解:(I)由
得
即
又
,∴
(Ⅱ)解法一:由(I)得,
依題意,
又
,故ω=3,∴
函數(shù)f(x)的圖象向左平移m個(gè)單位后所對(duì)應(yīng)的函數(shù)為
g(x)是偶函數(shù)當(dāng)且僅當(dāng)
即
從而,最小正實(shí)數(shù)
解法二:由(I)得,
,依題意,
又
,故ω=3,∴
函數(shù)f(x)的圖象向左平移m個(gè)單位后所對(duì)應(yīng)的函數(shù)為
,g(x)是偶函數(shù)當(dāng)且僅當(dāng)g(-x)=g(x)對(duì)x∈R恒成立
亦即
對(duì)x∈R恒成立.∴
=
即
對(duì)x∈R恒成立.∴
故
∴
從而,最小正實(shí)數(shù)
分析:(I)利用特殊角的三角函數(shù)值化簡(jiǎn)
,根據(jù)
直接求出φ的值;
(Ⅱ)解法一:在(I)的條件下,若函數(shù)f(x)的圖象的相鄰兩條對(duì)稱軸之間的距離等于
,求出周期,求出ω,得到函數(shù)f(x)的解析式;函數(shù)f(x)的圖象向左平移m個(gè)單位所對(duì)應(yīng)的函數(shù)是偶函數(shù).推出
,可求最小正實(shí)數(shù)m.
解法二:在(I)的條件下,若函數(shù)f(x)的圖象的相鄰兩條對(duì)稱軸之間的距離等于
,求出周期,求出ω,得到函數(shù)f(x)的解析式;利用g(x)是偶函數(shù)當(dāng)且僅當(dāng)g(-x)=g(x)對(duì)x∈R恒成立,使得函數(shù)f(x)的圖象向左平移m個(gè)單位所對(duì)應(yīng)的函數(shù)是偶函數(shù).化簡(jiǎn)
,然后再求最小正實(shí)數(shù)m.
點(diǎn)評(píng):本題是中檔題,考查三角函數(shù)的字母變量的求法,三角函數(shù)的圖象的平移,偶函數(shù)的性質(zhì),轉(zhuǎn)化思想的應(yīng)用,考查計(jì)算能力,是常考題.