考點(diǎn):利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性
專題:導(dǎo)數(shù)的綜合應(yīng)用
分析:根據(jù)函數(shù)f(x)為偶函數(shù),且其導(dǎo)數(shù)在[0,
]上大于0恒成立,可知f(x)在[-
,
]上的單調(diào)性,然后結(jié)合給出的四個(gè)條件逐一進(jìn)行判斷.
解答:
解:函數(shù)f(x)=x
2-cosx為偶函數(shù),f′(x)=2x+sinx,
當(dāng)0<x≤
時(shí),0<sinx≤1,0<2x≤π,
∴f′(x)>0,函數(shù)f(x)在[0,
]上為單調(diào)增函數(shù),
由偶函數(shù)性質(zhì)知函數(shù)在[-
,0]上為減函數(shù).
當(dāng)x
12>x
22時(shí),得|x
1|>|x
2|≥0,
∴f(|x
1|)>f(|x
2|),由函數(shù)f(x)在[-
,
]上為偶函數(shù)得f(x
1)>f(x
2),故①②成立;
當(dāng)
x1,x2∈[-,0]時(shí),由cosx
1>cosx
2,得x
1>x
2,此時(shí)f(x
1)<f(x
2),③不正確;
當(dāng)
x1,x2∈[-,0]時(shí),由sinx
1>sinx
2,得x
1>x
2,此時(shí)f(x
1)<f(x
2),④不正確.
∴能使f(x
1)>f(x
2)恒成立的條件序號(hào)是①②.
故選:B.
點(diǎn)評(píng):本題考查了利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性,主要考查導(dǎo)函數(shù)的正負(fù)與原函數(shù)的單調(diào)性之間的關(guān)系,即當(dāng)導(dǎo)函數(shù)大于0時(shí)原函數(shù)單調(diào)遞增,當(dāng)導(dǎo)函數(shù)小于0時(shí)原函數(shù)單調(diào)遞減,是中檔題.