5.4.3 正切函数的性质与图象
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【基础过关系列】2022-2023学年高一数学上学期同步知识点剖析精品讲义(人教A版2019)
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必修第一册同步巩固,难度2颗星!
基础知识
正切函数的定义域
根据正切函数的定义,\(\tan α=\dfrac{y}{x}\)(\(α\)的终边与单位圆交于点\(P(x,y)\),当交点\(P(x,y)\)在\(y\)轴上时,\(x=0\),
即当\(x=\dfrac{\pi}{2}+kπ\),\(k∈Z\),则\(\tan α\)没意义.
故\(y=\tan x\)的定义域是\(x≠\dfrac{\pi}{2}+kπ\),\(k∈Z\).
正切函数的周期性
由诱导公式\(\tan (x+π)=\tan x\),\(x∈R\),且\(x≠\dfrac{\pi}{2}+kπ\),\(k∈Z\),
可知道,正切函数是周期函数,周期是\(π\).
注 三角函数\(y=\tan (ωx+φ)\)的最小正周期 \(T=\dfrac{\pi}{|\omega|}\).
正切函数的奇偶性
由诱导公式\(\tan (-x)=-\tan x\),\(x∈R\),且\(x≠\dfrac{\pi}{2}+kπ\),\(k∈Z\),
可知,正切函数是奇函数.
正切函数的图象
设\(x∈\left[0,\dfrac{\pi}{2}\right)\),在直角坐标系中画出角\(x\)的终边与单位圆的角度\(B(x_0,y_0)\),过点\(B\)作\(x\)轴的垂线,垂足是\(M\);
过点\(A(1,0)\)作\(x\)轴的垂线与角\(x\)的终边交于点\(T\),则 \(\tan x=\dfrac{y_0}{x_0}=\dfrac{M B}{O M}=\dfrac{A T}{O A}=A T\).
由此可见,当\(x\in \left[0,\dfrac{\pi}{2}\right)\)时,线段\(AT\)的长度就是相应角\(x\)的正切值,
我们可以利用线段\(AT\)画出函数\(y=\tan x\),\(x\in \left[0,\dfrac{\pi}{2}\right)\)的图象,如下图;
当\(x\in \left[0,\dfrac{\pi}{2}\right)\)时,随着\(x\)的增大,线段\(AT\)的长度也在增大,而且当\(x\)趋向于\(\dfrac{\pi}{2}\)时,\(AT\)的长度趋向于无穷大.
相应地,函数\(y=\tan x\), \(x\in \left[0,\dfrac{\pi}{2}\right)\)的图象从左到右呈不断上升的趋势,且向右上方无限逼近直线\(x=\dfrac{\pi}{2}\).
又因为正切函数是奇函数,且周期为\(π\),
所以我们可以得到正切函数\(y=\tan x\),\(x\in R\),\(x≠\dfrac{\pi}{2}+kπ\),\(k∈Z\)的图象,把它叫做正切曲线.
正切函数的单调性
观察正切曲线可知,正切函数在\(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\)上单调递增.
由正切函数的周期性可得,正切函数在每一个区间\(\left(-\dfrac{\pi}{2}+kπ,\dfrac{\pi}{2}+kπ\right)(k\in Z)\)上都单调递增.
正切函数的值域
当\(x\in\left (-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\)时,\(\tan x\)在\((-∞,+∞)\)内可取到任意实数值, 但没有最大值、最小值.
因此,正切函数的值域是实数集\(R\).
正切函数的图像与性质汇总
注 表中的\(k∈Z\)
\(y=\tan x\) | |
---|---|
图像 | ![]() |
定义域 | \(\{x│x≠kπ+\dfrac{\pi}{2}\}\) |
值域 | \(R\) |
最值 | 既无最大值也无最小值 |
周期性 | \(π\) |
对称中心 | \(\left (\dfrac{k\pi}{2},0\right)\) |
对称轴 | 无对称轴 |
单调性 | 在\(\left(kπ-\dfrac{\pi}{2} ,kπ+\dfrac{\pi}{2}\right)\)上是增函数 |
基本方法
【题型1】正切函数的图象及应用
【典题1】 画出函数\(y=|\tan x|\)的图象,并根据图象判断其单调区间、奇偶性、周期性.
解析 由 \(y=|\tan x|\)得, \(y=\left\{\begin{array}{l}
\tan x, k \pi \leq x<k \pi+\dfrac{\pi}{2}(k \in \mathrm{Z}) \\
-\tan x,-\dfrac{\pi}{2}+k \pi<x<k \pi(k \in \mathrm{Z})
\end{array}\right.\),
其图象如图所示.
由图象可知,函数 \(y=|\tan x|\)是偶函数,单调递增区间为\(\left[kπ,kπ+\dfrac{\pi}{2}\right)(k\in Z)\),
单调递减区间为\(\left(-\dfrac{\pi}{2}+kπ,kπ\right](k\in Z)\),周期为\(π\).
【巩固练习】
1.作出函数\(f(x)=\tan\left(\dfrac{x}{2}-\dfrac{\pi}{3}\right)\)在一个周期内的简图.
参考答案
- 解析 令\(\dfrac{x}{2}-\dfrac{\pi}{3}=0\),则\(x=\dfrac{2\pi}{3}\);令\(\dfrac{x}{2}-\dfrac{\pi}{3}=\dfrac{\pi}{2}\),则\(x=\dfrac{5\pi}{3}\).
令\(\dfrac{x}{2}-\dfrac{\pi}{3}=-\dfrac{\pi}{2}\),则\(x=-\dfrac{\pi}{3}\).
\(\therefore\)函数\(f(x)=\tan\left(\dfrac{x}{2}-\dfrac{\pi}{3}\right)\)的图象与\(x\)轴的一个交点坐标是\(\left(\dfrac{2\pi}{3},0\right)\),
在这个交点左、右两侧相邻的两条渐近线方程分别是\(x=-\dfrac{\pi}{3}\),\(x=\dfrac{5\pi}{3}\),
从而得函数\(y=f(x)\)在一个周期\(\left(-\dfrac{\pi}{3},\dfrac{5\pi}{3}\right)\)内的简图(如图).
【题型2】正切函数的定义域
【典题1】 求下列函数的定义域:
(1) \(y=\dfrac{1}{1+\tan x}\);\(\qquad \qquad\) (2) \(y=\lg (\sqrt{3}-\tan x)\).
解析 (1)要使函数 \(y=\dfrac{1}{1+\tan x}\)有意义,必须且只需 \(\left\{\begin{array}{l}
1+\tan x \neq 0 \\
x \neq k \pi+\dfrac{\pi}{2}(k \in \mathrm{Z})
\end{array}\right.\)
所以函数的定义域为\(\{x∣x\in R\)且\(x≠kπ-\dfrac{\pi}{4},x≠kπ+\dfrac{\pi}{2},k\in Z\}\).
(2)因为\(\sqrt{3} -\tan x>0\),所以\(\tan x<\sqrt{3}\) .
又因为\(\tan x=\sqrt{3}\)时,\(x=\dfrac{\pi}{3}+kπ(k\in Z)\),
根据正切函数图象,得\(kπ-\dfrac{\pi}{2}<x<kπ+\dfrac{\pi}{3}(k\in Z)\),
所以函数的定义域是\(\{x∣kπ-\dfrac{\pi}{2}<x<kπ+\dfrac{\pi}{3},k\in Z\}\).
【巩固练习】
1.函数\(y=\tan \left(\dfrac{\pi}{4}-x\right)\)的定义域为( )
A.\(\{x∣x≠\dfrac{\pi}{4},x\in R\}\) \(\qquad \qquad \qquad \qquad \qquad\) B.\(\{x∣x≠-\dfrac{\pi}{4},x\in R\}\) \(\qquad \qquad \qquad \qquad\)
C.\(\{x∣x≠kπ+\dfrac{\pi}{4},x\in R,k\in Z\}\) \(\qquad \qquad \qquad\) D.\(\{x∣x≠kπ+\dfrac{3\pi}{4},x\in R,k\in Z\}\)
2.求函数 \(y=\sqrt{\tan x+1}+\lg (1-\tan x)\)的定义域.
参考答案
-
答案 \(D\)
解析 由\(\tan \left(\dfrac{\pi}{4}-x\right)=-\tan\left(x-\dfrac{\pi}{4}\right)\),
\(\therefore x-\dfrac{\pi}{4}≠kπ+\dfrac{\pi}{2}\),\(k∈Z\),从而\(x≠kπ+\dfrac{3\pi}{4}\),\(x\in R\),\(k∈Z\). -
答案 \(\left[kπ-\dfrac{\pi}{4},kπ+\dfrac{\pi}{4}\right)(k\in Z)\)
解析 由题意得 \(\left\{\begin{array}{l} \tan x+1 \geq 0 \\ 1-\tan x>0 \end{array}\right.\),即\(-1≤\tan x<1\).
在\(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\)内,满足上述不等式的\(x\)的取值范围是\(\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right)\),
又\(y=\tan x\)的周期为\(π\),
所以函数的定义域是\(\left[kπ-\dfrac{\pi}{4},kπ+\dfrac{\pi}{4}\right)(k\in Z)\).
【题型3】 正切函数的性质
【典题1】 已知函数\(f(x)=\tan\left(x+\dfrac{\pi}{3}\right)\),则下列关于\(f(x)\)的判断正确的是( )
A.在区间\(\left(\dfrac{\pi}{6},π\right)\)上单调递增 \(\qquad \qquad \qquad \quad \qquad\) B.最小正周期是\(π\)
C.图象关于直线\(x=\dfrac{\pi}{6}\)成轴对称 \(\qquad \qquad \qquad \qquad\) D.图象关于点\(\left(\dfrac{\pi}{6},0 \right)\)成中心对称
解析 \(A\).\(x\in \left(\dfrac{\pi}{6},π\right)⇒x+\dfrac{\pi}{3}\in \left(\dfrac{\pi}{2},\dfrac{4\pi}{3}\right)\);故单调递增, \(A\)正确;
\(B\).函数\(f(x)\)的最小正周期是\(π\),故\(B\)正确,
\(C\).正切函数没有对称轴,故\(C\)错误,
\(D\).令\(x+\dfrac{\pi}{3}=\dfrac{k\pi}{2}⇒x=\dfrac{k\pi}{2}-\dfrac{\pi}{3}\),\(k∈Z\);
则\(f(x)\)图象关于点\(\left(\dfrac{\pi}{6},0\right)\)成中心对称,故\(D\)正确,
故选:\(ABD\).
【典题2】 比较大小:\(\tan\left(-\dfrac{7\pi}{4}\right)\) \(\underline{\quad \quad}\) \(\tan \left(-\dfrac{9}{5} \pi\right)\).
解析 \(\because \tan \left(-\dfrac{7}{4} \pi\right)=-\tan \left(2 \pi-\dfrac{\pi}{4}\right)=\tan \dfrac{\pi}{4}\),
\(\tan \left(-\dfrac{9}{5} \pi\right)=-\tan \left(2 \pi-\dfrac{\pi}{5}\right)=\tan \dfrac{\pi}{5},\),
又\(0<\dfrac{\pi}{5}<\dfrac{\pi}{4}<\dfrac{\pi}{2}\),\(y=\tan x\)在\(\left(0,\dfrac{\pi}{2}\right)\)内单调递增,
\(\therefore \tan \dfrac{\pi}{5}<\tan \dfrac{\pi}{4}\),
\(\therefore \tan \left(-\dfrac{7}{4} \pi\right)>\tan \left(-\dfrac{9}{5} \pi\right)\).
【典题3】 若函数\(y=\tan \left(ωx+\dfrac{\pi}{6}\right)\)在\(\left[-\dfrac{\pi}{3},\dfrac{\pi}{3}\right]\)上单调递减,且在\(\left[-\dfrac{\pi}{3},\dfrac{\pi}{3}\right]\)上的最大值为\(\sqrt{3}\) ,
则\(ω\)的值为( )
A. \(-\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) C. \(-1\) \(\qquad \qquad \qquad \qquad\) D.\(1\)
解析 \(∵y=\tan \left(ωx+\dfrac{\pi}{6}\right)\)在\(\left[-\dfrac{\pi}{3},\dfrac{\pi}{3}\right]\)上单调递减,
\(\therefore T=\dfrac{\pi}{|\omega|} \geq \dfrac{2 \pi}{3}\),\(\therefore |ω|≤\dfrac{3}{2}\);
又\(y=\tan \left(ωx+\dfrac{\pi}{6}\right)\)在\(\left[-\dfrac{\pi}{3},\dfrac{\pi}{3}\right]\)上的最大值为\(\sqrt{3}\) ,
\(\therefore\)当\(x=-\dfrac{\pi}{3}\)时,\(y_{\max} =\sqrt{3}\),即\(\tan \left(-\dfrac{\pi}{3} ω+\dfrac{\pi}{6}\right)=\sqrt{3}\),
\(\therefore -\dfrac{\pi}{3} ω+\dfrac{\pi}{6}=kπ+\dfrac{\pi}{3}\),
\(\therefore-\dfrac{\pi}{3} \omega+\dfrac{\pi}{6}=k \pi+\dfrac{\pi}{3}\),
\(\therefore-\dfrac{1}{3} \omega=\dfrac{1}{6}+k(k \in Z)\),又\(|ω|≤\dfrac{3}{2}\),
\(\therefore ω=-\dfrac{1}{2}\).
故选:\(A\).
【巩固练习】
1.函数\(y=\tan \left(2x+\dfrac{\pi}{4}\right)\)的图象( )
A.关于原点对称 \(\qquad \qquad \qquad \qquad \qquad \qquad\) B.关于点\(\left (-\dfrac{\pi}{2},1\right)\)对称
C.关于直线\(x=-\dfrac{\pi}{8}\)对称 \(\qquad \qquad \qquad \qquad\) D.关于点\(\left(\dfrac{\pi}{8},0\right)\)对称
2.已知函数\(f(x)=\tan \left(2x-\dfrac{\pi}{3}\right)\),则下列说法错误的是( )
A. 函数\(f(x)\)的周期为\(\dfrac{\pi}{2}\)
B. 函数\(f(x)\)的值域为\(R\)
C. 点\(\left(\dfrac{\pi}{6},0\right)\)是函数\(f(x)\)的图象一个对称中心
D. \(f\left(\dfrac{2\pi}{5}\right)<f\left(\dfrac{3\pi}{5}\right)\)
3.方程\(\tan(2x+\dfrac{\pi}{3})=\sqrt{3}\)在区间\(\left[0,2π\right)\)上的解的个数是\(\underline{\quad \quad}\) .
4.已知函数\(f(x)=\tan(ωx+φ)(ω≠0,|φ|<\dfrac{\pi}{2})\),点\(\left(\dfrac{\pi}{3},0\right)\)和\(\left(\dfrac{5\pi}{6},0\right)\)是其相邻的两个对称中心,
且在区间\(\left(\dfrac{\pi}{3},\dfrac{2\pi}{3}\right)\)内单调递减,则\(φ=\) \(\underline{\quad \quad}\) .
5.设函数\(f(x)=\tan \left(2x-\dfrac{\pi}{3}\right)\).
(1)求\(f(x)\)的定义域、周期和单调区间;
(2)求不等式\(-1≤f(x)≤\sqrt{3}\)的解集;
(3)求\(f(x)\),\(x\in \left[0,π\right]\)的值域.
参考答案
-
答案 \(D\)
解析 函数\(y=\tan\left(2x+\dfrac{\pi}{4}\right)\)中,令\(2x+\dfrac{\pi}{4}=\dfrac{k\pi}{2}\),\(k∈Z\);
解得\(x=\dfrac{k\pi}{4}-\dfrac{\pi}{8}\),\(k∈Z\);
令\(k=1\),得\(x=\dfrac{\pi}{8}\),
所以\(y=\tan\left(2x+\dfrac{\pi}{4}\right)\)的图象关于原点\(\left(\dfrac{\pi}{8} ,0\right)\)对称,\(D\)正确.
故选:\(D\). -
答案 \(D\)
解析 \(∵f(x)=\tan \left(2x-\dfrac{\pi}{3}\right)\),
\(\therefore\) 函数\(f(x)\)的周期\(T=\dfrac{\pi}{2}\),故\(A\)正确;
由正切函数的图象和性质可知函数\(f(x)\)的值域为\(R\),故\(B\)正确;
由\(2x-\dfrac{\pi}{3}=kπ\),\(k∈Z\)可解得:\(x=\dfrac{k\pi}{2}+\dfrac{\pi}{6}\),\(k∈Z\),
则解得当\(k=0\)时,点\(\left(\dfrac{\pi}{6},0\right)\)是函数\(f(x)\)的图象一个对称中心,故\(C\)正确;
由 \(f\left(\dfrac{2 \pi}{5}\right)=\tan \left(2 \times \dfrac{2 \pi}{5}-\dfrac{\pi}{3}\right)=\tan \dfrac{7 \pi}{15}>0\); \(f\left(\dfrac{3 \pi}{5}\right)=\tan \left(2 \times \dfrac{3 \pi}{5}-\dfrac{\pi}{3}\right)=\tan \dfrac{13 \pi}{15}<0\),
从而\(f\left(\dfrac{2\pi}{5}\right)>f\left(\dfrac{3\pi}{5}\right)\),故\(D\)不正确.
故选:\(D\). -
答案 \(4\)
解析 方程\(\tan(2x+\dfrac{\pi}{3})=\sqrt{3}\) ,
\(\therefore 2x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+kπ\),\(k∈Z\),\(x=\dfrac{k\pi}{2}\),\(k∈Z\);
\(k=0\),\(k=1\),\(k=2\),\(k=3\),
求得方程在区间\([0,2π)\)上的解为\(0\),\(\dfrac{\pi}{2}\),\(π\),\(\dfrac{3\pi}{2}\)共\(4\)个. -
答案 \(\dfrac{\pi}{3}\)
解析 根据题意可得\(\left(\dfrac{\pi}{3},0\right)\)和\(\left(\dfrac{5\pi}{6},0\right)\)是其相邻的两个对称中心得 \(\dfrac{T}{2}=\dfrac{5 \pi}{6}-\dfrac{\pi}{3}=\dfrac{\pi}{2}\),\(\therefore T=π\);
又因为在区间\(\left(\dfrac{\pi}{3},\dfrac{2\pi}{3}\right)\)内单调递减,\(\therefore ω=-1\);
则\(f(x)=\tan(-x+φ)\);
当\(x=\dfrac{\pi}{3}\)时,\(f(\dfrac{\pi}{3})=0\),又\(∵|φ|<\dfrac{\pi}{2}⇒φ=\dfrac{\pi}{3}\). -
答案 (1)\(\left(\dfrac{k \pi}{2}-\dfrac{\pi}{12}, \dfrac{k \pi}{2}+\dfrac{5 \pi}{12}\right)\),\(k∈Z\);(2) \(\left[\dfrac{k\pi}{2}+\dfrac{\pi}{24},\dfrac{k\pi}{2}+\dfrac{\pi}{3}\right]\),\(k∈Z\);(3) \(R\).
解析 (1)\(\because\) 函数\(f(x)=\tan \left(2x-\dfrac{\pi}{3}\right)\),它的周期为\(\dfrac{\pi}{2}\).
根据函数的解析式可得\(2x-\dfrac{\pi}{3}≠kπ+\dfrac{\pi}{2}\),\(k∈Z\).求得 \(x \neq \dfrac{k \pi}{2}+\dfrac{5 \pi}{12}\),
故函数的定义域为\(\{x∣x≠\dfrac{k\pi}{2}+\dfrac{5 \pi}{12},k\in z\}\).
由\(kπ-\dfrac{\pi}{2}<2x-\dfrac{\pi}{3}<kπ+\dfrac{\pi}{2}\),\(k∈Z\)求得 \(\dfrac{k \pi}{2}-\dfrac{\pi}{12}<x<\dfrac{k \pi}{2}+\dfrac{5 \pi}{12}\),
故函数的增区间为 \(\left(\dfrac{k \pi}{2}-\dfrac{\pi}{12}, \dfrac{k \pi}{2}+\dfrac{5 \pi}{12}\right)\),\(k∈Z\).
(2)由不等式\(-1≤f(x)≤\sqrt{3}\),可得\(kπ-\dfrac{\pi}{4}≤2x-\dfrac{\pi}{3}≤kπ+\dfrac{\pi}{3}\),
求得\(\dfrac{k\pi}{2}+\dfrac{\pi}{24}≤x≤\dfrac{k\pi}{2}+\dfrac{\pi}{3}\),
故不等式的解集为\(\left[\dfrac{k\pi}{2}+\dfrac{\pi}{24},\dfrac{k\pi}{2}+\dfrac{\pi}{3}\right]\),\(k∈Z\).
(3)\(\because x\in \left[0,π\right]\),\(\therefore 2x-\dfrac{\pi}{3}\in \left[-\dfrac{\pi}{3},\dfrac{5\pi}{3}\right]\),
故\(\tan \left(2x-\dfrac{\pi}{3}\right)\in R\).
分层练习
【A组---基础题】
1.函数\(f(x)=\tan\left(ωx+\dfrac{\pi}{6}\right)\)的最小正周期为\(2π\),则\(f\left(\dfrac{\pi}{6}\right)=\) ( )
A.\(\dfrac{\sqrt{3}}{3}\) \(\qquad \qquad \qquad \qquad\) B.\(1\) \(\qquad \qquad \qquad \qquad\) C.\(\sqrt{3}\) \(\qquad \qquad \qquad \qquad\) D.\(0\)
2.现有下列四个命题:
①函数\(y=\tan x\)在定义域内是增函数;
②函数\(y=\tan(2x+1)\)的最小正周期是\(π\);
③函数\(y=\tan x\)的图象关于点\((π,0)\)成中心对称;
④函数\(y=\tan x\)的图象关于点\(\left(-\dfrac{\pi}{2},0\right)\)成中心对称.
其中正确命题的个数是( )
A.\(0\) \(\qquad \qquad \qquad \qquad\) B.\(1\) \(\qquad \qquad \qquad \qquad\) C.\(2\) \(\qquad \qquad \qquad \qquad\) D.\(3\)
3.函数\(f(x)=\tan \left(x+\dfrac{\pi}{4}\right)\)的单调区间为( )
A.\(\left(kπ-\dfrac{\pi}{2},kπ+\dfrac{\pi}{2}\right)\),\(k∈Z\) \(\qquad \qquad \qquad \qquad\) B.\(\left(kπ,(k+1)π\right)\),\(k∈Z\) \(\qquad \qquad \qquad \qquad\)
C.\(\left(kπ-\dfrac{3\pi}{4},kπ+\dfrac{\pi}{4}\right)\),\(k∈Z\) \(\qquad \qquad \qquad \qquad\) D.\(\left(kπ-\dfrac{\pi}{4},kπ+\dfrac{3\pi}{4}\right)\),\(k∈Z\)
4.下列关于函数\(y=\tan \left(x+\dfrac{\pi}{3}\right)\)的说法正确的是( )
A. 在区间\(\left(-\dfrac{\pi}{6},\dfrac{5\pi}{6}\right)\)上单调递增
B. 最小正周期是\(π\)
C. 图象关于点\(\left(\dfrac{\pi}{4},0\right)\)成中心对称
D. 图象关于直线\(x=\dfrac{\pi}{6}\)成轴对称
5.关于函数\(f(x)=|\tan x|\)的性质,下列叙述不正确的是( )
A.\(f(x)\)的最小正周期为\(\dfrac{\pi}{2}\)
B.\(f(x)\)是偶函数
C.\(f(x)\)的图象关于直线\(x=\dfrac{k\pi}{2}(k\in Z)\)对称
D.\(f(x)\)在每一个区间\(\left(kπ,kπ+\dfrac{\pi}{2}\right)(k\in Z)\)内单调递增
6.(多选)下列关于函数\(y=\tan \left(2x-\dfrac{\pi}{6}\right)\)的说法正确的是( )
A. 在区间\(\left(-\dfrac{\pi}{6},\dfrac{7\pi}{24}\right)\)上单调递增
B. 最小正周期是\(π\)
C. 图象关于\(\left(\dfrac{\pi}{3},0\right)\)成中心对称
D. 图象关于 \(\left(\dfrac{\pi}{12}, 0\right)\)成中心对称
7.函数\(y=\tan \left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)\),\(x\in \left(0,\dfrac{\pi}{6}\right]\)的值域是\(\underline{\quad \quad}\) .
8.已知函数 \(f(x)=\tan x+\dfrac{1}{\tan x}\),若\(f(a)=5\),则\(f(-a)=\) \(\underline{\quad \quad}\).
9.若\(f(n)=\tan \dfrac{n\pi}{3}\),\((n\in N^* )\),则\(f(1)+f(2)+⋯+f(100)=\) \(\underline{\quad \quad}\).
10.在\((0,2π)\)内,使\(\tan x>1\)成立的x的取值范围是\(\underline{\quad \quad}\).
11.已知函数\(y=3\tan ωx+1\)在\(\left(-\dfrac{\pi}{3},\dfrac{\pi}{4}\right)\)内是减函数,则\(ω\)的取值范围是\(\underline{\quad \quad}\).
12.已知函数\(f(x)=3\tan \left(\dfrac{1}{2} x-\dfrac{\pi}{3}\right)\).
(1)求\(f(x)\)的定义域和值域.(2)讨论\(f(x)\)的周期和单调区间.(3)求\(f(x)\)的对称中心.
13.设函数\(f(x)=\tan (ωx+φ)(ω>0,0<φ<\dfrac{\pi}{2})\),已知函数\(y=f(x)\)的图象与\(x\)轴相邻两个交点的距离为\(\dfrac{\pi}{2}\),
且图象关于点 \(M\left(-\dfrac{\pi}{8}, 0\right)\)对称.
(1)求\(f(x)\)的解析式;
(2)求\(f(x)\)的单调区间;
(3)求不等式\(-1≤f(x)≤\sqrt{3}\)的解集.
参考答案
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答案 \(B\)
解析 由已知 \(\dfrac{\pi}{\omega}=2 \pi\),\(\therefore ω=\dfrac{1}{2}\),\(\therefore f(x)=\tan \left(\dfrac{1}{2} x+\dfrac{\pi}{6}\right)\),
\(\therefore f(\dfrac{\pi}{6})=tan(\dfrac{1}{2}×\dfrac{\pi}{6}+\dfrac{\pi}{6})=\tan \dfrac{\pi}{4}=1\). -
答案 \(C\)
解析 ①函数\(y=\tan x\)在定义域内不是单调函数;故①错误,
②函数\(y=\tan(2x+1)\)的最小正周期是\(\dfrac{\pi}{2}\);故②错误
③函数\(y=\tan x\)的图象关于点\(\left(\dfrac{k\pi}{2},0\right)\)成中心对称,
当\(k=2\)时,对称中心为\((π,0)\);故③正确,
④函数\(y=\tan x\)的图象关于点\(\left(\dfrac{k\pi}{2},0\right)\)成中心对称,
当\(k=-1\)时,关于\(\left(-\dfrac{\pi}{2},0\right)\)成中心对称.故④正确,
故正确是③④,
故选:\(C\). -
答案 \(C\)
解析 由\(-\dfrac{\pi}{2}+kπ<x+\dfrac{\pi}{4}<\dfrac{\pi}{2}+kπ\),\(k∈Z\),
得\(-\dfrac{3\pi}{4}+kπ<x<\dfrac{\pi}{4}+kπ\),\(k∈Z\).故选\(C\). -
答案 \(B\)
解析 对于\(A\),由\(kπ-\dfrac{\pi}{2}<x+\dfrac{\pi}{3}<kπ+\dfrac{\pi}{2}\),\(k∈Z\),
即\(kπ-\dfrac{5\pi}{6}<x<kπ+\dfrac{\pi}{6}\),\(k∈Z\),
当\(k=0\)时,函数的单调递增区间为\(\left(-\dfrac{5\pi}{6},\dfrac{\pi}{6}\right)\),
当\(k=1\)时,函数的单调递增区间为\(\left(\dfrac{\pi}{6},\dfrac{7\pi}{6}\right)\),
故\(f(x)\)在区间\(\left(-\dfrac{\pi}{6},\dfrac{5\pi}{6}\right)\)上单调递增错误,\(A\)错误;
对于\(B\),函数\(f(x)\)的最小正周期为\(T=π\),命题正确;
对于\(C\),由\(x+\dfrac{\pi}{3}=\dfrac{k\pi}{2}\), 得\(x=-\dfrac{\pi}{3}+\dfrac{k\pi}{2}\),\(k∈Z\),
即函数\(f(x)\)的对称中心为\(\left(-\dfrac{\pi}{3}+\dfrac{k\pi}{2},0\right)\),
当\(k=1\)时,对称中心为\(\left(\dfrac{\pi}{6},0\right)\),\(f(x)\)图象不关于点\(\left(\dfrac{\pi}{4},0\right)\)成中心对称,\(C\)错误;
对于\(D\),正切函数是奇函数,图象没有对称轴,\(D\)错误.
故选:\(B\). -
答案 \(A\)
解析 对于函数\(f(x)=|\tan x|\)的性质,根据该函数的图象知,其最小正周期为\(π\),\(A\)错误;
又\(f(-x)=|\tan(-x)|=|\tan x|=f(x)\),所以\(f(x)\)是定义域上的偶函数,\(B\)正确;
根据函数\(f(x)\)的图象知,\(f(x)\)的图象关于直线\(x=\dfrac{k\pi}{2}(k\in Z)\)对称,\(C\)正确;
根据\(f(x)\)的图象知,\(f(x)\)在每一个区间\(\left(kπ,kπ+\dfrac{\pi}{2}\right)(k\in Z)\)内单调递增,\(D\)正确.
故选:\(A\). -
答案 \(ACD\)
解析 当\(x\in \left(-\dfrac{\pi}{6},\dfrac{7\pi}{24}\right)\)时,\(2x-\dfrac{\pi}{6}\in \left(-\dfrac{\pi}{2},\dfrac{5\pi}{12}\right)\),
所以\(y=\tan \left(2x-\dfrac{\pi}{6}\right)\)在区间\(\left(-\dfrac{\pi}{6},\dfrac{7\pi}{24}\right)\)上单调递增,故\(A\)正确;
函数\(y=\tan \left(2x-\dfrac{\pi}{6}\right)\)的最小正周期是\(\dfrac{\pi}{2}\),故\(B\)错误;
当\(x=\dfrac{\pi}{3}\)时,\(2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}\),所以函数\(y=\tan \left(2x-\dfrac{\pi}{6}\right)\)的图象关于\(\left(\dfrac{\pi}{3},0\right)\)成中心对称,故\(C\)正确;
当\(x=\dfrac{\pi}{12}\)时\(2x-\dfrac{\pi}{6}=0\),所以函数\(y=\tan \left(2x-\dfrac{\pi}{6}\right)\)的图象关于\(\left(\dfrac{\pi}{12},0\right)\)成中心对称,故\(D\)正确.
故选:\(ACD\). -
答案 \(\left(1,\sqrt{3} \right]\)
解析 由\(x\in \left(0,\dfrac{\pi}{6}\right]\),\(\therefore \dfrac{x}{2}+\dfrac{\pi}{4}\in \left(\dfrac{\pi}{4},\dfrac{\pi}{3}\right]\),结合正切函数的性质可得:\(1<y≤\sqrt{3}\).
故答案为:\(\left(1,\sqrt{3} \right]\). -
答案 \(-5\)
解析 \(f(x)\)的定义域为\(\left(kπ-\dfrac{\pi}{2},kπ\right)∪\left(kπ,kπ+\dfrac{\pi}{2}\right)(k\in Z)\).
可知\(f(x)\)的定义域关于原点对称.
又 \(f(-x)=\tan (-x)+\dfrac{1}{\tan (-x)}=-\left(\tan x+\dfrac{1}{\tan x}\right)=-f(x)\).
\(\therefore f(x)\)是奇函数.\(\therefore f(-a)=-f(a)=-5\). -
答案 \(\sqrt{3}\)
解析 \(\because f(n)=\tan \dfrac{n\pi}{3}\),\((n\in N^* )\),
根据正切函数的性质可得其周期 \(T=\dfrac{\pi}{\frac{\pi}{3}}=3\),
\(\therefore f(1)=\sqrt{3}\),\(f(2)=-\sqrt{3}\) ,\(f(3)=0\).
可得:\(f(1)=f(2)+f(3)=0\).
\(\therefore f(1)+f(2)+⋯+f(100)=33\left[f(1)+f(2)+f(3)\right]+f(1)=f(1)=\sqrt{3}\). -
答案 \(\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right)∪\left(\dfrac{5\pi}{4},\dfrac{3\pi}{2}\right)\)
解析 由\(\tan x>1\),可得 \(kπ+\dfrac{\pi}{2}>x>kπ+\dfrac{\pi}{4}\),\(k∈Z\).
再根据\(x\in (0,2π)\),求得\(x\in \left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right)∪\left(\dfrac{5\pi}{4},\dfrac{3\pi}{2}\right)\). -
答案 \(\left[-\dfrac{3}{2},0\right)\)
解析 \(\because\) 函数\(y=3\tan ωx+1\)在\(\left(-\dfrac{\pi}{3},\dfrac{\pi}{4}\right)\)内是减函数,
\(\therefore ω<0\)且函数\(y=3\tan ωx+1\)在\(\left(-\dfrac{\pi}{3},\dfrac{\pi}{3}\right)\)内也是减函数,
\(\therefore T=\left|\dfrac{\pi}{\omega}\right|≥\dfrac{\pi}{3}-(-\dfrac{\pi}{3})=\dfrac{2\pi}{3}\),
\(\therefore |ω|≤\dfrac{3}{2}\),\(\therefore -\dfrac{3}{2}≤ω≤\dfrac{3}{2}\),
又\(ω<0\),\(\therefore -\dfrac{3}{2}≤ω<0\).
故答案为:\(\left[-\dfrac{3}{2},0\right)\). -
答案 (1) 定义域为\(\{x∣x≠2kπ+\dfrac{5\pi}{3},k\in Z\}\),值域为\(R\);
(2) \(2π\),\(\left(-\dfrac{\pi}{3}+2kπ,\dfrac{5\pi}{3}+2kπ\right)\),\(k∈Z\);
(3)\(\left (\dfrac{2\pi}{3}+2kπ,0\right)\),\(k∈Z\).
解析 (1)\(\because\) 函数\(f(x)=3\tan \left(\dfrac{1}{2} x-\dfrac{\pi}{3}\right)\),
\(\therefore \dfrac{1}{2} x-\dfrac{\pi}{3}≠kπ+\dfrac{\pi}{2}\),\(k∈Z\),即\(x≠2kπ+\dfrac{5\pi}{3}\),\(k∈Z\),
\(\therefore f(x)\)的定义域为\(\{x∣x≠2kπ+\dfrac{5\pi}{3},k\in Z\}\),值域为\(R\);
(2) \(f(x)\)的最小正周期是 \(\dfrac{\pi}{\dfrac{1}{2}}=2 \pi\),
又令\(-\dfrac{\pi}{2}+kπ<\dfrac{1}{2} x-\dfrac{\pi}{3}<\dfrac{\pi}{2}+kπ\),\(k∈Z\),
\(\therefore -\dfrac{\pi}{3}+2kπ<x<\dfrac{5\pi}{3}+2kπ\),\(k∈Z\)
\(\therefore f(x)\)的单调增区间为\(\left(-\dfrac{\pi}{3}+2kπ,\dfrac{5\pi}{3}+2kπ\right)\),\(k∈Z\);
(3)令\(\dfrac{1}{2} x-\dfrac{\pi}{3}=kπ\),\(k∈Z\),解得\(x=\dfrac{2\pi}{3}+2kπ\),\(k∈Z\),
此时\(y=f(x)=0\),
\(\therefore\)函数\(f(x)\)的对称中心为\(\left(\dfrac{2\pi}{3}+2kπ,0\right)\),\(k∈Z\). -
答案 (1) \(f(x)=\tan \left(2x+\dfrac{\pi}{4}\right)\);
(2) 单调增区间为\(\left(-\dfrac{3 \pi}{8}+\dfrac{k \pi}{2}, \dfrac{\pi}{8}+\dfrac{k \pi}{2}\right)\),\(k∈Z\).无单调减区间;
(3) \(\left\{x \mid-\dfrac{\pi}{4}+\dfrac{k \pi}{2} \leq x \leq \dfrac{\pi}{24}+\dfrac{k \pi}{2}, \quad k \in Z\right\}\).
解析 (1)\(\because\)函数\(y=f(x)\)的图象与\(x\)轴相邻两交点的距离为\(\dfrac{\pi}{2}\),
即\(T=\dfrac{\pi}{2}\),\(\therefore ω=2\),
\(\therefore f(x)=\tan(2x+φ)\),
\(\because\) 图象关于点 \(M\left(-\dfrac{\pi}{8}, 0\right)\)对称.\(\therefore -2×\dfrac{\pi}{8}+φ=\dfrac{k\pi}{2}\),\(k∈Z\),
\(\because 0<φ<\dfrac{\pi}{2}\),\(\therefore φ=\dfrac{\pi}{4}\),
则\(f(x)=\tan (2x+\dfrac{\pi}{4})\).
(2)令\(-\dfrac{\pi}{2}+kπ<2x+\dfrac{\pi}{4}<\dfrac{\pi}{2}+kπ\),\(k∈Z\),
得\(-\dfrac{3\pi}{8}+\dfrac{k\pi}{2}<x<\dfrac{\pi}{8}+\dfrac{k\pi}{2}\),\(k∈Z\),
\(\therefore\)函数的单调增区间为 \(\left(-\dfrac{3 \pi}{8}+\dfrac{k \pi}{2}, \dfrac{\pi}{8}+\dfrac{k \pi}{2}\right)\),\(k∈Z\).无单调减区间.
(3)由(1)知,\(f(x)=\tan \left(2x+\dfrac{\pi}{4}\right)\).
由\(-1≤\tan (2x+\dfrac{\pi}{4})≤\sqrt{3}\),得\(-\dfrac{\pi}{4}+kπ≤2x+\dfrac{\pi}{4}≤\dfrac{\pi}{3}+kπ\),\(k∈Z\),
即\(-\dfrac{\pi}{4}+\dfrac{k\pi}{2}≤x≤\dfrac{\pi}{24}+\dfrac{k\pi}{2}\),\(k∈Z\),
\(\therefore\)不等式\(-1≤f(x)≤\sqrt{3}\)的解集为\(\{x∣-\dfrac{\pi}{4}+\dfrac{k\pi}{2}≤x≤\dfrac{\pi}{24}+\dfrac{k\pi}{2},k\in Z\}\).
【B组---提高题】
1.(多选)已知函数\(f(x)=\left |\tan \left(\dfrac{1}{2} x-\dfrac{\pi}{6}\right) \right|\),则下列说法正确的是( )
A. \(f(x)\)的周期是\(2π\);
B. \(f(x)\)的值域是\(\{y|y\in R,\)且\(y≠0\}\);
C. 直线\(x=\dfrac{5\pi}{3}\)是函数\(f(x)\)图象的一条对称轴;
D. \(f(x)\)的单调递减区间是\(\left(2kπ-\dfrac{2\pi}{3},2kπ+\dfrac{\pi}{3}\right]\),\(k∈Z\);
2.下列不等式中,正确的是( )
A. \(\tan \dfrac{4 \pi}{7}>\tan \dfrac{3 \pi}{7}\) \(\qquad \qquad \qquad \qquad \qquad \qquad\) B. \(\tan \dfrac{2 \pi}{5}<\tan \dfrac{3 \pi}{5}\)
C. \(\tan \left(-\dfrac{13 \pi}{7}\right)>\tan \left(-\dfrac{15 \pi}{8}\right)\) \(\qquad \qquad \qquad \qquad\) D. \(\tan \left(-\dfrac{13 \pi}{4}\right)<\tan \left(-\dfrac{12 \pi}{5}\right)\)
参考答案
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答案 \(AD\)
解析 \(A\).\(f(x)\)的周期和\(y=\tan \left(\dfrac{1}{2} x-\dfrac{\pi}{6}\right)\)周期相同,即\(T=\dfrac{\pi}{\dfrac{1}{2}}=2 \pi\),故\(A\)正确,
\(B\).\(\because y=\tan \left(\dfrac{1}{2} x-\dfrac{\pi}{6}\right)\)的值域为\(R\),
\(\therefore f(x)≥0\),即函数\(f(x)\)的值域为\(\left[0,+∞\right)\),故\(B\)错误,
\(C\).由绝对值的意义知当\(\dfrac{1}{2}-x-\dfrac{\pi}{6}=\dfrac{k\pi}{2}\),即对称轴为\(x=kπ+\dfrac{\pi}{3}\),\(k∈Z\),
则直线\(x=\dfrac{5\pi}{3}\)不是函数\(f(x)\)图象的一条对称轴,故\(C\)错误,
\(D\).由\(kπ-\dfrac{\pi}{2}<\dfrac{1}{2} x-\dfrac{\pi}{6}≤kπ\),\(k∈Z\) 得\(2kπ-\dfrac{2\pi}{3}<x≤2kπ+\dfrac{\pi}{3}\),\(k∈Z\),
即函数\(f(x)\)的单调递减区间是\(\left(2kπ-\dfrac{2\pi}{3},2kπ+\dfrac{\pi}{3}\right]\),\(k∈Z\),故\(D\)正确,
故选:\(AD\).
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答案 \(C\)
解析 根据正切函数的单调性与周期性,得
对于\(A\), \(\tan \dfrac{4 \pi}{7}<0<\tan \dfrac{3 \pi}{7}\),\(A\)错误;
对于\(B\), \(\tan \dfrac{2 \pi}{5}>0>\tan \dfrac{3 \pi}{5}\),\(B\)错误;
对于\(C\), \(\tan \left(-\dfrac{13 \pi}{7}\right)=\tan \left(-2 \pi+\dfrac{\pi}{7}\right)=\tan \dfrac{\pi}{7}\),
\(\tan \left(-\dfrac{15 \pi}{8}\right)=\tan \left(-2 \pi+\dfrac{\pi}{8}\right)=\tan \dfrac{\pi}{8}\),
\(\dfrac{\pi}{2}>\dfrac{\pi}{7}>\dfrac{\pi}{8}>0\), \(\therefore \tan \dfrac{\pi}{7}>\tan \dfrac{\pi}{8}\),\(C\)正确;
对于\(D\), \(\tan \left(-\dfrac{13 \pi}{4}\right)=\tan \left(-\dfrac{\pi}{4}\right)=-\tan \dfrac{\pi}{4}\),
且 \(\tan \left(-\dfrac{12 \pi}{5}\right)=\tan \left(-\dfrac{2 \pi}{5}\right)=-\tan \dfrac{2 \pi}{5}\),\(D\)错误.
故选:\(C\).