Question

当 $x$ 为实数时，求下式的最小可能值： $\begin{aligned} &(x+1)(x+2)(x+3)(x+4)\\ &\quad+2019, \end{aligned}$

What is the least possible value of $\begin{aligned} &(x+1)(x+2)(x+3)(x+4)\\ &\quad+2019, \end{aligned}$ where $x$ is a real number?

$2017$

$2018$

$2019$

$2020$

$2021$

先把前两个因子和后两个因子相乘，得到 $(x^2 + 5x + 4)(x^2 + 5x + 6).$

注意这两项相差 $2$ 。可把它写成平方差： $(x^2 + 5x + 5)^2 - 1.$

再加上 $2019$ ，得到 $(x^2 + 5x + 5)^2 + 2018.$

平方数非负，所以只要能让内部表达式为 $0$ ，就能让平方项为 $0$ 。

判别式为 $5^2 - 4 \cdot 5 = 5$ ，为正，因此确实存在使平方项为 $0$ 的取值。

所以正确答案是 B。 $0^2 + 2018 = 2018.$

Multiplying the first two terms and the last terms yields $(x^2 + 5x + 4)(x^2 + 5x + 6).$

Note that these two terms differ by $2.$ We can try to express this as a difference of squares, which is $(x^2 + 5x + 5)^2 - 1.$

Adding $2019$ to this gets us $(x^2 + 5x + 5)^2 + 2018.$

Squares are non-negative, so as long as we find a way to make the inner expression $0,$ we can make the square $0.$

The discriminant is $5^2 - 4 \cdot 5 = 5,$ which is positive meaning that there is a value that makes the square $0.$

This means that the minimum value would be $0^2 + 2018 = 2018.$ Thus, B is the correct answer.