D
分析:(1)若f(x)的圖象關(guān)于原點(diǎn)對(duì)稱則f(0)=0,因?yàn)閒(0)=1所以(1)錯(cuò).
(2)函數(shù)f(|x|)=f(|-x|)所以函數(shù)為偶函數(shù),當(dāng)x≥0時(shí)y=x
2-2ax+1此時(shí)其對(duì)稱軸為x=a.所以(2)錯(cuò).
(3)(4)根據(jù)函數(shù)的圖象可得若y=f(x)的圖象與直線y=2有兩個(gè)不同交點(diǎn),則a=1,若f(x)在R上是增函數(shù),則a≤0所以(3)(4)正確.
解答:(1)當(dāng)a=0時(shí),函數(shù)f(x)=x|x|+1由題得x∈R所以x=0有意義,所以所以當(dāng)a=0時(shí),f(x)的圖象不關(guān)于原點(diǎn)對(duì)稱.(1)錯(cuò).
(2)f(|x|)=x
2-2a|x|+1,所以函數(shù)f(|x|)=f(|-x|)所以函數(shù)為偶函數(shù).當(dāng)x≥0時(shí)y=x
2-2ax+1此時(shí)其對(duì)稱軸為x=a,當(dāng)a≤0時(shí)函數(shù)的最小值為1,由函數(shù)是偶函數(shù)得當(dāng)x≥0時(shí)函數(shù)的最小值也是1,所以f(|x|)有最小值1-a
2是錯(cuò)誤的.(2)錯(cuò).
(3)由題意得y=f(x)=
,
當(dāng)a<0時(shí)與a≥0時(shí)函數(shù)的圖象分別為
所以若y=f(x)的圖象與直線y=2有兩個(gè)不同交點(diǎn),則a=1.(3)正確.
(4)由(3)可得當(dāng)a≤0時(shí)函數(shù)函數(shù)f(x)=x|x|-2ax+1是增函數(shù).(4)正確.
故選D.
點(diǎn)評(píng):本題主要考查二次函數(shù)的奇偶性,單調(diào)性等性質(zhì),解決此類問題的方法是根據(jù)函數(shù)的解析式畫出函數(shù)的圖象,運(yùn)用數(shù)形結(jié)合的數(shù)學(xué)思想解決問題.