2009 AMC 12B 第 15 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

15.

假设 0<r<30 \lt r \lt 3。下面是五个关于 xx 的方程。哪个方程的解 xx 最大?

Assume 0<r<3.0 \lt r \lt 3. Below are five equations for x.x. Which equation has the largest solution x?x?

3(1+r)x=73(1 + r)^x = 7

3(1+r/10)x=73(1 + r/10)^x = 7

3(1+2r)x=73(1 + 2r)^x = 7

3(1+r)x=73(1 + \sqrt{r})^x = 7

3(1+1/r)x=73(1 + 1/r)^x = 7

答案:B
知识点:对数不等式
难度评级:1710
解答:

每个方程都给出 x=log(7/3)log(1+f(r)),x = \dfrac{\log(7/3)}{\log(1 + f(r))},所以正数 f(r)f(r) 越小,解越大。

0<r<3,0 \lt r \lt 3, 时,r10<r<2r\dfrac r{10}\lt r\lt2r,并且因为 r<100.r\lt100.r10<r\dfrac r{10}\lt\sqrt r。又因为 r2<9<10,r^2\lt9\lt10,所以 r10<1r.\dfrac r{10}\lt\dfrac1r. 因此五个量中 r/10r/10 最小,方程 (B) 的解最大。

所以正确答案是 B

Each equation gives x=log(7/3)log(1+f(r)),x = \dfrac{\log(7/3)}{\log(1 + f(r))}, which is largest when the positive quantity f(r)f(r) is smallest.

For 0<r<3,0 \lt r \lt 3, we have r10<r<2r\dfrac r{10}\lt r\lt2r and r10<r\dfrac r{10}\lt\sqrt r because r<100.r\lt100. Also r2<9<10,r^2\lt9\lt10, so r10<1r.\dfrac r{10}\lt\dfrac1r. Thus r/10r/10 is the smallest of the five quantities, and equation (B) has the largest solution.

Thus, the correct answer is B.

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