6. Let $P(x)$ be a quadratic polynomial with real coefficients such that $P(11)=181$ and $x^{2}-2 x+2 \leq P(x) \leq 2 x^{2}-4 x+3$ for any real number $x$. Find $P(21)$.
(1 mark)
設 $P(x)$ 為二次多項式, 其系數皆為實數。已知 $P(11)=181$, 且對任意實數 $x$ 皆有 $x^{2}-2 x+2 \leq P(x) \leq 2 x^{2}-4 x+3$, 求 $P(21)$ 。
(1 分)
6. 721
6. Note that $(x-1)^{2}+1 \leq P(x) \leq 2(x-1)^{2}+1$ for any real number $x$. Hence we must have $P(x)=a(x-1)^{2}+1$ for some $1 \leq a \leq 2$. Since $P(11)=181$, we have $181=a(11-1)^{2}+1$, or $a=\frac{9}{5}$. It follows that $P(21)=\left(\frac{9}{5}\right)(21-1)^{2}+1=721$.