1978 AMC 12 第 17 题

Question

若

k

为正数，且函数

f

满足：对于每个正数

x

，

\left[f(x^2+1)\right]^{\sqrt x}=k\text{，}

那么对于每个正数

y

，

\left[f\left(\frac{9+y^2}{y^2}\right)\right]^{\sqrt{\frac{12}{y}}}

等于

If

k

is a positive number and

f

is a function such that, for every positive number

x,

\left[f(x^2+1)\right]^{\sqrt x}=k,

then, for every positive number

y,

\left[f\left(\frac{9+y^2}{y^2}\right)\right]^{\sqrt{\frac{12}{y}}}

is equal to

\sqrt k

2k

k\sqrt k

k^2

y\sqrt k

选取

x

，使

x^2+1=\frac{9+y^2}{y^2}

Choose

x

so that

x^2+1=\frac{9+y^2}{y^2}

代入后比较

\sqrt{\frac{12}{y}}

与

\sqrt x

Compare

\sqrt{\frac{12}{y}}

with

\sqrt x

after making the substitution

令

x=\frac{3}{y}

，它是正数。于是

\begin{aligned} x^2+1&=\frac{9+y^2}{y^2},\\ \sqrt{\frac{12}{y}}&=2\sqrt x\text{。} \end{aligned}

因此所求表达式为

\left(\left[f(x^2+1)\right]^{\sqrt x}\right)^2=k^2\text{。}

因此，正确答案是 D。

Set

x=\frac{3}{y},

which is positive. Then

\begin{aligned} x^2+1&=\frac{9+y^2}{y^2},\\ \sqrt{\frac{12}{y}}&=2\sqrt x. \end{aligned}

Therefore the requested expression is

\left(\left[f(x^2+1)\right]^{\sqrt x}\right)^2=k^2.

Therefore, the correct answer is D.

其他年份的第 17 题