5.3.1 (3) 函数的单调性(运用)
${\color{Red}{欢迎到学科网下载资料学习 }}$
[ 【基础过关系列】高二数学同步精品讲义与分层练习(人教A版2019)]
( https://www.zxxk.com/docpack/2875423.html)
\({\color{Red}{ 跟贵哥学数学,so \quad easy!}}\)
选择性第二册同步巩固,难度2颗星!
基础知识
函数单调性与导数
在某个区间\((a ,b)\)内,若\(f'(x)>0\),则函数\(y=f(x)\)在这个区间内单调递增;
若\(f'(x)<0\) ,则函数\(y=f(x)\)在这个区间内单调递减.
解释
(1) 若函数\(y=f(x)\)在某个区间\((a ,b)\)内单调递增,则\(∀x∈(a ,b)\) ,\(f' (x)≥0\)(含等号)恒成立,但不存在一区间\((c ,d)⊆(a ,b)\)内使得\(f' (x)=0\);
假如存在一区间\((c ,d)⊆(a ,b)\)内使得\(f' (x)=0\),那原函数\(y=f(x)\)在区间\((c ,d)\)内恒等于一个常数,即\(f(x)=m\)(\(m\)是个常数),则原函数不可能在\((a ,b)\)内单调递增.
\(\qquad \qquad\)
函数\(y=f(x)\)在某个区间\((a ,b)\)内单调递减有类似结论!
(2)导函数“穿线图”与原函数“趋势图”
① 导函数“穿线图”关注导函数在各区间的正负,故特别注意函数与x轴的交点情况,
如\(f'(x)=x\)与\(f' (x)=e^x-1\)的“穿线图”视为一样的,它们在\((-∞,0)\)上为负,在\((0,+∞)\)上为正.
\(\qquad \qquad\)
② 原函数“趋势图”仅关注函数在各区间上的单调性,没顾及其最值或曲线形状等,
如由导函数\(f' (x)=x-1\)的“穿线图”易得原函数\(y=f(x)\)在\((-∞,0)\)上递减,在\((0,+∞)\)上为递增,趋势图可如下图,
③ 后面涉及到函数单调性均可通过分析导函数“穿线图”得出原函数的单调性.
基本方法
【题型1】已知函数单调性求参数范围
【典题1】 若函数\(f(x)=x^3-3kx+1\)在区间\((1,+∞)\)上单调递增,则实数\(k\)的取值范围是( )
A.\((-∞,1)\) \(\qquad \qquad \qquad\) B.\((-∞,1]\) \(\qquad \qquad \qquad\) C.\([-1,+∞)\) \(\qquad \qquad \qquad\) D.\([1,+∞)\)
解析 因为\(f(x)=x^3-3kx+1\),所以\(f'(x)=3x^2-3k\),
当\(k⩽0\)时,\(f'(x)⩾0\)恒成立,\(f(x)\)在\(R\)单调递增,满足题意;
当\(k>0\)时,令\(f'(x)=3x^2-3k⩾0\),
则 \(x \leqslant-\sqrt{k}\)或\(x⩾\sqrt{k}\),
因为\(f(x)\)在区间\((1,+∞)\)上单调递增,
所以\(\sqrt{k}⩽1\),即\(0<k⩽1\),
综上所述,实数\(k\)的取值范围是\((-∞,1]\).
故选:\(B\).
点拨 \(f(x)\)在\((a,b)\)上递增\(⇒f'(x)≥0\)在\((a,b)\)上恒成立,等号是否成立有时需要检验; \(f(x)\)在\((a,b)\)上递减\(⇒f'(x)≤0\)在\((a,b)\)上恒成立,等号是否成立有时需要检验.
【巩固练习】
1.若函数\(f(x)=(x^2-4ax+2) e^x\)在\(R\)上单调递增,则\(a\)的取值范围是\(\underline{\quad \quad}\).
2.已知函数\(f(x)=\sin x+a\cos x\)在区间\(\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)\)上是减函数,则实数\(a\)的取值范围为\(\underline{\quad \quad}\) .
3.已知函数\(f(x)=(x-a)\ln x\),\(a∈R\).若函数\(f(x)\)在\((0,+∞)\)上为增函数,则\(a\)的取值范围\(\underline{\quad \quad}\) .
参考答案
-
答案 \(\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]\)
解析 对函数求导:\(f'(x)=(x^2-4ax+2x+2-4a)⋅e^x\),
由已知有\(f'(x)≥0\)在\(R\)上恒成立,
又因为\(e^x>0\)恒成立,故仅需\(x^2+(2-4a)x+2-4a⩾0\)恒成立,
故\(△=(2-4a)^2-4(2-4a)⩽0\),解得\(-\dfrac{1}{2} ⩽a⩽\dfrac{1}{2}\) . -
答案 \([1,+∞)\)
解析 由题意得\(f'(x)=\cos x-a\sin x⩽0\)在区间\(\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)\)上恒成立,
所以 \(a \geqslant \dfrac{\cos x}{\sin x}=\dfrac{1}{\tan x}\)在区间\(\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)\)上恒成立,
因为当\(x \in\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)\)时,\(0<\dfrac{1}{\tan x}<1\),所以\(a⩾1\). -
答案 \(\left(-\infty,-e^{-2}\right]\)
解析 函数\(f(x)=(x-a)\ln x\),\(a∈R\),则\(f^{\prime}(x)=\ln x+1-\dfrac{a}{x}\),
函数\(f(x)\)在\((0,+∞)\)上为增函数,转化为\(f'(x)⩾0\)在\((0,+∞)\)上恒成立,
即\(a⩽x\ln x+x\)在\((0,+∞)\)上恒成立,
令\(g(x)=x\ln x+x\),\(x∈(0,+∞)\),则\(g'(x)=\ln x+2\),
由\(g'(x)=0\)得\(x=e^{-2}\),由\(g'(x)>0\)得\(x>e^{-2}\),由\(g'(x)<0\)得\(0<x<e^{-2}\),
\(\therefore g(x)\)在\((0,e^{-2})\)上单调递减,在\((e^{-2},+∞)\)上单调递增,
\(\therefore\)当\(x=e^{-2}\)时,\(g(x)\)取得极小值也是最小值,且\(g(e^{-2})=-e^{-2}\),
\(\therefore a⩽-e^{-2}\),
故实数\(a\)的取值范围为\((-∞,-e^{-2}]\).
【题型2】比较大小
【典题1】 若\(1<x<y<2\),则( )
A.\(e^x+3y<e^y+3x\) \(\qquad \qquad \qquad \qquad \qquad \qquad\) B.\(e^x+3y>e^y+3x\)
C.\(x^3+3y^2<y^3+3x^2\) \(\qquad \qquad \qquad \qquad \qquad \qquad\) D.\(x^3+3y^2>y^3+3x^2\)
解析 令\(f(x)=e^x-3x\),\(x∈(1,2)\),
则\(f'(x)=e^x-3\),由\(f'(x)=0\),可得\(x=\ln 3\),
所以当\(x∈(1,\ln 3)\)时,\(f'(x)<0\),\(f(x)\)单调递减,
当\(x∈(\ln 3,2)\)时,\(f'(x)>0\),\(f(x)\)单调递增,
所以当\(1<x<y<\ln 3\)时,\(e^x-3x>e^y-3y\),即\(e^x+3y>e^y+3x\),
当\(\ln 3<x<y<2\)时,\(e^x-3x<e^y-3y\),
即\(e^x+3y<e^y+3x\),故\(A\),\(B\)错误;
令\(g(x)=x^3-3x^2\),\(x∈(1,2)\),
则\(f'(x)=3x^2-6x=3x(x-2)<0\),
所以\(g(x)\)在\((1,2)\)上单调递减,
因为\(1<x<y<2\),所以\(x^3+3y^2>y^3+3x^2\),故\(C\)错误,\(D\)正确.
故选:\(D\).
点拨 需要根据不等式构造函数,再通过函数单调性判断大小.
【巩固练习】
1.若\(x, y \in\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\),且\(x\sin x-y\sin y>0\),则下列不等式一定成立的是( )
A.\(x<y\) \(\qquad \qquad \qquad \qquad\) B.\(x>y\) \(\qquad \qquad \qquad \qquad\) C.\(|x|<|y|\) \(\qquad \qquad \qquad \qquad\) D.\(|x|>|y|\)
2.若对于任意的\(0<x_1<x_2<a\),都有 \(\dfrac{\ln x_1}{x_1}-\dfrac{\ln x_2}{x_2}<\dfrac{1}{x_2}-\dfrac{1}{x_1}\),则\(a\)的最大值为( )
A.\(2e\) \(\qquad \qquad \qquad \qquad\) B.\(e\) \(\qquad \qquad \qquad \qquad\) C.\(1\) \(\qquad \qquad \qquad \qquad\) D.\(\dfrac{1}{2}\)
3.若\(0<x_1<x_2<1\),则下列结论正确的是( )
A.\(\ln \dfrac{x_1}{x_2}<e^{x_1}-e^{x_2}\) \(\qquad \qquad \qquad \qquad\) B. \(\ln \dfrac{x_1}{x_2}>e^{x_1}-e^{x_2}\)
C.\(\dfrac{x_1}{x_2}>\sqrt{e^{x_1-x_2}}\) \(\qquad \qquad \qquad \qquad \qquad\) D. \(\dfrac{x_1}{x_2}<\sqrt{e^{x_1-x_2}}\)
参考答案
-
答案 \(D\)
解析 令\(f(x)=x\sin x\),\(x \in\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\),则\(f(x)\)为偶函数,
当\(x>0\)时,\(f'(x)=\sin x+x\cos x>0\),即\(f(x)\)在\(\left[0, \dfrac{1}{2} \pi\right]\)上单调递增,
根据偶函数的对称性可知,\(f(x)\)在\(\left[-\dfrac{1}{2} \pi, 0\right)\)上单调递减,距离对称轴越远,函数值越大,
由\(x\sin x-y\sin y>0\),可得\(x\sin x>y\sin y\),即\(f(x)>f(y)\),
从而可得\(|x|>|y|\).
故选:\(D\). -
答案 \(C\)
解析 \(\because \dfrac{\ln x_1}{x_1}-\dfrac{\ln x_2}{x_2}<\dfrac{1}{x_2}-\dfrac{1}{x_1}\),\(\therefore \dfrac{\ln x_1+1}{x_1}<\dfrac{\ln x_2+1}{x_2}\),
据此可得函数\(f(x)=\dfrac{\ln x+1}{x}\)在定义域\((0,a)\)上单调递增,
其导函数:\(f^{\prime}(x)=\dfrac{1-(\ln x+1)}{x^2}=-\dfrac{\ln x}{x^2} \geq 0\)在\((0,a)\)上恒成立,
据此可得:\(0<x≤1\),
即实数\(a\)的最大值为\(1\).
故选:\(C\). -
答案 \(D\)
解析 对于\(A\):由\(\ln \dfrac{x_1}{x_2}<e^{x_1}-e^{x_2}\) ,
得\(\ln x_1-\ln x_2<e^{x_1}-e^{x_2}\),得:\(e^{x_1}-\ln x_1>e^{x_2}-\ln x_2\),
令\(f(x)=e^x-\ln x\),则\(f(x_1)>f(x_2)\),
由函数\(f(x)=e^x-\ln x\),得:\(f'(x)=e^x-\dfrac{1}{x} \),\(f'(x)\)在\((0,1)\)递增,
而\(f'(1)=e-1>0\),\(x→0\)时,\(f'(x)→-∞\),
故存在\(x_0∈(0,1)\),使得\(f'(x_0)=0\),
故\(f(x)\)在\((0,x_0)\)递减,在\((x_0,1)\)递增,
①若\(x_0<x_1<x_2<1\),则\(f(x_1 )<f(x_2 )\),
②若\(0<x_1<x_2<x_0\),则\(f(x_1 )>f(x_2 )\),
③若\(x_1<x_0<x_2\),则\(f(x_1 )\)和\(f(x_2 )\)无法比较大小,故\(A\)错误;
同理\(B\)错误;
对于\(C\):令\(g(x)=x^2 e^x\),\((0<x<1)\),
则\(g'(x)=e^x (x^2+2x)>0\),
故\(g(x)\)在\((0,1)\)递增,由\(0<x_1<x_2<1\),得\(g(x_1 )<g(x_2 )\),
即\(x_1^2 e^{x_1}<x_2^2 e^{x_2}\),即\(\left(\dfrac{x_1}{x_2}\right)^2<\dfrac{e^{x_2}}{e^{x_1}}\),
故\(\dfrac{x_1}{x_2}<\sqrt{e^{x_2-x_1}}\),
故\(C\)错误,\(D\)正确;
故选:\(D\).
【题型3】比较数值大小
【典题1】 若 \(a=\dfrac{\ln 4}{4}\), \(b=\dfrac{\ln 5.3}{5.3}\), \(c=\dfrac{\ln 6}{6}\),则\(a\)、\(b\)、\(c\)的大小是( )
A.\(a<b<c\) \(\qquad \qquad\) B.\(c<b<a\) \(\qquad \qquad\) C.\(c<a<b\) \(\qquad \qquad\) D.\(b<a<c\)
解析 设 \(f(x)=\dfrac{\ln x}{x}\), \(f^{\prime}(x)=\dfrac{1-\ln x}{x^2}\),
令\(f'(x)>0\),可得\(0<x<e\),令\(f'(x)<0\),可得\(x>e\),
所以\(f(x)\)在\((0,e)\)上单调递增,在\((e,+∞)\)上单调递减,
因为\(e<4<5.3<6\),
所以\(f(4)>f(5.3)>f(6)\),
即 \(\dfrac{\ln 4}{4}>\dfrac{\ln 5.3}{5.3}>\dfrac{\ln 6}{6}\),即\(a>b>c\).
故选:\(B\).
点拨 需要数值的结构特点构造函数,再通过函数单调性判断大小.
【典题2】 已知 \(a=e^{0.1}-1\),\(b=\sin 0.1\),\(c=\ln 1.1\),则( )
A.\(a<b<c\) \(\qquad \qquad\) B.\(b<c<a\) \(\qquad \qquad\) C.\(c<a<b\) \(\qquad \qquad\) D.\(c<b<a\)
解析 令\(f(x)=e^x-x-1\),则\(f'(x)=e^x-1\),
当\(x∈(0,+∞)\)时,\(f'(x)>0\),故\(f(x)\)在\((0,+∞)\)上是增函数,
故\(f(0.1)>f(0)\),即 \(e^{0.1}-0.1-1>0\),故\(a=e^{0.1}-1>0.1\),
令\(g(x)=\sin x-x\),则\(g'(x)=\cos x-1<0\)在\((0,1)\)上恒成立,
故\(g(x)=\sin x-x\)在\((0,1)\)上单调递减,
故\(g(0.1)<g(0)\),即\(\sin 0.1-0.1<0\),即\(b=\sin 0.1<0.1\),
令\(h(x)=\ln (x+1)-\sin x\),
则 \(h^{\prime}(x)=\dfrac{1}{x+1}-\cos x=\dfrac{1-(x+1) \cos x}{x+1}\),
令\(m(x)=1-(x+1)\cos x\),
则\(m'(x)=-\cos x+(x+1)\sin x\),
易知\(m'(x)\)在\(\left(0, \dfrac{\pi}{6}\right)\)上是增函数,
且\(m^{\prime}\left(\dfrac{\pi}{6}\right)=-\dfrac{\sqrt{3}}{2}+\left(1+\dfrac{\pi}{6}\right) \dfrac{1}{2}=\dfrac{-6 \sqrt{3}+6+\pi}{12}<0\),
故\(m'(x)<0\)在\(\left(0, \dfrac{\pi}{6}\right)\)上恒成立,故\(m(x)\)在\(\left(0, \dfrac{\pi}{6}\right)\)上是减函数,
又\(\because m(0)=1-1=0\),故\(m(x)<0\)在\(\left(0, \dfrac{\pi}{6}\right)\)上恒成立,
故\(h'(x)<0\)在\(\left(0, \dfrac{\pi}{6}\right)\)上恒成立,故\(h(x)\)在\(\left(0, \dfrac{\pi}{6}\right)\)上是减函数,
故\(h(0.1)<h(0)=0\),即\(\ln 1.1-\sin 0.1<0\),即\(c<b\),
故\(c<b<a\),
故选:\(D\).
【巩固练习】
1.已知 \(a=\dfrac{\ln \sqrt{2}}{4}\), \(b=\dfrac{1}{e^2}\), \(c=\dfrac{\ln \pi}{2 \pi}\),则\(a\)、\(b\)、\(c\)的大小关系为( )
A.\(a<c<b\) \(\qquad \qquad\) B. \(b<a<c\) \(\qquad \qquad\) C. \(a<b<c\) \(\qquad \qquad\) D. \(c<a<b\)
2.已知\(a=\dfrac{3}{\ln 3}\),\(b=e\), \(c=\dfrac{e^2}{2}\)(\(e\)为然对数的底数),则\(a\)、\(b\)、\(c\)的大小关系为( )
A.\(c>a>b\) \(\qquad \qquad\) B.\(c>b>a\) \(\qquad \qquad\) C.\(a>c>b\) \(\qquad \qquad\) D.\(b>c>a\)
3.设\(a=e^{-1}\) , \(b=\dfrac{1}{2} e^{-\frac{1}{2}}\),\(c=\ln 2\),则\(a\)、\(b\)、\(c\)的大小关系为( )
A.\(b<c<a\) \(\qquad \qquad\) B.\(a<b<c\) \(\qquad \qquad\) C. \(b<a<c\) \(\qquad \qquad\) D. \(c<a<b\)
4.设 \(a=\dfrac{10}{11}\), \(b=\dfrac{1}{e^{0.1}}\),\(c=0.9\),则( )
A.\(c<b<a\) \(\qquad \qquad\) B.\(c<a<b\) \(\qquad \qquad\) C.\(b<c<a\) \(\qquad \qquad\) D.\(a<c<b\)
参考答案
-
答案 \(C\)
解析 因为\(a=\dfrac{\ln \sqrt{2}}{4}=\dfrac{\ln 2}{8}=\dfrac{2 \ln 2}{16}=\dfrac{\ln 4}{16}\),\(b=\dfrac{1}{e^2}=\dfrac{\ln e}{e^2}\),\(c=\dfrac{\ln \pi}{2 \pi}=\dfrac{\ln \sqrt{\pi}}{\pi}\),
所以构造函数 \(f(x)=\dfrac{\ln x}{x^2}\),
则 \(f^{\prime}(x)=\dfrac{x-2 x \ln x}{x^4}=\dfrac{1-2 \ln x}{x^3}\),
令\(f'(x)>0\)得\(0<x<\sqrt{e}\);令\(f'(x)<0\)得\(x>\sqrt{e}\);
所以函数\(f(x)\)在\((0,\sqrt{e})\)上单调递增,在\((\sqrt{e},+∞)\)上单调递减,
因为\(\sqrt{e}<\sqrt{\pi}<e<4\),
所以\(f(\sqrt{\pi})>f(e)>f(4)\),即\(c>b>a\).
故选:\(C\). -
答案 \(A\)
解析 \(\because a=\dfrac{3}{\ln 3}\), \(b=e=\dfrac{e}{\ln e}\), \(c=\dfrac{e^2}{2}=\dfrac{e^2}{\ln e^2}\),
\(\therefore\)令\(f(x)=\dfrac{x}{\ln x}\),则 \(f^{\prime}(x)=\dfrac{\ln x-1}{(\ln x)^2}\),
\(\therefore\)当\(x∈(e,+∞)\)时,\(f'(x)>0\),
\(\therefore f(x)\)在\([e,+∞)\)上是增函数,
而\(a=\dfrac{3}{\ln 3}=f(3)\),\(b=e=\dfrac{e}{\ln e}=f(\mathrm{e})\),\(c=\dfrac{e^2}{2}=\dfrac{e^2}{\ln e^2}=f\left(e^2\right)\),
且\(e<3<e^2\),则\(c>a>b\),
故选:\(A\). -
答案 \(C\)
解析 设\(\text { 没 } f(x)=x e^{-x}\) ,则 \(f^{\prime}(x)=(1-x) e^{-x}\),
当\(x≤1\)时,\(f'(x)≥0\),\(f(x)\)单调递增,
当\(x>1\)时,\(f'(x)<0\),\(f(x)\)单调递减,
\(\therefore f(1)>f(\dfrac{1}{2} )\),即\(a>b\),
\(\because c=\ln 2>\ln \sqrt{e}=\dfrac{1}{2}>\dfrac{1}{e}=a\),\(\therefore b<a<c\),
故选:\(C\). -
答案 \(A\)
解析 令\(f(x)=e^x-(x+1)\),则\(f'(x)=e^x-1\),
当\(x>0\)时,\(f'(x)>0\),当\(x<0\)时,\(f'(x)<0\),
\(\therefore\)当\(x=0\)时,\(f(x)\)取得最小值,即\(f(x)⩾f(0)=0\),
\(\therefore e^x⩾x+1\),\(\therefore e^{0.1}>0.1+1=1.1\),
\(\therefore \dfrac{1}{e^{0.1}}<\dfrac{1}{1.1}=\dfrac{10}{11}\),即\(b<a\),
\(\therefore \dfrac{1}{e^{0.1}}=e^{-0.1}>-0.1+1=0.9\),即\(b>c\),
综上,可得\(c<b<a\).
故选:\(A\).
分层练习
【A组---基础题】
1.已知函数\(f(x)=(x^2-4x+a) e^x\)在区间\([-2,3]\)上单调递减,则实数\(a\)的取值范围是( )
A.\((-∞,-4]\) \(\qquad \qquad\) B.\((-∞,5]\) \(\qquad \qquad\) C.\((-∞,1]\) \(\qquad \qquad\) D.\((-∞,-8]\)
2.已知\(a=\dfrac{\ln 3}{3}\),\(b=\dfrac{1}{e}\),\(c=\dfrac{\ln 5}{5}\),则以下不等式正确的是( )
A.\(c>b>a\) \(\qquad \qquad\) B.\(a>b>c\) \(\qquad \qquad\) C.\(b>a>c\) \(\qquad \ \qquad\) D.\(b>c>a\)
3.已知\(a=\dfrac{1}{e}\),\(b=\dfrac{\ln 5}{5}\),\(c=\dfrac{2}{5}\),则\(a\)、\(b\)、\(c\)的大小关系为( )
A.\(b<c<a\)\(\qquad \qquad\) B.\(c<a<b\) \(\qquad \qquad\) C.\(c<b<a\) \(\qquad \qquad\) D.\(b<a<c\)
4.下列不等式正确的是( )
A.\(\dfrac{\ln 2}{2}>\dfrac{\ln 4}{4}\) \(\qquad \qquad\) B.\(\dfrac{2 \ln 3}{3}>\ln 2\) \(\qquad \qquad\) C.\(e\ln 10>10\) \(\qquad \qquad\) D. \(2^{\sqrt{6}}>6\)
5.设\(a=e^{0.5}\),\(b=\dfrac{3}{e}\),\(c=\ln 3\),其中\(e\)为自然对数的底数,则\(a\)、\(b\)、\(c\)的大小关系是( )
A.\(c<a<b\) \(\qquad \qquad\) B. \(b<c<a\) \(\qquad \qquad\) C. \(a<c<b\) \(\qquad \qquad\) D. \(c<b<a\)
6.已知\(a=\ln \sqrt[3]{3}\),\(b=e^{-1}\),\(c=\ln (e+1)-1\),则( )
A.\(a<b<c\) \(\qquad \qquad\) B. \(c<a<b\) \(\qquad \qquad\) C. \(a<c<b\) \(\qquad \qquad\) D. \(b<c<a\)
7.若\(\ln x-\ln y<\dfrac{1}{\ln x}-\dfrac{1}{\ln y}(x>1, y>1)\),则( )
A. \(e^{y-x}>1\) \(\qquad \qquad\) B. \(e^{y-x}<1\) \(\qquad \qquad\) C. \(e^{y-x-1}>1\) \(\qquad \qquad\) D. \(e^{y-x-1}<1\)
8.若\(\alpha, \beta \in\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\),且\(α\sin α-β\sin β>\cos α-\cos β\),则下列结论中必定成立的是( )
A.\(α>β\) \(\qquad \qquad \qquad\) B.\(α>-β\) \(\qquad \qquad \qquad\) C.\(α<β\) \(\qquad \qquad \qquad\) D.\(|α|>|β|\)
9.已知 \(0<\alpha<\beta<\dfrac{\pi}{2}\),则下列不等式中恒成立的是( )
A.\(α^α<β^β\)\(\qquad \qquad \qquad\) B.\(α^α≤β^β\) \(\qquad \qquad \qquad\) C.\(α^β>β^α\) \(\qquad \qquad \qquad\) D.\(α^β<β^α\)
10.已知函数\(f(x)=\dfrac{1}{2} x^2+2ax-\ln x\),若\(f(x)\)在区间\([1,3]\)上单调递增,则实数\(a\)的范围为\(\underline{\quad \quad}\) .
参考答案
-
答案 \(A\)
解析 因为函数\(f(x)=(x^2-4x+a) e^x\)在\([-2,3]\)上单调递减,
所以\(f'(x)=(x^2-2x+a-4) e^x⩽0\)在\([-2,3]\)上恒成立,
则\(x^2-2x+a-4⩽0\)在\([-2,3]\)上恒成立,
即\(a \leqslant\left(-x^2+2 x+4\right)_{\text {min }}\),
又\(-x^2+2x+4\)的最小值为\(-4\),所以\(a⩽-4\),
故选:\(A\). -
答案 \(C\)
解析 令 \(f(x)=\dfrac{\ln x}{x}\),则 \(f^{\prime}(x)=\dfrac{1-\ln x}{x^2}\),
当\(x>e\)时,\(f'(x)<0\),函数单调递减,
因为\(5>3>e\),所以\(f(5)<f(3)<f(e)\),
即 \(\dfrac{\ln 5}{5}<\dfrac{\ln 3}{3}<\dfrac{\ln e}{e}\),即\(b>a>c\).
故选:\(C\). -
答案 \(D\)
解析 \(\because a-c=\dfrac{1}{e}-\dfrac{2}{5}=\dfrac{5-2 e}{5 e}<0\),\(\therefore a<c\),
令 \(f(x)=\dfrac{\ln x}{x}\),则\(f^{\prime}(x)=\dfrac{1-\ln x}{x^2}\),
故当\(x∈(e,+∞)\)时, \(f^{\prime}(x)=\dfrac{1-\ln x}{x^2}<0\),
故\(f(x)\)在\([e,+∞)\)上是减函数,
而 \(a=\dfrac{1}{e}=f(\mathrm{e})\), \(b=\dfrac{\ln 5}{5}=f(5)\),
故\(b<a\),故\(b<a<c\),
故选:\(D\). -
答案 \(B\)
解析 因为 \(\dfrac{\ln 4}{4}=\dfrac{2 \ln 2}{4}=\dfrac{\ln 2}{2}\),\(A\)错误;
\(2\ln 3=\ln 9\),\(3\ln 2=\ln 8\),\(\ln 9>\ln 8\),故\(2\ln 3>3\ln 2\),
所以 \(\dfrac{2 \ln 3}{3}>\ln 2\),\(B\)正确;
令 \(f(x)=\dfrac{\ln x}{x}\),则 \(f^{\prime}(x)=\dfrac{1-\ln x}{x^2}\),
易得,当\(0<x<e\)时,\(f'(x)>0\),函数单调递增,
当\(x>e\)时,\(f'(x)<0\),函数单调递减,
故\(f(10)<f(e)\),
所以 \(\dfrac{\ln 10}{10}<\dfrac{1}{e}\),即\(e\ln 10<10\),\(C\)错误;
根据二次函数与幂函数性质可知,当\(2<x<4\)时,\(2^x<x^2\),
所以 \(2^{\sqrt{6}}<\sqrt{6}^2=6\),\(D\)错误.
故选:\(B\). -
答案 \(D\)
解析 易知\(a=e^{0.5}>\left(\dfrac{9}{4}\right)^{0.5}=1.5\),\(b=\dfrac{3}{e}<\dfrac{3}{2}=1.5\),\(c=\ln 3<\ln \sqrt{e^3}=1.5\),
令\(f(x)=\ln x-\dfrac{x}{e}\),则\(f^{\prime}(x)=\dfrac{1}{x}-\dfrac{1}{e}\),
故当\(x∈(e,+∞)\)时,\(f'(x)<0\),
故\(f(x)\)在\((e,+∞)\)上是减函数,
而\(f(e)=0\),故\(f(3)<0\),
即\(\ln 3-\dfrac{3}{e}<0\),即 \(\ln 3<\dfrac{3}{e}\),
故\(c<b<a\),
故选:\(D\). -
答案 \(B\)
解析 令 \(f(x)=\dfrac{\ln x}{x}(x \geq e)\),
则\(f^{\prime}(x)=\dfrac{1-\ln x}{x^2}<0\),即\(f(x)\)在\([e,+∞)\)上单调递减,
所以\(a=\ln \sqrt[3]{3}=\dfrac{1}{3} \ln 3=f(3)\),\(b=e^{-1}=\dfrac{1}{e}=f(e)\),所以\(a<b\),
因为\(c=\ln (e+1)-1=\ln \dfrac{e+1}{e}=\ln \left(1+\dfrac{1}{e}\right)\),
因为\(\sqrt[3]{3}>1+\dfrac{1}{e}\),所以 \(\ln \sqrt[3]{3}>\ln \left(1+\dfrac{1}{e}\right)\),即\(a>c\),
综上,\(b>a>c\).
故选:\(B\). -
答案 \(A\)
解析 依题意, \(\ln x-\dfrac{1}{\ln x}<\ln y-\dfrac{1}{\ln y}\),
令\(f(t)=t-\dfrac{1}{t}\),则 \(f^{\prime}(t)=1+\dfrac{1}{t^2}>0\),
\(\therefore\)函数\(f(t)\)在\(R\)上单调递增,
\(\because \ln x-\dfrac{1}{\ln x}<\ln y-\dfrac{1}{\ln y}\),即 \(f(\ln x)<f(\ln y)\),
\(\therefore \ln x<\ln y\),
\(\therefore 1<x<y\),
\(\therefore y-x>0\),
\(\therefore e^{y-x}>e^0=1\).
故选:\(A\). -
答案 \(D\)
解析 不等式\(α\sin α-β\sin β>\cos α-\cos β\),可整理为\(α\sin α-\cos α>β\sin β-\cos β\),
令\(f(x)=x\sin x-\cos x\), \(x \in\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\),
上述不等式等价于\(f(α)>f(β)\),
\(\because f(-x)=(-x)\sin (-x)-\cos (-x)=x\sin x-\cos x=f(x)\),
\(\therefore f(x)\)为偶函数.
又\(f'(x)=2\sin x+x\cos x\),
\(\therefore\)当\(0<x≤\dfrac{\pi}{2} \)时,\(\sin x>0\),\(x\cos x≥0\),\(\therefore f'(x)>0\),
\(\therefore f(x)\)在 \(\left(0, \dfrac{\pi}{2}\right]\)上单调递增,在 \(\left[-\dfrac{\pi}{2}, 0\right)\)上单调递减.
结合\(f(x)\)的单调性和奇偶性可作出函数\(f(x)\)的大致草图如下:
\(\because f(α)>f(β)\), \(\therefore |α|>|β|\).
故选:\(D\). -
答案 \(D\)
解析 构造函数 \(f(x)=\dfrac{\ln x}{x}\),则 \(f^{\prime}(x)=\dfrac{1-\ln x}{x}\),
令\(f'(x)>0\),解得\(0<x<e\),令\(f'(x)<0\),解得\(x>e\),
\(\therefore\)函数\(f(x)\)在(0,e)上单调递增,在(e,+∞)上单调递减,
\(\therefore\)函数\(f(x)\)在 \(\left(0, \dfrac{\pi}{2}\right)\)上单调递增,
\(\therefore f(α)<f(β)\),即 \(\dfrac{\ln \alpha}{\alpha}<\dfrac{\ln \beta}{\beta}\),
\(\therefore β\ln α<α\ln β\),即\(\ln α^β<\ln β^α\),
\(\therefore α^β<β^α\),
故选:\(D\). -
答案 \([0,+∞)\)
解析 由题意知,\(f'(x)=x+2a-\dfrac{1}{x} ⩾0\)在\([1,3]\)上恒成立,
即\(2a⩾\dfrac{1}{x} -x\),
又函数\(y=\dfrac{1}{x} -x\)在\([1,3]\)上单调递减,
所以当\(x=1\)时,函数\(y\)取得最大值,为\(0\),
所以\(2a≥0\),即\(a≥0\).
【B组---提高题】
1.若\(x,y∈(0,+∞)\),\(x+\ln x=e^y+\sin y\),则( )
A.\(\ln (x-y)<0\) \(\qquad \qquad\) B.\(\ln (y-x)>0\) \(\qquad \qquad\) C.\(x<e^y\) \(\qquad \qquad\) D.\(y<\ln x\)
2.已知\(a=8^{10}\),\(b=9^9\),\(c=10^8\),则\(a\),\(b\),\(c\)的大小关系为( )
A.\(b>c>a\) \(\qquad \qquad\) B.\(b>a>c\) \(\qquad \qquad\) C.\(a>c>b\) \(\qquad \qquad\) D.\(a>b>c\)
参考答案
-
答案 \(C\)
解析 设\(f(x)=x-\sin x(x>0)\),则\(f'(x)=1-\cos x⩾0\)(不恒为零),
故\(f(x)\)在\((0,+∞)\)上为增函数,故\(f(x)>f(0)=0\),
所以\(x>\sin x\),故\(y>\sin y\)在\((0,+∞)\)上恒成立,
所以\(x+\ln x<e^y+y=e^y+\ln e^y\),
但\(g(x)=x+\ln x\)为\((0,+∞)\)上的增函数,
故\(x<e^y\),即\(\ln x<y\),
所以\(C\)成立,\(D\)错误.
取\(x=e\),考虑\(1+e=e^y+\sin y\)的解,若\(y⩾e+1\),
则\(e^y⩾e^{e+1}>5>e+2⩾1+e-\sin y\),矛盾,
故\(y<e+1\),即\(y-x<1\),
此时\(\ln (y-x)<0\),故\(B\)错误.
取\(y=1\),考虑\(x+\ln x=e+\sin 1\),
若\(x⩽2\),则\(x+\ln x⩽2+\ln 2<3<e+\dfrac{1}{2}<e+\sin 1\),矛盾,
故\(x>2\),此时\(x-y>1\),此时\(\ln (x-y)>0\),故\(A\)错误,
故选:\(C\). -
答案 \(D\)
解析 令\(f(x)=(18-x)\ln x\),
则\(f^{\prime}(x)=-\ln x+\dfrac{18}{x}-1\)在\(x⩾8\)时单调递减,
又\(f^{\prime}(8)=\dfrac{5}{4}-\ln 8<\dfrac{5}{4}-\ln e^2<0\),
所以\(f'(x)<0\)在\(x⩾8\)时恒成立,
故\(f(x)\)在\(x⩾8\)时单调递减,所以\(f(8)>f(9)>f(10)\).
所以\(10\ln 8>9\ln 9>8\ln 10\),故 \(8^{10}>9^9>10^8\).
故\(a>b>c\).
故选:\(D\).
【C组---拓展题】
1.若 \(x+\dfrac{3}{2} y-2=e^x+3 \ln \dfrac{y}{2}\),其中\(x>2\),\(y>2\),则( )
A.\(e^x<y\) \(\qquad \qquad\) B.\(2x>y\) \(\qquad \qquad\) C. \(4 e^{\frac{x}{2}}>y\) \(\qquad \qquad\) D.\(2e^x>y\)
2.若实数\(a\),\(b\)满足 \(2 \ln a+\ln (2 b) \geq \dfrac{a^2}{2}+4 b-2\),则( )
A.\(a+b=\sqrt{2}+\dfrac{1}{4}\) \(\qquad \qquad\) B.\(a-2b=\sqrt{2}-\dfrac{1}{4} \) \(\qquad \qquad\) C.\(a^2+b>3\) \(\qquad \qquad\) D.\(a^2-4b<1\)
参考答案
-
答案 \(D\)
解析 因为 \(x+\dfrac{3}{2} y-2=e^x+3 \ln \dfrac{y}{2}\),
所以 \(e^x-x=\dfrac{3}{2} y-2-3 \ln \dfrac{y}{2}=2\left(\dfrac{y}{2}-1-\ln \dfrac{y}{2}\right)+\dfrac{y}{2}-\ln \dfrac{y}{2}\),
令\(f(x)=x-1-\ln x\),则 \(f^{\prime}(x)=1-\dfrac{1}{x}=\dfrac{x-1}{x}\),
当\(x>1\)时,\(f'(x)>0\),\(f(x)\)为增函数,
所以\(f(x)>f(1)=0\),
因为\(y>2\),所以 \(\dfrac{y}{2}-1-\ln \dfrac{y}{2}>0\),
所以 \(e^x-x>\dfrac{y}{2}-\ln \dfrac{y}{2}=e^{\ln \dfrac{y}{2}}-\ln \dfrac{y}{2}\),\(x>2\), \(\ln \dfrac{y}{2}>0\),
令\(g(x)=e^x-x\),则\(g'(x)=e^x-1\),
当\(x>0\)时,\(g'(x)>0\),\(g(x)\)单调递增,
所以 \(\text { 以 } e^x-x>\dfrac{y}{2}-\ln \dfrac{y}{2}=e^{\ln \dfrac{y}{2}}-\ln \dfrac{y}{2}\),等价于 \(x>\ln \dfrac{y}{2}\),
所以 \(e^x>\dfrac{y}{2}\),即\(2e^x>y\).
故选:\(D\). -
答案 \(A\)
解析 根据题意,若实数\(a\),\(b\)满足 \(2 \ln a+\ln (2 b) \geq \dfrac{a^2}{2}+4 b-2\),
则\(a>0\)且\(b>0\),
又由 \(\dfrac{a^2}{2}+4 b-2 \geq 2 \times \sqrt{\dfrac{a^2}{2} \times 4 b}-2=2 \sqrt{2 a^2 b}-2\),当且仅当\(a^2=8b\)时等号成立,
则有 \(2 \ln a+\ln (2 b) \geq 2 \sqrt{2 a^2 b}-2\),变形可得 \(\ln \left(2 a^2 b\right)-2 \sqrt{2 a^2 b}+2 \geq 0\),
设 \(g(x)=\ln x-2 \sqrt{x}+2\),
则其导数 \(g^{\prime}(x)=\dfrac{1}{x}-\dfrac{1}{\sqrt{x}}=\dfrac{1-\sqrt{x}}{x}\),
当\(0<x≤1\)时,\(g'(x)≥0\),则\(g(x)\)在区间\((0,1]\)上为增函数,
当\(x≥1\)时,\(g'(x)≤0\),则\(g(x)\)在区间\([1,+∞)\)上为减函数,
则有\(g(x)≤g(1)=0\),
若\(\ln \left(2 a^2 b\right)-2 \sqrt{2 a^2 b}+2 \geq 0\),
即\(g(2a^2 b)≥0\),必有\(2a^2 b=1\),
又\(a^2=8b\),所以\(a=\sqrt{2}\),\(b=\dfrac{1}{4} \) ,
据此分析选项:对于\(A\),\(a+b=\sqrt{2}+\dfrac{1}{4} \),\(A\)正确;
对于\(B\),\(a-2b=\sqrt{2}-\dfrac{1}{2} \),\(B\)错误;
对于\(C\), \(a^2+b=\dfrac{9}{4}\),\(C\)错误;
对于\(D\),\(a^2-4b=1\),\(D\)错误;
故选:\(A\).