第19讲 导数与不等式问题
课后自测诊断——及时查漏补缺·备考不留死角
1.(2019·启东中学检测)已知函数f(x)=1-x-1
e x,g(x)=x-ln x.
(1)证明:g(x)≥1.
(2)证明:(x-ln x)f(x)>1-1 e2.
证明:(1)g′(x)=x-1
x
,当0 当x>1时,g′(x)>0, 即g(x)在(0,1)上为减函数,在(1,+∞)上为增函数.所以g(x)≥g(1)=1,得证. (2)f(x)=1-x-1 e x ,f′(x)= x-2 e x , 所以当0 即f(x)在(0,2)上为减函数,在(2,+∞)上为增函数, 所以f(x)≥f(2)=1-1 e2 ,① 又由(1)知x-ln x≥1,②,且①②等号不同时取得. 所以(x-ln x)f(x)>1-1 e2. 2.设函数f(x)=1 x- e e x,g(x)=a(x 2-1)-ln x(a∈R,e为自然对数的底数). (1)证明:当x>1时,f(x)>0; (2)讨论g(x)的单调性; (3)若不等式f(x) 解:(1)证明:f(x)=e x-1-x x e x-1 , 令s(x)=e x-1-x,则s′(x)=e x-1-1, 当x>1时,s′(x)>0,所以s(x)在(1,+∞)上单调递增,又s(1)=0,所以s(x)>0, 从而当x >1时,f (x )>0. (2)g ′(x )=2ax -1x =2ax 2-1x (x >0), 当a ≤0时,g ′(x )<0,g (x )在(0,+∞)上单调递减. 当a >0时,由g ′(x )=0得x =12a . 当x ∈? ????0,12a 时,g ′(x )<0,g (x )单调递减; 当x ∈? ?? ??12a ,+∞时,g ′(x )>0,g (x )单调递增. 综上,当a ≤0时,g (x )在(0,+∞)上单调递减; 当a >0时,g (x )在? ????0,12a 上单调递减, 在? ?? ??12a ,+∞上单调递增. (3)由(1)知,当x >1时,f (x )>0. 当a ≤0,x >1时,g (x )=a (x 2-1)-ln x <0, 故当f (x ) 当0 >1时, g (x )在? ????1,12a 上单调递减,g ? ????12a ??12a >0, 所以此时f (x ) 当a ≥12时,令h (x )=g (x )-f (x )(x ≥1), 当x >1时,h ′(x )=2ax -1x +1x 2-e 1-x >x -1x +1x 2-1x =x 3-2x +1x 2>x 2-2x +1x 2 >0, 因此,h (x )在区间(1,+∞)上单调递增, 又h (1)=0,所以当x >1时,h (x )=g (x )-f (x )>0,即f (x ) 综上,a 的取值范围为???? ??12,+∞. 3.设函数f (x )=e x ln x (e 是自然对数的底数). (1)求曲线y =f (x )在点(1,f (1))处的切线方程; (2)令q (x )=1-2e x e x ,证明:当x >0时, f (x )>q (x )恒成立. 解:(1)函数f (x )的定义域为(0,+∞), f ′(x )=e x ln x +e x x , 所以f (1)=0,f ′(1)=e , 故曲线y =f (x )在点(1,f (1))处的切线方程为y =e(x -1). (2)证明:当x >0时,f (x )>q (x )恒成立, 等价于当x >0时,x ln x >x e x -2e 恒成立. 设函数g (x )=x ln x ,则g ′(x )=1+ln x , 所以当x ∈? ?? ??0,1e 时,g ′(x )<0; 当x ∈? ?? ??1e ,+∞时,g ′(x )>0. 故g (x )在? ????0,1e 上单调递减,在? ?? ??1e ,+∞上单调递增, 从而g (x )在(0,+∞)上的最小值为g ? ?? ??1e =-1e . 设函数h (x )=x e x -2e ,则h ′(x )=1-x e x , 所以当x ∈(0,1)时,h ′(x )>0; 当x ∈(1,+∞)时,h ′(x )<0. 故h (x )在(0,1)上单调递增,在(1,+∞)上单调递减, 从而h (x )在(0,+∞)上的最大值为h (1)=-1e . 因为g (x )min =g ? ?? ??1e =h (1)=h (x )max , 所以当x >0时,g (x )>h (x ), 所以当x >0时,f (x )>q (x )恒成立. 4.已知函数f (x )=x ln x -e x +1. (1)求曲线y =f (x )在点(1,f (1))处的切线方程; (2)证明:f (x ) 解:(1)依题意得f ′(x )=ln x +1-e x , 又f (1)=1-e ,f ′(1)=1-e ,故所求切线方程为y -1+e =(1-e)(x -1),即y =(1-e)x . (2)证明:依题意,要证f (x ) 即证x ln x -e x +1 即证x ln x 当0 故x ln x 当x >1时,令g (x )=e x +sin x -1-x ln x , 故g ′(x )=e x +cos x -ln x -1. 令h (x )=g ′(x )=e x +cos x -ln x -1, 则h ′(x )=e x -1x -sin x , 当x >1时,e x -1x >e -1>1, 所以h ′(x )=e x -1x -sin x >0, 故h (x )在(1,+∞)上单调递增. 故h (x )>h (1)=e +cos 1-1>0,即g ′(x )>0, 所以g (x )在(1,+∞)上单调递增, 所以g (x )>g (1)=e +sin 1-1>0, 即x ln x 综上所述,f (x ) 5.已知函数f (x )=x ln x -ax (a >0). (1)若函数f (x )在(1,+∞)上是减函数,求实数a 的最小值; (2)若?x 1,x 2∈[e ,e 2],使f (x 1)≤f ′(x 2)+a 成立,求实数a 的取值范围. 解:(1)因为f (x )在(1,+∞)上为减函数,所以f ′(x )=ln x -1ln x 2-a ≤0在(1,+ ∞)上恒成立. 所以当x ∈(1,+∞)时,f ′(x )max ≤0. 又f ′(x )=ln x -1ln x 2-a =-? ?? ??1ln x -122+14-a , 故当1ln x =12,即x =e 2时,f ′(x )max =14-a , 所以14-a ≤0,故a ≥14,所以a 的最小值为14. (2)“若?x 1,x 2∈[e ,e 2],使f (x 1)≤f ′(x 2)+a 成立”等价于当x ∈[e ,e 2]时,有f (x )min ≤f ′(x )max +a , 当x ∈[e ,e 2]时,有f ′(x )max +a =14, 问题等价于:“当x ∈[e ,e 2 ]时,有f (x )min ≤14”. ①当a ≥14时,f (x )在[e ,e 2]上为减函数, 则f (x )min =f (e 2 )=e 22-a e 2≤14,故a ≥12-14e 2. ②当0 ??1ln x -122+14-a 在[e ,e 2]上为增函数, 故f ′(x )的值域为[f ′(e),f ′(e 2)],即???? ??-a ,14-a . 由f ′(x )的单调性和值域知,存在唯一x 0∈(e ,e 2),使f ′(x 0)=0,且满足: 当x ∈(e ,x 0)时,f ′(x )<0,f (x )为减函数; 当x ∈(x 0,e 2)时,f ′(x )>0,f (x )为增函数. 所以f (x )min =f (x 0)=x 0ln x 0 -ax 0≤14,x 0∈(e ,e 2), 所以a ≥1ln x 0-14x 0>1ln e 2-14e >12-14=14