5.6 函数y=Asin(ωx+φ)的图像和性质
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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
A,ω,φ对f(x)=Asin (ωx+φ)(A>0,ω>0)的影响
\(A\)影响函数\(f(x)\)的最值,\(ω\)影响函数\(f(x)\)周期\((T=\dfrac{2\pi}{ω})\),\(φ\)影响函数\(f(x)\)水平位置.
![](https://cdn.nlark.com/yuque/0/2022/png/21447551/1670472149159-95bf7331-42f3-40d0-897c-fdfdc3d56242.png)
![](https://cdn.nlark.com/yuque/0/2022/png/21447551/1670472157960-8d68b137-eee9-4ce1-b4ee-1aebbae774ee.png)
![](https://cdn.nlark.com/yuque/0/2022/png/21447551/1670472165437-0ffc717b-8330-4cb2-a8bc-9748540ceaaf.png)
解释
对\(A\),\(ω\),\(φ\)的理解,想象下把\(y=\sin x\)看成一个人,
\(\left(1\right)\)\(A\)影响他的身高,\(A>1\)时就“个子长高”,\(0<A<1\)时就“浓缩精华”了;
\(\qquad\) \(\qquad\)
\(\left(2\right)\)\(ω\)影响他的“腰围”(周期),\(ω>1\)时就“减肥成功”,\(0<ω<1\)时就“减肥失败”了;
\(\qquad\) \(\qquad\)
\(\left(3\right)\)\(φ\)影响他的站位,\(φ>0\)时就“向左平移”,\(φ<0\)时就“向右平移”了.
\(\qquad\) \(\qquad\)
函数的变换
\(\left(1\right)\) 平移变换
① \(y=f(x)⟶ y=f(x±a)(a>0)\)将\(y=f(x)\)图像沿\(x\)轴向左(右)平移\(a\)个单位(左加右减);
② \(y=f(x)⟶y=f(x)± b (b>0)\)将\(y=f(x)\)图像沿\(x\)轴向上(下)平移\(b\)个单位(上加下减).
解释
\(f(x)=3\sin (2x+\dfrac{\pi}{3} )\)向左平移\(\dfrac{\pi}{4}\)个单位,得到的函数不是\(f(x)=3\sin (2x+\dfrac{\pi}{4} +\dfrac{\pi}{3} )\), 而是\(f(x)=3\sin \left[2(x+\dfrac{\pi}{4} )+\dfrac{\pi}{3} \right]\).
(2) 伸缩变换
① \(y=f(x)⟶ y=A f(x)(A>0)\)
将\(y=f(x)\)图像横坐标不变,纵坐标伸长到原来的\(A\)倍(\(A>1\)伸长,\(A<1\)缩短).
② \(y=f(x)⟶ y=f(ω x)(ω>0)\)
将\(y=f(x)\)图像纵坐标不变,横坐标缩到原来的 \(\dfrac{1}{\omega}\)倍( \(ω>1\)缩短,\(ω<1\)伸长);
问题 怎么理解呢?例:若将\(f(x)=3 \sin (x+\dfrac{\pi}{3} )\)图像纵坐标不变,横坐标缩到原来的\(\dfrac{1}{2}\)倍,那得到的函数是\(f(x)=3 \sin (2x+\dfrac{\pi}{3} )\)还是\(f(x)=3 \sin (\dfrac{1}{2} x+\dfrac{\pi}{3} )\)呢?
解释 我们把\(f(x)=3 \sin (x+\dfrac{\pi}{3} )\)的图象想象成一条弹簧,若纵坐标不变,横坐标缩到原来的\(\dfrac{1}{2}\)倍,那说明弹簧被压缩了,则周期变小,\(ω\)变大(\(T=\dfrac{2\pi}{ω}\),\(T\)与\(ω\)成反比),即变换后的函数应该是\(f(x)=3 \sin (2x+\dfrac{\pi}{3} )\).
基本方法
【题型1】 三角函数图象的变换
【典题1】 用“五点法”作函数\(y=2\sin (2x+\dfrac{\pi}{4} )\)在一个周期上的图象.
解析 \(\left(1\right)\)列出五个关键点如下:
$2x+\dfrac{\pi}{4}$ | $0$ | $\dfrac{\pi}{2}$ | $π$ | $\dfrac{3\pi}{2}$ | $2π$ |
$x$ | $-\dfrac{\pi}{8}$ | $\dfrac{\pi}{8}$ | $\dfrac{3\pi}{8}$ | $\dfrac{5\pi}{8}$ | $\dfrac{7\pi}{8}$ |
$y$ | $0$ | $2$ | $0$ | $-2$ | $0$ |
【典题2】 函数\(f(x)=3 \sin (2x-\dfrac{\pi}{6} )\)的图象可以看成由函数\(y=\sin x\)的图象如何变换得到的?
解析 方法1
\(y=\sin x \stackrel{\text { 向右平移 } \dfrac{\pi}{6} \text { 个单位 }}{\longrightarrow} y=\sin \left(x-\dfrac{\pi}{6}\right) \stackrel{\text { 纵坐标不变, 横坐标变为原来的 } \dfrac{1}{2} \text { 倍 }}{\longrightarrow}\)
\(y=\sin \left(2 x-\dfrac{\pi}{6}\right) \stackrel{\text { 横坐标不变, 纵坐标变为原来的 } 3 \text { 倍 }}{\longrightarrow} y=\sin \left(2 x-\dfrac{\pi}{6}\right)\).
具体图象变换显示如下
方法2
\(y=\sin x \stackrel{\text { 纵坐标不变, 横坐标变为原来的竩倍 }}{\longrightarrow} y=\sin 2 x \stackrel{\text { 向右平移 } \dfrac{\pi}{12} \text { 个单位 }}{\longrightarrow} y=\sin \left(2 x-\dfrac{\pi}{6}\right)\)
\(\stackrel{\text { 横坐标不变, 纵坐标变为原来的 } 3 \text { 倍 }}{\longrightarrow} y=\sin \left(2 x-\dfrac{\pi}{6}\right)\).
具体图象变换显示如下
点拨 函数的\(f(x)=A\sin (ωx+φ)(A>0,ω>0)\)的图象由其他三角函数变换得到,先变换\(A\)还是\(ω\),或\(φ\)均可以.
【典题3】 已知函数\(y=f(x)\),将\(y=f(x)\)的图象上的每一点的纵坐标保持不变,横坐标变为原来的\(2\)倍,然后把所得的图象沿着 轴向左平移\(\dfrac{\pi}{2}\)个单位,这样得到的是\(y=\dfrac{1}{2} \sin x\)的图象,那么函数\(y=f(x)\)的解析式是 ( )
A. \(f(x)=\dfrac{1}{2} \sin \left(\dfrac{x}{2} -\dfrac{\pi}{2} \right)\) \(\qquad \qquad \qquad \qquad\) B. \(f(x)=\dfrac{1}{2} \sin \left(2 x+\dfrac{\pi}{2} \right)\)
C. \(f(x)=\dfrac{1}{2} \sin \left(\dfrac{x}{2} +\dfrac{\pi}{2} \right)\) \(\qquad \qquad \qquad \qquad\) D. \(f(x)=\dfrac{1}{2} \sin \left(2 x-\dfrac{\pi}{2} \right)\)
解析 由题意曲线与\(y=\dfrac{1}{2} \sin x\)的图象沿\(x\)轴向右平移\(\dfrac{\pi}{2}\)个单位,再纵坐标不变,横坐标缩小为原来的一半即可得到\(y=f(x)\)的图形,故\(y=\dfrac{1}{2} \sin x\)的图形沿\(x\)轴向右平移\(\dfrac{\pi}{2}\)个单位所得图形对应的函数解析式为\(y=\dfrac{1}{2} \sin \left(x-\dfrac{\pi}{2} \right)\),然后再将所得的曲线上的点的纵坐标保持不变,横坐标缩小到原来的一半,所得的图形对应的解析式为\(f(x)=\dfrac{1}{2} \sin \left(2x-\dfrac{\pi}{2} \right)\).
【巩固练习】
1.用五点法画出函数\(f(x)=3\sin \left(\dfrac{x}{2} +\dfrac{\pi}{6} \right)+3\)在一个周期内的闭区间上的图象;
2.把函数\(y=\cos 2x+1\)的图象上所有点的横坐标伸长到原来的\(2\)倍(纵坐标不变),然后向左平移\(1\)个单位长度,再向下平移\(1\)个单位长度,得到的图象是( )
\(\qquad \qquad\)
3.为了得到函数\(f(x)=\sin \left(2x+\dfrac{3\pi}{4} \right)\)的图象,可以将函数\(g(x)=\cos 2x\)的图象( )
A.向右平移\(\dfrac{\pi}{4}\)个单位 \(\qquad \qquad \qquad \qquad\) B.向左平移\(\dfrac{\pi}{4}\)个单位
C.向右平移\(\dfrac{\pi}{8}\)个单位 \(\qquad \qquad \qquad \qquad\) D.向左平移\(\dfrac{\pi}{8}\)个单位
4.将函数\(y=\cos x\)的图象先左移\(\dfrac{\pi}{4}\),再纵坐标不变,横坐标缩为原来的\(\dfrac{1}{2}\),所得图象的解析式为 ( )
A.\(y=\sin \left(2x+\dfrac{\pi}{4} \right)\) \(\qquad \qquad \qquad \qquad\) B.\(y=\sin \left(\dfrac{1}{2} x+\dfrac{3\pi}{4} \right)\)
C.\(y=\sin \left(\dfrac{1}{2} x+\dfrac{\pi}{4} \right)\) \(\qquad \qquad \qquad \qquad\) D.\(y=\sin \left (2x+\dfrac{3\pi}{4} \right)\)
5.将函数\(f(x)=\sin \left(ωx+\dfrac{\pi}{6} \right)(ω>0)\)图象上所有点的横坐标伸长到原来的\(2\)倍,再向右平移\(\dfrac{\pi}{6}\)个单位长度,得到函数\(y=g(x)\)的图象,若\(y=g(x)\)为奇函数,则\(ω\)的最小值为( )
A. \(4\) \(\qquad \qquad \qquad \qquad\) B.\(3\) \(\qquad \qquad \qquad \qquad\) C. \(2\) \(\qquad \qquad \qquad \qquad\) D. \(1\)
参考答案
- 解析 列表
$x$ | $-\dfrac{\pi}{3}$ | $\dfrac{2\pi}{3}$ | $\dfrac{5\pi}{3}$ | $\dfrac{8\pi}{3}$ | $\dfrac{11\pi}{3}$ |
$\dfrac{x}{2} +\dfrac{\pi}{6}$ | $0$ | $\dfrac{\pi}{2}$ | $π$ | $\dfrac{3\pi}{2}$ | $2π$ |
$y$ | $3$ | $6$ | $3$ | $0$ | $3$ |
其函数图象如下,
![](https://cdn.nlark.com/yuque/0/2022/png/21447551/1669091927781-d3ffd27f-a25e-45d6-b384-df68ca7b3e86.png)
-
答案 \(A\)
解析 \(y=\cos 2x+1\)图象上所有点的横坐标伸长到原来的\(2\)倍得\(y_1=\cos x+1\),再向左平移\(1\)个单位长度得\(y_2=\cos (x+1)+1\),再向下平移\(1\)个单位长度得\(y_3=\cos (x+1)\),故相应图象为\(A\). -
答案 \(D\)
解析 为了得到函数\(f(x)=\sin \left(2x+\dfrac{3\pi}{4} \right)\)的图象,可以将函数\(g(x)=\cos 2x=\sin \left(2x+\dfrac{\pi}{2} \right)\)的图象向左平移\(\dfrac{\pi}{8}\)个单位,\(\sin \left[2\left(x+\dfrac{\pi}{8} \right)+\dfrac{\pi}{2} \right]=\sin \left(2x+\dfrac{3\pi}{4} \right)\).
故选:\(D\). -
答案 \(D\)
解析 函数\(y=\cos x=\sin \left(x+\dfrac{\pi}{2} \right)\),其图象先左移\(\dfrac{\pi}{4}\)个单位,得\(y=\sin \left(x+\dfrac{3\pi}{4} \right)\)的图象;
再纵坐标不变,横坐标缩为原来的\(\dfrac{1}{2}\),得函数\(y=\sin \left(2x+\dfrac{3\pi}{4}\right)\)的图象;
所以函数\(y\)的解析式为\(y=\sin \left(2x+\dfrac{3\pi}{4} \right)\).
故选:\(D\). -
答案 \(C\)
解析 由题意,将函数\(f(x)=\sin \left(ωx+\dfrac{\pi}{6} \right)(ω>0)\)图象上所有点的横坐标伸长到原来的\(2\)倍,再向右平移\(\dfrac{\pi}{6}\)个单位长度,
得到\(g(x)=\sin \left[\dfrac{\omega}{2}\left(x-\dfrac{\pi}{6}\right)+\dfrac{\pi}{6}\right]=\sin \left(\dfrac{\omega}{2} x+\dfrac{\pi}{6}-\dfrac{\omega \pi}{12}\right)\),
因为\(y=g(x)\)为奇函数,
所以 \(\dfrac{\pi}{6}-\dfrac{\omega \pi}{12}=k \pi(k \in Z)\),解得\(ω=2-12k(k∈Z)\),
又\(ω>0\),
所以当\(k=0\)时,\(ω\)取得最小值\(2\).
故选:\(C\).
【题型2】 求y=A\sin (ωx+φ)的解析式
【典题1】 已知函数\(f(x)=A\sin (ωx+φ)\left(A>0,ω>0,0<|φ|<\dfrac{\pi}{2} \right)\)的部分图象如图所示,下述四个结论:
①\(ω=2\);②\(φ=-\dfrac{\pi}{3}\);③\(f\left(x+\dfrac{\pi}{12} \right)\)是奇函数;④\(f\left(x-\dfrac{\pi}{12} \right)\)是偶函数中,其中所有正确结论的编号是\(\underline{\quad \quad}\) .
解析 由函数图象的最值可得\(A=1\),
由\(\dfrac{3}{4} T=\dfrac{\pi}{6} -\left(-\dfrac{7\pi}{12} \right)=\dfrac{3\pi}{4}\),解得\(T=π\),所以 \(\omega=\dfrac{2 \pi}{T}=2\),
此时\(f(x)=\sin (2x+φ)\),
代入\(\left(-\dfrac{7\pi}{12} ,1\right)\)得\(f\left(-\dfrac{7\pi}{12} \right)=\sin \left(-\dfrac{7\pi}{6} +φ\right)=1\),
\(\therefore -\dfrac{7\pi}{6} +φ=\dfrac{\pi}{2} +2kπ⇒φ=\dfrac{5\pi}{3} +2kπ\),
又\(\because 0<|φ|<\dfrac{\pi}{2}\) ,\(\therefore φ=-\dfrac{\pi}{3}\) ,
\(\therefore f(x)=\sin \left(2x-\dfrac{\pi}{3} \right)\),
\(\therefore\) ①、②正确;
\(\because f\left(x+\dfrac{\pi}{12} \right)=\sin \left[2\left(x+\dfrac{\pi}{12} \right)-\dfrac{\pi}{3} \right]=\sin \left(2x-\dfrac{\pi}{6} \right)\)不是奇函数,\(\therefore\)③错误;
\(\because f\left(x-\dfrac{\pi}{12} \right)=\sin \left[2\left(x-\dfrac{\pi}{12} \right)-\dfrac{\pi}{3} \right]=\sin \left(2x-\dfrac{\pi}{2} \right)=-\cos 2x\),
\(\therefore f(x-\dfrac{\pi}{12} )\)为偶函数,④正确.
综上知,正确的命题序号是①②④.
点拨 由函数\(y=A\sin (ωx+φ)+B(A>0,ω>0)\)的部分图象求解析式的方法
\(\left(1\right)\)求\(A\),\(B\):通过函数最值求解,由 \(\left\{\begin{array}{c}
f_{\max }=A+B \\
f_{\min }=-A+B
\end{array}\right.\)得 \(A=\dfrac{f_{\max }-f_{\min }}{2}\), \(B=\dfrac{f_{\max }+f_{\min }}{2}\);
\(\left(2\right)\)求\(ω\):根据图象求出周期\(T\),再利用\(T=\dfrac{2\pi}{ω}\)求出\(ω\);
\(\left(3\right)\)求\(φ\):求出\(A\),\(ω\)后代入函数图象一最值点,求出\(φ\).
【巩固练习】
1.函数\(f\left(x\right)=A\sin \left(ωx+φ\right)\)\((A,ω,φ\)为常数\(,A>0,ω>0 )\)的部分图象如图所示,则\(f\left(0\right)\)的值是\(\underline{\quad \quad}\).
2.若函数\(y=A\sin \left(ωx+φ\right)+B\left(A>0,ω>0\right)\)在其一个周期内的图象上有一个最高点\(\left(\dfrac{\pi}{12} ,3\right)\)和一个最低点\(\left(\dfrac{7\pi}{12} ,-5\right)\),求这个函数的解析式.
参考答案
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答案 \(\dfrac{\sqrt{6}}{2}\)
解析 由图可知:\(A=\sqrt{2}\), \(\dfrac{T}{4}=\dfrac{7 \pi}{12}-\dfrac{\pi}{3}=\dfrac{\pi}{4}\),
所以\(T=π\), \(\omega=\dfrac{2 \pi}{T}=2\),
又函数图象经过点\(\left(\dfrac{\pi}{3} ,0\right)\),所以\(2×\dfrac{\pi}{3} +φ=π\),则\(φ=\dfrac{\pi}{3}\) ,
故函数的解析式为\(f\left(x\right)=\sqrt{2} \sin \left(2x+\dfrac{\pi}{3} \right)\),
所以\(f\left(0\right)=\sqrt{2} \sin \dfrac{\pi}{3} = \dfrac{\sqrt{6}}{2}\). -
答案 \(y=4\sin \left(2x+\dfrac{\pi}{3} \right)-1\)
解析 由已知,\(y_{\max}=3\),\(y_{\min}=-5\),则
①\(A=\dfrac{y_{\max }-y_{\min }}{2}=\dfrac{3-(-5)}{2}=4\);
② \(B=\dfrac{y_{\max }+y_{\min }}{2}=\dfrac{3+(-5)}{2}=-1\);
③由 \(\dfrac{T}{2}=\dfrac{7 \pi}{12}-\dfrac{\pi}{12}=\dfrac{\pi}{2}\),\(\therefore T=π\),得 \(\omega=\dfrac{2 \pi}{T}=\dfrac{2 \pi}{\pi}=2\);
④函数的解析式\(y=A\sin \left(ωx+φ\right)+B=4\sin\left(2x+φ\right)-1\).
将点\(\left(\dfrac{\pi}{12} ,3\right)\)代入,得\(4\sin \left(2×\dfrac{\pi}{12} +φ\right)-1=3\),
即\(\sin \left(\dfrac{\pi}{6} +φ\right)=1\),
所以\(\dfrac{\pi}{6} +φ=2kπ+\dfrac{\pi}{2}\),\(k∈Z\),这里对\(φ\)没有限制,
应该说\(φ=2kπ+\dfrac{\pi}{3}\) ,\(k∈Z\)的任意一个解都满足题意,一 般取\(|φ|<\dfrac{\pi}{2}\) ,
故所求的函数解析式为\(y=4\sin \left(2x+\dfrac{\pi}{3} \right)-1\).
【题型3】函数\(y=A\sin \left(ωx+φ\right)\)性质的综合应用
【典题1】 已知函数\(f\left(x\right)=\sin ^2 ωx+\sqrt{3} \sin ωx\sin \left(ωx+\dfrac{\pi}{2} \right)\left(ω>0\right)\)的最小正周期为\(π\).
\(\left(1\right)\)求\(ω\)的值;
\(\left(2\right)\)求函数\(f(x)\)在区间\(\left[0 ,\dfrac{2\pi}{3} \right]\)上的取值范围.
\(\left(3\right)\)求函数\(f(x)\)的最大值,并且求使\(f(x)\)取得最大值的\(x\)的集合.
解析 \(\left(1\right)f\left(x\right)=\sin ^2 ωx+\sqrt{3} \sin ωx\sin \left(ωx+\dfrac{\pi}{2} \right)=\sin ^2 ωx+\sqrt{3} \sin ωx\cos ωx\)
\(=\dfrac{1-\cos 2 \omega x}{2}+\dfrac{\sqrt{3} \sin 2 \omega x}{2}=\dfrac{1}{2}+\dfrac{\sqrt{3} \sin 2 \omega x}{2}-\dfrac{\cos 2 \omega x}{2}\)
\(=\dfrac{1}{2} +\sin \left(2ωx-\dfrac{\pi}{6} \right)\),
\(\because\)函数\(f\left(x\right)=\sin ^2 ωx+\sqrt{3} \sin ωx\sin \left(ωx+\dfrac{\pi}{2} \right)\left(ω>0\right)\)的最小正周期为\(π\),
\(\therefore ω\)的值为\(1\);
\(\left(2\right)\)由\(\left(1\right)\)得\(f\left(x\right)=\dfrac{1}{2} +\sin \left(2x-\dfrac{\pi}{6} \right)\),
由\(x∈\left[0 ,\dfrac{2\pi}{3} \right]\)得\(2x-\dfrac{\pi}{6} ∈\left[-\dfrac{\pi}{6} ,\dfrac{7\pi}{6} \right]\) ,
所以\(\sin \left(2x-\dfrac{\pi}{6} \right)∈\left[-\dfrac{1}{2} ,1\right]\),
\(\therefore\)函数\(f(x)\)在区间\(\left[0 ,\dfrac{2\pi}{3} \right]\)上的取值范围是\(\left[0 ,\dfrac{3}{2} \right]\),
\(\left(3\right)\because f\left(x\right)=\dfrac{1}{2} +\sin \left(2x-\dfrac{\pi}{6} \right)\),
\(\therefore\)函数的最大值为\(\dfrac{3}{2}\),此时有\(2x-\dfrac{\pi}{6} =2kπ+\dfrac{\pi}{2}\) ,\(k∈Z\),
解得\(x=kπ+\dfrac{\pi}{3}\) ,\(k∈Z\).
即使\(f(x)\)取得最大值的x的集合是\(\{x|x=kπ+\dfrac{\pi}{3},k∈Z\}\).
点拨 通过各种公式(两角和差公式、倍角公式、积化和差公式等)转化,最终把函数的解析式转化为\(f\left(x\right)=A\sin \left(ωx+φ\right)+B\)或\(f\left(x\right)=A\cos \left(ωx+φ\right)+B\)的形式求解函数的各性质(单调性、对称性、周期、最值等).
【巩固练习】
1.已知函数\(f\left(x\right)=\sqrt{3} \sin x\cos x-\sin ^2 x\).
\(\left(1\right)\)求函数\(f(x)\)的最小正周期;
\(\left(2\right)\)求函数\(f(x)\)的单调增区间;
\(\left(3\right)\)求函数\(f(x)\)在区间\(\left[0 ,\dfrac{\pi}{2} \right]\)上的最大值.
2.已知函数\(f\left(x\right)=\sqrt{3} \sin ωx+\cos ωx\left(ω>0\right)\)图象的相邻对称轴与对称中心之间的距离为\(\dfrac{\pi}{4}\).
\(\left(1\right)\)求\(f(x)\)的单调递增区间;
\(\left(2\right)\)当\(x∈\left[-\dfrac{\pi}{6} ,\dfrac{7\pi}{12} \right]\)时,求\(f(x)\)的值域.
参考答案
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答案 \(\left(1\right) π\);\(\left(2\right) \left[-\dfrac{\pi}{3} +kπ ,\dfrac{\pi}{6} +kπ\right]\left(k∈Z\right)\);\(\left(3\right) \dfrac{1}{2}\) .
解析 \(\text { (1) } f(x)=\sqrt{3} \sin x \cos x-\sin ^2 x=\dfrac{\sqrt{3}}{2} \sin 2 x-\dfrac{1-\cos 2 x}{2}\)
\(=\dfrac{\sqrt{3}}{2} \sin 2 x+\dfrac{1}{2} \cos 2 x-\dfrac{1}{2}=\sin \left(2 x+\dfrac{\pi}{6}\right)-\dfrac{1}{2}\),
\(\left(1\right)\)最小正周期\(T=\dfrac{2\pi}{2} =π\);
\(\left(2\right)\)令\(-\dfrac{\pi}{2} +2kπ≤2x+\dfrac{\pi}{6} ≤\dfrac{\pi}{2} +2kπ\),\(k∈Z\),
则\(-\dfrac{\pi}{3} +kπ≤x≤\dfrac{\pi}{6} +kπ\),\(k∈Z\),
故单调增区间为:\(\left[-\dfrac{\pi}{3} +kπ ,\dfrac{\pi}{6} +kπ\right]\),\(\left(k∈Z\right)\),
\(\left(3\right)\)当\(x∈\left[0 ,\dfrac{\pi}{2} \right]\)时,\(2x+\dfrac{\pi}{6} ∈\left[\dfrac{\pi}{6},\dfrac{7\pi}{6} \right]\) ,
则\(f\left(x\right)=\sin \left(2x+\dfrac{\pi}{6} \right)-\dfrac{1}{2} ∈\left[-1 ,\dfrac{1}{2} \right]\),
所以函数\(f\left(x\right)\)在区间\(\left[0 ,\dfrac{\pi}{2} \right]\)上的最大值为\(\dfrac{1}{2}\) . -
答案 \(\left(1\right) \left[kπ-\dfrac{\pi}{3} ,kπ+\dfrac{\pi}{6} \right]\),\(k∈Z\) ;\(\left(2\right) \left[-\sqrt{3},2\right]\).
解析 \(\left(1\right)\because\) 函数\(f\left(x\right)=\sqrt{3} \sin ωx+\cos ωx=2\sin \left(ωx+\dfrac{\pi}{6} \right)\left(ω>0\right)\)
图象的相邻对称轴与对称中心之间的距离为 \(\dfrac{1}{4} ×\dfrac{2\pi}{ω}=\dfrac{\pi}{4}\),
\(\therefore ω=2\),\(f\left(x\right)=2\sin \left(2x+\dfrac{\pi}{6} \right)\).
令\(2kπ-\dfrac{\pi}{2} ≤2x+\dfrac{\pi}{6} ≤2kπ+\dfrac{\pi}{2}\) ,\(k∈Z\),
求得\(kπ-\dfrac{\pi}{3} ≤x≤kπ+\dfrac{\pi}{6}\),
可得函数的增区间为\(\left[kπ-\dfrac{\pi}{3} ,kπ+\dfrac{\pi}{6} \right]\),\(k∈Z\).
\(\left(2\right)\)当\(x∈\left[-\dfrac{\pi}{6} ,\dfrac{7\pi}{12} \right]\)时,\(2x+\dfrac{\pi}{6} ∈\left[-\dfrac{\pi}{6} ,\dfrac{4\pi}{3} \right]\),
\(\sin \left(2x+\dfrac{\pi}{6} \right)∈\left[-\dfrac{\sqrt{3}}{2} ,1\right]\),\(f\left(x\right)∈\left[-\sqrt{3},2\right]\),
故函数\(f(x)\)的值域为\(\left[-\sqrt{3},2\right]\).
分层练习
【A组---基础题】
1.为了得到函数\(y=\cos 3x\)的图象,只需把函数\(y=\cos \left(3x-\dfrac{\pi}{4} \right)\)的图象 ( )
A.向左平移\(\dfrac{\pi}{6}\)个单位长度 \(\qquad \qquad \qquad \qquad \qquad\) B.向右平移\(\dfrac{\pi}{6}\)个单位长度
C.向左平移\(\dfrac{\pi}{12}\)个单位长度 \(\qquad \qquad \qquad \qquad \qquad\) D.向右平移\(\dfrac{\pi}{12}\) 个单位长度
2.把函数\(y=\sin 2x\left(x∈R\right)\)的图象上所有的点向左平行移动\(\dfrac{\pi}{6}\)个单位长度,再把所得图象上所有点的横坐标缩短到原来的\(\dfrac{1}{2}\)倍(纵坐标不变),得到的图象所表示的函数是( )
A. \(y=\sin \left(4x+\dfrac{\pi}{6} \right),x∈R\) \(\qquad \qquad \qquad \qquad\) B. \(y=\sin \left(4x-\dfrac{\pi}{6} \right),x∈R\)
C. \(y=\sin \left(4x+\dfrac{\pi}{3} \right),x∈R\) \(\qquad \qquad \qquad \qquad\) D. \(y=\sin \left(4x-\dfrac{\pi}{3} \right),x∈R\)
3.下列函数中,图象的一部分如图所示的是( )
A.\(y=\sin \left(x+\dfrac{\pi}{6} \right)\) \(\qquad \qquad \qquad \qquad \qquad\) B.\(y=\sin \left(2x-\dfrac{\pi}{6} \right)\)
C.\(y=\cos \left(4x-\dfrac{\pi}{3} \right)\) \(\qquad \qquad \qquad \qquad \qquad\) D.\(y=\cos \left(2x-\dfrac{\pi}{6} \right)\)
4.将函数\(f\left(x\right)=A\sin \left(ωx+\dfrac{\pi}{6} \right)\left(A>0,ω>0\right)\)的图象上的点的横坐标缩短为原来的\(\dfrac{1}{2}\)倍,再向右平移\(\dfrac{\pi}{3}\) 个单位得到函数\(g\left(x\right)=2\cos \left(2x+φ\right)\)的图象,则下列说法正确的是( )
A.函数\(f(x)\)的最小正周期为\(π\)
B.函数\(f(x)\)的单调递增区间为\(\left[2kπ-\dfrac{2\pi}{3} ,2kπ+\dfrac{\pi}{3} \right]\left(k∈Z\right)\)
C.函数\(f(x)\)的图象有一条对称轴为\(x=\dfrac{2\pi}{3}\)
D.函数\(f(x)\)的图象有一个对称中心为\(\left(\dfrac{2\pi}{3} ,0\right)\)
5.如图,函数\(y=A\sin \left(ωx+φ\right)\left(A>0,ω>0,|φ|≤\dfrac{\pi}{2} \right)\)与坐标轴的三个交点\(P\)、\(Q\)、\(R\)满足\(P\left(1,0\right)\),\(∠PQR=\dfrac{\pi}{4}\) ,\(M\)为\(QR\)的中点, \(P M=\dfrac{\sqrt{34}}{2}\),则\(A\)的值为\(\underline{\quad \quad}\) .
6.给出下列六种图象变换的 方法:
①图象上所有点的纵坐标不变,横坐标缩短到原来的\(\dfrac{1}{2}\);
②图象上所有点的纵坐标不变,横坐标伸长为原来的\(2\)倍;
③图象向右平移\(\dfrac{\pi}{3}\) 个单位长度;
④图象向左平移\(\dfrac{\pi}{3}\)个单位长度;
⑤图象向右平移\(\dfrac{2\pi}{3}\)个单位长度;
⑥图象向左平移\(\dfrac{2\pi}{3}\)个单位长度.
请用上述变换中的两种变换,将函数\(y=\sin x\)的图象变换为函数\(y=\sin \left(\dfrac{x}{2} +\dfrac{\pi}{3} \right)\)的图象,那么这两种变换正确的标号是\(\underline{\quad \quad}\) (要求按变换先后顺序填上一种你认为正确的标号即可).
7.已知函数\(f\left(x\right)=\sin \left(ωx+φ\right)\left(ω>0,0≤φ≤π\right)\)是\(R\)上的偶函数,其图象关于点\(M\left(\dfrac{3\pi}{4} ,0\right)\)对称,且在区间\(\left[0,\dfrac{\pi}{2} \right]\)上是单调函数,求\(φ\)和\(ω\)的值.
8.函数\(f\left(x\right)=A\sin \left(ωx-\dfrac{\pi}{6} \right)+1\left(A>0,ω>0\right)\)的最大值为\(3\),其图象相邻两条对称轴之间的距离为\(\dfrac{\pi}{2}\).
\(\left(1\right)\)求函数\(f(x)\)的解析式;
\(\left(2\right)\)设\(α∈\left(0,\dfrac{\pi}{2} \right)\),\(f\left( \dfrac{α}{2}\right)=2\),求\(α\)的值.
9.已知函数\(f\left(x\right)=A\sin \left(ωx+φ\right)+B\left(A>0,ω>0,|φ|<\dfrac{\pi}{2} \right)\)的部分图象如图所示.
\(\left(1\right)\)求\(f(x)\)的解析式;
\(\left(2\right)\)求\(f(x)\)的单调递增区间和对称中心坐标;
\(\left(3\right)\)将\(f(x)\)的图象向左平移\(\dfrac{\pi}{6}\)个单位,再讲横坐标伸长到原来的\(2\)倍,纵坐标不变,最后将图象向上平移\(1\)个单位,得到函数\(g\left(x\right)\)的图象,求函数\(y=g\left(x\right)\)在\(x∈\left[0,\dfrac{7\pi}{6} \right]\)上的最大值和最小值.
10.函数\(f\left(x\right)=\sin \left(2x+\dfrac{\pi}{6} \right)+\cos 2x\).
\(\left(1\right)\)求\(f\left(0\right)\),\(f\left(\dfrac{\pi}{12} \right)\);
\(\left(2\right)\)求函数\(f(x)\)在\(\left[-\dfrac{\pi}{4} ,\dfrac{\pi}{4} \right]\)上的最大值与最小值.
参考答案
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答案 \(C\)
解析 函数\(y=\cos \left(3x-\dfrac{\pi}{4} \right)=\cos \left[3\left(x-\dfrac{\pi}{12} \right)\right]\),
所以只需把函数\(y=\cos \left(3x-\dfrac{\pi}{4} \right)\)的图象,向左平移\(\dfrac{\pi}{12}\)个长度单位,
即可得到函数\(y=\cos \left[3\left(x+\dfrac{\pi}{12} -\dfrac{\pi}{12} \right)\right]=\cos 3x\)的图象.
故选:\(C\). -
答案 \(C\)
解析 函数\(y=\sin 2x\left(x∈R\right)\)的图象上所有的点向左平行移动\(\dfrac{\pi}{6}\)个单位长度得到\(y=\sin \left(2x+\dfrac{\pi}{3} \right)\)的图象,再把所得图象上所有点的横坐标缩短到原来的\(\dfrac{1}{2}\)倍(纵坐标不变),得到\(y=\sin \left(4x+\dfrac{\pi}{3} \right),x∈R\).
故选:\(C\). -
答案 \(D\)
解析 “五点法”对应解方程.设\(y=A\sin \left(ωx+φ\right)\),显然\(A=1\),
又图象过点\(\left(-\dfrac{\pi}{6} ,0\right),\left(\dfrac{\pi}{12} ,1\right)\)
所以\(ω×\left(-\dfrac{\pi}{6} \right)+φ=0\), \(ω×\dfrac{\pi}{12} +φ=\dfrac{\pi}{2}\)解得\(ω=2\),\(φ=\dfrac{\pi}{3}\) .
所以函数解析式为\(y=\sin \left(2x+\dfrac{\pi}{3} \right)=\cos \left(2x-\dfrac{\pi}{6} \right)\).故选\(D\). -
答案 \(B\)
解析 函数\(f\left(x\right)=A\sin \left(ωx+\dfrac{\pi}{6} \right)\left(A>0,ω>0\right)\)的图象上的点的横坐标缩短为原来的\(\dfrac{1}{2}\)倍,再向右平移\(\dfrac{\pi}{3}\)个单位得到:\(g\left(x\right)=A\sin \left(2ωx+\dfrac{\pi}{6} -\dfrac{2πω}{3}\right)\)的图象.
与\(g\left(x\right)=2\cos \left(2x+φ\right)=2\sin \left(2x+φ+\dfrac{\pi}{2} \right)\)比较,
又由于\(A>0\),\(ω>0\),所以\(A=2\),\(ω=1\).
故\(\sin \left(2x-\dfrac{\pi}{2} \right)=\cos \left(2x-π\right)=\cos \left(2x+φ\right)\),
得到:\(φ=2kπ-π\),\(k∈Z\),
所以:\(f\left(x\right)=2\sin \left(x+\dfrac{\pi}{6} \right)\),\(g\left(x\right)=-2\cos 2x\).
故函数\(f(x)\)的周期为\(2π\),\(A\)错误;
令\(2kπ-\dfrac{\pi}{2} ≤x+\dfrac{\pi}{6} ≤2kπ+\dfrac{\pi}{2}\),\(k∈Z\),
解得\(2kπ-\dfrac{2\pi}{3} ≤x≤2kπ+\dfrac{\pi}{3}\),\(k∈Z\),
函数\(f(x)\)单调递增区间为\(\left[2kπ-\dfrac{2\pi}{3} ,2kπ+\dfrac{\pi}{3} \right]\left(k∈Z\right)\),故\(B\)正确;
由于\(f\left(\dfrac{2\pi}{3} \right)=2\sin \dfrac{5\pi}{6} =1\),可得\(C\),\(D\)错误.
故选:\(B\). -
答案 \(5\sqrt{2}\)
解析 由\(∠PQR=\dfrac{\pi}{4}\),所以\(OQ=OR\),设\(Q\left(m,0\right)\),则\(R\left(0,-m\right)\),
又\(M\)为\(QR\)的中点,所以\(M\left(\dfrac{m}{2},-\dfrac{m}{2}\right)\);
又 \(P M=\dfrac{\sqrt{34}}{2}\),即 \(\sqrt{\left(1-\dfrac{m}{2}\right)^2+\left(0+\dfrac{m}{2}\right)^2}=\dfrac{\sqrt{34}}{2}\);
整理得\(m^2-2m-15=0\),解得\(m=5\)或\(m=-3\)(不合题意,舍去);
所以\(R\left(0,-5\right)\),\(Q\left(5,0\right)\);
所以\(\dfrac{1}{2} T=4\),解得\(T=8\),所以\(\dfrac{2\pi}{ω}=8\),解得\(ω=\dfrac{\pi}{4}\);
把\(P\left(1,0\right)\)代入\(f\left(x\right)=A\sin \left(\dfrac{\pi}{4} x+φ\right)\),即\(A\sin \left(\dfrac{\pi}{4} +φ\right)=0\),
由\(|φ|≤\dfrac{\pi}{2}\),得\(φ=-\dfrac{\pi}{4}\);
把\(R\left(0,-5\right)\),代入\(f\left(x\right)=A\sin \left(\dfrac{\pi}{4} x-\dfrac{\pi}{4} \right)\)
得\(A\sin\left (-\dfrac{\pi}{4} \right)=-5\),解得\(A=5\sqrt{2}\). -
答案 ④②或②⑥
解析 \(y=\sin x \stackrel{④}{\rightarrow} y=\sin \left(x+\dfrac{\pi}{3}\right) \stackrel{②}{\rightarrow} y=\sin \left(\dfrac{1}{2} x+\dfrac{\pi}{3}\right)\)
或 \(y=\sin x \stackrel{②}{\rightarrow} y=\sin \dfrac{x}{2} \stackrel{⑥}{\rightarrow} y=\sin \left(\dfrac{x}{2}+\dfrac{\pi}{3}\right)\). -
答案 \(φ=\dfrac{\pi}{2}\),\(ω=2\)或\(\dfrac{2}{3}\)
解析 由\(f\left(x\right)\)是偶函数,得\(f\left(-x\right)=f\left(x\right)\),即函数\(f(x)\)的图象关于\(y\)轴对称,
\(\therefore f\left(x\right)\)当\(x=0\)时取得最值,即\(\sin φ=1\)或\(-1\).
依题设\(0≤φ≤π\),解得\(φ=\dfrac{\pi}{2}\) .
由\(f(x)\)的图象关于点\(M\)对称,
可知\(\sin \left(\dfrac{3\pi}{4} ω+\dfrac{\pi}{2} \right)=0\),解得 \(\omega=\dfrac{4 k}{3}-\dfrac{2}{3}\) ,\(k∈Z\).
又\(f(x)\)在\(\left[0,\dfrac{\pi}{2} \right]\)上是单调函数,\(\therefore T≥π\),即\(\dfrac{2\pi}{ω}≥π\),\(\therefore ω≤2\).
又\(ω>0\),
\(\therefore\)当\(k=1\)时,\(ω=\dfrac{2}{3}\);当\(k=2\)时,\(ω=2\).
\(\therefore φ=\dfrac{\pi}{2}\),\(ω=2\)或\(\dfrac{2}{3}\). -
答案 \(\left(1\right) y=2\sin \left(2x-\dfrac{\pi}{6} \right)+1\);\(\left(2\right) \dfrac{\pi}{3}\).
解析 \(\left(1\right)\because\) 函数\(f(x)\)的最大值为\(3\),\(\therefore A+1=3\),即\(A=2\).
\(\because\) 函数图象的相邻两条对称轴之间的距离为\(\dfrac{\pi}{2}\) ,
\(\therefore\) 最小正周期\(T=π\).
\(\therefore ω=2\).
故函数\(f(x)\)的解析式为\(y=2\sin \left(2x-\dfrac{\pi}{6} \right)+1\).
\(\left(2\right) \therefore f\left( \dfrac{α}{2}\right)=2\sin \left(α-\dfrac{\pi}{6} \right)+1=2\),
即\(\sin \left(α-\dfrac{\pi}{6} \right)=\dfrac{1}{2}\),
\(\because 0<α<\dfrac{\pi}{2}\) ,\(\therefore -\dfrac{\pi}{6} <α-\dfrac{\pi}{6} <\dfrac{\pi}{3}\) .
\(\therefore α-\dfrac{\pi}{6} =\dfrac{\pi}{6}\).故\(α=\dfrac{\pi}{3}\). -
答案 \(\left(1\right)f\left(x\right)=2\sin \left(2x+\dfrac{\pi}{3} \right)-1\);\(\left(2\right)\left(\dfrac{k\pi}{2} -\dfrac{\pi}{6} ,-1\right)\),\(k∈Z\);\(\left(3\right)\)最小值\(-2\),最大值\(\sqrt{3}\) .
解析 \(\left(1\right)\)由图象可知 \(\left\{\begin{array}{l} A+B=1 \\ -A+B=-3 \end{array}\right.\),可得:\(A=2\),\(B=-1\),
又由于 \(\dfrac{T}{2}=\dfrac{7 \pi}{12}-\dfrac{\pi}{12}\) ,可得:\(T=π\),所以 \(\omega=\dfrac{2 \pi}{T}=2\),
由图象及五点法作图可知: \(2 \times \dfrac{\pi}{12}+\varphi=\dfrac{\pi}{2}\) ,所以\(φ=\dfrac{\pi}{3}\),
所以\(f\left(x\right)=2\sin \left(2x+\dfrac{\pi}{3} \right)-1\).
\(\left(2\right)\)由\(\left(1\right)\)知,\(f\left(x\right)=2\sin \left(2x+\dfrac{\pi}{3} \right)-1\),
令\(2kπ-\dfrac{\pi}{2} ≤2x+\dfrac{\pi}{3} ≤2kπ+\dfrac{\pi}{2}\),\(k∈Z\),
得\(kπ-\dfrac{5\pi}{12} ≤x≤kπ+\dfrac{\pi}{12}\),\(k∈Z\),
所以\(f(x)\)的单调递增区间为\(\left[kπ-\dfrac{5\pi}{12} ,kπ+\dfrac{\pi}{12} \right]\),\(k∈Z\),
令\(2x+\dfrac{\pi}{3} =kπ\),\(k∈Z\),得\(x=\dfrac{k\pi}{2} -\dfrac{\pi}{6}\) ,\(k∈Z\),
所以\(f(x)\)的对称中心的坐标为\(\left(\dfrac{k\pi}{2} -\dfrac{\pi}{6} ,-1\right)\),\(k∈Z\).
\(\left(3\right)\)由已知的图象变换过程可得:\(g\left(x\right)=2\sin \left(x+\dfrac{2\pi}{3} \right)\),
因为\(0≤x≤\dfrac{7\pi}{6}\),所以\(\dfrac{2\pi}{3} ≤x+\dfrac{2\pi}{3} ≤\dfrac{11\pi}{6}\),
所以当\(x+\dfrac{2\pi}{3} =\dfrac{3\pi}{2}\),得\(x=\dfrac{5\pi}{6}\)时,\(g\left(x\right)\)取得最小值\(g\left(\dfrac{5\pi}{6} \right)=-2\),
当\(x+\dfrac{2\pi}{3} =\dfrac{2\pi}{3}\),即\(x=0\)时,\(g\left(x\right)\)取得最大值\(g\left(0\right)=\sqrt{3}\). -
答案 \(\left(1\right) f\left(0\right)=\dfrac{3}{2}\) ,\(f\left(\dfrac{\pi}{12} \right)=\sqrt{3}\);\(\left(2\right)\)最大值为\(\sqrt{3}\),最小值为 \(-\dfrac{\sqrt{3}}{2}\) .
解析 \(\left(1\right)\)因为 \(f(x)=\sin \left(2 x+\dfrac{\pi}{6}\right)+\cos 2 x=\dfrac{\sqrt{3}}{2} \sin 2 x+\dfrac{1}{2} \cos 2 x+\cos 2 x\)
\(=\dfrac{\sqrt{3}}{2} \sin 2 x+\dfrac{3}{2} \cos 2 x=\sqrt{3} \sin \left(2 x+\dfrac{\pi}{3}\right)\),
所以\(f\left(0\right)=\sqrt{3} \sin \left(2×0+\dfrac{\pi}{3} \right)=\dfrac{3}{2}\),\(f\left(\dfrac{\pi}{12} \right)=\sqrt{3} \sin \left(2×\dfrac{\pi}{12} +\dfrac{\pi}{3} \right)=\sqrt{3}\);
\(\left(2\right)f\left(x\right)=\sqrt{3} \sin \left(2x+\dfrac{\pi}{3} \right)\),
因为\(x∈\left[-\dfrac{\pi}{4} ,\dfrac{\pi}{4} \right]\),所以\(-\dfrac{\pi}{6} ≤2x+\dfrac{\pi}{3} ≤\dfrac{5\pi}{6}\),
故\(-\dfrac{1}{2} ≤\sin \left(2x+\dfrac{\pi}{3} \right)≤1\),所以 \(-\dfrac{\sqrt{3}}{2} \leq f(x) \leq \sqrt{3}\),
所以函数\(f(x)\)在\(\left[-\dfrac{\pi}{4} ,\dfrac{\pi}{4} \right]\)上的最大值为\(\sqrt{3}\),最小值为 \(-\dfrac{\sqrt{3}}{2}\) .
【B组---提高题】
1.已知函数\(f\left(x\right)=\sin \left(ωx+φ\right)\left(ω>0,0<φ<π\right)\),\(f\left(0\right)=f\left(\dfrac{2 \pi}{9}\right)=-f\left(\dfrac{\pi}{3} \right)\),且\(f\left(x\right)\)在\(\left(\dfrac{\pi}{6} ,\dfrac{4\pi}{9}\right)\)上单调,则函数\(y=f(x)\)的解析式是\(\underline{\quad \quad}\) .
2.如图是函数\(f\left(x\right)=A \sin \left(ωx+φ\right)\left(A>0,ω>0,0<φ<\dfrac{\pi}{2} \right)\)的部分图象,\(M\)、\(N\)是它与\(x\)轴的两个不同交点,\(D\)是\(M\)、\(N\)之间的最高点且横坐标为\(\dfrac{\pi}{4}\),点\(F\left(0,1\right)\)是线段\(DM\)的中点.
\(\left(1\right)\)求函数\(f(x)\)的解析式及\(\left[π,2π\right]\)上的单调增区间;
\(\left(2\right)\)若\(x∈\left[-\dfrac{\pi}{12} ,\dfrac{5\pi}{12} \right]\)时,函数\(h\left(x\right)=f^2 \left(x\right)-a f\left(x\right)+1\)的最小值为\(\dfrac{1}{2}\) ,求实数\(a\)的值.
参考答案
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答案 \(f\left(x\right)=\sin \left(3x+\dfrac{\pi}{6} \right)\)
解析 对于函数\(f\left(x\right)=\sin \left(ωx+φ\right) \left (ω>0,0<φ<π\right)\),
由\(f\left(0\right)=f\left(\dfrac{2\pi}{9}\right)\),可得函数的图象关于直线\(x=\dfrac{1}{2} \left(0+\dfrac{2\pi}{9}\right)=\dfrac{\pi}{9}\)对称;
又\(f\left(\dfrac{2\pi}{9}\right)=-f\left(\dfrac{\pi}{3} \right)\),可得函数的图象关于点\(\left(\dfrac{\dfrac{2 \pi}{9}+\dfrac{\pi}{3}}{2},0\right).\)对称,即\(\left(\dfrac{5 \pi}{18},0\right)\);
\(\therefore \dfrac{T}{4}+k T=\dfrac{5 \pi}{18}-\dfrac{\pi}{9}=\dfrac{\pi}{6}\) ,\(k∈Z\), 解得 \(T=\dfrac{2 \pi}{3(4 k+1)}\),
\(\therefore \omega=\dfrac{2 \pi}{T}=3(4 k+1)\);
\(\because f\left(x\right)\)在\(\left(\dfrac{\pi}{6} ,\dfrac{4\pi}{9}\right)\)上单调,\(\therefore \dfrac{T}{2} \geq \dfrac{4 \pi}{9}-\dfrac{\pi}{6}\),解得\(T>\dfrac{5\pi}{9}\),
\(\therefore 0<\omega \leq \dfrac{18}{5} \text {, }\),
又 \(\omega=\dfrac{2 \pi}{T}=3(4 k+1)\), \(\therefore ω=3\),
\(\because \left(\dfrac{5 \pi}{18},0\right)\)是对称中心,\(\therefore f\left(\dfrac{5 \pi}{18}\right)=0\),
即 \(\sin \left(3 \times \dfrac{5 \pi}{18}+\varphi\right)=0\),
又\(\because 0<φ<π\) , \(\therefore φ=\dfrac{\pi}{6}\),
\(\therefore f\left(x\right)=\sin \left(3x+\dfrac{\pi}{6} \right)\). -
答案 \(\left(1\right) f\left(x\right)=2\sin \left(x+\dfrac{\pi}{4} \right)\),\(\left[\dfrac{5\pi}{4} ,2π\right]\);\(\left(2\right) \dfrac{3}{2}\).
解析 \(\left(1\right)\)取\(MN\)的中点为\(H\),则\(DH⊥MN\),
因为\(F\)为\(DM\)的中点,且\(F\)在\(y\)轴上,
则\(OF//DH\)且\(OF=\dfrac{1}{2} DH\),则\(OM=OH\),
所以\(D\left(\dfrac{\pi}{4} ,2\right)\),\(M\left(-\dfrac{\pi}{4} ,0\right)\),则\(A=2\),
\(T=\dfrac{2 \pi}{\omega}=4\left[\dfrac{\pi}{4}-\left(-\dfrac{\pi}{4}\right)\right]=2 \pi\),所以\(ω=1\),
所以\(f\left(x\right)=2\sin \left(x+φ\right)\),
由\(f\left(\dfrac{\pi}{4} \right)=2\),解得\(φ=2kπ+\dfrac{\pi}{4}\),\(k∈Z\),
由\(0<φ<\dfrac{\pi}{2}\) ,所以\(φ=\dfrac{\pi}{4}\) ,
即\(f\left(x\right)=2\sin \left(x+\dfrac{\pi}{4} \right)\),
令\(-\dfrac{\pi}{2} +2kπ≤x+\dfrac{\pi}{4} ≤\dfrac{\pi}{2} +2kπ\),解得\(-\dfrac{3\pi}{4} +2kπ≤x≤\dfrac{\pi}{4} +2kπ\),
又\(x∈\left[π,2π\right]\),
所以函数\(f\left(x\right)\)在\(\left[π,2π\right]\)上的单调增区间为:\(\left[\dfrac{5\pi}{4} ,2π\right]\);
\(\left(2\right)\)因为\(-\dfrac{\pi}{12} ≤x≤\dfrac{5\pi}{12}\),所以\(\dfrac{\pi}{6} ≤x+\dfrac{\pi}{4} ≤\dfrac{2\pi}{3}\),
所以\(\dfrac{1}{2} ≤\sin \left(x+\dfrac{\pi}{4} \right)≤1\),所以\(1≤f\left(x\right)≤2\),
令\(t=f\left(x\right)\),则\(t∈\left[1,2\right]\),
则 \(g(t)=t^2-a t+1=\left(t-\dfrac{a}{2}\right)^2+1-\dfrac{a^2}{4}\),
①当\(\dfrac{a}{2} ≤1\),即\(a≤2\)时, \(g(t)_{\min }=g(1)=\dfrac{1}{2}\),解得:\(a=\dfrac{3}{2}\) ,
②当\(1<\dfrac{a}{2} <2\),即\(2<a<4\)时, \(g(t)_{\min }=g\left(\dfrac{a}{2}\right)=1-\dfrac{a^2}{4}=\dfrac{1}{2}\),解得:\(a=±\sqrt{2}\)(舍),
③当\(\dfrac{a}{2} ≥2\)即\(a≥4\)时, \(g(t)_{\min }=g(2)=\dfrac{1}{2}\),解得 \(a=\dfrac{9}{4}\)(舍),
综合①②③得实数\(a\)的值为\(\dfrac{3}{2}\) .
【C组---拓展题】
1.已知函数\(f\left(x\right)=\sqrt{3} \sin \left(2ωx+φ\right)+1\left(ω>0,-\dfrac{\pi}{2} <φ<\dfrac{\pi}{2} \right)\),函数\(f(x)\)的图象经过点\(\left(-\dfrac{\pi}{12} ,1\right)\)且\(f(x)\)的最小正周期为\(\dfrac{\pi}{2}\).
\(\left(1\right)\)求函数\(f(x)\)的解析式;
\(\left(2\right)\)将函数\(y=f(x)\)图象上所有的点向下平移\(1\)个单位长度,再函数图象上所有点的横坐标变为原来的\(2\)倍,纵坐标不变,再将图象上所有的点的横坐标不变,纵坐标变为原来的 \(\dfrac{2 \sqrt{3}}{3}\)倍,得到函数\(y=h\left(x\right)\)图象,令函数\(g\left(x\right)=h\left(x\right)+1\),区间\([a ,b ] (a ,b∈R\)且\(a<b )\)满足:\(y=g\left(x\right)\)在\(\left[a ,b\right]\)上至少有\(30\)个零点,在所有满足上述条件的\(\left[a ,b\right]\)中,求\(b-a\)的最小值.
\(\left(3\right)\)若\(m\left[1+\sqrt{3}\left(f\left(\dfrac{x}{8}-\dfrac{\pi}{12} \right)-1\right)\right]+\dfrac{1}{2} +\dfrac{3}{2} \cos x≤0\)对任意\(x∈\left[0 ,2π\right]\)恒成立,求实数\(m\)的取值范围.
参考答案
- 答案 \(\left(1\right) f\left(x\right)=\sqrt{3} \sin \left(4x+\dfrac{\pi}{3} \right)+1\);\(\left(2\right) \dfrac{43\pi}{3}\);\(\left(3\right) \left(-∞ ,-2\right]\).
解析 \(\left(1\right)\because f\left(x\right)=\sqrt{3} \sin \left(2ωx+φ\right)+1\),
又函数\(f\left(x\right)\)的最小正周期为\(\dfrac{\pi}{2}\),\(\therefore \dfrac{2\pi}{2} ω=\dfrac{\pi}{2}\),\(\therefore ω=2\).
\(\therefore f\left(x\right)=\sqrt{3} \sin \left(4x+φ\right)+1\).
又函数\(f(x)\)经过点\(\left(-\dfrac{\pi}{12} ,1\right)\),
所以\(f\left(-\dfrac{\pi}{12} \right)=\sqrt{3} \sin \left(-\dfrac{\pi}{3} +φ\right)+1=1\),
于是\(\left(4×\left(-\dfrac{\pi}{12} \right)+φ\right)=kπ\),\(k∈Z\)
因为\(-\dfrac{\pi}{2} <ϕ<\dfrac{\pi}{2}\),所以\(ϕ=\dfrac{\pi}{3}\).
故\(f\left(x\right)=\sqrt{3} \sin \left(4x+\dfrac{\pi}{3} \right)+1\).
\(\left(2\right)\)由题意,\(h(x)=2 \sin \left(2 x+\dfrac{\pi}{3}\right) g(x)=2 \sin \left(2 x+\dfrac{\pi}{3}\right)+1\).
令\(g\left(x\right)=0\)得:\(\sin \left(2x+\dfrac{\pi}{3} \right)=-\dfrac{1}{2}\),
\(\therefore 2x+\dfrac{\pi}{3} =2kπ+\dfrac{7\pi}{6}\)或\(2x+\dfrac{\pi}{3} =2kπ+\dfrac{11\pi}{6}\) ,\(k∈Z\)
解得:\(x=kπ+\dfrac{5\pi}{12}\)或\(x=kπ+\dfrac{3\pi}{4}\) ,\(k∈Z\)
\(\therefore\)相邻两个零点之间的距离为\(\dfrac{\pi}{3}\)或\(\dfrac{2\pi}{3}\).
若\(b-a\)最小,则\(a\) ,\(b\)均为\(g\left(x\right)\)的零点,
此时在区间\(\left[a ,π+a\right]\),\(\left[a ,2π+a\right]\),… ,\(\left[a ,mπ+a\right]\left(m∈N^*\right)\)分别恰有\(3\),\(5\),…,\(2m+1\)个零点.
\(\therefore\)在区间\(\left[a ,14π+a\right]\)恰有\(2×14+1=29\)个零点.
\(\therefore \left(14π+a ,b\right]\)至少有一个零点.
\(\therefore b-\left(14π+a\right)≥\dfrac{\pi}{3}\),即\(b-a≥14π+\dfrac{\pi}{3} =\dfrac{43\pi}{3}\).
检验可知,在\(\left[\dfrac{5\pi}{12} ,\dfrac{5\pi}{12} +\dfrac{43\pi}{4} \right]\)恰有\(30\)个零点,满足题意(可有可无)
\(\therefore b-a\)的最小值为\(\dfrac{43\pi}{3}\) .
\(\left(3\right)\)由题意得\(m\left(3\sin \dfrac{x}{2} +1\right)≤3\sin ^2 \dfrac{x}{2} -2\).
\(\because x∈\left[0 ,2π\right]\),\(\therefore \dfrac{x}{2} ∈\left[0 ,π\right]\),
\(\therefore \sin \dfrac{x}{2} ∈\left[0 ,1\right]\),\(m \leq \dfrac{3 \sin ^2 \dfrac{x}{2}-2}{3 \sin \dfrac{x}{2}+1}\).
设\(t=3\sin \dfrac{x}{2} +1\),\(t∈\left[1 ,4\right]\).则 \(\sin \dfrac{x}{2}=\dfrac{t-1}{3}\).
设 \(y=\dfrac{3 \sin ^2 \dfrac{x}{2}-2}{3 \sin \dfrac{x}{2}+1}\).
则 \(y=\dfrac{3 \cdot \dfrac{1}{9}(t-1)^2-2}{t}=\dfrac{t^2-2 t-5}{3 t}=\dfrac{1}{3}\left(t-\dfrac{5}{t}-2\right)\)在\(t∈\left[1 ,4\right]\)上是增函数.
\(\therefore\)当\(t=1\)时, \(y_{\min }=-2\),\(\therefore m≤-2\).
故实数\(m\)的取值范围是\(\left(-∞ ,-2\right]\).