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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! 

yasbib555  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
 #1
avatar+3190 
0

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

 

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

Rom  Oct 12, 2018
 #2
avatar+217 
0

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

yasbib555  Oct 12, 2018
 #3
avatar+316 
0

And for b we do not know anything?

Dimitristhym  Oct 13, 2018
 #5
avatar+94110 
0

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

Melody  Oct 13, 2018
 #4
avatar+94110 
+1

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. 

 

f(0)=0

f(1)=3a+2b+1

 

 

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

 

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

 

Melody  Oct 13, 2018
 #6
avatar+94110 
0

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
 #7
avatar+217 
0

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

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