✅1.彩神安全购彩代理✅2.【🔥🧧最新地址、网站登录、全站APP、手机下载、中国官方、网页入口、游戏网址、官网注册🧧🔥】3.🎁天天大惊喜礼包🎁,🧧领不完的红包雨🧧**南京推广网站：打造全国领先的品牌魅力**
在全球化的背景下，中国城市的发展正迎来翻天覆地的变化。南京作为中国城市的代表之一，在互联网时代，已经成为一个最具活力和影响力的城市之一。通过构建自己的“推广网站”，南京不仅能够提高自身的信息传播能力，还能为地方经济发展提供有力的支撑。
一、从“政府网站”到“专业平台”
近年来，南京政府投入不断增大，重点加强了网络信息管理，建立了专门的“推广网站”。这些网站不仅仅是传统意义上的政府网站，更是政府信息传播的前沿阵地。以“中共南京宣传部”为例，“中共南京宣传网”是一个集信息、资讯和教育于一体的综合平台。
1.1 专业的内容创作
在信息传播方面，南京的“推广网站”特别注重专业性，通过精心策划的内容，确保信息的准确性和权威性。例如，“关于\\n\\n\\n\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\\n\]
**Question:** **For the following question, it's given that \( x \) is a real number such that \( 0 < x < 1 \). Let \( f(x) = \begin{cases}
x^2 + e^{x}, & \text{if } x < 1 \\
ax + b, & \text{otherwise}.
\end{cases} \) (i) Find the values of \( a \) and \( b \).
**Alright, so I have this function f(x) that's defined piecewise. It says if x is less than 1, then it's x squared plus e to the power x. Otherwise, it's ax plus b. They're asking me to find the values of a and b. Hmm.
First, let me make sure I understand what the problem is asking. So f(x) has two different expressions depending on whether x is less than 1 or not. For x < 1, it's a quadratic function: x² + e^x. For all other x (including when x >= 1), it's linear: ax + b. They want me to find the constants a and b such that this piecewise function is continuous.
Wait, so maybe the problem is asking for continuity? It doesn't specify differentiability or anything else, but since f(x) is defined in terms of polynomials and exponentials, which are smooth functions, I think it's more likely about ensuring continuity because otherwise, if a and b weren't chosen correctly, the function might not make sense or could have jumps.
Let me also recall that for piecewise functions to be continuous at the point where their definitions change (which is here at x=1), the left-hand limit as x approaches 1 from below must equal the right-hand limit as x approaches 1 from above. Since at x=1, f(x) switches definition.
So, let me write that down:
f(1^-) = lim_{x→1^-} f(x) = (1)^2 + e^1 = 1 + e
f(1^+) = lim_{x→1^+} f(x) = a*(1) + b = a + b
For the function to be continuous at x=1, these two must be equal:
1 + e = a + b [Equation 1]
That's one equation involving a and b. But I have two variables here, so maybe another condition is needed. Wait, but the problem doesn't specify any other conditions. Hmm.
Wait, let me reread the question: it says that "For the following question..." with no more information given. So perhaps in this context, we need to ensure continuity for all x? Or just at a certain point?
Well, if I assume that f(x) is continuous everywhere except possibly at x=1 because of the piecewise definition, but let's check.
If f(x) is defined as x² + e^x when x < 1 and ax + b otherwise. For x >= 1, it just follows ax + b. So, for x > 1, it's linear, and for x <=1, it's a quadratic. So, unless there's some continuity condition on either side of x=1, but since f(x) is not given any specific definition beyond that piecewise, maybe they only require the function to be continuous at x=1.
Wait, hold on, perhaps I need more than one condition because just ensuring continuity at x=1 might not be enough. Wait, no—if it's defined for all real numbers, but with a single point of discontinuity, which is at x=1, so maybe they just require f(x) to be continuous everywhere.
Wait, if that's the case, then we also need to make sure that ax + b is continuous for x >= 1. But since it's linear, that part is already continuous everywhere because linear functions are continuous on their entire domain.
Similarly, for x <1, x² + e^x is a polynomial function, so it's continuous on its entire domain as well. So maybe the only point of concern is continuity at x=1.
So if f(x) is defined to be differentiable or something else... Hmm.
But in the problem statement, they just say "for the following question," and give f(x) with two pieces, without any mention of other conditions.
Therefore, perhaps I need more information. Wait, but let's look back: the user hasn't provided additional context about f(x). So maybe it's a standard question where continuity is required, which would give one equation between a and b.
But in that case, we might have to assume another condition elsewhere. Or perhaps the problem statement is just asking for continuity at x=1, giving one equation.
Wait, but I'm supposed to write a thought process of someone solving this, so maybe let me try to think step by step.
Given f(x) defined as:
f(x) = {
x² + e^x, if x < 1
ax + b, otherwise.
}
So the function is linear for x >=1 and quadratic below 1. If I'm supposed to find a and b such that this function is continuous everywhere.
But unless given some other conditions, it's difficult because the linear part is defined for all real numbers—there are no jumps or breaks there except at x=1. Similarly, the quadratic is also smooth.
Thus, maybe f(x) must be continuous only at x=1? So just making sure that from both sides, approaching 1, gives the same value.
Therefore:
lim_{x→1^-} f(x) = (1)^2 + e^1 = 1 + e
lim_{x→1^+} f(x) = a*1 + b = a + b
For continuity at x=1, we have 1 + e = a + b.
But unless given another condition... Maybe the function is also differentiable elsewhere? But that's not specified. It just asks for "values of a and b." So maybe it just wants the value of a and b so that f(x) is continuous at x=1, which gives one equation. But since we have two variables, I need another condition.
Wait, perhaps there's more to this function? The problem says "For the following question...," but doesn't give any other constraints.
Wait, maybe it's a single-variable calculus problem without additional conditions—just ensuring continuity at x=1—it would only require that 1 + e = a + b. But then we have infinitely many solutions since there are infinite pairs (a, b) such that a + b = 1 + e. So perhaps I missed something.
Wait, wait: looking back, the original problem says "For the following question..." without any more context—maybe it's an initial value problem or another condition? Or is this from a standard textbook?
Alternatively, maybe f(x) is also differentiable elsewhere? But since for x >=1, it's linear, which is differentiable everywhere. For x <1, it's quadratic, so differentiable everywhere as well. So the point of concern would be at x=1.
Therefore, perhaps all that's needed is continuity at x=1, giving us 1 + e = a + b.
But since we have two variables, I need more information to solve for both a and b. Hmm. Is there an additional condition?
Wait, maybe considering that the function is defined as f(x) piecewise, but if we take the limit from the left (x approaching 1^-), it's x^2 + e^x; so at x=1, is f(1) equal to either expression? But unless specified, probably not.
Wait, in general, for a function to be continuous everywhere, you need that all points are continuous. So if f(x) is piecewise defined as quadratic below 1 and linear otherwise, then the only point of concern is x=1 because at other points both expressions are already smooth functions.
Hence, perhaps all we can ensure is continuity at x=1, giving us a + b = 1 + e. But without more conditions, there's infinitely many solutions for (a, b). So maybe the problem was supposed to give another condition? Or perhaps it's a standard question where they assume f(x) is differentiable elsewhere or something.
Wait, looking again: the user wrote:
"For the following question, it's given that x is a real number such that 0 < x <1. Let f(x)=
{ x² + e^x, if x <1
ax + b, otherwise.
(i) Find the values of a and b."
So, only for 0