Apologize problem for Comrade Otto B

Algebra Level 5

Suppose a quadratic function $ax^2+bx+c$ where $a,b,c \in \mathbb{R}$ and $a\neq 0$ , satisfies the following conditions:

1: When $x\in \mathbb{R}$ , $f(x-4)=f(2-x)$ and $f(x)\geq x$ .

2: When $x \in (0,2)$ , $f(x) \leq \left(\dfrac{x+1}{2}\right)^2$ .

3: The minimum value of $f(x)$ on $\mathbb{R}$ is 0.

Find the maximum value of $m$ ( $m>1$ ) such that there exists $t\in \mathbb{R}$ such that $f(x+t) \leq x$ holds so long as $x\in [1,m]$ .

The answer is 9.0.

1 solution

Otto Bretscher
Feb 27, 2016

No need to apologize at all (for what??), but I appreciate and enjoy the lovely problem, Comrade!

I thought about this problem largely in terms of the graphs of $f(x)$ and $y=x$ .

With all the information we have, we can see that $f(1)=1,f'(1)=1$ and $f'(-1)=0$ , so that $f(x)=\frac{1}{4}x^2+\frac{1}{2}x+\frac{1}{4}$ . Note that $f(x)\leq 1$ on the interval $[-3,1]$ .

If we make $t<-4$ or $t>0$ , then $f(1+t)>1$ , violating the condition $f(x+t)\leq x$ for $x=1$ .

If we make $-4\leq t\leq 0$ , then the larger solution of $f(x+t)=x$ is $(\sqrt{-t}+1)^2\leq 9$ . The maximum value of $\boxed{9}$ is attained when $t=-4$ .

Again, this all becomes much clearer if you take a look at the graphs: We are shifting the graph of $f(x)$ four units to the right.

I apologize you from the "obnoxious guys" that caused you trouble.

Anyways , a great standard solution. Yes , Graphical way is the best approach for this problem.

Nihar Mahajan - 5 years, 3 months ago

May I ask where did you get this question from?

Nihar: From a Chinese Mathematical Constest Book.

Pi Han Goh - 5 years, 3 months ago

Apologize problem for Comrade Otto B

The answer is 9.0.

1 solution

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