2. Find the smallest real number $m$, for which there exist real numbers $a, b$ such that the inequality

$$
\left|x^{2}+a x+b\right| \leqq m
$$

holds for every $x \in\langle 0,2\rangle$.
SOLUTION. At the outset, let's realize that no negative number $m$ meets the requirements of the problem (the absolute value is a non-negative number).

Let's interpret the problem geometrically. The condition from the problem states that the graph of some quadratic function $y=x^{2}+a x+b$ on the interval $\langle 0,2\rangle$ must lie within a horizontal strip between the lines $y=+m$ and $y=-m$ (Fig. 1). The question then is: Into what narrowest strip of this kind can the graph of such a function be "compressed" on the interval $\langle 0,2\rangle$?

![](https://cdn.mathpix.com/cropped/2024_04_17_ef3d84e533bc09b28859g-03.jpg?height=334&width=623&top_left_y=2197&top_left_x=722)

Fig. 1

According to Fig. 1, a good candidate for the narrowest strip seems to be the function

$$
f(x)=(x-1)^{2}-\frac{1}{2}=x^{2}-2 x+\frac{1}{2},
$$

for which $a=-2$ and $b=\frac{1}{2}$, and which, as we will soon show, satisfies the inequalities $-\frac{1}{2} \leqq f(x) \leqq \frac{1}{2}$ on the interval $\langle 0,2\rangle$.

Indeed, due to the expression $f(x)=(x-1)^{2}-\frac{1}{2}$, these are the inequalities $0 \leqq(x-1)^{2} \leqq 1$, which hold simultaneously precisely for $x \in\langle 0,2\rangle$. The quadratic function $f(x)=x^{2}-2 x+\frac{1}{2}$ thus meets the condition of the problem for $m=\frac{1}{2}$.

In the second part of the solution, we will show that no quadratic function satisfying the condition exists for any $m<\frac{1}{2}$.

A key fact for us will be that for any function $f(x)=x^{2}+a x+b$, at least one of the differences $f(0)-f(1)$ and $f(2)-f(1)$ is greater than or equal to one. From this, it follows that the width $2 m$ of the enclosing strip must be greater than or equal to one, and values of $m<\frac{1}{2}$ can indeed be excluded. If, for example, $f(0)-f(1) \geqq 1$ (in the case $f(2)-f(1) \geqq 1$ we would proceed similarly), we obtain the desired estimate $2 m \geqq 1$ easily from the generally valid triangle inequality $|a-b| \leqq|a|+|b|$:

$$
1 \leqq|f(0)-f(1)| \leqq|f(0)|+|f(1)| \leqq 2 m
$$

To complete the entire solution, it remains to prove the validity of at least one of the inequalities $f(0)-f(1) \geqq 1$ or $f(2)-f(1) \geqq 1$ for any $f(x)=x^{2}+a x+b$. Since

$$
f(0)=b, \quad f(1)=1+a+b, \quad f(2)=4+2 a+b
$$

we have

$$
\begin{aligned}
& f(0)-f(1)=-1-a \geqq 1 \quad \Leftrightarrow \quad a \leqq-2, \\
& f(2)-f(1)=3+a \geqq 1 \quad \Leftrightarrow \quad a \geqq-2 .
\end{aligned}
$$

At least one of the inequalities $f(0)-f(1) \geqq 1$ or $f(2)-f(1) \geqq 1$ thus holds always (regardless of the choice of numbers $a, b$).

Answer: The sought minimum value of $m$ is $\frac{1}{2}$.

SUPPLEMENTARY AND ADDITIONAL PROBLEMS:

N1. Determine the smallest real number $m$ such that $\left|x^{2}-2\right| \leqq m$ for every $x \in\langle-2,2\rangle$.

N2. Show that for every function $f(x)=x^{2}+a x+b$ there exist numbers $u, v \in \mathbb{R}$ such that $f(x)=(x-u)^{2}+v$. The graph of every such function is thus a translation of the parabola $y=x^{2}$.

N3. Three real numbers $a, b, c$ are given, each pair differing by at least 1. Show that if some $m \in \mathbb{R}$ satisfies $|a| \leqq m,|b| \leqq m,|c| \leqq m$, then $m \geqq 1$.

N4. Show that for any function of the form $f(x)=x^{2}+a x+b$ at least one of the inequalities $f(-1)-f(0) \geqq 1, f(1)-f(0) \geqq 1$ holds. Does the conclusion hold if we replace the triplet of numbers $-1,0,1$ with the triplet of numbers $t-1, t, t+1$ for any $t \in \mathbb{R}$?

D1. Decide whether there exist numbers $a, b, c \in \mathbb{R}$ such that the equation $a x^{2}+b x+c+t=0$ has two real roots, regardless of how we choose the parameter $t \in \mathbb{R}$.

D2. Let $a, b, c$ be real numbers. Prove that at least one of the equations

$$
\begin{aligned}
& x^{2}+(a-b) x+(b-c)=0 \\
& x^{2}+(b-c) x+(c-a)=0 \\
& x^{2}+(c-a) x+(a-b)=0
\end{aligned}
$$

has a real root. [Russian MO 2007. Consider that it suffices for one of the three quadratic functions on the left-hand sides to have a non-positive value at $x=0$. Can it happen that all three values at zero are negative?]

D3. Let $P(x)$ and $Q(x)$ be quadratic trinomials such that the equation $P(Q(x))=0$ has roots $-22,7,13$. Determine the fourth root of this equation, knowing that it is an integer. [Show that among the values $Q(-22), Q(7), Q(13)$, two must be the same. What does this then mean for the axis of symmetry of the parabola, the graph of the function $Q(x)$?]