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I was given the following:

Given that $P(x)$ is a second degree polynomial, and that $P(11)=151$, and that $$\forall x \in \mathbb{R}, \,\, x^2-2x+2 \le P(x) \le 2x^2-4x+3$$ determine $P(21)$.

Where do I even start from?


Is there a source (online or a book) where I can find similar testing questions on polynomials? (Testing not in the sense that Putnam-like questions, but in the sense that they are somewhat different than those found in textbooks.)

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The graph helps conceptualise why this is sufficient information; the bounding quadratics given share a minimum at $(1,1)$. For the inequality to hold, $P(x)$ must also have its minimum at $(1,1)$, and there is only one parabola passing through $(11,151)$ with that minimum.

Since the minimum is at $(1,1)$, $P(x)=a(x-1)^2+1$.

$$P(11)=151$$ $$100a+1=151$$ $$a=1.5$$

Two parabolas and the point (11,151).

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  • $\begingroup$ How does one show that if $P(x)$ have a minimum other than $(1,1)$, then the inequality is broken? $\endgroup$ – joeblack May 13 at 13:11
  • $\begingroup$ Call the bounding polynomials $A(x)$ and $B(x)$. We know that $A(1)\leq P(1) \leq B(1)$ so $P(1)=1$. $(1,1)$ is the minimum of $A(x)$ so $A(x)>1$ everywhere except $x=1$. But then $P(x)\geq A(x)>1$ everywhere except $x=1$, so $(1,1)$ is the minimum of $P(x)$. $\endgroup$ – dbmag9 May 13 at 13:17
  • $\begingroup$ @joeblack You can write any parabola with minimum at $(m,n)$ as $P(x) = a(x-m)^2 + n$ (with $a>0$). The term $(x-m)^2$ is never smaller than $0$, so $x=m$ must be the minimum, and $P(m) = a\cdot 0 + n = n$. But then, $P(x)$ only has one variable parameter left: $a$. So defining a single other point fixes the parabola. $\endgroup$ – Thern May 13 at 13:18
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    $\begingroup$ This is a valuable lesson: if you can draw it, it's usefull to do this. You immediately see that the function goes through (1,1). Nice answer! $\endgroup$ – Student May 13 at 13:42
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    $\begingroup$ @joeblack It needs to pass through $(1,1)$ because $A(x)$ and $B(x)$ go through that point (if $1\leq P(x) \leq 1$ then $P(x)=1$). Given that it passes through $(1,1)$, that must be its minimum because $A(x)$ has its minimum there. $\endgroup$ – dbmag9 May 13 at 14:39
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We are given that $(1-x)^{2}+1 \leq P(x) \leq 2(1-x)^{2}+1$. Let $Q(x)=P(x)-1-(1-x)^{2}$. Then $0 \leq Q(x) \leq (1-x)^{2}$ and $Q$ is also a polynomial of degree $2$. This implies that $Q(x)=c(1-x)^{2}$ for some constant $c$. Can you finish from here?

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Note that the given condition can be written as $$ (x-1)^2\leq P(x)-1\leq 2(x-1)^2 $$ So this automatically forces $P(x)-1$ to have a repeated root at $x=1$, thus $P(x)=\lambda (x-1)^2+1$ (since $\deg P=2$) for some $1\leq\lambda\leq 2$. Now use $P(11)$ to determine $\lambda$, hence $P(21)$.

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  • $\begingroup$ I am sorry if this is a stupid question: Why does $(x-1)^2\leq P(x)-1\leq 2(x-1)^2$ forces $P(x)-1$ to have a repeated root at $x=1$? (Or, had it been $(x-2)^2\leq P(x)-1\leq 2(x-2)^2$, would $P(x)$ be forced to have a repeated root at $x=2$? $\endgroup$ – joeblack May 13 at 13:00
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    $\begingroup$ I do have a similar concern as @joeblack. $P(x)$ could be of the form $P(x) = \lambda(x-1)^2 + \kappa(x-1) + 1$ and still fulfill the inequalities. You still have to prove why $\kappa$ is zero. $\endgroup$ – Thern May 13 at 13:08
  • $\begingroup$ Yes. Dividing $(x-1)^2\leq P(x)-1\leq 2(x-1)^2$ by $\lvert x-1\rvert$ and take limit $x\to 1$ gives $0\leq\lvert P'(1)\rvert\leq 0$, so $P'(1)=0$ and $P(1)-1=0$. $P(x)-1$ has.a repeat root. Similar with your $(x-2)$ case. $\endgroup$ – user10354138 May 13 at 13:08
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$$ P(x)=ax^2+bx+c$$ so $$151 = 121a +11b+c$$

$$(x-1)^2+1\leq P(x)\leq 2(x-1)^2+1\implies P(1) = 1$$

so $$ a+b+c=1$$

so $$120a+10b= 150\implies 12a+b=15$$ and $$P(x) = (x-1)(ax+a+b)+1$$

so $$x-1\leq ax+a+b\leq 2(x-1)$$ $$\implies x=1:\;\;\;2a+b=0$$

So $a={3\over 2}$, $b=-3$ and $c={5\over 2}$ so $P(21) = 601$.

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    $\begingroup$ You lost $200a$ between the left and right sides of $440a+20b+1=20\cdot15+1.$ Notice that $(1,1),$ $(11,151),$ and $(21,301)$ are collinear, so you've concluded that $P(x)=15x+1.$ This is incorrect. $\endgroup$ – David K May 13 at 10:18
  • $\begingroup$ Now it is corrected. $\endgroup$ – Tarzan May 14 at 8:40

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