Question

17．下列函數(shù)既是奇函數(shù)又是偶函數(shù)的是（　　）

• A． $f（x）=x+\frac{1}{x}$
• B． $f（x）=\frac{1}{x^2}$
• C． $f（x）=\sqrt{{x^2}-1}+\sqrt{1-{x^2}}$
• D． $f（x）=\left\{\begin{array}{l}\frac{1}{2}{x^2}+1，x＞0\\-\frac{1}{2}{x^2}-1，x＜0\end{array}\right.$

Answer 1

分析 根據(jù)奇偶性的定義和函數(shù)定義域必須關于原點對稱判斷即可．

解答 解：對于A：$f（x）=x+\frac{1}{x}$，則f（-x）=$-x-\frac{1}{x}$=-f（x），是奇函數(shù)．
對于B：$f（x）=\frac{1}{x^2}$，則f（-x）=$\frac{1}{（-x）^{2}}=\frac{1}{{x}^{2}}=f（x）$是偶函數(shù)．
對于C：$f（x）=\sqrt{{x^2}-1}+\sqrt{1-{x^2}}$，∵定義域為{-1，1}，則f（-x）=f（x）=0，f（-x）=-f（x）=0，∴既是奇函數(shù)又是偶函數(shù)
對于D：$f（x）=\left\{\begin{array}{l}\frac{1}{2}{x^2}+1，x＞0\\-\frac{1}{2}{x^2}-1，x＜0\end{array}\right.$，則f（-x）=$\left\{\begin{array}{l}{\frac{1}{2}{x}^{2}+1，x＜0}\\{-\frac{1}{2}{x}^{2}-1，x＞0}\end{array}\right.$⇒f（-x）=-f（x）是奇函數(shù)．
故選C．

點評 本題考查了函數(shù)的奇偶性的定義判斷，注意奇偶性判斷的前提條件是函數(shù)定義域必須關于原點對稱．屬于基礎題．