5.4.3 正切函式的性質與圖象
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基礎知識
正切函式的定義域
根據正切函式的定義,\(\tan α=\dfrac{y}{x}\)(\(α\)的終邊與單位圓交於點\(P(x,y)\),當交點\(P(x,y)\)在\(y\)軸上時,\(x=0\),即當\(x=\dfrac{\pi}{2}+kπ\),\(k∈Z\),則\(\tan α\)沒意義.
故\(y=\tan x\)
正切函式的週期性
由誘導公式\(\tan (x+π)=\tan x\),\(x∈R\),且\(x≠\dfrac{\pi}{2}+kπ\),\(k∈Z\),
可知道,正切函式是周期函式,週期是\(π\).
注 三角函式\(y=\tan (ωx+φ)\)的最小正週期 \(T=\dfrac{\pi}{|\omega|}\).
正切函式的奇偶性
由誘導公式\(\tan (-x)=-\tan x\),\(x∈R\),且\(x≠\dfrac{\pi}{2}+kπ\),\(k∈Z\),
可知,正切函式是奇函式.
正切函式的圖象
設\(x∈\left[0,\dfrac{\pi}{2}\right)\),在直角座標系中畫出角\(x\)的終邊與單位圓的角度\(B(x_0,y_0)\),過點\(B\)作\(x\)軸的垂線,垂足是\(M\);過點\(A(1,0)\)作\(x\)軸的垂線與角\(x\)的終邊交於點\(T\),則 \(\tan x=\dfrac{y_0}{x_0}=\dfrac{M B}{O M}=\dfrac{A T}{O A}=A T\).
由此可見,當\(x\in \left[0,\dfrac{\pi}{2}\right)\)時,線段\(AT\)的長度就是相應角\(x\)的正切值,我們可以利用線段\(AT\)畫出函式\(y=\tan x\)
當\(x\in \left[0,\dfrac{\pi}{2}\right)\)時,隨著\(x\)的增大,線段\(AT\)的長度也在增大,而且當\(x\)趨向於\(\dfrac{\pi}{2}\)時,\(AT\)的長度趨向於無窮大.相應地,函式\(y=\tan x\), \(x\in \left[0,\dfrac{\pi}{2}\right)\)的圖象從左到右呈不斷上升的趨勢,且向右上方無限逼近直線\(x=\dfrac{\pi}{2}\).
又因為正切函式是奇函式,且週期為\(π\),所以我們可以得到正切函式\(y=\tan x\),\(x\in R\),\(x≠\dfrac{\pi}{2}+kπ\),\(k∈Z\)的圖象,把它叫做正切曲線.
正切函式的單調性
觀察正切曲線可知,正切函式在\(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\)上單調遞增.
由正切函式的週期性可得,正切函式在每一個區間\(\left(-\dfrac{\pi}{2}+kπ,\dfrac{\pi}{2}+kπ\right)(k\in Z)\)上都單調遞增.
正切函式的值域
當\(x\in\left (-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\)時,\(\tan x\)在\((-∞,+∞)\)內可取到任意實數值, 但沒有最大值、最小值.
因此,正切函式的值域是實數集\(R\).
正切函式的影象與性質彙總
注 表中的k\(k∈Z\)
\(y=\tan x\) |
---|
| 影象| |
| 定義域 | \(\{x│x≠kπ+\dfrac{\pi}{2}\}\) |
| 值域 | \(R\) |
| 最值 | 既無最大值也無最小值 |
| 週期性 | \(π\) |
| 對稱中心 | \(\left (\dfrac{k\pi}{2},0\right)\) |
| 對稱軸 | 無對稱軸 |
| 單調性 | 在\(\left(kπ-\dfrac{\pi}{2} ,kπ+\dfrac{\pi}{2}\right)\)上是增函式 |
基本方法
【題型1】正切函式的圖象及應用
【典題1】 畫出函式\(y=|\tan x|\)的圖象,並根據圖象判斷其單調區間、奇偶性、週期性.
解析 由 \(y=|\tan x|\)得, \(y=\left\{\begin{array}{l}
\tan x, k \pi \leq x<k \pi+\dfrac{\pi}{2}(k \in \mathrm{Z}) \\
-\tan x,-\dfrac{\pi}{2}+k \pi<x<k \pi(k \in \mathrm{Z})
\end{array}\right.\),
其圖象如圖所示.
由圖象可知,函式 \(y=|\tan x|\)是偶函式,單調遞增區間為\(\left[kπ,kπ+\dfrac{\pi}{2}\right)(k\in Z)\),
單調遞減區間為\(\left(-\dfrac{\pi}{2}+kπ,kπ\right](k\in Z)\),週期為\(π\).
【鞏固練習】
1.作出函式\(f(x)=\tan\left(\dfrac{x}{2}-\dfrac{\pi}{3}\right)\)在一個週期內的簡圖.
參考答案
-
解析 令\(\dfrac{x}{2}-\dfrac{\pi}{3}=0\),則\(x=\dfrac{2\pi}{3}\);令\(\dfrac{x}{2}-\dfrac{\pi}{3}=\dfrac{\pi}{2}\),則\(x=\dfrac{5\pi}{3}\).
令\(\dfrac{x}{2}-\dfrac{\pi}{3}=-\dfrac{\pi}{2}\),則\(x=-\dfrac{\pi}{3}\).
\(\therefore\)函式\(f(x)=\tan\left(\dfrac{x}{2}-\dfrac{\pi}{3}\right)\)的圖象與\(x\)軸的一個交點座標是\(\left(\dfrac{2\pi}{3},0\right)\),在這個交點左、右兩側相鄰的兩條漸近線方程分別是\(x=-\dfrac{\pi}{3}\),\(x=\dfrac{5\pi}{3}\),從而得函式\(y=f(x)\)在一個週期\(\left(-\dfrac{\pi}{3},\dfrac{5\pi}{3}\right)\)內的簡圖(如圖).
【題型2】正切函式的定義域
【典題1】 求下列函式的定義域:
(1) \(y=\dfrac{1}{1+\tan x}\);\(\qquad \qquad\) (2) \(y=\lg (\sqrt{3}-\tan x)\).
解析 (1)要使函式 \(y=\dfrac{1}{1+\tan x}\)有意義,必須且只需 \(\left\{\begin{array}{l}
1+\tan x \neq 0 \\
x \neq k \pi+\dfrac{\pi}{2}(k \in \mathrm{Z})
\end{array}\right.\)
所以函式的定義域為\(\{x∣x\in R\)且\(x≠kπ-\dfrac{\pi}{4},x≠kπ+\dfrac{\pi}{2},k\in Z\}\).
(2)因為\(\sqrt{3} -\tan x>0\),所以\(\tan x<\sqrt{3}\) .
又因為\(\tan x=\sqrt{3}\)時,\(x=\dfrac{\pi}{3}+kπ(k\in Z)\),
根據正切函式圖象,得\(kπ-\dfrac{\pi}{2}<x<kπ+\dfrac{\pi}{3}(k\in Z)\),
所以函式的定義域是\(\{x∣kπ-\dfrac{\pi}{2}<x<kπ+\dfrac{\pi}{3},k\in Z\}\).
【鞏固練習】
1.函式\(y=\tan \left(\dfrac{\pi}{4}-x\right)\)的定義域為( )
A.\(\{x∣x≠\dfrac{\pi}{4},x\in R\}\) \(\qquad \qquad \qquad \qquad\) 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}$ \(\qquad \qquad \qquad \qquad\) D.\(\{x∣x≠kπ+\dfrac{3\pi}{4},x\in R,k\in Z\}\)
2.求函式 \(y=\sqrt{\tan x+1}+\lg (1-\tan x)\)的定義域.
參考答案
-
答案 \(D\)
解析 由\(\tan \left(\dfrac{\pi}{4}-x\right)=-\tan\left(x-\dfrac{\pi}{4}\right)\),
\(\therefore x-\dfrac{\pi}{4}≠kπ+\dfrac{\pi}{2}\),\(k∈Z\),從而\(x≠kπ+\dfrac{3\pi}{4}\),\(x\in R\),\(k∈Z\). -
答案 \(\left[kπ-\dfrac{\pi}{4},kπ+\dfrac{\pi}{4}\right)(k\in Z)\)
解析 由題意得 \(\left\{\begin{array}{l} \tan x+1 \geq 0 \\ 1-\tan x>0 \end{array}\right.\),即\(-1≤\tan x<1\).
在\(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\)內,滿足上述不等式的\(x\)的取值範圍是\(\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right)\),
又\(y=\tan x\)的週期為\(π\),
所以函式的定義域是\(\left[kπ-\dfrac{\pi}{4},kπ+\dfrac{\pi}{4}\right)(k\in Z)\).
【題型3】 正切函式的性質
【典題1】 已知函式\(f(x)=\tan\left(x+\dfrac{\pi}{3}\right)\),則下列關於\(f(x)\)的判斷正確的是( )
A.在區間\(\left(\dfrac{\pi}{6},π\right)\)上單調遞增 \(\qquad \qquad \qquad \qquad\) B.最小正週期是\(π\)
C.圖象關於直線\(x=\dfrac{\pi}{6}\)成軸對稱 \(\qquad \qquad \qquad \qquad\) D.圖象關於點\(\left(\dfrac{\pi}{6},0 \right)\)成中心對稱
解析 \(A\).\(x\in \left(\dfrac{\pi}{6},π\right)⇒x+\dfrac{\pi}{3}\in \left(\dfrac{\pi}{2},\dfrac{4\pi}{3}\right)\);故單調遞增, \(A\)正確;
\(B\).函式\(f(x)\)的最小正週期是\(π\),故\(B\)正確,
\(C\).正切函式沒有對稱軸,故\(C\)錯誤,
\(D\).令\(x+\dfrac{\pi}{3}=\dfrac{k\pi}{2}⇒x=\dfrac{k\pi}{2}-\dfrac{\pi}{3}\),\(k∈Z\);
則\(f(x)\)圖象關於點\(\left(\dfrac{\pi}{6},0\right)\)成中心對稱,故\(D\)正確,
故選:\(ABD\).
【典題2】比較大小:\(\tan\left(-\dfrac{7\pi}{4}\right)\) \(\underline{\quad \quad}\) \(\tan \left(-\dfrac{9}{5} \pi\right)\).
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\(\left(0,\dfrac{\pi}{2}\right)\)內單調遞增,\(\therefore \tan \dfrac{\pi}{5}<\tan \dfrac{\pi}{4}\),
\(\therefore \tan \left(-\dfrac{7}{4} \pi\right)>\tan \left(-\dfrac{9}{5} \pi\right)\).
【典題3】 若函式\(y=\tan \left(ωx+\dfrac{\pi}{6}\right)\)在\(\left[-\dfrac{\pi}{3},\dfrac{\pi}{3}\right]\)上單調遞減,且在\(\left[-\dfrac{\pi}{3},\dfrac{\pi}{3}\right]\)上的最大值為\(\sqrt{3}\) ,則\(ω\)的值為( )
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\(\left[-\dfrac{\pi}{3},\dfrac{\pi}{3}\right]\)上單調遞減,
\(\therefore T=\dfrac{\pi}{|\omega|} \geq \dfrac{2 \pi}{3}\),\(\therefore |ω|≤\dfrac{3}{2}\);
又\(y=\tan \left(ωx+\dfrac{\pi}{6}\right)\)在\(\left[-\dfrac{\pi}{3},\dfrac{\pi}{3}\right]\)上的最大值為\(\sqrt{3}\) ,
\(\therefore\)當\(x=-\dfrac{\pi}{3}\)時,\(y_{\max} =\sqrt{3}\),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\(|ω|≤\dfrac{3}{2}\),
\(\therefore ω=-\dfrac{1}{2}\).
故選:\(A\).
【鞏固練習】
1.函式\(y=\tan \left(2x+\dfrac{\pi}{4}\right)\)的圖象( )
A.關於原點對稱 \(\qquad \qquad \qquad \qquad\) B.關於點\(\left (-\dfrac{\pi}{2},1\right)\)對稱 \(\qquad \qquad \qquad \qquad\) C.關於直線\(x=-\dfrac{\pi}{8}\)對稱 \(\qquad \qquad \qquad \qquad\) D.關於點\(\left(\dfrac{\pi}{8},0\right)\)對稱
2.已知函式\(f(x)=\tan \left(2x-\dfrac{\pi}{3}\right)\),則下列說法錯誤的是( )
A. 函式\(f(x)\)的週期為\(\dfrac{\pi}{2}\)
B. 函式\(f(x)\)的值域為\(R\)
C. 點\(\left(\dfrac{\pi}{6},0\right)\)是函式\(f(x)\)的圖象一個對稱中心
D. \(f\left(\dfrac{2\pi}{5}\right)<f\left(\dfrac{3\pi}{5}\right)\)
- 方程\(\tan(2x+\dfrac{\pi}{3})=\sqrt{3}\)在區間\(\left[0,2π\right)\)上的解的個數是\(\underline{\quad \quad}\) .
4.已知函式\(f(x)=\tan(ωx+φ)(ω≠0,|φ|<\dfrac{\pi}{2})\),點\(\left(\dfrac{\pi}{3},0\right)\)和\(\left(\dfrac{5\pi}{6},0\right)\)是其相鄰的兩個對稱中心,且在區間\(\left(\dfrac{\pi}{3},\dfrac{2\pi}{3}\right)\)內單調遞減,則\(φ=\) \(\underline{\quad \quad}\) .
5.設函式\(f(x)=\tan \left(2x-\dfrac{\pi}{3}\right)\).
(1)求\(f(x)\)的定義域、週期和單調區間;
(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.
參考答案
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答案 \(D\)
解析 函式\(y=\tan\left(2x+\dfrac{\pi}{4}\right)\)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\(y=\tan\left(2x+\dfrac{\pi}{4}\right)\)的圖象關於原點\(\left(\dfrac{\pi}{8} ,0\right)\)對稱,\(D\)正確.
故選:\(D\). -
答案 \(D\)
解析 \(∵f(x)=\tan \left(2x-\dfrac{\pi}{3}\right)\),
\(\therefore\) 函式\(f(x)\)的週期\(T=\dfrac{\pi}{2}\),故\(A\)正確;
由正切函式的圖象和性質可知函式\(f(x)\)的值域為\(R\),故\(B\)正確;
由\(2x-\dfrac{\pi}{3}=kπ\),\(k∈Z\)可解得:\(x=\dfrac{k\pi}{2}+\dfrac{\pi}{6}\),\(k∈Z\),
則解得當\(k=0\)時,點\(\left(\dfrac{\pi}{6},0\right)\)是函式\(f(x)\)的圖象一個對稱中心,故\(C\)正確;
由 \(f\left(\dfrac{2 \pi}{5}\right)=\tan \left(2 \times \dfrac{2 \pi}{5}-\dfrac{\pi}{3}\right)=\tan \dfrac{7 \pi}{15}>0\); \(f\left(\dfrac{3 \pi}{5}\right)=\tan \left(2 \times \dfrac{3 \pi}{5}-\dfrac{\pi}{3}\right)=\tan \dfrac{13 \pi}{15}<0\),
從而\(f\left(\dfrac{2\pi}{5}\right)>f\left(\dfrac{3\pi}{5}\right)\),故\(D\)不正確.
故選:\(D\). -
答案 \(4\)
解析 方程\(\tan(2x+\dfrac{\pi}{3})=\sqrt{3}\) ,
\(\therefore 2x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+kπ\),\(k∈Z\),\(x=\dfrac{k\pi}{2}\),\(k∈Z\);
\(k=0\),\(k=1\),\(k=2\),\(k=3\),
求得方程在區間\([0,2π)\)上的解為\(0\),\(\dfrac{\pi}{2}\),\(π\),\(\dfrac{3\pi}{2}\)共\(4\)個. -
答案 \(\dfrac{\pi}{3}\)
解析 根據題意可得\(\left(\dfrac{\pi}{3},0\right)\)和\(\left(\dfrac{5\pi}{6},0\right)\)是其相鄰的兩個對稱中心得 \(\dfrac{T}{2}=\dfrac{5 \pi}{6}-\dfrac{\pi}{3}=\dfrac{\pi}{2}\),\(\therefore T=π\);
又因為在區間\(\left(\dfrac{\pi}{3},\dfrac{2\pi}{3}\right)\)內單調遞減,\(\therefore ω=-1\);
則\(f(x)=\tan(-x+φ)\);
當\(x=\dfrac{\pi}{3}\)時,\(f(\dfrac{\pi}{3})=0\),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)\(\because\) 函式\(f(x)=\tan \left(2x-\dfrac{\pi}{3}\right)\),它的週期為\(\dfrac{\pi}{2}\).
根據函式的解析 式可得\(2x-\dfrac{\pi}{3}≠kπ+\dfrac{\pi}{2}\),\(k∈Z\).求得 \(x \neq \dfrac{k \pi}{2}+\dfrac{5 \pi}{12}\),
故函式的定義域為\(\{x∣x≠\dfrac{k\pi}{2}+\dfrac{5 \pi}{12},k\in z\}\).
由\(kπ-\dfrac{\pi}{2}<2x-\dfrac{\pi}{3}<kπ+\dfrac{\pi}{2}\),\(k∈Z\)求得 \(\dfrac{k \pi}{2}-\dfrac{\pi}{12}<x<\dfrac{k \pi}{2}+\dfrac{5 \pi}{12}\),
故函式的增區間為 \(\left(\dfrac{k \pi}{2}-\dfrac{\pi}{12}, \dfrac{k \pi}{2}+\dfrac{5 \pi}{12}\right)\),\(k∈Z\).
(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\(\dfrac{k\pi}{2}+\dfrac{\pi}{24}≤x≤\dfrac{k\pi}{2}+\dfrac{\pi}{3}\),
故不等式的解集為\(\left[\dfrac{k\pi}{2}+\dfrac{\pi}{24},\dfrac{k\pi}{2}+\dfrac{\pi}{3}\right]\),\(k∈Z\).
(3)\(\because x\in \left[0,π\right]\),\(\therefore 2x-\dfrac{\pi}{3}\in \left[-\dfrac{\pi}{3},\dfrac{5\pi}{3}\right]\),
故\(\tan \left(2x-\dfrac{\pi}{3}\right)\in R\).
分層練習
【A組---基礎題】
1.函式\(f(x)=\tan\left(ωx+\dfrac{\pi}{6}\right)\)的最小正週期為\(2π\),則\(f\left(\dfrac{\pi}{6}\right)=\) ( )
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.現有下列四個命題:
①函式\(y=\tan x\)在定義域內是增函式;
②函式\(y=\tan(2x+1)\)的最小正週期是\(π\);
③函式\(y=\tan x\)的圖象關於點\((π,0)\)成中心對稱;
④函式\(y=\tan x\)的圖象關於點\(\left(-\dfrac{\pi}{2},0\right)\)成中心對稱.
其中正確命題的個數是( )
A.\(0\) \(\qquad \qquad \qquad \qquad\) B.\(1\) \(\qquad \qquad \qquad \qquad\) C.\(2\) \(\qquad \qquad \qquad \qquad\) D.\(3\)
3.函式\(f(x)=\tan \left(x+\dfrac{\pi}{4}\right)\)的單調區間為( )
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.下列關於函式\(y=\tan \left(x+\dfrac{\pi}{3}\right)\)的說法正確的是( )
A. 在區間\(\left(-\dfrac{\pi}{6},\dfrac{5\pi}{6}\right)\)上單調遞增
B. 最小正週期是\(π\)
C. 圖象關於點\(\left(\dfrac{\pi}{4},0\right)\)成中心對稱
D. 圖象關於直線\(x=\dfrac{\pi}{6}\)成軸對稱
5.關於函式\(f(x)=|\tan x|\)的性質,下列敘述不正確的是( )
A.\(f(x)\)的最小正週期為\(\dfrac{\pi}{2}\)
B.\(f(x)\)是偶函式
C.\(f(x)\)的圖象關於直線\(x=\dfrac{k\pi}{2}(k\in Z)\)對稱
D.\(f(x)\)在每一個區間\(\left(kπ,kπ+\dfrac{\pi}{2}\right)(k\in Z)\)內單調遞增
6.(多選)下列關於函式\(y=\tan \left(2x-\dfrac{\pi}{6}\right)\)的說法正確的是( )
A. 在區間\(\left(-\dfrac{\pi}{6},\dfrac{7\pi}{24}\right)\)上單調遞增
B. 最小正週期是\(π\)
C. 圖象關於\(\left(\dfrac{\pi}{3},0\right)\)成中心對稱
D. 圖象關於 \(\left(\dfrac{\pi}{12}, 0\right)\)成中心對稱
7.函式\(y=\tan \left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)\),\(x\in \left(0,\dfrac{\pi}{6}\right]\)的值域是 .
8.已知函式 \(f(x)=\tan x+\dfrac{1}{\tan x}\),若\(f(a)=5\),則\(f(-a)=\) \(\underline{\quad \quad}\).
9.若\(f(n)=\tan \dfrac{n\pi}{3}\),\((n\in N^* )\),則\(f(1)+f(2)+⋯+f(100)=\) \(\underline{\quad \quad}\).
10.在\((0,2π)\)內,使\(\tan x>1\)成立的x的取值範圍是\(\underline{\quad \quad}\).
11.已知函式\(y=3\tan ωx+1\)在\(\left(-\dfrac{\pi}{3},\dfrac{\pi}{4}\right)\)內是減函式,則\(ω\)的取值範圍是\(\underline{\quad \quad}\).
12.已知函式\(f(x)=3\tan \left(\dfrac{1}{2} x-\dfrac{\pi}{3}\right)\).
(1)求\(f(x)\)的定義域和值域.(2)討論\(f(x)\)的週期和單調區間.(3)求\(f(x)\)的對稱中心.
13.設函式\(f(x)=\tan (ωx+φ)(ω>0,0<φ<\dfrac{\pi}{2})\),已知函式\(y=f(x)\)的圖象與\(x\)軸相鄰兩個交點的距離為\(\dfrac{\pi}{2}\),且圖象關於點 \(M\left(-\dfrac{\pi}{8}, 0\right)\)對稱.
(1)求\(f(x)\)的解析 式;
(2)求\(f(x)\)的單調區間;
(3)求不等式\(-1≤f(x)≤\sqrt{3}\)的解集.
參考答案
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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函式\(y=\tan x\)在定義域內不是單調函式;故①錯誤,
②函式\(y=\tan(2x+1)\)的最小正週期是\(\dfrac{\pi}{2}\);故②錯誤
③函式\(y=\tan x\)的圖象關於點\(\left(\dfrac{k\pi}{2},0\right)\)成中心對稱,
當\(k=2\)時,對稱中心為\((π,0)\);故③正確,
④函式\(y=\tan x\)的圖象關於點\(\left(\dfrac{k\pi}{2},0\right)\)成中心對稱,
當\(k=-1\)時,關於\(\left(-\dfrac{\pi}{2},0\right)\)成中心對稱.故④正確,
故正確是③④,
故選:\(C\). -
答案 \(C\)
解析 由\(-\dfrac{\pi}{2}+kπ<x+\dfrac{\pi}{4}<\dfrac{\pi}{2}+kπ\),\(k∈Z\),
得\(-\dfrac{3\pi}{4}+kπ<x<\dfrac{\pi}{4}+kπ\),\(k∈Z\).故選\(C\). -
答案 \(B\)
解析 對於\(A\),由\(kπ-\dfrac{\pi}{2}<x+\dfrac{\pi}{3}<kπ+\dfrac{\pi}{2}\),\(k∈Z\),
即\(kπ-\dfrac{5\pi}{6}<x<kπ+\dfrac{\pi}{6}\),\(k∈Z\),
當\(k=0\)時,函式的單調遞增區間為\(\left(-\dfrac{5\pi}{6},\dfrac{\pi}{6}\right)\),
當\(k=1\)時,函式的單調遞增區間為\(\left(\dfrac{\pi}{6},\dfrac{7\pi}{6}\right)\),
故\(f(x)\)在區間\(\left(-\dfrac{\pi}{6},\dfrac{5\pi}{6}\right)\)上單調遞增錯誤,\(A\)錯誤;
對於\(B\),函式\(f(x)\)的最小正週期為\(T=π\),命題正確;
對於\(C\),由\(x+\dfrac{\pi}{3}=\dfrac{k\pi}{2}\), 得\(x=-\dfrac{\pi}{3}+\dfrac{k\pi}{2}\),\(k∈Z\),
即函式\(f(x)\)的對稱中心為\(\left(-\dfrac{\pi}{3}+\dfrac{k\pi}{2},0\right)\),
當\(k=1\)時,對稱中心為\(\left(\dfrac{\pi}{6},0\right)\),\(f(x)\)圖象不關於點\(\left(\dfrac{\pi}{4},0\right)\)成中心對稱,\(C\)錯誤;
對於\(D\),正切函式是奇函式,圖象沒有對稱軸,\(D\)錯誤.
故選:\(B\). -
答案 \(A\)
解析 對於函式\(f(x)=|\tan x|\)的性質,根據該函式的圖象知,其最小正週期為\(π\),\(A\)錯誤;
又\(f(-x)=|\tan(-x)|=|\tan x|=f(x)\),所以\(f(x)\)是定義域上的偶函式,\(B\)正確;
根據函式\(f(x)\)的圖象知,\(f(x)\)的圖象關於直線\(x=\dfrac{k\pi}{2}(k\in Z)\)對稱,\(C\)正確;
根據\(f(x)\)的圖象知,\(f(x)\)在每一個區間\(\left(kπ,kπ+\dfrac{\pi}{2}\right)(k\in Z)\)內單調遞增,\(D\)正確.
故選:\(A\). -
答案 \(ACD\)
解析 當\(x\in \left(-\dfrac{\pi}{6},\dfrac{7\pi}{24}\right)\)時,\(2x-\dfrac{\pi}{6}\in \left(-\dfrac{\pi}{2},\dfrac{5\pi}{12}\right)\),
所以\(y=\tan \left(2x-\dfrac{\pi}{6}\right)\)在區間\(\left(-\dfrac{\pi}{6},\dfrac{7\pi}{24}\right)\)上單調遞增,故\(A\)正確;
函式\(y=\tan \left(2x-\dfrac{\pi}{6}\right)\)的最小正週期是\(\dfrac{\pi}{2}\),故\(B\)錯誤;
當\(x=\dfrac{\pi}{3}\)時,\(2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}\),所以函式\(y=\tan \left(2x-\dfrac{\pi}{6}\right)\)的圖象關於\(\left(\dfrac{\pi}{3},0\right)\)成中心對稱,故\(C\)正確;
當\(x=\dfrac{\pi}{12}\)時\(2x-\dfrac{\pi}{6}=0\),所以函式\(y=\tan \left(2x-\dfrac{\pi}{6}\right)\)的圖象關於\(\left(\dfrac{\pi}{12},0\right)\)成中心對稱,故\(D\)正確.
故選:\(ACD\). -
答案 \(\left(1,\sqrt{3} \right]\)
解析 由\(x\in \left(0,\dfrac{\pi}{6}\right]\),\(\therefore \dfrac{x}{2}+\dfrac{\pi}{4}\in \left(\dfrac{\pi}{4},\dfrac{\pi}{3}\right]\),結合正切函式的性質可得:\(1<y≤\sqrt{3}\).
故答案為:\(\left(1,\sqrt{3} \right]\). -
答案 \(-5\)
解析 \(f(x)\)的定義域為\(\left(kπ-\dfrac{\pi}{2},kπ\right)∪\left(kπ,kπ+\dfrac{\pi}{2}\right)(k\in Z)\).
可知\(f(x)\)的定義域關於原點對稱.
又 \(f(-x)=\tan (-x)+\dfrac{1}{\tan (-x)}=-\left(\tan x+\dfrac{1}{\tan x}\right)=-f(x)\).
\(\therefore f(x)\)是奇函式.\(\therefore f(-a)=-f(a)=-5\). -
答案 \(\sqrt{3}\)
解析 \(\because f(n)=\tan \dfrac{n\pi}{3}\),\((n\in N^* )\),
根據正切函式的性質可得其週期 \(T=\dfrac{\pi}{\dfrac{\pi}{3}}=3\),
\(\therefore f(1)=\sqrt{3}\),\(f(2)=-\sqrt{3}\) ,\(f(3)=0\).
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 \(kπ+\dfrac{\pi}{2}>x>kπ+\dfrac{\pi}{4}\),\(k∈Z\).
再根據\(x\in (0,2π)\),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 \(\because\) 函式\(y=3\tan ωx+1\)在\(\left(-\dfrac{\pi}{3},\dfrac{\pi}{4}\right)\)內是減函式,
\(\therefore ω<0\)且函式\(y=3\tan ωx+1\)在\(\left(-\dfrac{\pi}{3},\dfrac{\pi}{3}\right)\)內也是減函式,
\(\therefore T=\left|\dfrac{\pi}{\omega}\right|≥\dfrac{\pi}{3}-(-\dfrac{\pi}{3})=\dfrac{2\pi}{3}\),
\(\therefore |ω|≤\dfrac{3}{2}\),\(\therefore -\dfrac{3}{2}≤ω≤\dfrac{3}{2}\),
又\(ω<0\),\(\therefore -\dfrac{3}{2}≤ω<0\).
故答案為:\(\left[-\dfrac{3}{2},0\right)\). -
答案 (1) 定義域為\(\{x∣x≠2kπ+\dfrac{5\pi}{3},k\in Z\}\),值域為\(R\);
(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)\(\because\) 函式\(f(x)=3\tan \left(\dfrac{1}{2} x-\dfrac{\pi}{3}\right)\),
\(\therefore \dfrac{1}{2} x-\dfrac{\pi}{3}≠kπ+\dfrac{\pi}{2}\),\(k∈Z\),即\(x≠2kπ+\dfrac{5\pi}{3}\),\(k∈Z\),
\(\therefore f(x)\)的定義域為\(\{x∣x≠2kπ+\dfrac{5\pi}{3},k\in Z\}\),值域為\(R\);
(2) \(f(x)\)的最小正週期是 \(\dfrac{\pi}{\dfrac{1}{2}}=2 \pi\),
又令\(-\dfrac{\pi}{2}+kπ<\dfrac{1}{2} x-\dfrac{\pi}{3}<\dfrac{\pi}{2}+kπ\),\(k∈Z\),
\(\therefore -\dfrac{\pi}{3}+2kπ<x<\dfrac{5\pi}{3}+2kπ\),\(k∈Z\)
\(\therefore f(x)\)的單調增區間為\(\left(-\dfrac{\pi}{3}+2kπ,\dfrac{5\pi}{3}+2kπ\right)\),\(k∈Z\);
(3)令\(\dfrac{1}{2} x-\dfrac{\pi}{3}=kπ\),\(k∈Z\),解得\(x=\dfrac{2\pi}{3}+2kπ\),\(k∈Z\),
此時\(y=f(x)=0\),
\(\therefore\)函式\(f(x)\)的對稱中心為\(\left(\dfrac{2\pi}{3}+2kπ,0\right)\),\(k∈Z\). -
答案 (1) \(f(x)=\tan \left(2x+\dfrac{\pi}{4}\right)\);
(2) 單調增區間為\(\left(-\dfrac{3 \pi}{8}+\dfrac{k \pi}{2}, \dfrac{\pi}{8}+\dfrac{k \pi}{2}\right)\),\(k∈Z\).無單調減區間;
(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)\(\because\)函式\(y=f(x)\)的圖象與\(x\)軸相鄰兩交點的距離為\(\dfrac{\pi}{2}\),
即\(T=\dfrac{\pi}{2}\),\(\therefore ω=2\),
\(\therefore f(x)=\tan(2x+φ)\),
\(\because\) 圖象關於點 \(M\left(-\dfrac{\pi}{8}, 0\right)\)對稱.\(\therefore -2×\dfrac{\pi}{8}+φ=\dfrac{k\pi}{2}\),\(k∈Z\),
\(\because 0<φ<\dfrac{\pi}{2}\),\(\therefore φ=\dfrac{\pi}{4}\),
則\(f(x)=\tan (2x+\dfrac{\pi}{4})\).
(2)令\(-\dfrac{\pi}{2}+kπ<2x+\dfrac{\pi}{4}<\dfrac{\pi}{2}+kπ\),\(k∈Z\),
得\(-\dfrac{3\pi}{8}+\dfrac{k\pi}{2}<x<\dfrac{\pi}{8}+\dfrac{k\pi}{2}\),\(k∈Z\),
\(\therefore\)函式的單調增區間為 \(\left(-\dfrac{3 \pi}{8}+\dfrac{k \pi}{2}, \dfrac{\pi}{8}+\dfrac{k \pi}{2}\right)\),\(k∈Z\).無單調減區間.
(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\(-1≤f(x)≤\sqrt{3}\)的解集為\(\{x∣-\dfrac{\pi}{4}+\dfrac{k\pi}{2}≤x≤\dfrac{\pi}{24}+\dfrac{k\pi}{2},k\in Z\}\).
【B組---提高題】
1.(多選)已知函式\(f(x)=\left |\tan \left(\dfrac{1}{2} x-\dfrac{\pi}{6}\right) \right|\),則下列說法正確的是( )
A. \(f(x)\)的週期是\(2π\);
B. \(f(x)\)的值域是\(\{y|y\in R,\)且\(y≠0\}\);
C. 直線\(x=\dfrac{5\pi}{3}\)是函式\(f(x)\)圖象的一條對稱軸;
D. \(f(x)\)的單調遞減區間是\(\left(2kπ-\dfrac{2\pi}{3},2kπ+\dfrac{\pi}{3}\right]\),\(k∈Z\);
2.下列不等式中,正確的是( )
A. \(\tan \dfrac{4 \pi}{7}>\tan \dfrac{3 \pi}{7}\) \(\qquad \qquad \qquad \qquad\) B. \(\tan \dfrac{2 \pi}{5}<\tan \dfrac{3 \pi}{5}\) \(\qquad \qquad \qquad \qquad\) C. \(\tan \left(-\dfrac{13 \pi}{7}\right)>\tan \left(-\dfrac{15 \pi}{8}\right)\) \(\qquad \qquad \qquad \qquad\) D. \(\tan \left(-\dfrac{13 \pi}{4}\right)<\tan \left(-\dfrac{12 \pi}{5}\right)\)
參考答案
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答案 \(AD\)
解析 \(A\).\(f(x)\)的週期和\(y=\tan \left(\dfrac{1}{2} x-\dfrac{\pi}{6}\right)\)週期相同,即\(T=\dfrac{\pi}{\dfrac{1}{2}}=2 \pi\),故\(A\)正確,
\(B\).\(\because y=\tan \left(\dfrac{1}{2} x-\dfrac{\pi}{6}\right)\)的值域為\(R\),
\(\therefore f(x)≥0\),即函式\(f(x)\)的值域為\(\left[0,+∞\right)\),故\(B\)錯誤,
\(C\).由絕對值的意義知當\(\dfrac{1}{2}-x-\dfrac{\pi}{6}=\dfrac{k\pi}{2}\),即對稱軸為\(x=kπ+\dfrac{\pi}{3}\),\(k∈Z\),
則直線\(x=\dfrac{5\pi}{3}\)不是函式\(f(x)\)圖象的一條對稱軸,故\(C\)錯誤,
\(D\).由\(kπ-\dfrac{\pi}{2}<\dfrac{1}{2} x-\dfrac{\pi}{6}≤kπ\),\(k∈Z\) 得\(2kπ-\dfrac{2\pi}{3}<x≤2kπ+\dfrac{\pi}{3}\),\(k∈Z\),
即函式\(f(x)\)的單調遞減區間是\(\left(2kπ-\dfrac{2\pi}{3},2kπ+\dfrac{\pi}{3}\right]\),\(k∈Z\),故\(D\)正確,
故選:\(AD\). -
答案 \(C\)
解析 根據正切函式的單調性與週期性,得
對於\(A\), \(\tan \dfrac{4 \pi}{7}<0<\tan \dfrac{3 \pi}{7}\),\(A\)錯誤;
對於\(B\), \(\tan \dfrac{2 \pi}{5}>0>\tan \dfrac{3 \pi}{5}\),\(B\)錯誤;
對於\(C\), \(\tan \left(-\dfrac{13 \pi}{7}\right)=\tan \left(-2 \pi+\dfrac{\pi}{7}\right)=\tan \dfrac{\pi}{7}\),
\(\tan \left(-\dfrac{15 \pi}{8}\right)=\tan \left(-2 \pi+\dfrac{\pi}{8}\right)=\tan \dfrac{\pi}{8}\),
\(\dfrac{\pi}{2}>\dfrac{\pi}{7}>\dfrac{\pi}{8}>0\), \(\therefore \tan \dfrac{\pi}{7}>\tan \dfrac{\pi}{8}\),\(C\)正確;
對於\(D\), \(\tan \left(-\dfrac{13 \pi}{4}\right)=\tan \left(-\dfrac{\pi}{4}\right)=-\tan \dfrac{\pi}{4}\),
且 \(\tan \left(-\dfrac{12 \pi}{5}\right)=\tan \left(-\dfrac{2 \pi}{5}\right)=-\tan \dfrac{2 \pi}{5}\),\(D\)錯誤.
故選:\(C\).