1. Given the function f (x) = ln (x + 1) + ax ^ 2-x, a ∈ R (1) When a = 1 / 4, find the extremum of function y = f (x); (2) Is there a real number B ∈ (1,2) such that when x ∈ (- 1, b], the maximum value of function f (x) is f (b)? If there is a range of real number a, if not, please explain the reason I'd like to ask teachers and students to help me solve some problems. We have a holiday these days. First, I'll take a good look at the 20 and 21 questions in my exams. There are more than 70 questions. If I do them one by one, I don't have enough time to take a day off. So I want to take a serious look at the solutions. I hope the process can be more detailed,

1. Given the function f (x) = ln (x + 1) + ax ^ 2-x, a ∈ R (1) When a = 1 / 4, find the extremum of function y = f (x); (2) Is there a real number B ∈ (1,2) such that when x ∈ (- 1, b], the maximum value of function f (x) is f (b)? If there is a range of real number a, if not, please explain the reason I'd like to ask teachers and students to help me solve some problems. We have a holiday these days. First, I'll take a good look at the 20 and 21 questions in my exams. There are more than 70 questions. If I do them one by one, I don't have enough time to take a day off. So I want to take a serious look at the solutions. I hope the process can be more detailed,

So in 0 is the maximum and in 1 is the minimum, the second problem is a ratio of classification calculation. Through the boundary, the two poles are defined as a & gt; = 0.5 f (0) min, f (the point solved in the graph, denoted as n) max, - 1 & lt; n & lt; 00 & lt; a & lt; 0.5, & nbsp; f (0) max & nbsp; f (n) min, n & gt; 0-0.5 & lt; a & lt; 0 & nbsp; & nbsp; f (0) max & nbsp; f (n) max n & lt; - 2, which is impossible, so there is no a = & lt; N pole; -When 0.5 & nbsp; f (0) max & nbsp; f (n) max - 1 & lt; n & lt; 0A = 0, as a special case, we can see that the maximum can only be taken at 0, not ∈ (1,2). In the first case, we must take x = n, then B = n, n ∈ (1,2) or x = B, we should also satisfy f (b) & gt; = f (n), then f (2) & gt; = f (n), we can solve a & gt; = 1 / 6. In the second case, we must take x = 0, then B = 0, not conforming to the meaning of the problem or x = B, we should also satisfy f (b) & gt; = f (n), then f (2) & gt; =In the last case, X can not go to B to get the maximum value, X can only take the extreme value, and the extreme value is not ∈ (1,2). To sum up, a & gt; = 1 / 6, I'm a senior. Ha ha, today I'm addicted to high school life. I'm going to study microelectronics, and I'm going to study nano in the United States, When you go to college, you will find that college is not as full as high school