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# pls help ASAP, I appreciate it so much, you don't even know!! tysm <33

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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!

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

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

Oct 12, 2018
#2
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sorry, i meant 3ax^2+2bx+c

yasbib555  Oct 12, 2018
#3
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And for b we do not know anything?

Dimitristhym  Oct 13, 2018
#5
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Rom would have answered you long ago if you had proofed your question properly in the first place.

Melody  Oct 13, 2018
#4
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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 Oct 13, 2018
#6
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You know Yasbib555, I spent hours on this question, it would be nice to get a response from you.