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Algebra

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10 Jan 2009 20:30:22 IST

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A problem concerning a quadratic trinomial

None

Suppose a quadratic function $f(x) = ax^2 + bx + c$ where $a,\, b,\, c \in \mathbb{ R}$ and $a\neq 0$ , satisfies the following

conditions:

(i) $f(x-4)=f(2-x)$ and $f(x)\geq x\ \forall \, x\in \mathbb{R}$

(ii) $f(x)\leq \left(\dfrac{x+1}{2}\right)^2\ \forall\, x\in (0,\,2)$

(iii)The minimum value of $f(x)$ in $\mathbb{R}$ is 0.

Find the maximal $m$ (where $m>1$ ) such that there exists some $t\in \mathbb{R}$ such that $f(x+t)\leq x$ for all $x\in [1,\,m]$ .

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Comments (4)

aqswdefr

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Joined: 11 Oct 2008

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11 Jan 2009 02:44:25 IST

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f(x) = ( x+ b/2a)^2 !!!

thes can hlep or not ???

Anant Kumar

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11 Jan 2009 09:05:19 IST

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And how exactly are they going to help?

Anant Kumar

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13 Jan 2009 13:18:38 IST

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No one to take it up? Or is it too difficult?

akki ~~ unlucky forever ~~

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13 Jan 2009 16:13:42 IST

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from (ii) $f(x)\leq \left(\dfrac{x+1}{2}\right)^2\ \forall\, x\in (0,\,2)$ put x=1, we get $F(1)\le1$

from (i), $F(x)\ge\ x$ , put x=1, we get $F(1)\ge\ 1$

so, F(1)=1 => a+b+c=1

from(i), we can say, F(-3)=F(1)=1 => 9a-3b+c=1

similarly we can see from (i) that F(x) is symmetrical to x+1=0, & a must be +ve

as $a,\, b,\, c \in \mathbb{ R}$ , & min. value is 0, so , F(-1)=0 => a-b+c = 0

solving all 3 equn. $F(x)=(\frac{x+1}{2})^2$

now,

F(x+t)<=x

=> $(x+t-1)^2\le0$

=> x=1-t

as $1\le\ x\le\ m$

=> t<0

so, m > 1-t

hence no such maxima exist for m

do correct me if nything is wrong

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