# Is it possible to determine $P(21)$, if $P(x)$ is a 2nd degree polynomial, $P(11)=151$, and for all $x\in\Bbb R$, $x^2-2x+2\le P(x)\le2x^2-4x+3$?

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.)

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 $$(1,1)$$, $$P(x)=a(x-1)^2+1$$.

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

• How does one show that if $P(x)$ have a minimum other than $(1,1)$, then the inequality is broken? – joeblack May 13 at 13:11
• 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 $P(x)$. – dbmag9 May 13 at 13:17
• @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. – Thern May 13 at 13:18
• 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! – Student May 13 at 13:42
• @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. – dbmag9 May 13 at 14:39

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?

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 $$P(21)$$.

• 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 $x=2$? – joeblack May 13 at 13:00
• 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. – Thern May 13 at 13:08
• 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. – user10354138 May 13 at 13:08

$$P(x)=ax^2+bx+c$$ 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 $$P(21) = 601$$.

• 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. – David K May 13 at 10:18
• Now it is corrected. – Tarzan May 14 at 8:40