Let f(x) = ax^2+bx+c be such that |f(x)| \le 1 \ \forall \ x \in \[-1,1]
Prove that |2ax+b| \le 4 \ \forall \ x \in [-1,1]
This topic is current in mathlinks.ro, and in any case no borrowed solutions please.
39lol
and sir its a famous question
also in many old posts of mathlinks
62I can think of 2 different cases and two graphs in each...
BUt I guess this is better left for the students!
341I am familiar with the question. I have seen many solutions with some nice manipulations to get the solution. I am looking to see if someone can get a more JEE accessible solution [in fact a problem of this kind had come in an old JEE paper (perhaps soon after I was born. That old :D] and preceded by an IMO shortlist problem.
Nishant sir has indicated the direction.
11Well I did not need graphs for this.....
Beleive me - a funny thing happened 2day, while attempting this - i shall disclose that later - time for some work now...
f(1)+f(-1)=2(a+c)....(1)
|2ax+b|\le |2a+b|\\ \ |2a+b|=|f(1)+f(-1)-2f(0)+\frac{f(1)-f(-1)}{2}|
Now triangles inequality, and put max value for each i.e
\boxed{f(1)=f(0)=f(-1)=1}
Done!