高考數(shù)學(xué)專(zhuān)項(xiàng)研究：導(dǎo)數(shù)（25）三角函數(shù)與導(dǎo)數(shù)壓軸第五章函數(shù)模型第二節(jié)三角函數(shù)與導(dǎo)數(shù)壓軸一、逐段分析(一)人為分段:①以?π2②尋找比邊界值大或者小的值進(jìn)行分段。1.已知函數(shù)fx=sinx(1)f′x在區(qū)間(2)fx解:(1)設(shè)gx=f當(dāng)x∈?1,π2時(shí),g′設(shè)為α.則當(dāng)x∈?1,α?xí)r,g所以gx在?1,α單調(diào)遞增,在α,π2單調(diào)遞減,故g(2)fx的定義域?yàn)?(I)當(dāng)x∈(?1,0]時(shí),由(1)知,f′x在?1,0單調(diào)遞增,而f′0=0,所以當(dāng)x∈(II)當(dāng)x∈0,π2時(shí),由(1)知,f′x在0,α單調(diào)遞增,在α,π2單調(diào)遞減,而f′0=0,f′π2又f0=0,fπ2=1(III)當(dāng)x∈π2,π時(shí),f′x<0,所以f(Iv)當(dāng)x∈π,+∞時(shí),lnx+1>1,所以fx<2.已知函數(shù)fx=axsin（1）求fx的解析式(2)判斷函數(shù)fx在解:(1)f′x=asin當(dāng)a=0時(shí),f當(dāng)a<0時(shí),x∈0,π2有又最大值為f0=?32當(dāng)a>0時(shí),x∈0,又最大值為fπ2（2）f′x=sin又f0當(dāng)x∈πg(shù)′x故存在x0∈故而fx在π2,fx在x0,綜上,fx在03.已知函數(shù)fx=cosx解:當(dāng)x∈?π2當(dāng)x∈[當(dāng)x∈π所以f′x?當(dāng)x∈當(dāng)x∈f所以fx在x∈π又fx為偶函數(shù),在(?∞,3]綜上,fx4.設(shè)函數(shù)fx(I)討論fx(II)設(shè)fx≤1解:(I)f′(i)當(dāng)a≥1時(shí),f′x≥0,當(dāng)且僅當(dāng)a=(ii)當(dāng)a≤0時(shí),f′x≤0,當(dāng)且僅當(dāng)a=0,x=0(iii)當(dāng)0<a<1時(shí),由當(dāng)x∈0,當(dāng)x∈x1當(dāng)x∈x2(II)由fx≤1+sinx令gx=sinx當(dāng)x∈0,arccos2π時(shí),g又g0=gπ2當(dāng)a≤2π(i)當(dāng)0≤x≤π2(ii)當(dāng)π2≤綜上,a的取值范圍是?∞,2(二)自然分段:以題目中的明顯提示為單位進(jìn)行分段1.對(duì)?x∈0ff令gx=cosx?sin當(dāng)gxmax≤0即g0≤0,a當(dāng)gxmin≥0即gπ2≥0,當(dāng)gx=0有解時(shí),存在唯一解使得gx綜上,a2.求fx解:fff當(dāng)x=?f(三)導(dǎo)數(shù)異號(hào):一階導(dǎo)和二階導(dǎo)在某點(diǎn)處異號(hào)已知函數(shù)f（1）討論fx在0（2）證明:fx>0在π解:(1)f①在x∈0,π2故存在x1∈0,π2使gx②在x∈π有xsinx≥minfx在x∈π③在x∈5π6,π所以f′x在x∈5πfx在5π6,x2綜上,fx在0(2)f①若x∈π2,f′x=1?xsinf②若x∈5π當(dāng)x∈5π6,π,f故存在唯一x2∈5π6,π使由1=x0sin令?x二、放縮消三角(一)正弦函數(shù):x>01.證明:x證明:xlnx+2.a∈0證明:πx+a3.證明:4解:x3x?令??x在0,1所以?xmin=4.sinax≥asinx解:當(dāng)a=?1,0易知該函數(shù)單調(diào)遞減,所以ax≤x當(dāng)?1<則sinxx<sin5.當(dāng)x∈0解:x(二)余弦函數(shù):cos1.證明:x法一:解:即證x令令易知?′x單調(diào)遞減且?′1=0,故易知gx在x=故g法二:當(dāng)x∈(0,當(dāng)x>1即證f又f1(三)函數(shù)邊界:sin1.證明:x解:lnx+g′又g02.是否存在m>0使得函數(shù)fx解:當(dāng)a≤0當(dāng)x∈1,+∞當(dāng)a>0是否存在m>0使得gxg′x=2ax?g12a=ln當(dāng)g12a=ln2ag所以,?m=e1a使得(四)正切函數(shù):x1.已知x∈0,π解:得fx單調(diào)遞增,所以三、三角函數(shù)在一起常見(jiàn)模型:x1.x>0證明:易知?易知分子恒小于0,那么?