1 is too simple, so I won't do it;
Question 2:
Let y=f(x)-x2+x, and f (y) = y
Moreover, because there is only one real number that makes f(x)=x, y is a unique constant, and it is set to n[ if the value range of y=f(x)-x2+x is not unique, that is, y can have multiple values, isn't it a violation of the condition of "only one real number makes f(x)=x" in the title? Note that here n is a fixed value and f(n)=n]
That is to say, for any real number x, there is f (x)-x2+x = n. Of course, when x is equal to real number n, this equation also holds, so substituting x = n into this equation gives:
f(n)-n2+n=n
And because f(n)=n, so:
n-n2+n=n
-n2+n=0
Get two candidate values of n: 0 and 1, and then check which candidate value meets the conditions listed in the question.
When n=0, f(x)-x2+x=n=0.
f(x)=x2-x
There are two real numbers (real number 0 and real number 2) that satisfy f(x)=x, which do not meet the meaning of the question and are excluded.
F (x)-x2+x = n= 1 when n =1.
f(x)=x2-x+ 1
Only one real number (real number 1) satisfies f(x)=x, which is consistent with the meaning of the question.
Therefore, the solution of the original function equation is
F(x)=x2-x+ 1 (like your notation, x2 in this question represents the square of x).
You put forward that you think F (x) = X. I think you habitually regard f(x)-x2+x as a variable Y, so that f(y)=y, that is, F (x) = X. If the range of f(x)-x2+x is all real numbers, it will definitely lead to F (x) = X. However, when you think this is ridiculous, you should subconsciously realize that it can't be a compound variable, but a constant.
In addition, in response to my friend's answer, I think this question 1 and question 2 are two different questions of the same function equation, and the conditions cannot be mixed. The national college entrance examination questions, how serious things are, can't afford to make mistakes.