若函數(shù)f(x)定義域?yàn)镽,滿足對(duì)任意x1,x2∈R,有f(x1+x2)≤f(x1)+f(x2),則稱f(x)為“V形函數(shù)”;若函數(shù)g(x)定義域?yàn)镽,g(x)恒大于0,且對(duì)任意x1,x2∈R,有l(wèi)gg(x1+x2)≤lgg(x1)+lgg(x2),則稱g(x)為“對(duì)數(shù)V形函數(shù)”.
(1)當(dāng)f(x)=x2時(shí),判斷f(x)是否為V形函數(shù),并說明理由;
(2)當(dāng)g(x)=x2+2時(shí),證明:g(x)是對(duì)數(shù)V形函數(shù);
(3)若f(x)是V形函數(shù),且滿足對(duì)任意x∈R,有f(x)≥2,問f(x)是否為對(duì)數(shù)V形函數(shù)?證明你的結(jié)論.
【答案】
分析:(1)由f(x
1+x
2)-[f(x
1)+f(x
2)]=(x
1+x
2)
2-(
+
)=2x
1x
2,可得2x
1x
2符號(hào)不定,從而可得結(jié)論;
(2)利用反證法證明.假設(shè)對(duì)任意x
1,x
2∈R,有l(wèi)gg(x
1+x
2)≤lgg(x
1)+lgg(x
2),則可得(x
1+x
2)
2+2≤(x
12+2)(x
22+2),即證x
12x
22+(x
1-x
2)
2+2≥0,顯然成立;
(3)f(x)是對(duì)數(shù)V形函數(shù),根據(jù)f(x)是V形函數(shù),利用對(duì)任意x∈R,有f(x)≥2,證明f(x
1+x
2)≤f(x
1)f(x
2),從而可得f(x)是對(duì)數(shù)V形函數(shù).
解答:(1)解:f(x
1+x
2)-[f(x
1)+f(x
2)]=(x
1+x
2)
2-(
+
)=2x
1x
2∵x
1,x
2∈R,∴2x
1x
2符號(hào)不定,∴當(dāng)2x
1x
2≤0時(shí),f(x)是V形函數(shù);當(dāng)2x
1x
2>0時(shí),f(x)不是V形函數(shù);
(2)證明:假設(shè)對(duì)任意x
1,x
2∈R,有l(wèi)gg(x
1+x
2)≤lgg(x
1)+lgg(x
2),
則lgg(x
1+x
2)-lgg(x
1)-lgg(x
2)=lg[(x
1+x
2)
2+2]-lg(x
12+2)-lg(x
22+2)≤0,
∴(x
1+x
2)
2+2≤(x
12+2)(x
22+2),
∴x
12x
22+(x
1-x
2)
2+2≥0,顯然成立,
∴假設(shè)正確,g(x)是對(duì)數(shù)V形函數(shù);
(3)解:f(x)是對(duì)數(shù)V形函數(shù)
證明:∵f(x)是V形函數(shù),∴對(duì)任意x
1,x
2∈R,有f(x
1+x
2)≤f(x
1)+f(x
2),
∵對(duì)任意x∈R,有f(x)≥2,∴
+
≤1,∴0<f(x
1)+f(x
2)≤f(x
1)f(x
2),
∴f(x
1+x
2)≤f(x
1)f(x
2),
∴l(xiāng)gf(x
1+x
2)≤lgf(x
1)+lgf(x
2),
∴f(x)是對(duì)數(shù)V形函數(shù).
點(diǎn)評(píng):本題考查新定義,考查學(xué)生分析解決問題的能力,解題的關(guān)鍵是正確理解新定義.