2024-2025學(xué)年高中數(shù)學(xué)同步備課試題 選擇性必修二（人教A版2019）第五章 5-2-2 導(dǎo)數(shù)的四則運(yùn)算法則 含解析

5.2.2導(dǎo)數(shù)的四則運(yùn)算法則課后訓(xùn)練鞏固提升A組1.下列運(yùn)算中正確的是()A.(ax2+bx+c)'=a(x2)'+b(x)'B.(sinx-2x2)'=(sinx)'-2'(x2)'C.sinxxD.(cosx·sinx)'=(sinx)'cosx+(cosx)'cosx答案:A2.若函數(shù)f(x)=ax4+bx2+c滿足f'(1)=2,則f'(-1)等于()A.-1 B.-2 C.2 D.0解析:∵f'(x)=4ax3+2bx為奇函數(shù),∴f'(-1)=-f'(1)=-2.答案:B3.已知函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),且滿足f(x)=2xf'(e)+lnx,則f'(e)=()A.e-1 B.-1 C.-e-1 D.-e解析:∵f(x)=2xf'(e)+lnx,∴f'(x)=2f'(e)+1x∴f'(e)=2f'(e)+1e,解得f'(e)=-1e.故選答案:C4.若函數(shù)f(x)=x2-2x-4lnx,則f'(x)>0的解集為()A.(0,+∞) B.(-1,0)∪(2,+∞)C.(2,+∞) D.(-1,0)解析:∵f(x)=x2-2x-4lnx,∴f'(x)=2x-2-4x由f'(x)>0,得(x+1)(∵f(x)的定義域?yàn)?0,+∞),∴x>2.故選C.答案:C5.已知一物體沿直線運(yùn)動(dòng),位移s(單位:m)與時(shí)間t(單位:s)之間的關(guān)系為s(t)=t2+3t,則該物體在t=2s時(shí)的瞬時(shí)速度為(A.194 B.17C.154 D.解析:∵s'(t)=2t-3t2,∴s'(2)=4-答案:D6.已知曲線y=aex+xlnx在點(diǎn)(1,ae)處的切線方程為y=2x+b,則()A.a=e,b=-1 B.a=e,b=1C.a=1e,b=1 D.a=1e,解析:由題意得y'=aex+lnx+1,則y'|x=1=ae+1=2,因而ae=1,故a=1e將點(diǎn)(1,1)的坐標(biāo)代入y=2x+b,得b=-1.答案:D7.(多選題)若函數(shù)f(x)的導(dǎo)函數(shù)f'(x)的圖象關(guān)于y軸對稱,則f(x)的解析式可能為()A.f(x)=3cosx B.f(x)=x3+xC.f(x)=x+1x D.f(x)=ex解析:對于A,f(x)=3cosx,其導(dǎo)數(shù)f'(x)=-3sinx為奇函數(shù),圖象不關(guān)于y軸對稱,不符合題意;對于B,f(x)=x3+x,其導(dǎo)數(shù)f'(x)=3x2+1為偶函數(shù),圖象關(guān)于y軸對稱,符合題意;對于C,f(x)=x+1x,其導(dǎo)數(shù)f'(x)=1-1x2為偶函數(shù),圖象關(guān)于y軸對稱對于D,f(x)=ex+x,其導(dǎo)數(shù)f'(x)=ex+1不是偶函數(shù),圖象不關(guān)于y軸對稱,不符合題意.故選BC.答案:BC8.已知函數(shù)f(x)=f'π4cosx+sinx,則fπ4的值為解析:∵f'(x)=-f'π4sinx+cosx∴f'π4=-f'π4×22+2∴f(x)=(2-1)cosx+sinx.∴fπ4=1答案:19.已知函數(shù)f(x)=ax+bex的圖象在點(diǎn)P(-1,2)處的切線與直線y=-3x平行,則函數(shù)f(x)的解析式是.

解析:由題意知f'(-1)=-3,即a+be-1=-3.①由題意知f(-1)=2,即-a+be-1=2.②由①②聯(lián)立,解得a=-52,b=-12故f(x)=-52x-12ex+答案:f(x)=-52x-12e10.已知曲線y=x4+ax2+1在點(diǎn)(-1,a+2)處切線的斜率為8,則a=.

解析:y'=4x3+2ax,因?yàn)榍€在點(diǎn)(-1,a+2)處切線的斜率為8,所以y'|x=-1=-4-2a=8,解得a=-6.答案:-611.求下列函數(shù)的導(dǎo)數(shù):(1)y=xex;(2)y=(x+1)(x+2)(x+3);(3)y=2x(4)y=xsinx-2cos解:(1)y'=x'·ex+x·(ex)'=ex+xex=(1+x)ex.(2)因?yàn)?x+1)(x+2)(x+3)=(x2+3x+2)(x+3)=x3+6x2+11x+6,所以y'=[(x+1)(x+2)(x+3)]'=(x3+6x2+11x+6)'=3x2+12x+11.(3)y'=2xx2(4)y'=(xsinx)'-2cosx'=sinx+xcosx-B組1.在一次降雨過程中,某地的降雨量y(單位:mm)與時(shí)間t(單位:min)的函數(shù)關(guān)系可近似地表示為y=f(t)=10t,則在t=40min時(shí)的瞬時(shí)降雨強(qiáng)度(降雨量的導(dǎo)數(shù))為(A.20mm B.400mmC.12mm/min D.1解析:∵f'(t)=102t,∴f'(40)=答案:D2.設(shè)函數(shù)f(x)=x3+(a-1)x2+ax,若f(x)為奇函數(shù),則曲線y=f(x)在點(diǎn)(0,0)處的切線方程為()A.y=-2x B.y=-x C.y=2x D.y=x解析:∵f(x)為奇函數(shù),∴f(-x)=-f(x),得a=1.∴f(x)=x3+x.∴f'(x)=3x2+1.∴f'(0)=1.∴切線方程為y=x.答案:D3.已知函數(shù)f(x)的圖象如圖所示,直線y=kx+2是曲線y=f(x)在點(diǎn)(3,f(3))處的切線,令g(x)=xf(x),g'(x)是g(x)的導(dǎo)函數(shù),則g'(3)=()A.-1 B.0 C.2 D.4解析:由題圖可知曲線y=f(x)在點(diǎn)(3,f(3))處切線的斜率為2-10-3=-13,因?yàn)間(x)=xf(x),所以g'(x)=f(x)+xf'(x).所以g'(3)=f(3)+3f'(3).又由題圖可知f(3)=1,所以g'(3)=1+3×-13=答案:B4.設(shè)函數(shù)f(x)=sinθ3x3+3cosθ2x2+tanθ,其中θ∈0,5πA.[-2,2] B.[2,3C.[3,2] D.[2,2]解析:由題意得f'(x)=sinθ·x2+3cosθ·x,∴f'(1)=sinθ+3cosθ=2sinθ+∵θ∈0,5π12,∴2sinθ+π3∈[答案:D5.設(shè)點(diǎn)P是曲線y=ex-3x+23上的任意一點(diǎn),曲線在點(diǎn)P處的切線的傾斜角為α,則角α的取值范圍為(A.2π3,C.0,π2解析:y=ex-3x+23的導(dǎo)數(shù)為y'=ex-3由ex>0,可得切線的斜率k>-3.由tanα>-3,且0≤α<π,得0≤α<π2或2π3答案:D6.曲線y=3(x2+x)ex在點(diǎn)(0,0)處的切線方程為.

解析:∵y'=3(2x+1)ex+3(x2+x)ex=3(x2+3x+1)ex,∴k=y'|x=0=3.∴曲線y=3(x2+x)ex在點(diǎn)(0,0)處的切線方程為y=3x.答案:y=3x7.若曲線y=xα+1(α∈Q,且α≠0)在點(diǎn)(1,2)處的切線經(jīng)過坐標(biāo)原點(diǎn),則α=.

解析:由題意得y'=αxα-1,則曲線在點(diǎn)(1,2)處的切線的斜率k=y'|x=1=α.因?yàn)榍芯€經(jīng)過坐標(biāo)原點(diǎn),所以k=α=2-01答案:28.曲線y=cosx-x2在點(diǎn)(0,1)處的切線方程為.解析:由y'=-sinx-12,得y'|x=0=-12.故切線方程為y-1=-12x,即x+2y-2答案:x+2y-2=09.已知曲線y=f(x)=x3+ax+b在點(diǎn)P(2,-6)處的切線方程是13x-y-32=0.(1)求a,b的值;(2)如果曲線y=f(x)在某點(diǎn)處的切線與直線l:y=-14x+3垂直,求切點(diǎn)坐標(biāo)與切線的方程解:(1)f(x)=x3+ax+b的導(dǎo)數(shù)f'(x)=3x2+a.由題意可得f'(2)=12+a=13,f(2)=8+2a+b=-6,從而可得a=1,b=-16.(2)因?yàn)榍芯€與直線y=-14x+3垂直所以切線的斜率k=4.設(shè)切點(diǎn)的坐標(biāo)為(x0,y0),則f'(x0)=3x02+1=4,解得x0=±當(dāng)x0=1時(shí),y0=1+1-16=-14,切點(diǎn)的坐標(biāo)為(1,-14),切線方程為y=4x-18;當(dāng)x0=-1時(shí),y0=-1-1-16=-18,切點(diǎn)坐標(biāo)為(-1,-18),切線方程為y=4x-14.10.已知曲線C:y=x2-2x+3,直線l:x-y-4=