2023年高考數(shù)學(xué)主觀題專題六 函數(shù)與導(dǎo)數(shù)

專題六函數(shù)與導(dǎo)數(shù)以函數(shù)為載體,以導(dǎo)數(shù)為工具,考查函數(shù)性質(zhì)及導(dǎo)數(shù)極值理論,單調(diào)性及其應(yīng)用為目標(biāo),是最近幾年函數(shù)與導(dǎo)數(shù)交匯試題的顯著特點(diǎn)和命題趨向,高考導(dǎo)數(shù)問(wèn)題命題的五大熱點(diǎn)如下:熱點(diǎn)一、在導(dǎo)數(shù)與函數(shù)性質(zhì)的交匯點(diǎn)命題:主要考查導(dǎo)數(shù)的簡(jiǎn)單應(yīng)用,包括求函數(shù)的極值,求函數(shù)的單調(diào)區(qū)間,證明函數(shù)的單調(diào)性等.命題的熱點(diǎn):三次函數(shù)求導(dǎo)后為二次函數(shù),結(jié)合一元二次方程根的分布,考查代數(shù)推理能力、語(yǔ)言轉(zhuǎn)化能力和待定系數(shù)法等數(shù)學(xué)思想.歷年高考命題分析熱點(diǎn)二、在導(dǎo)數(shù)與含參數(shù)函數(shù)的交匯點(diǎn)命題:主要考查含參數(shù)函數(shù)的極值問(wèn)題,分類討論思想及解不等式的能力,利用分離變量法求參數(shù)的取值范圍等問(wèn)題.熱點(diǎn)三、在導(dǎo)數(shù)與解析幾何交匯點(diǎn)命題:主要考查對(duì)導(dǎo)數(shù)的幾何意義,切線的斜率,導(dǎo)數(shù)與函數(shù)單調(diào)性,最(極)值等綜合運(yùn)用知識(shí)的能力.熱點(diǎn)四、在導(dǎo)數(shù)與向量問(wèn)題交匯點(diǎn)命題:依托向量把函數(shù)單調(diào)性,奇偶性,解不等式等知識(shí)融合在一起.即考查了向量的有關(guān)知識(shí),又考查了函數(shù)性質(zhì)及解不等式等內(nèi)容.熱點(diǎn)五、在導(dǎo)數(shù)與函數(shù)模型構(gòu)建交匯點(diǎn)命題:主要考查考生將實(shí)際問(wèn)題轉(zhuǎn)化為數(shù)學(xué)問(wèn)題,運(yùn)用導(dǎo)數(shù)工具和不等式知識(shí)去解決最優(yōu)化問(wèn)題的數(shù)學(xué)應(yīng)用意識(shí)和實(shí)踐能力.這部分的題目難度較大,特別是對(duì)藝術(shù)類考生而言.因此,考生在復(fù)習(xí)時(shí)可以酌情選做.【近7年全國(guó)卷考點(diǎn)統(tǒng)計(jì)】試卷類型2016201720182019202020212022全國(guó)卷(甲卷)12121212121212全國(guó)卷(乙卷)12121212121212新高考全國(guó)Ⅰ卷

1212新高考全國(guó)Ⅱ卷

1212典例解析【例】設(shè)函數(shù)f(x)=x3-kx2+x(k∈R).(1)當(dāng)k=1時(shí),求函數(shù)f(x)的單調(diào)區(qū)間;【解析】∵f(x)=x3-kx2+x,∴f'(x)=3x2-2kx+1.(1)∵當(dāng)k=1時(shí),f'(x)=3x2-2x+1,Δ=4-12=-8<0,∴f'(x)>0,f(x)在R上單調(diào)遞增.【例】設(shè)函數(shù)f(x)=x3-kx2+x(k∈R).(2)當(dāng)k<0時(shí),求函數(shù)f(x)在[k,-k]上的最小值m和最大值M.【解析】(2)

當(dāng)k<0時(shí),?x∈[k,-k],都有f(x)-f(k)=x3-kx2+x-k3+k3-k=(x2+1)(x-k)≥0,故f(x)≥f(k);

f(x)-f(-k)=x3-kx2+x+k3+k3+k=(x+k)(x2-2kx+2k2+1)=(x+k)[(x-k)2+k2+1]≤0.故f(x)≤f(-k),而f(k)=k<0,f(-k)=-2k3-k>0,所以f(x)max=f(-k)=-2k3-k,f(x)min=f(k)=k.考點(diǎn)訓(xùn)練

2.已知函數(shù)f(x)=ex(ax+b)-x2-4x,曲線y=f(x)在點(diǎn)(0,f(0))處切線方程為y=4x+4.(1)求a,b的值;解:(1)f'(x)=ex(ax+a+b)-2x-4,由已知得f(0)=4,f'(0)=4,故b=4,a+b=8,從而a=b=4.

2.已知函數(shù)f(x)=ex(ax+b)-x2-4x,曲線y=f(x)在點(diǎn)(0,f(0))處切線方程為y=4x+4.(2)討論f(x)的單調(diào)性,并求f(x)的極大值.

x(-∞,-2)-2(-2,-ln2)-ln2(-ln2,+∞)f'(x)+0-0+f(x)↗極大↘極小↗

4.已知函數(shù)f(x)=(x+1)lnx-a(x-1).(1)當(dāng)a=4時(shí),求曲線y=f(x)在(1,f(1))處的切線方程;

4.已知函數(shù)f(x)=(x+1)lnx-a(x-1).(2)若當(dāng)x∈(1,+∞)時(shí),f(x)>0,求a的取值范圍.

5.已知函數(shù)f(x)=x2e-x.(1)求f(x)的極小值和極大值;解:(1)f(x)的定義域?yàn)?-∞,+∞),f'(x)=-e-xx(x-2).當(dāng)x∈(-∞,0)或x∈(2,+∞)時(shí),f'(x)<0;當(dāng)x∈(0,2)時(shí),f'(x)>0.所以f(x)在(-∞,0),(2,+∞)單調(diào)遞減,在(0,2)單調(diào)遞增.故當(dāng)x=0時(shí),f(x)取得極小值,極小值為f(0)=0;當(dāng)x=2時(shí),f(x)取得極大值,極大值為f(2)=4e-2.5.已知函數(shù)f(x)=x2e-x.(2)當(dāng)曲線y=f(x)的切線l的斜率為負(fù)數(shù)時(shí),求l在x軸上的截距的取值范圍.

7.已知函數(shù)f(x)=x3-3x2+ax+2,曲線y=f(x)在點(diǎn)(0,2)處的切線與x軸交點(diǎn)的橫坐標(biāo)為-2.(1)求a的值;

7.已知函數(shù)f(x)=x3-3x2+ax+2,曲線y=f(x)在點(diǎn)(0,2)處的切線與x軸交點(diǎn)的橫坐標(biāo)為-2.(2)證明:當(dāng)k<1時(shí),曲線y=f(x)與直線y=kx-2只有一個(gè)交點(diǎn).

(2)證明:由(1)知,f(x)=x3-3x2+x+2.

設(shè)g(x)=f(x)-kx+2=x3-3x2+(1-k)x+4,由題設(shè)知1-k>0,當(dāng)x≤0時(shí),g'(x)=3x2-6x+1-k>0,g(x)在(-∞,0]上單調(diào)遞增,又g(-1)=k-1<0,g(0)=4>0,所以g(x)=0在(-∞,0]有唯一實(shí)根.當(dāng)x>0時(shí),令h(x)=x3-3x2+4,則g(x)=h(x)+(1-k)x>h(x).h'(x)=3x2-6x=3x(x-2),h(x)在(0,2)上單調(diào)遞減,在(2,+∞)上單調(diào)遞增,所以g(x)>h(x)≥h(2)=0,所以g(x)=0在(0,+∞)沒(méi)有實(shí)根.綜上,g(x)=0在R上有唯一實(shí)根,即曲線y=f(x)與直線y=kx-2只有一個(gè)交點(diǎn).

9.已知函數(shù)f(x)=x2-ax-alnx(a∈R).(1)若函數(shù)f(x)在x=1處取得極值,求a的值;

9.已知函數(shù)f(x)=x2-ax-alnx(a∈R).(3)當(dāng)x∈[e,+∞)時(shí),f(x)≥0恒成立,求a的取值范圍.

x(0,1)1(1,+∞)f'(x)+0-f(x)遞增遞減

12.設(shè)函數(shù)f(x)=e2x-alnx.(1)討論f(x)的導(dǎo)函數(shù)f'(x)的零點(diǎn)的個(gè)數(shù);

13.已知函數(shù)f(x)=(x-2)ex+a(x-1)2.(1)討論f(x)的單調(diào)性;解:(1)f'(x)=(x-1)ex+2a(x-1)=(x-1)(ex+2a).①設(shè)a≥