AMC12 2007 B

AMC12 2007 B · Q17

AMC12 2007 B · Q17. It mainly tests Absolute value, Logarithms (rare).

If $a$ is a nonzero integer and $b$ is a positive number such that $ab^2=\log_{10}b$, what is the median of the set $\{0,1,a,b,1/b\}$?

若$a$为非零整数，$b$为正数，且满足 $ab^2=\log_{10}b$，则集合 $\{0,1,a,b,1/b\}$ 的中位数是多少？

(A) 0 0

(B) 1 1

(D) $b$ $b$

(E) $\frac{1}{b}$ $\frac{1}{b}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Note that if $a$ is positive, then, the equation will have no solutions for $b$. This becomes more obvious by noting that at $a=1$, $ab^2 > \log_{10} b$. The LHS quadratic function will increase faster than the RHS logarithmic function, so they will never intersect. This puts $a$ as the smallest in the set since it must be negative. Checking the new equation: $-b^2 = \log_{10}b$ Near $b=0$, $-b^2 > \log_{10} b$ but at $b=1$, $-b^2 < \log_{10} b$ This implies that the solution occurs somewhere in between: $0 < b < 1$ This also implies that $\frac{1}{b} > 1$ This makes our set (ordered) $\{a,0,b,1,1/b\}$ The median is $b \Rightarrow \mathrm {(D)}$

注意如果$a$为正，则方程对$b$没有解。这一点在$a=1$时更明显：此时$ab^2 > \log_{10} b$。左边是二次函数，增长速度快于右边的对数函数，因此它们不会相交。因此$a$必须为负，从而在集合中最小。检验新的方程：$-b^2 = \log_{10}b$ 当$b$接近$0$时，$-b^2 > \log_{10} b$，但当$b=1$时，$-b^2 < \log_{10} b$ 这表明解在两者之间：$0 < b < 1$ 这也意味着$\frac{1}{b} > 1$ 于是集合按从小到大排列为 $\{a,0,b,1,1/b\}$ 中位数是$b \Rightarrow \mathrm {(D)}$

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