(x) is linear, can I derive f(2) directly from the given information? Let's try that.We know that f(x + y) = f(x) + f(y). Let's set x = y = 4. Then, f(4 + 4) = f(4) + f(4), which simplifies to f(8) = 2f(4). We know f(8) is 3, so 3 = 2f(4). Solving for f(4), we get f(4) = 3/2.Now, let's set x = y = 2. Then, f(2 + 2) = f(2) + f(2), which simplifies to f(4) = 2f(2). We already found that f(4) is 3/2, so 3/2 = 2f(2). Solving for f(2), we get f(2) = (3/2)/2 = 3/4.Okay, so that confirms it. Whether I assume linearity or break it down step by step using the given functional equation, I end up with f(2) = 3/4. That seems solid.Just to be thorough, let me check if there's another angle to approach this problem. Maybe using induction or some other method? Well, since the function is defined for all positive real numbers, and we're dealing with addition, induction might not be the most straightforward approach here. The step-by-step substitution method I used earlier seems more appropriate.Another thought: could there be a non-linear function that satisfies f(x + y) = f(x) + f(y)? I recall that without additional constraints like continuity, there are pathological solutions to Cauchy's equation. However, since the domain is ℝ₊ and we're given a specific value at 8, it's likely that the function is linear in this context. So, I think my initial assumption holds.Let me recap:1. Given f(x + y) = f(x) + f(y) for all x, y ∈ ℝ₊.2. Given f(8) = 3.3. Assume f is linear: f(x) = kx.4. Solve for k using f(8) = 3: k = 3/8.5. Therefore, f(2) = (3/8)*2 = 3/4.Alternatively:1. Start with f(8) = 3.2. Express f(8) as f(4 + 4) = f(4) + f(4) = 2f(4).3. So, 2f(4) = 3 ⇒ f(4) = 3/2.4. Then, express f(4) as f(2 + 2) = f(2) + f(2) = 2f(2).5. So, 2f(2) = 3/2 ⇒ f(2) = 3/4.Both methods lead to the same result, which gives me confidence that f(2) is indeed 3/4.I think I've covered all the bases here. I don't see any mistakes in my reasoning, and both approaches confirm the same answer. So, I feel pretty confident that the correct value of f(2) is 3/4.**Final Answer**The value of ( f(2) ) is boxed{B}.

answer：First, I need to determine the fraction of the total length that each part represents when the wire is divided into 3 equal parts.The total length of the wire is (frac{4}{5}) meters. Dividing this length into 3 equal parts means each part will be (frac{1}{3}) of the total length.To find the fraction, I divide the total length by the number of parts:[frac{4}{5} div 3 = frac{4}{5} times frac{1}{3} = frac{4}{15}]However, the question asks for the fraction of the total length, not the exact length of each part. Since the wire is divided into 3 equal parts, each part is (frac{1}{3}) of the total length.Therefore, the correct answer is:

answer：Alright, so I've got this problem here about functions and their domains. It says that the domain of the function ( y = f(x + 1) ) is ([1, 2]), and I need to find the domain of the function ( y = f(2x - 1) ). The options are A: ([1/2, 1]), B: ([3/2, 2]), C: ([-1, 1]), and D: ([3, 5]). Hmm, okay, let's break this down step by step.First, I remember that the domain of a function is the set of all possible input values (x-values) for which the function is defined. So, if the domain of ( y = f(x + 1) ) is ([1, 2]), that means ( x ) can be any value from 1 to 2, inclusive. Now, ( f(x + 1) ) is a transformation of the original function ( f(x) ). Specifically, it's a horizontal shift. Since it's ( x + 1 ), that means the graph of ( f(x) ) is shifted one unit to the left. But how does this affect the domain?Well, if the domain of ( f(x + 1) ) is ([1, 2]), that tells me something about the domain of ( f(x) ) itself. Let's think about it: when you replace ( x ) with ( x + 1 ) in the function, you're effectively shifting the input. So, for ( f(x + 1) ) to be defined, ( x + 1 ) must lie within the domain of ( f ). Given that ( x ) is in ([1, 2]), then ( x + 1 ) would be in ([2, 3]). So, this implies that the original function ( f(x) ) has a domain of ([2, 3]). That makes sense because if ( x + 1 ) must be within ([2, 3]) for ( f(x + 1) ) to be defined, then ( f(x) ) itself must be defined for inputs from 2 to 3.Okay, so now I know that ( f(x) ) has a domain of ([2, 3]). The next part of the problem asks for the domain of ( y = f(2x - 1) ). Again, this is a transformation of ( f(x) ), but this time it's a combination of a horizontal shift and a horizontal scaling.Let me recall how transformations affect the domain. If you have a function ( f(kx + c) ), the domain is affected by the scaling factor ( k ) and the shift ( c ). Specifically, to find the domain of ( f(kx + c) ), you need to solve the inequality that ensures ( kx + c ) lies within the domain of ( f ).In this case, ( f(2x - 1) ) means ( k = 2 ) and ( c = -1 ). So, we need to find all ( x ) such that ( 2x - 1 ) is within the domain of ( f ), which we've established is ([2, 3]).So, let's set up the inequality:[2 leq 2x - 1 leq 3]Now, I need to solve this inequality for ( x ).First, let's add 1 to all parts of the inequality to isolate the term with ( x ):[2 + 1 leq 2x - 1 + 1 leq 3 + 1]Simplifying that:[3 leq 2x leq 4]Now, divide all parts by 2 to solve for ( x ):[frac{3}{2} leq x leq 2]So, ( x ) must be between ( frac{3}{2} ) and 2, inclusive. That means the domain of ( y = f(2x - 1) ) is ([ frac{3}{2}, 2 ]).Looking back at the options, that's option B. So, I think the answer is B.But wait, let me double-check to make sure I didn't make any mistakes. Sometimes when dealing with transformations, it's easy to mix up the direction of the shift or the scaling factor.Starting again, the domain of ( f(x + 1) ) is ([1, 2]). That means ( x + 1 ) must be in ([2, 3]), so ( f(x) ) has a domain of ([2, 3]). For ( f(2x - 1) ), we set ( 2x - 1 ) to be within ([2, 3]). Solving ( 2x - 1 geq 2 ) gives ( 2x geq 3 ), so ( x geq frac{3}{2} ). Similarly, ( 2x - 1 leq 3 ) gives ( 2x leq 4 ), so ( x leq 2 ). Therefore, ( x ) is between ( frac{3}{2} ) and 2, which confirms my earlier result.I think I'm confident that the domain is ([ frac{3}{2}, 2 ]), so option B is correct.**Final Answer**boxed{B}

answer：Okay, so I have this problem here: the function is y equals (k minus 3) times x plus 2. And the question is asking, if y increases as x increases, what are the possible values of k? The options are A:1, B:2, C:3, D:4.Alright, let me try to figure this out step by step. First, I remember that in a linear function like y = mx + b, the slope m determines whether the function is increasing or decreasing. If m is positive, the function increases as x increases, and if m is negative, it decreases. So in this case, the slope is (k - 3). So, for y to increase as x increases, the slope (k - 3) must be greater than zero. That means I need to solve the inequality k - 3 > 0. If I add 3 to both sides, I get k > 3. So k has to be greater than 3.Now, looking at the options, A is 1, B is 2, C is 3, and D is 4. So which of these are greater than 3? Well, 1, 2, and 3 are not greater than 3; they're equal to or less than 3. Only 4 is greater than 3. So that means k must be 4.Wait, let me double-check that. If k is 4, then the slope becomes 4 - 3, which is 1. So y = 1x + 2, which is y = x + 2. Yeah, that definitely increases as x increases because the slope is positive.What if k was 3? Then the slope would be 0, making the function y = 0x + 2, which is just y = 2. That's a horizontal line, so y doesn't increase or decrease; it stays constant. So k can't be 3 because we need y to increase.Similarly, if k was 2, the slope would be 2 - 3 = -1, so y = -x + 2, which is a decreasing function. And if k was 1, the slope would be 1 - 3 = -2, so y = -2x + 2, which also decreases. So none of those work.Therefore, the only value that makes y increase as x increases is k = 4, which is option D.

answer：Alright, so I've got this problem here about a sequence of positive terms. The first term is 1, and there's this recurrence relation: (n+1)a_{n+1}^2 - n a_n^2 + a_n a_{n+1} = 0 for n = 1, 2, 3, and so on. I need to find the general term a_n. Hmm, okay, let's try to figure this out step by step.First, I should probably write down the recurrence relation to get a better look at it. So, it's (n+1)a_{n+1}^2 - n a_n^2 + a_n a_{n+1} = 0. That's a quadratic in terms of a_{n+1}, right? Maybe I can rearrange it to make it easier to solve.Let me try to rearrange the equation. If I move the n a_n^2 term to the other side, I get (n+1)a_{n+1}^2 + a_n a_{n+1} = n a_n^2. Hmm, that looks a bit better. Maybe I can factor this equation somehow. Let me see if I can factor it into something like (something with a_{n+1}) times (something else) equals zero.Looking at the left side, (n+1)a_{n+1}^2 + a_n a_{n+1}, I notice that both terms have a_{n+1} in them. So, I can factor out an a_{n+1}, which gives me a_{n+1}[(n+1)a_{n+1} + a_n] = n a_n^2. Hmm, not sure if that helps directly, but maybe I can set up the equation differently.Wait, another thought: since this is a quadratic in a_{n+1}, maybe I can use the quadratic formula to solve for a_{n+1}. Let's try that. The standard quadratic form is ax^2 + bx + c = 0. Comparing that to our equation, (n+1)a_{n+1}^2 + a_n a_{n+1} - n a_n^2 = 0. So, a = (n+1), b = a_n, and c = -n a_n^2.Applying the quadratic formula, a_{n+1} = [-b ± sqrt(b^2 - 4ac)] / (2a). Plugging in the values, that becomes a_{n+1} = [-a_n ± sqrt(a_n^2 - 4*(n+1)*(-n a_n^2))]/(2*(n+1)). Let's simplify the discriminant inside the square root.So, the discriminant D = a_n^2 - 4*(n+1)*(-n a_n^2) = a_n^2 + 4n(n+1)a_n^2. Factor out a_n^2: D = a_n^2[1 + 4n(n+1)]. Let's compute 1 + 4n(n+1): that's 1 + 4n^2 + 4n. Hmm, that looks like a perfect square. Let me check: (2n + 1)^2 = 4n^2 + 4n + 1. Yep, exactly! So, D = a_n^2*(2n + 1)^2.Therefore, sqrt(D) = a_n*(2n + 1). So, plugging back into the quadratic formula, a_{n+1} = [-a_n ± a_n*(2n + 1)] / [2*(n+1)]. Let's consider both the plus and minus cases.First, the plus case: [-a_n + a_n*(2n + 1)] / [2*(n+1)] = [a_n*(-1 + 2n + 1)] / [2*(n+1)] = [a_n*(2n)] / [2*(n+1)] = [2n a_n] / [2(n+1)] = n a_n / (n+1).Second, the minus case: [-a_n - a_n*(2n + 1)] / [2*(n+1)] = [ -a_n*(1 + 2n + 1) ] / [2*(n+1)] = [ -a_n*(2n + 2) ] / [2*(n+1)] = [ -2(n + 1)a_n ] / [2(n + 1)] = -a_n.But wait, the problem states that the sequence consists of positive terms. So, a_{n+1} must be positive. In the minus case, we get a_{n+1} = -a_n, which would be negative since a_n is positive. That's not allowed. So, we discard the minus case.Therefore, the only valid solution is a_{n+1} = (n a_n) / (n + 1). Okay, so that's a simpler recurrence relation. Now, let's see if we can find a pattern or solve this recurrence.Given that a_{n+1} = (n / (n + 1)) a_n, and we know that a_1 = 1. Let's compute the first few terms to see if we can spot a pattern.a_1 = 1.a_2 = (1 / 2) a_1 = (1 / 2)*1 = 1/2.a_3 = (2 / 3) a_2 = (2 / 3)*(1/2) = (2 / 3)*(1/2) = 1/3.a_4 = (3 / 4) a_3 = (3 / 4)*(1/3) = 1/4.a_5 = (4 / 5) a_4 = (4 / 5)*(1/4) = 1/5.Hmm, I see a pattern here. It seems like a_n = 1/n. Let's test this hypothesis.Assume that a_n = 1/n. Then, a_{n+1} should be (n / (n + 1)) * a_n = (n / (n + 1))*(1/n) = 1/(n + 1). Which matches our assumption. So, by induction, if a_n = 1/n, then a_{n+1} = 1/(n + 1). Since a_1 = 1, which is 1/1, the formula holds for n=1. Therefore, by mathematical induction, a_n = 1/n for all positive integers n.Wait, but just to be thorough, let's plug a_n = 1/n back into the original recurrence relation to make sure it satisfies the equation.Original recurrence: (n+1)a_{n+1}^2 - n a_n^2 + a_n a_{n+1} = 0.Substitute a_n = 1/n and a_{n+1} = 1/(n + 1):Left side: (n + 1)*(1/(n + 1))^2 - n*(1/n)^2 + (1/n)*(1/(n + 1)).Simplify each term:First term: (n + 1)*(1/(n + 1)^2) = (n + 1)/(n + 1)^2 = 1/(n + 1).Second term: n*(1/n^2) = n/n^2 = 1/n.Third term: (1/n)*(1/(n + 1)) = 1/(n(n + 1)).So, putting it all together:1/(n + 1) - 1/n + 1/(n(n + 1)).Let's combine these terms. Let's find a common denominator, which is n(n + 1).First term: 1/(n + 1) = n / [n(n + 1)].Second term: -1/n = -(n + 1) / [n(n + 1)].Third term: 1 / [n(n + 1)].So, adding them up:[n - (n + 1) + 1] / [n(n + 1)] = [n - n - 1 + 1] / [n(n + 1)] = [0] / [n(n + 1)] = 0.Which matches the right side of the equation. So, yes, a_n = 1/n satisfies the recurrence relation.Therefore, the general term of the sequence is a_n = 1/n.I think that's it. It makes sense, and I've checked it by plugging it back into the original equation. So, I'm confident that a_n = 1/n is the correct solution.