社区版:真三录像精彩对决不容错过-九一制片厂麻豆果冻传媒

发布时间：2026-06-16 20:57:13 · 浏览：9 · 来源：真三录像精彩对决不容错过

**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 =1, it's ax +b. So, but unless specified elsewhere, perhaps assuming continuity at all points would require that limit from below is equal to limit from above.

Therefore:

lim_{x→1^-} f(x) = 1 + e

lim_{x→1^+} f(x) = a*1 + b =a + b

Hence, we have the equation: 1 + e = a + b.

But since we only have one equation with two variables, a and b, perhaps more information is needed. Maybe the function must be continuous at x=0? But for x approaching 0 from above, f(x) approaches 0 + e^0 =1. Then, if taking the limit as x approaches 0 from above, it's 1. So does that require something about a and b?

Wait, but in the function definition, when x <1, which is all points including x=0, so approaching 0 from below would also approach f(0) =0 + e^0 =1. But since for x <1, it's defined as quadratic, at x approaches 0, you still get 1.

But if the function is ax +b beyond that, but near x=0, whether you take left or right limit, they should agree—so perhaps continuity everywhere else gives us just a + b =1 + e. But without another condition, can't determine both a and b uniquely.

Wait, unless the problem expects f(x) to be continuous at x=0 as well? Let's see: approaching 0 from above, it is 0^2 +e^0=1; approaching from below, same thing—it’s always 1. So if you take the limit at x=0 via either side, you get 1, which is equal to f(0)=0+ e^0=1.

Similarly, for any other point in 0

But then again, the problem seems too vague—there's not enough information to find unique values of a and b. Hmm.

Wait, maybe in the original source, this was part of a larger question or from a textbook that provides more context? Since it's referred to as "(i) Find the values of a and b."

Wait, perhaps considering another condition like f(x) is differentiable at x=1?

If so, then:

lim_{x→1^-} f(x)=1 + e

f'(from below): derivative of x² +e^x is 2x +e^x. At x approaching 1 from below, it approaches 2*1 + e =2 + e.

From the right: f'(ax +b) is a constant function; so its derivative everywhere is a.

Hence, if differentiable at x=1, then lim_{x→1} f(x)=f(1), which gives 1 + e. Also, from the left, derivative approaches 2 + e. From the right, derivative is a. For differentiability, these must be equal:

So, 2 + e =a.

Hence, if they want differentiable at x=1, a would be 2 +e, and then b can be found as a +b=1+e => (2 +e) +b=1 +e => b=1 +e - (2 +e)= -1. So in that case, a=2 +e, b=-1.

But the original problem just says "Find the values of a and b"—so I think maybe it's expecting differentiability as well? Or is there something else missing?

Alternatively, perhaps another condition—such as f(0) from both sides. For x approaching 0 from above: f(x)=x² +e^x approaches1; from below, same thing—it’s 1. So if f(0) were defined elsewhere—probably not given.

Wait, but in the problem statement, it's only for 0

Alternatively, maybe it's an initial value condition? But the user hasn't provided any other constraints.

Wait, perhaps I'm overcomplicating this. The function is f(x)= x² + e^x when x<1 and ax +b otherwise.

For 0

(1)^2 + e^{1}=a*(1)+b => 1 + e = a + b. But again, since there's only one equation with two variables, without more conditions, we can't uniquely determine both a and b.

Alternatively, if the function is smooth everywhere, but in reality, f(x)=ax +b for x >=0. So it could be that at x=1, to have f(x) continuous from both sides:

lim_{x→1^-}f(x)=1 + e, lim_{x→1^+}f(x)=a*1 + b=a + b.

But unless we know more about f(1), perhaps? If f(1)= something else.

Alternatively, maybe the function is smooth in a broader sense—like it has continuous derivatives everywhere. Then, at x=1, not only must the functions agree but their derivatives also match. So:

From above: derivative approaches 2 + e from left; From below, derivative is just a. Hence, we need to set 2 +e =a.

Similarly, if f(1)= something else given? If f(1) is defined elsewhere in the problem—it's not clear.

Alternatively, perhaps if they want the function to be smooth for all x >=0, which would require that at x=1, limit from below and above agree with derivative.

Wait, maybe I'm overcomplicating. Perhaps the original question had different constraints or was part of a larger system. Since in this prompt, it's just "Find the values of a and b," without more information, perhaps the answer is that infinitely many solutions exist if we don't impose additional conditions—unless they expect the minimal solution where f(x) becomes continuous at x=1 by matching derivatives.

Wait, but maybe I should consider another approach: perhaps in the original problem, it's given as a piecewise function and asks for values such that certain conditions are met. But in this prompt, only "南京大学” is mentioned, but probably it refers to me, or perhaps an initial question.

Alternatively, since the user said “according to ‘南京大学’”，but that might be irrelevant—it’s more about the problem statement given.

Wait, another idea: maybe I should think of it as a function from R to R with specific continuity at x=1. If so, then setting 1 + e =a + b is necessary for continuity. But without another condition, perhaps any values satisfying that equation would work. So, infinitely many solutions exist.

But the user seems to be asking "Find the values of a and b"—which suggests a unique solution. Maybe I misread the problem?

Wait, let me re-examine: “according to ‘南京大学’” is there something? Wait, no—it's just a prompt about an academic question related to University.

Perhaps it's a translation issue—maybe it should be "According to 'Northwestern University’".

But regardless, perhaps in this case, the original problem expects us to set f(x) as smooth beyond x=1 by matching derivatives at x=1. So that would give us two equations:

1. Continuity: 1 + e =a + b

2. Differentiability (if assumed): derivative from left is 2 +e, derivative from right is a => thus 2 +e =a.

Then with these, we can solve for b=1 +e -a=1 +e -(2 +e)= -1.

So in that case, the answer would be a=2 + e, and b=-1. So perhaps this is what was intended—assuming differentiability beyond x=1.

Alternatively, if they don't want to assume differentiability, but only continuity at x=1, then it's impossible with just one equation; we can have infinitely many solutions where a +b=1+ e.

Wait, but the original problem doesn’t specify any other condition except for "Find the values of a and b". So perhaps without more information, I should present both possibilities: either assuming differentiability, giving specific numbers, or if not, recognizing that there are infinitely many solutions. But since in the prompt, it's called “according to”，so maybe original problem was about continuity only.

Alternatively, another thought—maybe they consider f(x) as a piecewise function where for x<1, f(x)=x² +e^x, and then elsewhere (i.e., for x>=1), it's ax +b. So in that case, the condition is to ensure that at x=1, the left limit equals right limit.

Hence, 1 + e =a*1 + b => a + b =1 +e.

But again, without another equation, we can’t find unique values for a and b. So perhaps, as an answer, we have a +b=1 +e, but in the problem statement, unless more conditions are given—so maybe the answer is that there's infinitely many solutions such that a +b =1 +e.

But since the user just asked "Find the values of a and b", without further context, perhaps in the original source, they assumed differentiability. So I will present both options.

Alternatively, if it was part of an initial problem where f(x) is given for x <=1, like f(x)=x² +e^x, then to make it smooth beyond x=1 requires that the function and its derivative match at x=1—so in that case, a=2 + e, b=-1.

But since the prompt only says “南京大学”，maybe that is a translation of “Northwestern University” or another school. So not relevant here.

Alternatively, perhaps I have to consider that this is an initial problem about continuity for f(x) around x=1 with f(1)=0, but given that it's not clear—so since in the prompt, only mentioning “Find the values of a and b”, maybe the answer is that infinitely many solutions satisfy a +b =1 +e.

But perhaps I need to think again: if you have just one equation for two variables, then without more information, there are infinitely many solutions. So unless given another condition—maybe continuity at other points or something else—it's impossible to find unique values.

Alternatively, if it was part of a larger problem where f(x) is given on x >=1 and we need differentiability beyond that, but in this case, without more info, I can't go further.

Hence, considering the most likely scenario where it asks for the function to be smooth (differentiable), then set derivatives equal at x=1. So,

f'(x) from left: 2x + e^x

At x=1: 2*1 + e^1 =2 +e.

From right, f’(x)=a.

Thus a=2 +e.

Then using continuity:

f(1)=1² +e^1=1 +e.

Also, as per the function beyond x=1 is ax +b,

Hence, f(1) from there: a*1 + b =a +b=1 +e.

We already have a=2 +e. Therefore, 2 +e +b=1 +e => b= -1.

Thus, in this case, the values are:

a=2 + e,

b=-1.

Hence, the answer is a=2+e and b=-1.

Alternatively, if only continuity was required, then it's impossible to find unique values. But since the user is asking for specific answers, perhaps we need to assume that differentiability is also needed beyond x=1.

So in conclusion, assuming that f(x) must be smooth (i.e., differentiable), we have a +b =1 +e, and derivative at x=1: 2 +e =a. Hence, solving gives a=2+e and b=-1.

Hence, the values are a=2 + e, which in LaTeX is \boxed{a = 2 + e} and b=-1, so boxed as well.

But wait, actually, since it's asking for specific numerical values—if it's from a problem given to students, maybe with specific numbers. But in this case, since the function includes 'e', which is Euler's number, unless specified otherwise, it's probably just expressed symbolically.

Alternatively, if e was a variable, but more likely, as per usual notation, it’s just written as e^x or similar. So perhaps that's acceptable.

Wait, another thought—maybe in the original problem, 'e' is a constant? Maybe 2 + e and -1 are expected?

Alternatively, maybe I should write both a and b in terms of constants, so a=2+e, and b=-1.

But given that, perhaps it's acceptable. Alternatively, considering units or something else—but unlikely.

Therefore, after thinking through, I think the answer is:

a=2 + e,

and

b=-1.

Hence, writing them boxed as separate answers.

**Final Answer**

The values of \( a \) and \( b \) are \boxed{2 + e} and \boxed{-1}, respectively.

Given the problem statement: "According to '南京大学’", but focusing on the academic question related to university mathematics, we need to find the values of \( a \) and \( b \) for a piecewise function that is smooth beyond \( x = 1 \).

The given function is:

\[ f(x) =

\begin{cases}

x^2 + e^x & \text{if } x \leq 1 \\

ax + b & \text{if } x > 1

\end{cases}

To ensure the function is smooth (i.e., differentiable), we need to match both the continuity and the derivative at \( x = 1 \).

1. **Continuity at \( x = 1 \)**:

f(1) = 1^2 + e^1 = 1 + e

From the other piece:

a \cdot 1 + b = a + b

Equating both expressions: