di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Trigonometric Functions


Question: $已知函数f(x)=\sqrt{2}\\sin  (\omega x+\phi )\left(\omega >0,|\varphi |\leq \frac{\pi }{2}\right)的部分图象如图所示.$<img_2144>
$将函数f(x)的图象向左平移\frac{\pi }{4}个单位长度，再将图象上各点的横坐标伸长到原来的2倍(纵坐标不变)，得到函数g(x)的图象，若关于x的方程g(x)-m=0在区间\left[0,\frac{3\pi }{4}\right]上有两个不同的实数解，求实数m的取值范围.$

Solution:
$由题意得g(x)=\sqrt{2}\\sin  \left(x+\frac{\pi }{4}\right)，g(x)-m=0在区间\left[0,\frac{3\pi }{4}\right]上有两个不同的实数解，即直线y=m与函数g(x)=\sqrt{2}\\sin  \left(x+\frac{\pi }{4}\right)的图象有两个不同的交点，画出y=g(x)在\left[0,\frac{3\pi }{4}\right]上的图象及直线y=m,如图.$<img_2722>$由图可知，若g(x)-m=0在\left[0,\frac{3\pi }{4}\right]上有两个不同的实数解，则m\in [1,\sqrt{2}).$

Final answer: $[1,\sqrt 2)$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(\right.A>0,\omega >0,|\phi |<\frac{\pi }{2}\left)\right.的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1664456008530206721/img/compression/M_5832cbf7c122cc168968a9bec677338f.jpg)
$求A，\omega 和\phi 的值；$

Solution:
$由题图可得A=1，T=2\times \left(\frac{4}{3}-\frac{1}{3}\right)=2，则\omega =\frac{2\pi }{T}=\pi .易知当x=\frac{\frac{1}{3}+\frac{4}{3}}{2}=\frac{5}{6}时， f(x)取得最大值，则\frac{5\pi }{6}+\phi =\frac{\pi }{2}+2k\pi (k\in Z)，所以\phi =-\frac{\pi }{3}+2k\pi (k\in Z)，又因为|\phi |<\frac{\pi }{2}，所以\phi =-\frac{\pi }{3}.$

Final answer: $A=1$, $\omega =\pi , $\phi = -\pi /3$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(\right.A>0,\omega >0,|\phi |<\frac{\pi }{2}\left)\right.的部分图象如图所示.$<img_1899>
$求A，\omega 和\phi 的值；$

Solution:
$由题图可得A=1，T=2\times \left(\frac{4}{3}-\frac{1}{3}\right)=2，则\omega =\frac{2\pi }{T}=\pi .易知当x=\frac{\frac{1}{3}+\frac{4}{3}}{2}=\frac{5}{6}时， f(x)取得最大值，则\frac{5\pi }{6}+\phi =\frac{\pi }{2}+2k\pi (k\in Z)，所以\phi =-\frac{\pi }{3}+2k\pi (k\in Z)，又因为|\phi |<\frac{\pi }{2}，所以\phi =-\frac{\pi }{3}.$



Final answer: $A=1$, $\omega =\pi , $\phi = -\pi /3$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(\right.A>0,\omega >0,|\phi |<\frac{\pi }{2}\left)\right.的部分图象如图所示.$<img_1899>
$求函数y=f(x)在[1,2]上的单调递减区间；$

Solution:
$由(1)可知f(x)=\\sin  \left(\pi x-\frac{\pi }{3}\right)，令\frac{\pi }{2}+2k\pi \leq \pi x-\frac{\pi }{3}\leq \frac{3\pi }{2}+2k\pi ，k\in Z，解得\frac{5}{6}+2k\leq x\leq \frac{11}{6}+2k，k\in Z.故f(x)的单调递减区间为\left[\frac{5}{6}+2k,\frac{11}{6}+2k\right](k\in Z).取k=0，可得f(x)在[1，2]上的单调递减区间为\left[1,\frac{11}{6}\right].$

Final answer: $[1,\frac{11}6]$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(\right.A>0,\omega >0,|\phi |<\frac{\pi }{2}\left)\right.的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1664456008530206721/img/compression/M_5832cbf7c122cc168968a9bec677338f.jpg)
$求函数y=f(x)在[1,2]上的单调递减区间；$

Solution:
$由(1)可知f(x)=\\sin  \left(\pi x-\frac{\pi }{3}\right)，令\frac{\pi }{2}+2k\pi \leq \pi x-\frac{\pi }{3}\leq \frac{3\pi }{2}+2k\pi ，k\in Z，解得\frac{5}{6}+2k\leq x\leq \frac{11}{6}+2k，k\in Z.故f(x)的单调递减区间为\left[\frac{5}{6}+2k,\frac{11}{6}+2k\right](k\in Z).取k=0，可得f(x)在[1，2]上的单调递减区间为\left[1,\frac{11}{6}\right].$

Final answer: $[1,\frac{11}6]$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(\right.A>0,\omega >0,|\phi |<\frac{\pi }{2}\left)\right.的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1664456008530206721/img/compression/M_5832cbf7c122cc168968a9bec677338f.jpg)
$若函数y=f(x)在区间[a,b]上恰有2 020个零点，求b-a的取值范围.$

Solution:
$令f(x)=\\sin  \left(\pi x-\frac{\pi }{3}\right)=0，得\pi x-\frac{\pi }{3}=k\pi ，k\in Z，解得x=k+\frac{1}{3}，k\in Z，所以f(x)在\left[\frac{1}{3},\frac{7}{3}\right)上有两个零点.因为f(x)的最小正周期为2，所以若函数y=f(x)在区间[a，b]上恰有2 020个零点，则1 009\times 2+1\leq b-a<1 010\times 2+1，故b-a的取值范围为[2 019,2 021).$

Final answer: $[2 019, 2 021]$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(\right.A>0,\omega >0,|\phi |<\frac{\pi }{2}\left)\right.的部分图象如图所示.$<img_1899>
$若函数y=f(x)在区间[a,b]上恰有2 020个零点，求b-a的取值范围.$

Solution:
$令f(x)=\\sin  \left(\pi x-\frac{\pi }{3}\right)=0，得\pi x-\frac{\pi }{3}=k\pi ，k\in Z，解得x=k+\frac{1}{3}，k\in Z，所以f(x)在\left[\frac{1}{3},\frac{7}{3}\right)上有两个零点.因为f(x)的最小正周期为2，所以若函数y=f(x)在区间[a，b]上恰有2 020个零点，则1 009\times 2+1\leq b-a<1 010\times 2+1，故b-a的取值范围为[2 019,2 021).$

Final answer: $[2 019, 2 021]$
Subfield: Plane Geometry


Question: $某港口水深y(单位:m)是时间t(0\leq t\leq 24,单位:h,凌晨0:00时，t=0)的函数,下面是水深数据:
t(h)
0
3
6
9
12
15
18
21
24
y(m)
10.0
13.0
9.9
7.0
10.0
13.0
10.1
7.0
10.0
据上述数据得到的曲线如图所示,该曲线可近似地看成函数y=A\\sin   \omega t+b(A>0,\omega >0)的图象.$![Image](https://cdntk.53zaixian.com/tq/1664456018887827458/img/compression/M_1c2319590377ed26dade09ed47fe0485.jpg)$
$
$试根据数据表和曲线,求y=A\\sin   \omega t+b(A>0,\omega >0)的解析式;$

Solution:
$根据题意可得\left\{\begin{matrix}A+b=13,\\ -A+b=7,\end{matrix}\right.得\left\{\begin{matrix}A=3,\\ b=10,\end{matrix}\right.T=15-3=12,\therefore \omega =\frac{2\pi }{T}=\frac{\pi }{6},\therefore 函数的解析式为y=3\\sin  \frac{\pi }{6}t+10(0\leq t\leq 24).$

Final answer: $y=3\sin (\frac{\pi}{6}t)+10$
Subfield: Plane Geometry


Question: $某港口水深y(单位:m)是时间t(0\leq t\leq 24,单位:h,凌晨0:00时，t=0)的函数,下面是水深数据:
t(h)
0
3
6
9
12
15
18
21
24
y(m)
10.0
13.0
9.9
7.0
10.0
13.0
10.1
7.0
10.0
据上述数据得到的曲线如图所示,该曲线可近似地看成函数y=A\\sin   \omega t+b(A>0,\omega >0)的图象.$<img_1992>$
$
$试根据数据表和曲线,求y=A\\sin   \omega t+b(A>0,\omega >0)的解析式;$

Solution:
$根据题意可得\left\{\begin{matrix}A+b=13,\\ -A+b=7,\end{matrix}\right.得\left\{\begin{matrix}A=3,\\ b=10,\end{matrix}\right.T=15-3=12,\therefore \omega =\frac{2\pi }{T}=\frac{\pi }{6},\therefore 函数的解析式为y=3\\sin  \frac{\pi }{6}t+10(0\leq t\leq 24).$

Final answer: $y=3\sin (\frac{\pi}{6}t)+10$
Subfield: Plane Geometry


Question: $某港口水深y(单位:m)是时间t(0\leq t\leq 24,单位:h,凌晨0:00时，t=0)的函数,下面是水深数据:
t(h)
0
3
6
9
12
15
18
21
24
y(m)
10.0
13.0
9.9
7.0
10.0
13.0
10.1
7.0
10.0
据上述数据得到的曲线如图所示,该曲线可近似地看成函数y=A\\sin   \omega t+b(A>0,\omega >0)的图象.$![Image](https://cdntk.53zaixian.com/tq/1664456018887827458/img/compression/M_1c2319590377ed26dade09ed47fe0485.jpg)$
$
$一般情况下,船舶航行时船底与海底的距离不小于4.5 m是安全的,如果某船的吃水深度(船底与水面的距离)为7 m,那么该船在什么时间段能够安全进港?$

Solution:
$由题意得y\geq 4.5+7,即3\\sin  \frac{\pi }{6}t+10\geq 11.5,\therefore \\sin  \frac{\pi }{6}t\geq \frac{1}{2},\therefore \frac{\pi }{6}t\in \left[2k\pi +\frac{\pi }{6},2k\pi +\frac{5\pi }{6}\right],k\in Z,又0\leq t\leq 24,\therefore k=0或k=1,\therefore t\in [1,5]或t\in [13,17].故该船在1:00至5:00或13:00至17:00能安全进港.$

Final answer: $该船在1:00至5:00或13:00至17:00能安全进港.$
Subfield: Plane Geometry


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(\right.A>0,\omega >0,|\phi |<\frac{\pi }{2}\left)\right.的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1664456042237517825/img/compression/M_fdb8cb5e664105eae0f8f1368c3b3404.jpg)
$求函数f(x)的解析式；$

Solution:
$由题图可知A=2，T=2\times \left(\frac{7}{6}-\frac{1}{6}\right)=2，所以\omega =\frac{2\pi }{T}=\pi ，所以f(x)=2\\sin  (\pi x+\phi ).又点\left(\frac{1}{6},2\right)在f(x)的图象上，所以2\\sin  \left(\frac{\pi }{6}+\varphi \right)=2，所以\frac{\pi }{6}+\phi =\frac{\pi }{2}+2k\pi ，k\in Z,即\phi =\frac{\pi }{3}+2k\pi ，k\in Z，又|\phi |<\frac{\pi }{2}，所以\phi =\frac{\pi }{3}.故f(x)=2\\sin  \left(\pi x+\frac{\pi }{3}\right).$

Final answer: $f(x) = 2\sin (\pi x + \pi /3)$
Subfield: Plane Geometry


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(\right.A>0,\omega >0,|\phi |<\frac{\pi }{2}\left)\right.的部分图象如图所示.$<img_2096>
$求函数f(x)的解析式；$

Solution:
$由题图可知A=2，T=2\times \left(\frac{7}{6}-\frac{1}{6}\right)=2，所以\omega =\frac{2\pi }{T}=\pi ，所以f(x)=2\\sin  (\pi x+\phi ).又点\left(\frac{1}{6},2\right)在f(x)的图象上，所以2\\sin  \left(\frac{\pi }{6}+\varphi \right)=2，所以\frac{\pi }{6}+\phi =\frac{\pi }{2}+2k\pi ，k\in Z,即\phi =\frac{\pi }{3}+2k\pi ，k\in Z，又|\phi |<\frac{\pi }{2}，所以\phi =\frac{\pi }{3}.故f(x)=2\\sin  \left(\pi x+\frac{\pi }{3}\right).$

Final answer: $f(x) = 2\sin (\pi x + \pi /3)$
Subfield: Plane Geometry


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(\right.A>0,\omega >0,|\phi |<\frac{\pi }{2}\left)\right.的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1664456042237517825/img/compression/M_fdb8cb5e664105eae0f8f1368c3b3404.jpg)
$求方程f(x)=-\frac{3}{2}在区间[0,4]内的所有实数根之和.$

Solution:
$如图，易得f(x)在[0,4]上的图象与直线y=-\frac{3}{2}有4个交点，则方程f(x)=-\frac{3}{2}在[0,4]上有4个实数根，设这4个实数根分别为x_{1}，x_{2}，x_{3}，x_{4}(x_{1}$![Image](https://cdntk.53zaixian.com/tq/1664456042497564674/img/compression/M_4aa7a8f46c50816173796aa00f5b904a.jpg)$将方程的根的问题转化为对应函数的图象的交点问题，结合图象分析解决问题.由图可得x_{1}，x_{2}关于直线x=\frac{7}{6}对称，x_{3}，x_{4}关于直线x=\frac{19}{6}对称，所以x_{1}+x_{2}=\frac{7}{3}，x_{3}+x_{4}=\frac{19}{3}，所以x_{1}+x_{2}+x_{3}+x_{4}=\frac{26}{3}.$

Final answer: $\frac{26}{3}$
Subfield: Plane Geometry


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(\right.A>0,\omega >0,|\phi |<\frac{\pi }{2}\left)\right.的部分图象如图所示.$<img_2096>
$求方程f(x)=-\frac{3}{2}在区间[0,4]内的所有实数根之和.$

Solution:
$如图，易得f(x)在[0,4]上的图象与直线y=-\frac{3}{2}有4个交点，则方程f(x)=-\frac{3}{2}在[0,4]上有4个实数根，设这4个实数根分别为x_{1}，x_{2}，x_{3}，x_{4}(x_{1}$<img_2315>$将方程的根的问题转化为对应函数的图象的交点问题，结合图象分析解决问题.由图可得x_{1}，x_{2}关于直线x=\frac{7}{6}对称，x_{3}，x_{4}关于直线x=\frac{19}{6}对称，所以x_{1}+x_{2}=\frac{7}{3}，x_{3}+x_{4}=\frac{19}{3}，所以x_{1}+x_{2}+x_{3}+x_{4}=\frac{26}{3}.$

Final answer: $\frac{26}{3}$
Subfield: Plane Geometry


Question: $如图,点A,B是圆心在原点,半径分别为1和2的圆上的动点.动点A从初始位置A_{0}\left(\\\cos   \frac{\pi }{3},\\\sin   \frac{\pi }{3}\right)开始,按逆时针方向以2 rad/s的角速度做圆周运动,同时点B从初始位置B_{0}(2,0)开始,按顺时针方向以2 rad/s的角速度做圆周运动.记t(s)时刻,点A,B的纵坐标分别为y_{1},y_{2}.$![Image](https://cdntk.53zaixian.com/tq/1664456043135098883/img/compression/M_de8c612acf85570b22839e8a70fa3aec.jpg)
$求t=\frac{\pi }{4}时,A,B两点间的距离;$

Solution:
$当t=\frac{\pi }{4}时,\angle xOA=\frac{\pi }{3}+\frac{\pi }{2}=\frac{5\pi }{6},\angle xOB=\frac{\pi }{2},过点A作AD\perp BO，交BO的延长线于点D,则\angle AOD=\frac{\pi }{3}.\because OA=1,\therefore OD=\frac{1}{2},AD=\frac{\sqrt{3}}{2},\because OB=2,\therefore BD=2+\frac{1}{2}=\frac{5}{2},\therefore AB=\sqrt{AD^2+BD^2}=\sqrt{\frac{3}{4}+\frac{25}{4}}=\sqrt{7},即A,B两点间的距离为\sqrt{7}.$

Final answer: $\sqrt{7}$
Subfield: Plane Geometry


Question: $如图,点A,B是圆心在原点,半径分别为1和2的圆上的动点.动点A从初始位置A_{0}\left(\\\cos   \frac{\pi }{3},\\\sin   \frac{\pi }{3}\right)开始,按逆时针方向以2 rad/s的角速度做圆周运动,同时点B从初始位置B_{0}(2,0)开始,按顺时针方向以2 rad/s的角速度做圆周运动.记t(s)时刻,点A,B的纵坐标分别为y_{1},y_{2}.$<img_2253>
$求t=\frac{\pi }{4}时,A,B两点间的距离;$

Solution:
$当t=\frac{\pi }{4}时,\angle xOA=\frac{\pi }{3}+\frac{\pi }{2}=\frac{5\pi }{6},\angle xOB=\frac{\pi }{2},过点A作AD\perp BO，交BO的延长线于点D,则\angle AOD=\frac{\pi }{3}.\because OA=1,\therefore OD=\frac{1}{2},AD=\frac{\sqrt{3}}{2},\because OB=2,\therefore BD=2+\frac{1}{2}=\frac{5}{2},\therefore AB=\sqrt{AD^2+BD^2}=\sqrt{\frac{3}{4}+\frac{25}{4}}=\sqrt{7},即A,B两点间的距离为\sqrt{7}.$

Final answer: $\sqrt{7}$
Subfield: Plane Geometry


Question: $如图,点A,B是圆心在原点,半径分别为1和2的圆上的动点.动点A从初始位置A_{0}\left(\\\cos   \frac{\pi }{3},\\\sin   \frac{\pi }{3}\right)开始,按逆时针方向以2 rad/s的角速度做圆周运动,同时点B从初始位置B_{0}(2,0)开始,按顺时针方向以2 rad/s的角速度做圆周运动.记t(s)时刻,点A,B的纵坐标分别为y_{1},y_{2}.$![Image](https://cdntk.53zaixian.com/tq/1664456043135098883/img/compression/M_de8c612acf85570b22839e8a70fa3aec.jpg)
$求y=y_{1}+y_{2}关于时间t(t>0)的函数关系式,并求当t\in \left(0,\frac{\pi }{2}\right]时,这个函数的值域.$

Solution:
$依题意知y_{1}=\\sin  \left(2t+\frac{\pi }{3}\right),y_{2}=-2\\sin   2t,所以y=y_{1}+y_{2}=\\sin  \left(2t+\frac{\pi }{3}\right)-2\\sin   2t=\frac{\sqrt{3}}{2}\\cos   2t-\frac{3}{2}\\sin   2t=\sqrt{3}\\cos  \left(2t+\frac{\pi }{3}\right),结合已知条件，构造三角函数模型，利用三角函数知识解决相关问题.即函数关系式为y=\sqrt{3}\\cos  \left(2t+\frac{\pi }{3}\right)(t>0).当t\in \left(0,\frac{\pi }{2}\right]时,2t+\frac{\pi }{3}\in \left(\frac{\pi }{3},\frac{4\pi }{3}\right],\therefore \\cos  \left(2t+\frac{\pi }{3}\right)\in \left[-1,\frac{1}{2}\right),\therefore y\in \left[-\sqrt{3},\frac{\sqrt{3}}{2}\right).$

Final answer: $y=\sqrt{3}\cos  \left(2t+\frac{\pi }{3}\right)$, $[-\sqrt{3},\frac{\sqrt{3}}{2}]$
Subfield: Plane Geometry


Question: $如图,点A,B是圆心在原点,半径分别为1和2的圆上的动点.动点A从初始位置A_{0}\left(\\\cos   \frac{\pi }{3},\\\sin   \frac{\pi }{3}\right)开始,按逆时针方向以2 rad/s的角速度做圆周运动,同时点B从初始位置B_{0}(2,0)开始,按顺时针方向以2 rad/s的角速度做圆周运动.记t(s)时刻,点A,B的纵坐标分别为y_{1},y_{2}.$<img_2253>
$求y=y_{1}+y_{2}关于时间t(t>0)的函数关系式$

Solution:
$依题意知y_{1}=\\sin  \left(2t+\frac{\pi }{3}\right),y_{2}=-2\\sin   2t,所以y=y_{1}+y_{2}=\\sin  \left(2t+\frac{\pi }{3}\right)-2\\sin   2t=\frac{\sqrt{3}}{2}\\cos   2t-\frac{3}{2}\\sin   2t=\sqrt{3}\\cos  \left(2t+\frac{\pi }{3}\right),结合已知条件，构造三角函数模型，利用三角函数知识解决相关问题.即函数关系式为y=\sqrt{3}\\cos  \left(2t+\frac{\pi }{3}\right)(t>0).当t\in \left(0,\frac{\pi }{2}\right]时,2t+\frac{\pi }{3}\in \left(\frac{\pi }{3},\frac{4\pi }{3}\right],\therefore \\cos  \left(2t+\frac{\pi }{3}\right)\in \left[-1,\frac{1}{2}\right),\therefore y\in \left[-\sqrt{3},\frac{\sqrt{3}}{2}\right).$



Final answer: $y=\sqrt{3}\cos  \left(2t+\frac{\pi }{3}\right)$
Subfield: Plane Geometry


Question: $已知函数f(x)=2\\sin  (\omega x+\phi )(\omega >0，0<\phi <\pi )在一个周期内的图象如图所示．$![Image](https://cdntk.53zaixian.com/tq/1669581747885907969/img/compression/M_497e24bb3cee682d6c8405d116c2fac4.jpg)
$求f(x)的解析式；$

Solution:
$由题图可知，\frac{T}{2}=\frac{5\pi }{12}-\left(-\frac{\pi }{12}\right)=\frac{\pi }{2}，所以T=\pi ，所以\omega =2．(2分)因为f(x)的图象过点\left(-\frac{\pi }{12},2\right)，所以2\\sin  \left[2 \times \left(-\frac{\pi }{12}\right)+\varphi \right]=2，即\\sin  \left(-\frac{\pi }{6}+\varphi \right)=1，所以-\frac{\pi }{6}+\phi =2k\pi +\frac{\pi }{2}，k\in Z，解得\phi =2k\pi +\frac{2\pi }{3}，k\in Z．又因为0<\phi <\pi ，所以\phi =\frac{2\pi }{3}．(4分)所以f(x)=2\\sin  \left(2x+\frac{2\pi }{3}\right)．(5分)$

Final answer: $f(x)=2\sin (2x+\frac{2\pi} {3})$
Subfield: Plane Geometry


Question: $已知函数f(x)=2\\sin  (\omega x+\phi )(\omega >0，0<\phi <\pi )在一个周期内的图象如图所示．$<img_2358>
$求f(x)的解析式；$

Solution:
$由题图可知，\frac{T}{2}=\frac{5\pi }{12}-\left(-\frac{\pi }{12}\right)=\frac{\pi }{2}，所以T=\pi ，所以\omega =2．(2分)因为f(x)的图象过点\left(-\frac{\pi }{12},2\right)，所以2\\sin  \left[2 \times \left(-\frac{\pi }{12}\right)+\varphi \right]=2，即\\sin  \left(-\frac{\pi }{6}+\varphi \right)=1，所以-\frac{\pi }{6}+\phi =2k\pi +\frac{\pi }{2}，k\in Z，解得\phi =2k\pi +\frac{2\pi }{3}，k\in Z．又因为0<\phi <\pi ，所以\phi =\frac{2\pi }{3}．(4分)所以f(x)=2\\sin  \left(2x+\frac{2\pi }{3}\right)．(5分)$

Final answer: $f(x)=2\sin (2x+\frac{2\pi} {3})$
Subfield: Plane Geometry


Question: $已知函数f(x)=2\\sin  (\omega x+\phi )(\omega >0，0<\phi <\pi )在一个周期内的图象如图所示．$![Image](https://cdntk.53zaixian.com/tq/1669581747885907969/img/compression/M_497e24bb3cee682d6c8405d116c2fac4.jpg)
$直接写出f(x)在区间[0，\pi ]上的单调区间；$

Solution:
$在区间[0，\pi ]上，函数f(x)的单调递增区间为\left[\frac{5\pi }{12},\frac{11\pi }{12}\right]，(7分)单调递减区间为\left[0,\frac{5\pi }{12}\right]，\left[\frac{11\pi }{12},\pi \right]．(9分)$

Final answer: $f(x)的单调区间：[0, \pi ]上的单调递增区间为[5\pi /12, 11\pi /12]，单调递减区间为[0, 5\pi /12], [11\pi /12, \pi ]。$
Subfield: Plane Geometry


Question: $已知函数f(x)=2\\sin  (\omega x+\phi )(\omega >0，0<\phi <\pi )在一个周期内的图象如图所示．$<img_2358>
$直接写出f(x)在区间[0，\pi ]上的单调递增区间；$

Solution:
$在区间[0，\pi ]上，函数f(x)的单调递增区间为\left[\frac{5\pi }{12},\frac{11\pi }{12}\right]，(7分)单调递减区间为\left[0,\frac{5\pi }{12}\right]，\left[\frac{11\pi }{12},\pi \right]．(9分)$



Final answer: [5\pi /12, 11\pi /12]
Subfield: Plane Geometry


Question: $已知函数f(x)=2\\sin  (\omega x+\phi )(\omega >0，0<\phi <\pi )在一个周期内的图象如图所示．$<img_2358>
$已知∀x\in R， f(a-x)=f(a+x)都成立，直接写出一个满足题意的a的值．$

Solution:
$因为∀x\in R,f(a-x)=f(a+x)恒成立，所以函数f(x)的图象关于直线x=a对称，(11分)由题图可知，a=\frac{5\pi }{12}符合题意(答案不唯一).(12分)$

Final answer: $\frac{5\pi }{12}$
Subfield: Plane Geometry


Question: $已知函数f(x)=2\\sin  (\omega x+\phi )(\omega >0，0<\phi <\pi )在一个周期内的图象如图所示．$![Image](https://cdntk.53zaixian.com/tq/1669581747885907969/img/compression/M_497e24bb3cee682d6c8405d116c2fac4.jpg)
$已知∀x\in R， f(a-x)=f(a+x)都成立，直接写出一个满足题意的a的值．$

Solution:
$因为∀x\in R,f(a-x)=f(a+x)恒成立，所以函数f(x)的图象关于直线x=a对称，(11分)由题图可知，a=\frac{5\pi }{12}符合题意(答案不唯一).(12分)$

Final answer: $\frac{5\pi }{12}$
Subfield: Plane Geometry


Question: $已知函数f(x)=\\cos  (\omega x+\phi )(\omega >0,0<\phi <\pi )的部分图象如图所示.$<img_2212>
$求函数f(x)的单调递减区间；可用k表示任意整数$

Solution:
$由题图得最小正周期T=2\times \left(\frac{5}{4}-\frac{1}{4}\right)=2，\therefore \omega =\frac{2\pi }{T}=\pi ，\therefore f(x)=\\cos  (\pi x+\phi ).\because f \left(\frac{3}{4}\right)=\\cos  \left(\frac{3\pi }{4}+\varphi \right)=-1，\therefore \frac{3\pi }{4}+\phi =\pi +2k\pi ，k\in Z，\therefore \phi =\frac{\pi }{4}+2k\pi ，k\in Z，又0<\phi <\pi ，\therefore \phi =\frac{\pi }{4}，\therefore f(x)=\\cos  \left(\pi x+\frac{\pi }{4}\right).令2k\pi \leq \pi x+\frac{\pi }{4}\leq 2k\pi +\pi ，k\in Z，得2k-\frac{1}{4}\leq x\leq 2k+\frac{3}{4}，k\in Z，\therefore 函数f(x)的单调递减区间为\left[2k-\frac{1}{4},2k+\frac{3}{4}\right]，k\in Z.$



Final answer: $[2k-\frac14, 2k+\frac34]$
Subfield: Plane Geometry


Question: $已知函数f(x)=\\cos  (\omega x+\phi )(\omega >0,0<\phi <\pi )的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1665979845117558786/img/compression/M_257f588b16d3a5194c955075729de747.jpg)
$求函数f(x)的单调递减区间；$

Solution:
$由题图得最小正周期T=2\times \left(\frac{5}{4}-\frac{1}{4}\right)=2，\therefore \omega =\frac{2\pi }{T}=\pi ，\therefore f(x)=\\cos  (\pi x+\phi ).\because f \left(\frac{3}{4}\right)=\\cos  \left(\frac{3\pi }{4}+\varphi \right)=-1，\therefore \frac{3\pi }{4}+\phi =\pi +2k\pi ，k\in Z，\therefore \phi =\frac{\pi }{4}+2k\pi ，k\in Z，又0<\phi <\pi ，\therefore \phi =\frac{\pi }{4}，\therefore f(x)=\\cos  \left(\pi x+\frac{\pi }{4}\right).令2k\pi \leq \pi x+\frac{\pi }{4}\leq 2k\pi +\pi ，k\in Z，得2k-\frac{1}{4}\leq x\leq 2k+\frac{3}{4}，k\in Z，\therefore 函数f(x)的单调递减区间为\left[2k-\frac{1}{4},2k+\frac{3}{4}\right]，k\in Z.$

Final answer: $[2k-\frac14, 2k+\frac34] (k\in Z)$
Subfield: Plane Geometry


Question: $已知函数f(x)=\\cos  (\omega x+\phi )(\omega >0,0<\phi <\pi )的部分图象如图所示.$<img_2212>
$将f(x)的图象向右平移\frac{1}{2}个单位长度，再将所得图象上所有点的横坐标变为原来的\pi 倍(纵坐标不变)，得到函数g(x)的图象，若g(x)=a-1在\left[0,\frac{7\pi }{4}\right]上有两个解，求实数a的取值范围.$

Solution:
$易得g(x)=\\sin  \left(x+\frac{\pi }{4}\right).若方程g(x)=a-1在\left[0,\frac{7\pi }{4}\right]上有两个解，则函数y=\\sin  \left(x+\frac{\pi }{4}\right)的图象与直线y=a-1在\left[0,\frac{7\pi }{4}\right]上有两个不同的交点.利用转化思想,将方程的解的个数问题转化为函数图象与直线的交点问题.令t=x+\frac{\pi }{4},x\in \left[0,\frac{7\pi }{4}\right]，则t\in \left[\frac{\pi }{4},2\pi \right]，y=\\sin   t.作出函数y=\\sin   t在\left[\frac{\pi }{4},2\pi \right]上的图象，如图所示：$<img_2988>$由图可得，\frac{\sqrt{2}}{2}\leq a-1<1或-1<a-1\leq 0，解得1+\frac{\sqrt{2}}{2}\leq a<2或0<a\leq 1，故实数a的取值范围为0<a\leq 1或1+\frac{\sqrt{2}}{2}\leq a<2.$

Final answer: $[0,1] \cup [1+\frac{\sqrt 2}{2}2]$
Subfield: Plane Geometry


Question: $已知函数f(x)=\\cos  (\omega x+\phi )(\omega >0,0<\phi <\pi )的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1665979845117558786/img/compression/M_257f588b16d3a5194c955075729de747.jpg)
$将f(x)的图象向右平移\frac{1}{2}个单位长度，再将所得图象上所有点的横坐标变为原来的\pi 倍(纵坐标不变)，得到函数g(x)的图象，若g(x)=a-1在\left[0,\frac{7\pi }{4}\right]上有两个解，求实数a的取值范围.$

Solution:
$易得g(x)=\\sin  \left(x+\frac{\pi }{4}\right).若方程g(x)=a-1在\left[0,\frac{7\pi }{4}\right]上有两个解，则函数y=\\sin  \left(x+\frac{\pi }{4}\right)的图象与直线y=a-1在\left[0,\frac{7\pi }{4}\right]上有两个不同的交点.利用转化思想,将方程的解的个数问题转化为函数图象与直线的交点问题.令t=x+\frac{\pi }{4},x\in \left[0,\frac{7\pi }{4}\right]，则t\in \left[\frac{\pi }{4},2\pi \right]，y=\\sin   t.作出函数y=\\sin   t在\left[\frac{\pi }{4},2\pi \right]上的图象，如图所示：$![Image](https://cdntk.53zaixian.com/tq/1665979845465686017/img/compression/M_fcd4f69440bccf4917b2667731c28879.jpg)$由图可得，\frac{\sqrt{2}}{2}\leq a-1<1或-1<a-1\leq 0，解得1+\frac{\sqrt{2}}{2}\leq a<2或0<a\leq 1，故实数a的取值范围为0<a\leq 1或1+\frac{\sqrt{2}}{2}\leq a<2.$

Final answer: $[0,1] \cup [1+\frac{\sqrt 2}{2}2]$
Subfield: Plane Geometry


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )(A>0,\omega >0,|\phi |<\pi )的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1669583595185586177/img/compression/M_4a7d60a0cef2754330353163723d018c.jpg)
$求f(x)的解析式及图象的对称中心；$

Solution:
$由题图可知，A=2，\frac{3}{4}T=\frac{5\pi }{12}-\left(-\frac{\pi }{3}\right)=\frac{3\pi }{4}，因为T=\frac{2\pi }{\omega }=\pi ，所以\omega =2，所以f(x)=2\\sin  (2x+\phi ).又函数f(x)的图象过点\left(\frac{5\pi }{12},2\right)，所以2=2\\sin  \left(\frac{5\pi }{6}+\varphi \right)，即\frac{5\pi }{6}+\phi =\frac{\pi }{2}+2k\pi ,k\in Z，又|\phi |<\pi ，所以\phi =-\frac{\pi }{3}，所以f(x)=2\\sin  \left(2x-\frac{\pi }{3}\right).(3分)令2x-\frac{\pi }{3}=k\pi ,k\in Z，得x=\frac{\pi }{6}+\frac{k\pi }{2},k\in Z，所以f(x)图象的对称中心为\left(\frac{\pi }{6}+\frac{k\pi }{2},0\right),k\in Z.(5分)$

Final answer: $f(x)=2\sin (2x-\pi /3)$, $(\pi /6+k\pi /2,0)，其中k\in Z$
Subfield: Plane Geometry


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )(A>0,\omega >0,|\phi |<\pi )的部分图象如图所示.$<img_2680>
$求f(x)的解析式；$

Solution:
$由题图可知，A=2，\frac{3}{4}T=\frac{5\pi }{12}-\left(-\frac{\pi }{3}\right)=\frac{3\pi }{4}，因为T=\frac{2\pi }{\omega }=\pi ，所以\omega =2，所以f(x)=2\\sin  (2x+\phi ).又函数f(x)的图象过点\left(\frac{5\pi }{12},2\right)，所以2=2\\sin  \left(\frac{5\pi }{6}+\varphi \right)，即\frac{5\pi }{6}+\phi =\frac{\pi }{2}+2k\pi ,k\in Z，又|\phi |<\pi ，所以\phi =-\frac{\pi }{3}，所以f(x)=2\\sin  \left(2x-\frac{\pi }{3}\right).(3分)令2x-\frac{\pi }{3}=k\pi ,k\in Z，得x=\frac{\pi }{6}+\frac{k\pi }{2},k\in Z，所以f(x)图象的对称中心为\left(\frac{\pi }{6}+\frac{k\pi }{2},0\right),k\in Z.(5分)$



Final answer: $f(x)=2\sin (2x-\pi /3)$$
Subfield: Plane Geometry


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )(A>0,\omega >0,|\phi |<\pi )的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1669583595185586177/img/compression/M_4a7d60a0cef2754330353163723d018c.jpg)
$若f \left(\frac{\alpha }{2}\right)=\frac{4}{3}，求\\cos  \left(\alpha +\frac{\pi }{6}\right)的值；$

Solution:
$易得f \left(\frac{\alpha }{2}\right)=2\\sin  \left(\alpha -\frac{\pi }{3}\right)=\frac{4}{3}，即\\sin  \left(\alpha -\frac{\pi }{3}\right)=\frac{2}{3}.因为\left(\alpha +\frac{\pi }{6}\right)-\left(\alpha -\frac{\pi }{3}\right)=\frac{\pi }{2}，所以\\cos  \left(\alpha +\frac{\pi }{6}\right)=\\cos  \left[\frac{\pi }{2}+\left(\alpha -\frac{\pi }{3}\right)\right]=-\\sin  \left(\alpha -\frac{\pi }{3}\right)=-\frac{2}{3}.(8分)$

Final answer: $-\frac{2}{3}$
Subfield: Plane Geometry


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )(A>0,\omega >0,|\phi |<\pi )的部分图象如图所示.$<img_2680>
$若f \left(\frac{\alpha }{2}\right)=\frac{4}{3}，求\\cos  \left(\alpha +\frac{\pi }{6}\right)的值；$

Solution:
$易得f \left(\frac{\alpha }{2}\right)=2\\sin  \left(\alpha -\frac{\pi }{3}\right)=\frac{4}{3}，即\\sin  \left(\alpha -\frac{\pi }{3}\right)=\frac{2}{3}.因为\left(\alpha +\frac{\pi }{6}\right)-\left(\alpha -\frac{\pi }{3}\right)=\frac{\pi }{2}，所以\\cos  \left(\alpha +\frac{\pi }{6}\right)=\\cos  \left[\frac{\pi }{2}+\left(\alpha -\frac{\pi }{3}\right)\right]=-\\sin  \left(\alpha -\frac{\pi }{3}\right)=-\frac{2}{3}.(8分)$

Final answer: $-\frac{2}{3}$
Subfield: Plane Geometry


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )(A>0,\omega >0,|\phi |<\pi )的部分图象如图所示.$<img_2680>
$先将f(x)的图象的横坐标不变，纵坐标缩短为原来的\frac{1}{2}，得到函数g(x)的图象，再将g(x)的图象向右平移\frac{\pi }{12}个单位长度，得到h(x)的图象，求函数y=h(x)在x\in \left[\frac{\pi }{12},\frac{3\pi }{4}\right]上的单调递减区间.$

Solution:
$易得h(x)=\\sin  \left(2x-\frac{\pi }{6}-\frac{\pi }{3}\right)=-\\cos   2x.(10分)因为x\in \left[\frac{\pi }{12},\frac{3\pi }{4}\right]，所以2x\in \left[\frac{\pi }{6},\frac{3\pi }{2}\right]，令\pi \leq 2x\leq \frac{3\pi }{2}，解得\frac{\pi }{2}\leq x\leq \frac{3\pi }{4}，所以函数y=h(x)在x\in \left[\frac{\pi }{12},\frac{3\pi }{4}\right]上的单调递减区间为\left[\frac{\pi }{2},\frac{3\pi }{4}\right].(12分)$

Final answer: $[\frac{\pi}{2},\frac{3\pi}{4}]$
Subfield: Plane Geometry


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )(A>0,\omega >0,|\phi |<\pi )的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1669583595185586177/img/compression/M_4a7d60a0cef2754330353163723d018c.jpg)
$先将f(x)的图象的横坐标不变，纵坐标缩短为原来的\frac{1}{2}，得到函数g(x)的图象，再将g(x)的图象向右平移\frac{\pi }{12}个单位长度，得到h(x)的图象，求函数y=h(x)在x\in \left[\frac{\pi }{12},\frac{3\pi }{4}\right]上的单调递减区间.$

Solution:
$易得h(x)=\\sin  \left(2x-\frac{\pi }{6}-\frac{\pi }{3}\right)=-\\cos   2x.(10分)因为x\in \left[\frac{\pi }{12},\frac{3\pi }{4}\right]，所以2x\in \left[\frac{\pi }{6},\frac{3\pi }{2}\right]，令\pi \leq 2x\leq \frac{3\pi }{2}，解得\frac{\pi }{2}\leq x\leq \frac{3\pi }{4}，所以函数y=h(x)在x\in \left[\frac{\pi }{12},\frac{3\pi }{4}\right]上的单调递减区间为\left[\frac{\pi }{2},\frac{3\pi }{4}\right].(12分)$

Final answer: $[\frac{\pi}{2},\frac{3\pi}{4}]$