2018届二轮 题型专项训练8 专题卷(全国通用)

题型专项训练8函数与导数(解答题专项)

1.已知函数f(x)=x ln x+ax(a∈R).

(1)当a=0时,求f(x)的最小值;

(2)若函数g(x)=f(x)+ln x在区间[1,+∞)上为增函数,求实数a的取值范围.

2.已知函数f(x)=a ln x+x2+bx(a,b∈R)在x1=2,x2=3处取得极值.

(1)求a,b的值;

(2)求f(x)在点P(1,f(1))处的切线方程.

3.(2017浙江绍兴鲁迅中学模拟)已知函数f(x)=ln x-

(1)若函数f(x)在(0,+∞)上单调递增,求实数a的取值范围;

(2)设m>n>0,求证:ln m-ln n>

4.(2017浙江湖州、丽水、衢州三地市4月联考)已知函数f(x)=lo x-m log2x+a,g(x)=x2+1.

(1)当a=1时,求f(x)在x∈[1,4]上的最小值;

(2)当a>0,m=2时,若对任意的实数t∈[1,4],均存在x i∈[1,8](i=1,2),且x1≠x2,使得=f(t)成立,求实数a 的取值范围.

5.已知二次函数f(x)=x2+bx+c,其中常数b,c∈R.

(1)若任意的x∈[-1,1],f(x)≥0,f(2+x)≤0,试求实数c的取值范围;

(2)若对任意的x1,x2∈[-1,1],有|f(x1)-f(x2)|≤4,试求实数b的取值范围.

6.已知a∈R,函数f(x)=+a ln x.

(1)若函数f(x)在(0,2)上递减,求实数a的取值范围;

(2)当a>0时,求f(x)的最小值g(a)的最大值;

(3)设h(x)=f(x)+|(a-2)x|,x∈[1,+∞),求证:h(x)≥2.

参考答案

题型专项训练8函数与导数(解答题专项)

1.解(1)f(x)的定义域为(0,+∞),f'(x)=ln x+1,令f'(x)=0,得x=.

当x∈(0,+∞)时,f'(x),f(x)的变化的情况如下:

x

∴f(x)的最小值是f=-.

(2)由题意得g'(x)=ln x+a+1+.

∵函数g(x)在区间[1,+∞)上为增函数,

∴当x∈[1,+∞)时,g'(x)≥0,即ln x+≥-(a+1)在[1,+∞)上恒成立,

设h(x)=ln x+,

∴h'(x)=,

∴h(x)=ln x+在[1,+∞)上递增,

∴-(a+1)≤h(x)min=h(1)=1,

∴a≥-2.

2.解(1)f'(x)=+x+b=,

令f'(x)==0,

据题意,得2,3是方程x2+bx+a=0的两根,

则有所以

(2)f(x)=6ln x+x2-5x,

则f(1)=-5=-,得P.

又由f'(x)=,得f'(1)=1-5+6=2.

从而,得所求切线方程为l:y+=2(x-1),即4x-2y-13=0.

3.(1)解f',

因为f(x)在(0,+∞)上单调递增,所以f'(x)≥0在(0,+∞)上恒成立,

即x2+(2-2a)x+1≥0在(0,+∞)上恒成立,所以2a-2≤x+在(0,+∞)上恒成立, 因为x+≥2,当且仅当x=1时等号成立,所以2a-2≤2,解得a≤2.

(2)证明要证ln m-ln n>,只需证ln,只需证ln>0.

设h(x)=ln x-,由(1)可知h(x)在(0,+∞)上单调递增,

因为>1,所以h>h(1)=0,

即ln>0,所以原等式成立.

4.解(1)当a=1时,f(x)=lo x-m log2x+1=+1-,其中0≤log2x≤2.

因此,①当≤0,即m≤0时,f(x)min=f(1)=1;

②当≥2,即m≥4时,f(x)min=f(4)=5-2m;

③当0

综上,f(x)min=

(2)令log2t=u(0≤u≤2),则f(t)=u2-2u+a的值域是[a-1,a].

因为y==x+-2a(1≤x≤8),

利用图形可知

即

解得3

故实数a的取值范围是(3,11-2].

5.解(1)因为-1≤x≤1,所以1≤2+x≤3.

由已知,可得对任意的-1≤x≤1,f(x)≥0恒成立;

对任意的1≤x≤3,f(x)≤0恒成立,故f(1)≥0,且f(1)≤0,

即f(1)=0,也即1为函数y=f(x)的一个零点.

因此可设f(x)=(x-1)(x-c).

所以,任意的1≤x≤3,f(x)≤0恒成立,则[1,3]?[1,c],即c的取值范围为c≥3.

(2)函数f(x)=x2+bx+c对?x1,x2∈[-1,1],有|f(x1)-f(x2)|≤4恒成立,

即f(x)max-f(x)min≤4,记f(x)max-f(x)min=M,则M≤4.

当>1,即|b|>2时,

M=|f(1)-f(-1)|=|2b|>4,与M≤4矛盾;

当≤1,即-2≤b≤2时,

M=max{f(1),f(-1)}-f

=-f

=≤4,即-2≤b≤2.

综上,实数b的取值范围为-2≤b≤2.

6.(1)解函数定义域为(0,+∞),函数f(x)在(0,2)上递减??x∈(0,2),恒有f'(x)≤0成立,而f'(x)=≤0??x∈(0,2),恒有a≤成立,

而>1,则a≤1.

(2)解当a>0时,f'(x)==0?x=.

当x∈(0,+∞)时,f'(x),f(x)的变化情况如下:

所以f(x)的最小值g(a)=f=a+a ln ,

g'(a)=ln 2-ln a=0?a=2,

当a∈(0,+∞)时,g'(a),g(a)的变化情况如下:

所以g(a)的最大值为g(2)=2.

(3)证明当a≥2时,h(x)=f(x)+(a-2)x=+a ln x+(a-2)x,

h'(x)=+a-2≥0,

所以h(x)在[1,+∞)上是增函数,故h(x)≥h(1)=a≥2.

当a<2时,h(x)=f(x)-(a-2)x=+a ln x-(a-2)x,

h'(x)=-a+2==0,

解得x=-<0或x=1,所以h(x)在[1,+∞)上单调递增,即h(x)≥h(1)=4-a>2.

综上所述:h(x)≥2.

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