their nervous systems and ability to produce baby birds, and can lead to kidney(肾) failures and death. So condors with high levels of lead are sent to Los Angeles Zoo, where they are treated with calcium EDTA, a chemical that removes lead from the blood over several days. This work is starting to pay off. The annual death rate for adult condors has dropped from 38% in 2000 to 5.4% in 2011. 题组训练18 定积分与微积分基本定理
132
1.若a>2,则函数f(x)=x-ax+1在区间(0,2)上恰好有( )
3A.0个零点 C.2个零点 答案 B
解析 ∵f′(x)=x-2ax,且a>2, ∴当x∈(0,2)时,f′(x)<0, 即f(x)在(0,2)上是单调减函数. 11
又∵f(0)=1>0,f(2)=-4a<0,
3∴f(x)在(0,2)上恰好有1个零点.故选B. 2.函数y=xe的图像大致为( )
2x
2
B.1个零点 D.3个零点
答案 A
解析 因为y′=2xe+xe=x(x+2)e,所以当x<-2或x>0时,y′>0,函数y=xe为增函数;当-2
1xπ
3.函数f(x)=e(sinx+cosx)在区间[0,]上的值域为( )
22π
11
A.[,e2]
22π
C.[1,e2] 答案 A
1x1xπx
解析 f′(x)=e(sinx+cosx)+e(cosx-sinx)=ecosx,当0≤x≤时,f′(x)≥0.
222
π
11
B.(,e2)
22π
D.(1,e2)
2x
2x
x
2x
x
2x
their nervous systems and ability to produce baby birds, and can lead to kidney(肾) failures and death. So condors with high levels of lead are sent to Los Angeles Zoo, where they are treated with calcium EDTA, a chemical that removes lead from the blood over several days. This work is starting to pay off. The annual death rate for adult condors has dropped from 38% in 2000 to 5.4% in 2011. π
∴f(x)是[0,]上的增函数.
2π
π1
∴f(x)的最大值为f()=e2,
221
f(x)的最小值为f(0)=.
2
4.(2018·山东陵县一中月考)已知函数f(x)=xe,当x∈[-1,1]时,不等式f(x) A.[,+∞) eC.[e,+∞) 答案 D 解析 由f′(x)=e(2x+x)=x(x+2)e,得当-1 5.(2014·课标全国Ⅰ)已知函数f(x)=ax-3x+1,若f(x)存在唯一的零点x0,且x0>0,则a的取值范围是( ) A.(2,+∞) C.(-∞,-2) 答案 C 解析 当a=0时,显然f(x)有两个零点,不符合题意. 22 当a≠0时,f′(x)=3ax-6x,令f′(x)=0,解得x1=0,x2=. a 2232 当a>0时,>0,所以函数f(x)=ax-3x+1在(-∞,0)与(,+∞)上为增函数,在(0, aa2 )上为减函数,因为f(x)存在唯一零点x0,且x0>0,则f(0)<0,即1<0,不成立. a 22232 当a<0时,<0,所以函数f(x)=ax-3x+1在(-∞,)和(0,+∞)上为减函数,在(, aaa284 0)上为增函数,因为f(x)存在唯一零点x0,且x0>0,则f()>0,即a·3-3·2+1>0,解 aaa得a>2或a<-2,又因为a<0,故a的取值范围为(-∞,-2).选C. 6.f(x)是定义在R上的偶函数,当x<0时,f(x)+xf′(x)<0,且f(-4)=0,则不等式xf(x)>0的解集为( ) A.(-4,0)∪(4,+∞) C.(-∞,-4)∪(4,+∞) B.(-4,0)∪(0,4) D.(-∞,-4)∪(0,4) B.(1,+∞) D.(-∞,-1) 3 2 x 2 x 2x 1 B.(,+∞) eD.(e,+∞) their nervous systems and ability to produce baby birds, and can lead to kidney(肾) failures and death. So condors with high levels of lead are sent to Los Angeles Zoo, where they are treated with calcium EDTA, a chemical that removes lead from the blood over several days. This work is starting to pay off. The annual death rate for adult condors has dropped from 38% in 2000 to 5.4% in 2011. 答案 D 解析 设g(x)=xf(x),则当x<0时,g′(x)=[xf(x)]′=x′f(x)+xf′(x)=xf′(x)+f(x)<0,所以函数g(x)在区间(-∞,0)上是减函数.因为f(x)是定义在R上的偶函数.所以g(x)=xf(x)是R上的奇函数,所以函数g(x)在区间(0,+∞)上是减函数.因为f(-4)=0,所以f(4)=0,即g(4)=0,g(-4)=0,所以xf(x)>0化为g(x)>0.设x>0,不等式为g(x)>g(4),即0 7.(2018·衡水调研卷)已知函数f(x)=alnx-bx,a,b∈R.若不等式f(x)≥x对所有的b∈(-∞,0],x∈(e,e]都成立,则实数a的取值范围是( ) A.[e,+∞) e2 C.[,e) 2答案 B alnx-x2 解析 由题意可得bx≤alnx-x,所以b≤.由b∈(-∞,0],故对任意的x∈(e,2 xalnx-xx222 e],都有≥0,即alnx≥x对一切x∈(e,e]恒成立,即a≥对一切x∈(e,e]2 xlnxxlnx-1e2 恒成立.令h(x)=,则h′(x)=2>0在x∈(e,e]上恒成立,故h(x)max=,所 lnx(lnx)2e 以a≥.故选B. 2 32x3x 8.(2018·湖南衡阳期末)设函数f(x)=e(x+x-6x+2)-2ae-x,若不等式f(x)≤0 2在[-2,+∞)上有解,则实数a的最小值为( ) 31A.-- 2e31C.-- 42e答案 C 321332xx3x 解析 由f(x)=e(x+x-6x+2)-2ae-x≤0,得a≥x+x-3x+1-x. 2242e1332x 令g(x)=x+x-3x+1-x,则 242e 323x-131 g′(x)=x+x-3+x=(x-1)(x+3+x). 222e22e 当x∈[-2,1)时,g′(x)<0,当x∈(1,+∞)时,g′(x)>0, 32 B.-- 2e1 D.-1- e 2 2 2 2 2 e B.[,+∞) 2D.[e,+∞) 2 2
2019版高考数学一轮总复习第三章导数及应用题组训练18定积分与微积分基本定理理



