di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Plane Geometry


Question: $建设生态文明是关系人民福祉、关乎民族未来的长远大计.某市通宵营业的大型商场为响应国家节能减排的号召，在气温低于0 ^\circ C时，才开放中央空调，否则关闭中央空调.下图是该市冬季某一天的气温y(单位：^\circ C)随时间t(0\leq t\leq 24，单位：时)的大致变化曲线，已知该曲线近似满足y=f(t)=A\\sin  \left(\omega t-\frac{3\pi }{4}\right)+b(A>0,\omega >0)的关系.$<img_2060>
$请根据(1)的结论，求该商场的中央空调在一天内开启的时长.$

Solution:
$根据题意，令8\\sin  \left(\frac{\pi }{12}t-\frac{3\pi }{4}\right)+4<0，即\\sin  \left(\frac{\pi }{12}t-\frac{3\pi }{4}\right)<-\frac{1}{2}，(6分)由正弦函数的图象得\frac{7\pi }{6}+2k\pi <\frac{\pi }{12}t-\frac{3\pi }{4}<\frac{11\pi }{6}+2k\pi ,k\in Z，(8分)解得23+24k<t<31+24k,k\in Z，(9分)又因为0\leq t\leq 24，所以当k=-1时，0\leq t<7，当k=0时，23<t\leq 24，所以0\leq t<7或23<t\leq 24,(11分)所以该商场的中央空调在一天内开启的时长为8小时.(12分)$

Final answer: $8$ 小时
Subfield: Plane Geometry


Question: $建设生态文明是关系人民福祉、关乎民族未来的长远大计.某市通宵营业的大型商场为响应国家节能减排的号召，在气温低于0 ^\circ C时，才开放中央空调，否则关闭中央空调.下图是该市冬季某一天的气温y(单位：^\circ C)随时间t(0\leq t\leq 24，单位：时)的大致变化曲线，已知该曲线近似满足y=f(t)=A\\sin  \left(\omega t-\frac{3\pi }{4}\right)+b(A>0,\omega >0)的关系.$![Image](https://cdntk.53zaixian.com/tq/1706200452651651073/img/compression/M_52db569a52ba523d6dfc8abebeb52c0f.jpg)
$请根据(1)的结论，求该商场的中央空调在一天内开启的时长.$

Solution:
$根据题意，令8\\sin  \left(\frac{\pi }{12}t-\frac{3\pi }{4}\right)+4<0，即\\sin  \left(\frac{\pi }{12}t-\frac{3\pi }{4}\right)<-\frac{1}{2}，(6分)由正弦函数的图象得\frac{7\pi }{6}+2k\pi <\frac{\pi }{12}t-\frac{3\pi }{4}<\frac{11\pi }{6}+2k\pi ,k\in Z，(8分)解得23+24k<t<31+24k,k\in Z，(9分)又因为0\leq t\leq 24，所以当k=-1时，0\leq t<7，当k=0时，23<t\leq 24，所以0\leq t<7或23<t\leq 24,(11分)所以该商场的中央空调在一天内开启的时长为8小时.(12分)$

Final answer: $8$ 小时
Subfield: Trigonometric Functions


Question: $函数f(x)=A\\sin  (\omega x+\phi )\left(\right.其中A>0,\omega >0,|\phi |<\frac{\pi }{2}\left)\right.的部分图象如图所示,把函数f(x)的图象向右平移\frac{\pi }{4}个单位长度,再向下平移1个单位长度后得到函数g(x)的图象.$<img_2458>
$当x\in \left[\frac{\pi }{4},\frac{17\pi }{24}\right]时,求g(x)的值域;$

Solution:
$由题图知A=1,\frac{1}{4}T=\frac{7\pi }{12}-\frac{\pi }{3},即T=\pi ,\therefore \omega =\frac{2\pi }{T}=2,\therefore f(x)=\\sin  (2x+\phi ).把\left(\frac{7\pi }{12},-1\right)代入f(x)的解析式,得\\sin  \left(\frac{7\pi }{6}+\varphi \right)=-1,\therefore \frac{7\pi }{6}+\phi =\frac{3\pi }{2}+2k\pi ，k\in Z,\therefore \phi =2k\pi +\frac{\pi }{3},k\in Z.\because |\phi |<\frac{\pi }{2},\therefore \phi =\frac{\pi }{3}.\therefore f(x)=\\sin  \left(2x+\frac{\pi }{3}\right).(3分)由题意得g(x)=\\sin  \left[2\left(x-\frac{\pi }{4}\right)+\frac{\pi }{3}\right]-1=\\sin  \left(2x-\frac{\pi }{6}\right)-1.(4分)设\alpha =2x-\frac{\pi }{6},则当x\in \left[\frac{\pi }{4},\frac{17\pi }{24}\right]时，\alpha \in \left[\frac{\pi }{3},\frac{5\pi }{4}\right],\therefore \\sin   \alpha \in \left[-\frac{\sqrt{2}}{2},1\right],\therefore g(x)的值域为\left[-\frac{\sqrt{2}}{2}-1,0\right].(6分)$



Final answer: $\left[-\frac{\sqrt{2}}{2}-1,0\right]$
Subfield: Trigonometric Functions


Question: $函数f(x)=A\\sin  (\omega x+\phi )\left(\right.其中A>0,\omega >0,|\phi |<\frac{\pi }{2}\left)\right.的部分图象如图所示,把函数f(x)的图象向右平移\frac{\pi }{4}个单位长度,再向下平移1个单位长度后得到函数g(x)的图象.$![Image](https://cdntk.53zaixian.com/tq/1706200453008166914/img/compression/M_74a35475bc97fc1ceb20fd045d29d2e2.jpg)
$当x\in \left[\frac{\pi }{4},\frac{17\pi }{24}\right]时,求g(x)的值域;$

Solution:
$由题图知A=1,\frac{1}{4}T=\frac{7\pi }{12}-\frac{\pi }{3},即T=\pi ,\therefore \omega =\frac{2\pi }{T}=2,\therefore f(x)=\\sin  (2x+\phi ).把\left(\frac{7\pi }{12},-1\right)代入f(x)的解析式,得\\sin  \left(\frac{7\pi }{6}+\varphi \right)=-1,\therefore \frac{7\pi }{6}+\phi =\frac{3\pi }{2}+2k\pi ，k\in Z,\therefore \phi =2k\pi +\frac{\pi }{3},k\in Z.\because |\phi |<\frac{\pi }{2},\therefore \phi =\frac{\pi }{3}.\therefore f(x)=\\sin  \left(2x+\frac{\pi }{3}\right).(3分)由题意得g(x)=\\sin  \left[2\left(x-\frac{\pi }{4}\right)+\frac{\pi }{3}\right]-1=\\sin  \left(2x-\frac{\pi }{6}\right)-1.(4分)设\alpha =2x-\frac{\pi }{6},则当x\in \left[\frac{\pi }{4},\frac{17\pi }{24}\right]时，\alpha \in \left[\frac{\pi }{3},\frac{5\pi }{4}\right],\therefore \\sin   \alpha \in \left[-\frac{\sqrt{2}}{2},1\right],\therefore g(x)的值域为\left[-\frac{\sqrt{2}}{2}-1,0\right].(6分)$

Final answer: $\left[-\frac{\sqrt{2}}{2}-1,0\right]$
Subfield: Trigonometric Functions


Question: $函数f(x)=A\\sin  (\omega x+\phi )\left(\right.其中A>0,\omega >0,|\phi |<\frac{\pi }{2}\left)\right.的部分图象如图所示,把函数f(x)的图象向右平移\frac{\pi }{4}个单位长度,再向下平移1个单位长度后得到函数g(x)的图象.$![Image](https://cdntk.53zaixian.com/tq/1706200453008166914/img/compression/M_74a35475bc97fc1ceb20fd045d29d2e2.jpg)
$令F(x)=f(x)-3,若对任意实数x，[F(x)]^{2}-(2+m)F(x)+2+m\leq 0恒成立,求m的最大值.$

Solution:
$\because f(x)=\\sin  \left(2x+\frac{\pi }{3}\right),\therefore F(x)=f(x)-3=\\sin  \left(2x+\frac{\pi }{3}\right)-3.易知F(x)\in [-4,-2].(8分)令t=F(x),则t\in [-4,-2],设h(t)=t^{2}-(2+m)t+2+m,由题意得,在[-4,-2]上,h(t)_{max}\leq 0.(10分)易知当t\in [-4,-2]时，h(t)的最大值在t=-4或t=-2处取得,\therefore \left\{\begin{matrix}h(-2)\leq 0,\\ h(-4)\leq 0,\end{matrix}\right.即\left\{\begin{matrix}4-(2+m)\times (-2)+2+m\leq 0,\\ 16-(2+m)\times (-4)+2+m\leq 0,\end{matrix}\right.解得\left\{\begin{matrix}m\leq -\frac{10}{3},\\ m\leq -\frac{26}{5},\end{matrix}\right.\therefore m\leq -\frac{26}{5}.\therefore m的最大值为-\frac{26}{5}.(12分)$

Final answer: $-\frac{26}{5}$
Subfield: Trigonometric Functions


Question: $函数f(x)=A\\sin  (\omega x+\phi )\left(\right.其中A>0,\omega >0,|\phi |<\frac{\pi }{2}\left)\right.的部分图象如图所示,把函数f(x)的图象向右平移\frac{\pi }{4}个单位长度,再向下平移1个单位长度后得到函数g(x)的图象.$<img_2458>
$令F(x)=f(x)-3,若对任意实数x，[F(x)]^{2}-(2+m)F(x)+2+m\leq 0恒成立,求m的最大值.$

Solution:
$\because f(x)=\\sin  \left(2x+\frac{\pi }{3}\right),\therefore F(x)=f(x)-3=\\sin  \left(2x+\frac{\pi }{3}\right)-3.易知F(x)\in [-4,-2].(8分)令t=F(x),则t\in [-4,-2],设h(t)=t^{2}-(2+m)t+2+m,由题意得,在[-4,-2]上,h(t)_{max}\leq 0.(10分)易知当t\in [-4,-2]时，h(t)的最大值在t=-4或t=-2处取得,\therefore \left\{\begin{matrix}h(-2)\leq 0,\\ h(-4)\leq 0,\end{matrix}\right.即\left\{\begin{matrix}4-(2+m)\times (-2)+2+m\leq 0,\\ 16-(2+m)\times (-4)+2+m\leq 0,\end{matrix}\right.解得\left\{\begin{matrix}m\leq -\frac{10}{3},\\ m\leq -\frac{26}{5},\end{matrix}\right.\therefore m\leq -\frac{26}{5}.\therefore m的最大值为-\frac{26}{5}.(12分)$

Final answer: $-\frac{26}{5}$
Subfield: Trigonometric Functions


Question: $如图所示，一条河的两岸互相平行，河的宽度d=500 m,一艘船从A点出发航行到河对岸，船航行速度的大小为|v_{1}|=10 km/h,水流速度的大小为|v_{2}|=4 km/h,设v_{1}和v_{2}的夹角为\theta (0^\circ <\theta <180^\circ ).$![Image](https://cdntk.53zaixian.com/tq/1706236243067744259/img/compression/M_ea0a25713a4bcd311d48951ed2106e55.jpg)
$当\\cos   \theta 多大时，船能垂直到达对岸？$

Solution:
$船垂直到达对岸，即v_{1}+v_{2}与v_{2}垂直,即(v_{1}+v_{2})\cdot v_{2}=0,所以v_{1}\cdot v_{2}+v^2_2=0,即|v_{1}||v_{2}|\\cos   \theta +|v_{2}|^{2}=0，所以40\\cos   \theta +16=0，解得\\cos   \theta =-\frac{2}{5}.$

Final answer: $-\frac{2}{5}$
Subfield: Trigonometric Functions


Question: $如图所示，一条河的两岸互相平行，河的宽度d=500 m,一艘船从A点出发航行到河对岸，船航行速度的大小为|v_{1}|=10 km/h,水流速度的大小为|v_{2}|=4 km/h,设v_{1}和v_{2}的夹角为\theta (0^\circ <\theta <180^\circ ).$<img_1576>
$当\\cos   \theta 多大时，船能垂直到达对岸？$

Solution:
$船垂直到达对岸，即v_{1}+v_{2}与v_{2}垂直,即(v_{1}+v_{2})\cdot v_{2}=0,所以v_{1}\cdot v_{2}+v^2_2=0,即|v_{1}||v_{2}|\\cos   \theta +|v_{2}|^{2}=0，所以40\\cos   \theta +16=0，解得\\cos   \theta =-\frac{2}{5}.$

Final answer: $-\frac{2}{5}$
Subfield: Trigonometric Functions


Question: $如图所示，一条河的两岸互相平行，河的宽度d=500 m,一艘船从A点出发航行到河对岸，船航行速度的大小为|v_{1}|=10 km/h,水流速度的大小为|v_{2}|=4 km/h,设v_{1}和v_{2}的夹角为\theta (0^\circ <\theta <180^\circ ).$![Image](https://cdntk.53zaixian.com/tq/1706236243067744259/img/compression/M_ea0a25713a4bcd311d48951ed2106e55.jpg)
$当船垂直到达对岸时，航行所需时间是否最短？为什么？$

Solution:
$设船航行到对岸所需的时间为t h,则t=\frac{d}{|\boldsymbol{v}_1|\\\sin    \theta }=\frac{0.5}{10\\\sin    \theta }=\frac{1}{20\\\sin    \theta }.故当\theta =90^\circ 时，船的航行时间最短，为\frac{1}{20} h.故当船垂直到达对岸时，航行所需时间不是最短的.$

Final answer: $船垂直到达对岸时，航行所需时间不是最短的。$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(A>0,\omega >0,|\varphi |<\frac{\pi }{2}\right)的部分图象如图所示.$<img_2168>
$求函数f(x)的解析式；$

Solution:
$由题图可知，A=2，且T=2\left(\frac{11}{12}\pi -\frac{5}{12}\pi \right)=\pi ，所以\omega =\frac{2\pi }{T}=2，所以f(x)=2\\sin  (2x+\phi ).因为f\left(\frac{5}{12}\pi \right)=2\\sin  \left(\frac{5}{6}\pi +\varphi \right)=2，所以\\sin  \left(\frac{5}{6}\pi +\varphi \right)=1，所以\frac{5}{6}\pi +\phi =2k\pi +\frac{\pi }{2}(k\in Z),所以\phi =2k\pi －\frac{\pi }{3}(k\in Z).因为|\phi |<\frac{\pi }{2},所以\phi =－\frac{\pi }{3}，所以f(x)=2\\sin  \left(2x-\frac{\pi }{3}\right).令2k\pi －\frac{\pi }{2}\leq 2x－\frac{\pi }{3}\leq 2k\pi +\frac{\pi }{2}(k\in Z)，解得k\pi －\frac{\pi }{12}\leq x\leq k\pi +\frac{5\pi }{12}(k\in Z).故f(x)的解析式为f(x)=2\\sin  \left(2x-\frac{\pi }{3}\right)，f(x)的单调递增区间为\left[k\pi -\frac{\pi }{12},k\pi +\frac{5\pi }{12}\right](k\in Z).$



Final answer: $f(x)=2\sin (2x-\frac{\pi} 3)$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(A>0,\omega >0,|\varphi |<\frac{\pi }{2}\right)的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1706236342003367938/img/compression/M_3b1c85e7484deeac1f204f5d184a5170.jpg)
$求函数f(x)的解析式及单调递增区间；$

Solution:
$由题图可知，A=2，且T=2\left(\frac{11}{12}\pi -\frac{5}{12}\pi \right)=\pi ，所以\omega =\frac{2\pi }{T}=2，所以f(x)=2\\sin  (2x+\phi ).因为f\left(\frac{5}{12}\pi \right)=2\\sin  \left(\frac{5}{6}\pi +\varphi \right)=2，所以\\sin  \left(\frac{5}{6}\pi +\varphi \right)=1，所以\frac{5}{6}\pi +\phi =2k\pi +\frac{\pi }{2}(k\in Z),所以\phi =2k\pi －\frac{\pi }{3}(k\in Z).因为|\phi |<\frac{\pi }{2},所以\phi =－\frac{\pi }{3}，所以f(x)=2\\sin  \left(2x-\frac{\pi }{3}\right).令2k\pi －\frac{\pi }{2}\leq 2x－\frac{\pi }{3}\leq 2k\pi +\frac{\pi }{2}(k\in Z)，解得k\pi －\frac{\pi }{12}\leq x\leq k\pi +\frac{5\pi }{12}(k\in Z).故f(x)的解析式为f(x)=2\\sin  \left(2x-\frac{\pi }{3}\right)，f(x)的单调递增区间为\left[k\pi -\frac{\pi }{12},k\pi +\frac{5\pi }{12}\right](k\in Z).$

Final answer: $f(x)=2\sin (2x-\frac{\pi} 3)$，$单调递增区间：[k\pi -\pi /12, k\pi +5\pi /12](k\in Z)$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(A>0,\omega >0,|\varphi |<\frac{\pi }{2}\right)的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1706236342003367938/img/compression/M_3b1c85e7484deeac1f204f5d184a5170.jpg)
$若\alpha \in \left(0,\frac{\pi }{2}\right)，且f\left(\frac{\alpha }{2}+\frac{\pi }{6}\right)=\frac{6}{5}，求f\left(\frac{\alpha }{2}-\frac{\pi }{6}\right)的值.$

Solution:
$由(1)知，f(x)=2\\sin  \left(2x-\frac{\pi }{3}\right)，所以f\left(\frac{\alpha }{2}+\frac{\pi }{6}\right)=2\\sin   \alpha =\frac{6}{5}，所以\\sin   \alpha =\frac{3}{5}，又\alpha \in \left(0,\frac{\pi }{2}\right)，所以\\cos   \alpha =\sqrt{1-\\\sin   ^2\alpha }=\frac{4}{5},所以f\left(\frac{\alpha }{2}-\frac{\pi }{6}\right)=2\\sin  \left(\alpha -\frac{2\pi }{3}\right)=2\\sin   \alpha \\cos  \frac{2\pi }{3}－2\\cos   \alpha \\sin  \frac{2\pi }{3}=－\frac{4\sqrt{3}+3}{5}.$

Final answer: $－\frac{4\sqrt{3}+3}{5}$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(A>0,\omega >0,|\varphi |<\frac{\pi }{2}\right)的部分图象如图所示.$<img_2168>
$若\alpha \in \left(0,\frac{\pi }{2}\right)，且f\left(\frac{\alpha }{2}+\frac{\pi }{6}\right)=\frac{6}{5}，求f\left(\frac{\alpha }{2}-\frac{\pi }{6}\right)的值.$

Solution:
$由(1)知，f(x)=2\\sin  \left(2x-\frac{\pi }{3}\right)，所以f\left(\frac{\alpha }{2}+\frac{\pi }{6}\right)=2\\sin   \alpha =\frac{6}{5}，所以\\sin   \alpha =\frac{3}{5}，又\alpha \in \left(0,\frac{\pi }{2}\right)，所以\\cos   \alpha =\sqrt{1-\\\sin   ^2\alpha }=\frac{4}{5},所以f\left(\frac{\alpha }{2}-\frac{\pi }{6}\right)=2\\sin  \left(\alpha -\frac{2\pi }{3}\right)=2\\sin   \alpha \\cos  \frac{2\pi }{3}－2\\cos   \alpha \\sin  \frac{2\pi }{3}=－\frac{4\sqrt{3}+3}{5}.$

Final answer: $－\frac{4\sqrt{3}+3}{5}$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\cos  (\omega x+\phi )\left(A>0,\omega >0,|\varphi |\leq \frac{\pi }{2}\right)的部分图象如图所示.$<img_2974>
$求f(x)的解析式；$

Solution:
$由题图可知A=1，最小正周期T=4\times \left(\frac{\pi }{3}-\frac{\pi }{12}\right)=\pi ，\therefore \omega =\frac{2\pi }{T}=2，\therefore f\left(\frac{\pi }{12}\right)=\\cos  \left(\frac{\pi }{6}+\varphi \right)=1，\therefore \frac{\pi }{6}+\phi =2k\pi (k\in Z)，解得\phi =-\frac{\pi }{6}+2k\pi (k\in Z)，又|\phi |\leq \frac{\pi }{2}，\therefore \phi =-\frac{\pi }{6}，\therefore f(x)=\\cos  \left(2x-\frac{\pi }{6}\right).$

Final answer: $f(x)=\cos (2x-\frac{\pi} 6)$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\cos  (\omega x+\phi )\left(A>0,\omega >0,|\varphi |\leq \frac{\pi }{2}\right)的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1702594484084342787/img/compression/M_09fcde146cab5ee6c94a1a399d61f774.jpg)
$求f(x)的解析式；$

Solution:
$由题图可知A=1，最小正周期T=4\times \left(\frac{\pi }{3}-\frac{\pi }{12}\right)=\pi ，\therefore \omega =\frac{2\pi }{T}=2，\therefore f\left(\frac{\pi }{12}\right)=\\cos  \left(\frac{\pi }{6}+\varphi \right)=1，\therefore \frac{\pi }{6}+\phi =2k\pi (k\in Z)，解得\phi =-\frac{\pi }{6}+2k\pi (k\in Z)，又|\phi |\leq \frac{\pi }{2}，\therefore \phi =-\frac{\pi }{6}，\therefore f(x)=\\cos  \left(2x-\frac{\pi }{6}\right).$

Final answer: $f(x)=\cos (2x-\frac{\pi} 6)$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\cos  (\omega x+\phi )\left(A>0,\omega >0,|\varphi |\leq \frac{\pi }{2}\right)的部分图象如图所示.$<img_2974>
$将函数f(x)的图象向右平移\frac{\pi }{4}个单位后得到曲线C，把C上各点的横坐标保持不变，纵坐标变为原来的2倍，得到函数g(x)的图象，若关于x的方程g(x)-m=0在\left[0,\frac{\pi }{2}\right]上有两个不同的实数解\alpha ,\beta ，求实数m的取值范围及\alpha +\beta 的值.$

Solution:
$由题意得g(x)=2f\left(x-\frac{\pi }{4}\right)=2\\cos  \left(2x-\frac{\pi }{2}-\frac{\pi }{6}\right)=2\\sin  \left(2x-\frac{\pi }{6}\right).作出g(x)=2\\sin  \left(2x-\frac{\pi }{6}\right)，x\in \left[0,\frac{\pi }{2}\right]的图象及直线y=m，如图所示.$<img_1513>$由图可知，当1\leq m<2时，函数g(x)的图象与直线y=m在\left[0,\frac{\pi }{2}\right]上有两个不同的交点，即方程g(x)-m=0在\left[0,\frac{\pi }{2}\right]上有两个不同的实数解，且\alpha +\beta =\frac{2\pi }{3}.$

Final answer: $[1,2)$, $\frac{2\pi }{3}$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\cos  (\omega x+\phi )\left(A>0,\omega >0,|\varphi |\leq \frac{\pi }{2}\right)的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1702594484084342787/img/compression/M_09fcde146cab5ee6c94a1a399d61f774.jpg)
$将函数f(x)的图象向右平移\frac{\pi }{4}个单位后得到曲线C，把C上各点的横坐标保持不变，纵坐标变为原来的2倍，得到函数g(x)的图象，若关于x的方程g(x)-m=0在\left[0,\frac{\pi }{2}\right]上有两个不同的实数解\alpha ,\beta ，求实数m的取值范围及\alpha +\beta 的值.$

Solution:
$由题意得g(x)=2f\left(x-\frac{\pi }{4}\right)=2\\cos  \left(2x-\frac{\pi }{2}-\frac{\pi }{6}\right)=2\\sin  \left(2x-\frac{\pi }{6}\right).作出g(x)=2\\sin  \left(2x-\frac{\pi }{6}\right)，x\in \left[0,\frac{\pi }{2}\right]的图象及直线y=m，如图所示.$![Image](https://cdntk.53zaixian.com/tq/1702594484541521922/img/compression/M_d42f435ef3287c54d1bcba98ab1a5fb2.jpg)$由图可知，当1\leq m<2时，函数g(x)的图象与直线y=m在\left[0,\frac{\pi }{2}\right]上有两个不同的交点，即方程g(x)-m=0在\left[0,\frac{\pi }{2}\right]上有两个不同的实数解，且\alpha +\beta =\frac{2\pi }{3}.$

Final answer: $[1,2)$, $\frac{2\pi }{3}$
Subfield: Trigonometric Functions


Question: $函数f(x)=A\\sin  (\omega x+\theta )\left(A>0,\omega >0,\left|\theta \right|<\frac{\pi }{2}\right)的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1702594510663647233/img/compression/M_87a4f0fdc0a81fa3299b8b0cc1ba3bdf.jpg)
$求函数f(x)的解析式；$

Solution:
$由题图知，A=\sqrt{2}，\frac{3}{4}T=\frac{5\pi }{6}－\frac{\pi }{12}=\frac{3\pi }{4},\therefore T=\pi ，\therefore \omega =2.又点\left(\frac{\pi }{12},\sqrt{2}\right)在f(x)的图象上，\therefore 2\times \frac{\pi }{12}+\theta =\frac{\pi }{2}+2k\pi (k\in Z),解得\theta =\frac{\pi }{3}+2k\pi (k\in Z)，\because |\theta |<\frac{\pi }{2}，\therefore \theta =\frac{\pi }{3}.\therefore f(x)=\sqrt{2}\\sin  \left(2x+\frac{\pi }{3}\right).$

Final answer: $f(x)=\sqrt 2\sin (2x+\frac{\pi }3)$
Subfield: Trigonometric Functions


Question: $函数f(x)=A\\sin  (\omega x+\theta )\left(A>0,\omega >0,\left|\theta \right|<\frac{\pi }{2}\right)的部分图象如图所示.$<img_2601>
$求函数f(x)的解析式；$

Solution:
$由题图知，A=\sqrt{2}，\frac{3}{4}T=\frac{5\pi }{6}－\frac{\pi }{12}=\frac{3\pi }{4},\therefore T=\pi ，\therefore \omega =2.又点\left(\frac{\pi }{12},\sqrt{2}\right)在f(x)的图象上，\therefore 2\times \frac{\pi }{12}+\theta =\frac{\pi }{2}+2k\pi (k\in Z),解得\theta =\frac{\pi }{3}+2k\pi (k\in Z)，\because |\theta |<\frac{\pi }{2}，\therefore \theta =\frac{\pi }{3}.\therefore f(x)=\sqrt{2}\\sin  \left(2x+\frac{\pi }{3}\right).$

Final answer: $f(x)=\sqrt 2\sin (2x+\frac{\pi }3)$
Subfield: Trigonometric Functions


Question: $函数f(x)=A\\sin  (\omega x+\theta )\left(A>0,\omega >0,\left|\theta \right|<\frac{\pi }{2}\right)的部分图象如图所示.$<img_2601>
$若对任意实数x\in \left[0,\frac{\pi }{2}\right]，\left|f(x)-m\right|<2，求实数m的取值范围.$

Solution:
$\because x\in \left[0,\frac{\pi }{2}\right]，\therefore 2x+\frac{\pi }{3}\in \left[\frac{\pi }{3},\frac{4\pi }{3}\right]，\therefore \\sin  \left(2x+\frac{\pi }{3}\right)\in \left[-\frac{\sqrt{3}}{2},1\right]，\therefore f(x)\in \left[-\frac{\sqrt{6}}{2},\sqrt{2}\right].由|f(x)-m|<2，得m－2将f(x)的范围问题，转化为与f(x)的最值有关的问题.\therefore \left\{\begin{matrix}mf(x)_{\max }-2,\end{matrix}\right.解得\sqrt{2}－2\therefore 实数m的取值范围为\left(\sqrt{2}-2,2-\frac{\sqrt{6}}{2}\right).$



Final answer: $\left(\sqrt{2}-2,2-\frac{\sqrt{6}}{2}\right)$
Subfield: Trigonometric Functions


Question: $函数f(x)=A\\sin  (\omega x+\theta )\left(A>0,\omega >0,\left|\theta \right|<\frac{\pi }{2}\right)的部分图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1702594510663647233/img/compression/M_87a4f0fdc0a81fa3299b8b0cc1ba3bdf.jpg)
$若对任意实数x\in \left[0,\frac{\pi }{2}\right]，\left|f(x)-m\right|<2，求实数m的取值范围.$

Solution:
$\because x\in \left[0,\frac{\pi }{2}\right]，\therefore 2x+\frac{\pi }{3}\in \left[\frac{\pi }{3},\frac{4\pi }{3}\right]，\therefore \\sin  \left(2x+\frac{\pi }{3}\right)\in \left[-\frac{\sqrt{3}}{2},1\right]，\therefore f(x)\in \left[-\frac{\sqrt{6}}{2},\sqrt{2}\right].由|f(x)-m|<2，得m－2将f(x)的范围问题，转化为与f(x)的最值有关的问题.\therefore \left\{\begin{matrix}mf(x)_{\max }-2,\end{matrix}\right.解得\sqrt{2}－2\therefore 实数m的取值范围为\left(\sqrt{2}-2,2-\frac{\sqrt{6}}{2}\right).$

Final answer: $\left(\sqrt{2}-2,2-\frac{\sqrt{6}}{2}\right)$
Subfield: Plane Geometry


Question: $如图，在平面直角坐标系中，角\alpha ，\beta 的始边均为x轴正半轴，终边分别与圆O交于A，B两点，若\alpha \in \left(\frac{7\pi }{12},\pi \right)，\beta =\frac{\pi }{12}，且点A的坐标为(-2,m).$![Image](https://cdntk.53zaixian.com/tq/1702594560928186370/img/compression/M_12fa740fed48f6760368ef755021cd91.jpg)
$若\\tan   2\alpha =-\frac{4}{3}，求实数m的值；$

Solution:
$由题意得\\tan   \alpha =-\frac{m}{2}，所以\\tan   2\alpha =\frac{2\\\tan    \alpha }{1-{\\\tan    }^2\alpha }=\frac{-m}{1-\frac{m^2}{4}}=\frac{4m}{m^2-4}=-\frac{4}{3}，解得m=1或m=-4.因为\alpha \in \left(\frac{7\pi }{12},\pi \right)，所以\\tan   \alpha <0，所以m=1.$

Final answer: $m=1$
Subfield: Plane Geometry


Question: $如图，在平面直角坐标系中，角\alpha ，\beta 的始边均为x轴正半轴，终边分别与圆O交于A，B两点，若\alpha \in \left(\frac{7\pi }{12},\pi \right)，\beta =\frac{\pi }{12}，且点A的坐标为(-2,m).$<img_3064>
$若\\tan   2\alpha =-\frac{4}{3}，求实数m的值；$

Solution:
$由题意得\\tan   \alpha =-\frac{m}{2}，所以\\tan   2\alpha =\frac{2\\\tan    \alpha }{1-{\\\tan    }^2\alpha }=\frac{-m}{1-\frac{m^2}{4}}=\frac{4m}{m^2-4}=-\frac{4}{3}，解得m=1或m=-4.因为\alpha \in \left(\frac{7\pi }{12},\pi \right)，所以\\tan   \alpha <0，所以m=1.$

Final answer: $m=1$
Subfield: Plane Geometry


Question: $如图，在平面直角坐标系中，角\alpha ，\beta 的始边均为x轴正半轴，终边分别与圆O交于A，B两点，若\alpha \in \left(\frac{7\pi }{12},\pi \right)，\beta =\frac{\pi }{12}，且点A的坐标为(-2,m).$<img_3064>
$若\\tan  \angle AOB=-\frac{3}{4}，求\\sin   2\alpha 的值.$

Solution:
$由题意得\\tan  \angle AOB=\\tan  (\alpha -\beta )=\frac{\\\sin   \left(\alpha -\frac{\pi }{12}\right)}{\\\cos   \left(\alpha -\frac{\pi }{12}\right)}=-\frac{3}{4}.因为\alpha \in \left(\frac{7\pi }{12},\pi \right)，所以\alpha -\frac{\pi }{12}\in \left(\frac{\pi }{2},\frac{11\pi }{12}\right)，所以\\sin  \left(\alpha -\frac{\pi }{12}\right)=\frac{3}{5}，\\cos  \left(\alpha -\frac{\pi }{12}\right)=-\frac{4}{5}，所以\\sin  \left(2\alpha -\frac{\pi }{6}\right)=2\\sin  \left(\alpha -\frac{\pi }{12}\right)\\cos  \left(\alpha -\frac{\pi }{12}\right)=-\frac{24}{25}，\\cos  \left(2\alpha -\frac{\pi }{6}\right)=2\\cos  ^{2}\left(\alpha -\frac{\pi }{12}\right)-1=\frac{7}{25}，所以\\sin   2\alpha =\\sin  \left[\left(2\alpha -\frac{\pi }{6}\right)+\frac{\pi }{6}\right]=\frac{\sqrt{3}}{2}\\sin  \left(2\alpha -\frac{\pi }{6}\right)+\frac{1}{2}\cdot \\cos  \left(2\alpha -\frac{\pi }{6}\right)=\frac{7-24\sqrt{3}}{50}.$

Final answer: $\frac{7-24\sqrt{3}}{50}$
Subfield: Plane Geometry


Question: $如图，在平面直角坐标系中，角\alpha ，\beta 的始边均为x轴正半轴，终边分别与圆O交于A，B两点，若\alpha \in \left(\frac{7\pi }{12},\pi \right)，\beta =\frac{\pi }{12}，且点A的坐标为(-2,m).$![Image](https://cdntk.53zaixian.com/tq/1702594560928186370/img/compression/M_12fa740fed48f6760368ef755021cd91.jpg)
$若\\tan  \angle AOB=-\frac{3}{4}，求\\sin   2\alpha 的值.$

Solution:
$由题意得\\tan  \angle AOB=\\tan  (\alpha -\beta )=\frac{\\\sin   \left(\alpha -\frac{\pi }{12}\right)}{\\\cos   \left(\alpha -\frac{\pi }{12}\right)}=-\frac{3}{4}.因为\alpha \in \left(\frac{7\pi }{12},\pi \right)，所以\alpha -\frac{\pi }{12}\in \left(\frac{\pi }{2},\frac{11\pi }{12}\right)，所以\\sin  \left(\alpha -\frac{\pi }{12}\right)=\frac{3}{5}，\\cos  \left(\alpha -\frac{\pi }{12}\right)=-\frac{4}{5}，所以\\sin  \left(2\alpha -\frac{\pi }{6}\right)=2\\sin  \left(\alpha -\frac{\pi }{12}\right)\\cos  \left(\alpha -\frac{\pi }{12}\right)=-\frac{24}{25}，\\cos  \left(2\alpha -\frac{\pi }{6}\right)=2\\cos  ^{2}\left(\alpha -\frac{\pi }{12}\right)-1=\frac{7}{25}，所以\\sin   2\alpha =\\sin  \left[\left(2\alpha -\frac{\pi }{6}\right)+\frac{\pi }{6}\right]=\frac{\sqrt{3}}{2}\\sin  \left(2\alpha -\frac{\pi }{6}\right)+\frac{1}{2}\cdot \\cos  \left(2\alpha -\frac{\pi }{6}\right)=\frac{7-24\sqrt{3}}{50}.$

Final answer: $\frac{7-24\sqrt{3}}{50}$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(\right.A>0,\omega >0,-\frac{\pi }{2}<\phi <\frac{\pi }{2}\left)\right.在一个周期内的图象如图所示.$<img_1830>
$求函数f(x)的最小正周期T及f(x)的解析式；$

Solution:
$由题图可得A=1，\frac{T}{4}=\frac{\pi }{12}+\frac{\pi }{6}=\frac{\pi }{4}，所以T=\pi ，则\omega =\frac{2\pi }{T}=2，所以f(x)=\\sin  (2x+\phi ).(2分)因为f(x)的图象过点\left(\frac{\pi }{12},1\right)，所以2\times \frac{\pi }{12}+\phi =\frac{\pi }{2}+2k\pi (k\in Z)，即\phi =\frac{\pi }{3}+2k\pi (k\in Z)，又-\frac{\pi }{2}<\phi <\frac{\pi }{2}，所以\phi =\frac{\pi }{3}，(4分)故f(x)=\\sin  \left(2x+\frac{\pi }{3}\right).(5分)$

Final answer: $\pi$, $f(x) = \sin (2x + \frac{\pi} 3)$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(\right.A>0,\omega >0,-\frac{\pi }{2}<\phi <\frac{\pi }{2}\left)\right.在一个周期内的图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1702591227042074626/img/compression/M_62ab783c8d4230a0d20666fd2e60cb8c.jpg)
$求函数f(x)的最小正周期T及f(x)的解析式；$

Solution:
$由题图可得A=1，\frac{T}{4}=\frac{\pi }{12}+\frac{\pi }{6}=\frac{\pi }{4}，所以T=\pi ，则\omega =\frac{2\pi }{T}=2，所以f(x)=\\sin  (2x+\phi ).(2分)因为f(x)的图象过点\left(\frac{\pi }{12},1\right)，所以2\times \frac{\pi }{12}+\phi =\frac{\pi }{2}+2k\pi (k\in Z)，即\phi =\frac{\pi }{3}+2k\pi (k\in Z)，又-\frac{\pi }{2}<\phi <\frac{\pi }{2}，所以\phi =\frac{\pi }{3}，(4分)故f(x)=\\sin  \left(2x+\frac{\pi }{3}\right).(5分)$

Final answer: $\pi$, $f(x) = \sin (2x + \frac{\pi} 3)$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(\right.A>0,\omega >0,-\frac{\pi }{2}<\phi <\frac{\pi }{2}\left)\right.在一个周期内的图象如图所示.$![Image](https://cdntk.53zaixian.com/tq/1702591227042074626/img/compression/M_62ab783c8d4230a0d20666fd2e60cb8c.jpg)
$求函数f(x)的图象的对称轴方程及单调递增区间；$

Solution:
$结合题中图象及函数的最小正周期可知，函数f(x)的图象的对称轴方程为x=\frac{\pi }{12}+\frac{k\pi }{2}(k\in Z).(7分)令-\frac{\pi }{2}+2k\pi \leq 2x+\frac{\pi }{3}\leq \frac{\pi }{2}+2k\pi ,k\in Z，得-\frac{5\pi }{12}+k\pi \leq x\leq \frac{\pi }{12}+k\pi ,k\in Z，故函数f(x)的单调递增区间为\left[-\frac{5\pi }{12}+k\pi ,\frac{\pi }{12}+k\pi \right](k\in Z).(9分)$

Final answer: $图象的对称轴方程：x = \pi /12 + k\pi /2 (k\in Z)
单调递增区间：[-5\pi /12 + k\pi , \pi /12 + k\pi ] (k\in Z)$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=A\\sin  (\omega x+\phi )\left(\right.A>0,\omega >0,-\frac{\pi }{2}<\phi <\frac{\pi }{2}\left)\right.在一个周期内的图象如图所示.$<img_1830>
$将f(x)的图象向右平移\frac{\pi }{3}个单位，再将所得图象上所有点的横坐标伸长为原来的2倍(纵坐标不变)，得到函数g(x)的图象，若g(x)=a-1在x\in \left[\frac{\pi }{2},\frac{3\pi }{2}\right]上有两个不同的解，求实数a的取值范围.$

Solution:
$f(x)的图象向右平移\frac{\pi }{3}个单位得到y=\\sin  \left[2\left(x-\frac{\pi }{3}\right)+\frac{\pi }{3}\right]=\\sin  \left(2x-\frac{\pi }{3}\right)的图象，故g(x)=\\sin  \left(x-\frac{\pi }{3}\right).(10分)作y=g(x)在\left[\frac{\pi }{2},\frac{3\pi }{2}\right]上的图象，如图所示：$<img_1618>$因为g(x)=a-1在x\in \left[\frac{\pi }{2},\frac{3\pi }{2}\right]上有两个不同的解，所以\frac{1}{2}\leq a-1<1，即\frac{3}{2}\leq a<2.故实数a的取值范围是\left[\frac{3}{2},2\right).(12分)$



Final answer: $\left[\frac{3}{2},2\right)$