We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

1. Given f(x)=3ax^2+2bx+c with a≠0, when 0≤x≤1 |f(x)|≤1, please determine the maximum value of a. 


2. Given

- m is the greater root of the equation (1999x)^2-1998x2000x-1=0 

- n is the lesser root of the equation x^2+1998x-1999=0, 

please determine the value of m-n! 

 Oct 12, 2018
edited by yasbib555  Oct 12, 2018
edited by yasbib555  Oct 12, 2018
edited by yasbib555  Oct 13, 2018
edited by yasbib555  Oct 13, 2018

do you mean f(x) = 3a2 + 2bx + x


or do you mean f(x) = 3a2 + 2bx + x2 ?

 Oct 12, 2018

sorry, i meant 3ax^2+2bx+c

yasbib555  Oct 12, 2018

And for b we do not know anything?

Dimitristhym  Oct 13, 2018

Rom would have answered you long ago if you had proofed your question properly in the first place.

Melody  Oct 13, 2018

1. Given f(x)=3ax^2+2bx+x with a≠0, when 0≤x≤1 |f(x)|≤1, please determine the maximum value of a. 






\(-1\le3a+2b+1\qquad and \qquad 3a+2b+1\le+1\\ -2\le3a+2b\qquad and \qquad 3a+2b\le0\\ -2-2b\le3a\qquad and \qquad 3a\le-2b\\ \frac{-2-2b}{3}\le a\qquad and \qquad \qquad a\le\frac{-2b}{3}\\ \frac{-2b}{3}-\frac{2}{3}\le a\qquad and \qquad \qquad a\le\frac{-2b}{3}\\ \frac{-2b}{3}-\frac{2}{3}\le a\le\frac{-2b}{3}\\ \)


I want the biggest 'a' so it follows that  

\(a=\frac{-2b}{3}\\ b=-\frac{3a}{2}\)

It also follows that a is positive and b is negative


so the question becomes


\( f(x)=3ax^2+2bx+x \\  f(x)=3ax^2+2*\frac{-3a}{2}x+x \\  f(x)=3ax^2-3ax+x \\ or\\  f(x)=x(3ax-3a+1) \)


with a≠0, when 0≤x≤1 |f(x)|≤1, please determine the maximum value of a. 


and f(1)= 3a+2b+1 = 3a-3a+1 = 1


So f(x) is a concave up parabola passing through (0,0) and (1,1) 


\(f'(x)=6ax-3a+1\\ stat\;pt \;is\;when\; f'(x)=0\\ 6ax-3a+1=0\\ 6ax=3a-1\\ x=\frac{3a-1}{6a}\\ or\\ x=\frac{1}{2}-\frac{1}{6a}\\ \)


Now a is positive and as big as possible so I am going to assume that an 'a' exists such that

the stationary point will lie between    x=0 and x=0.5  

This means that the stationary point (minimim) lies in the designated region of the graph.

So the minimum value must be greater than or equal to -1   (we already know it is less than 1)


The minimum value of f(x) is


\(f(\frac{3a-1}{6a})=\frac{3a-1}{6a}[(3a*\frac{3a-1}{6a})-3a+1]\ge-1\\ \frac{3a-1}{6a}[(\frac{3a-1-6a+2}{2})]\ge-1\\ remembering \;that\;a>0\\ (3a-1)(3a-1-6a+2)\ge-12a\\ (3a-1)(-3a+1)\ge-12a\\ (3a-1)(3a-1)\le12a\\ 9a^2-6a+1\le12a\\ 9a^2-18a+1\le0\\ \text{If this was set to y it would be a concave up parabola so} \\\text{ the function will be less than zero between the roots. } \\\text{Find the roots.}\\ a=\frac{18\pm\sqrt{324-36}}{18}\\ a=1\pm\frac{2\sqrt{2}}{3}\\ \text{So the biggest value of a is}\\ a=1+\frac{2\sqrt{2}}{3}\\ a\approx 1.9428 \)


Here is the graph


 Oct 13, 2018

You know Yasbib555, I spent hours on this question, it would be nice to get a response from you.

I know that you have seen my answer already.

Melody  Oct 13, 2018

Omg im so sorry, things have been really busy with school and curricular activities but i know that's no excuse. Thank you so much! I really do appreciate it, you've helped me a lot. I am sorry if I did p**s you off in some way, I didn't mean it. 

yasbib555  Oct 19, 2018

12 Online Users