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11. 设 $f(x)$ 是定义域为 $R$ 的偶函数, 且在 $(0,+\infty)$ 单调递减, 则 $(\quad)$

Gaokao · Math · previous-year question

  1. A.$f\left(\log _{5} \frac{1}{4}\right)>f\left(2^{-\frac{3}{2}}\right)>f\left(2^{-\frac{2}{3}}\right)$
  2. B.$f\left(\log _{8} \frac{1}{4}\right)>f\left(2^{-\frac{2}{3}}\right)>f\left(2^{-\frac{3}{2}}\right)$
  3. C.$f\left(2^{-\frac{3}{2}}\right)>f\left(2^{-\frac{2}{3}}\right)>f\left(\log _{5} \frac{1}{4}\right)$correct
  4. D.$f\left(2^{-\frac{2}{3}}\right)>f\left(2^{-\frac{3}{2}}\right)>f\left(\log _{5} \frac{1}{4}\right)$

Answer

C. $f\left(2^{-\frac{3}{2}}\right)>f\left(2^{-\frac{2}{3}}\right)>f\left(\log _{5} \frac{1}{4}\right)$

Explanation

【详解】 $\because f(x)$ 是 $\mathrm{R}$ 的偶函数, $\therefore f\left(\log _{3} \frac{1}{4}\right)=f\left(\log _{3} 4\right)$. $\therefore \log _{3} 4>1=2^{0}>2^{-\frac{3}{2}}$, 又 $f(x)$ 在 $(0,+\infty)$ 单调递减, $f\left(\log _{3} 4\right)<f\left(2^{-\frac{2}{3}}\right)<f\left(2^{-\frac{3}{2}}\right)$, $\therefore f\left(2^{-\frac{3}{2}}\right)>f\left(2^{-\frac{2}{3}}\right)>f\left(\log _{3} \frac{1}{4}\right)$, 故选 C.

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