《2025年经纶学典学霸黑白题高中数学必修第一册苏教版》

1. (2025·江苏常州高一期末)为了得到函数$ y = \sin\left(2x + \frac{\pi}{6}\right) $的图象,可以将函数$ y = \cos 2x $的图象 (A)

A.向右平移$\frac{\pi}{6}$个单位长度
B.向右平移$\frac{\pi}{3}$个单位长度
C.向左平移$\frac{\pi}{6}$个单位长度
D.向左平移$\frac{\pi}{3}$个单位长度

答案： 1.A解析：$y = \sin \left( 2x + \frac {\pi}{6} \right) = \cos \left( \frac {\pi}{2} - 2x - \frac {\pi}{6} \right) = \cos \left( \frac {\pi}{3} - 2x \right) = \cos \left( 2x - \frac {\pi}{3} \right)$，设$x \to x + \varphi$，$y = \cos 2(x + \varphi) = \cos (2x + 2\varphi)$，令$2\varphi = - \frac {\pi}{3}$，$\varphi = - \frac {\pi}{6}$，把函数$y = \cos 2x$的图象向右平移$\frac {\pi}{6}$个单位长度得到函数$y = \sin \left( 2x + \frac {\pi}{6} \right)$的图象。故选A。

2. 如图是周期为$\pi$的函数$ y = \cos(\omega x + \varphi) $ ($\omega ＞ 0, 0 ＜ \varphi ＜ \pi$)的部分图象,则$\varphi =$ (A)


A.$\frac{\pi}{6}$
B.$\frac{\pi}{3}$
C.$\frac{2\pi}{3}$
D.$\frac{5\pi}{6}$

答案： 2.A解析：由周期$T = \frac {2\pi}{\omega} = \pi$，得$\omega = 2$，由五点对应法得$2 × \frac {\pi}{6} + \varphi = \frac {\pi}{2}$，得$\varphi = \frac {\pi}{6}$。故选A。

3. * (多选)(2025·山西太原高一期末)已知函数$ f(x) = \sin 2x $,$ g(x) = \tan 2x $,则下列结论正确的是 (BC)

A.$ f(x) $,$ g(x) $的最小正周期都是$\pi$
B.$ f(x) $,$ g(x) $都是奇函数
C.$ f(x) $,$ g(x) $在$\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$上都单调递增
D.$ f(x) $,$ g(x) $的对称中心相同

答案： 3.BC解析：对于A，易知$g(x) = \tan 2x$的最小正周期是$\frac {\pi}{2}$，即A错误；对于B，易知$f(-x) = \sin(-2x) = - \sin 2x = -f(x)$，$g(-x) = \tan(-2x) = -\tan 2x = -g(x)$，且$f(x)$，$g(x)$的定义域都为$\mathbf {R}$，满足奇函数定义，因此它们都是奇函数，可得B正确；对于C，当$x \in \left( - \frac {\pi}{4},\frac {\pi}{4} \right)$时，$2x \in \left( - \frac {\pi}{2},\frac {\pi}{2} \right)$，再由正弦函数和正切函数单调性可得C正确；对于D，$f(x)$的对称中心的横坐标满足$2x = k\pi$，$k \in \mathbf {Z}$，即$x = \frac {k}{2}\pi$，$k \in \mathbf {Z}$，可得$f(x)$的对称中心为$\left( \frac {k}{2}\pi,0 \right)$，$k \in \mathbf {Z}$，而对于$g(x) = \tan 2x$，其对称中心的横坐标满足$2x = \frac {k}{2}\pi$，$k \in \mathbf {Z}$，可得$x = \frac {k}{4}\pi$，$k \in \mathbf {Z}$，即$g(x)$的对称中心为$\left( \frac {k}{4}\pi,0 \right)$，$k \in \mathbf {Z}$，因此$f(x)$，$g(x)$的对称中心不完全相同，故D错误。故选BC。

4. * (多选)下列不等式中一定成立的是 (AD)

A.$ \sin 470° ＞ \sin 115° $
B.$ \cos \frac{16\pi}{7} ＞ \cos \frac{17\pi}{8} $
C.$ \cos 226° ＞ \sin 224° $
D.$ \tan 1600° ＞ \tan 1415° $

答案： 4.AD解析：$\sin 470° = \sin 110° ＞ \sin 115°$，故A正确；因为$\cos \frac {16\pi}{7} = \cos \frac {2\pi}{7}$，$\cos \frac {17\pi}{8} = \cos \frac {\pi}{8}$，又$\frac {2\pi}{7} ＞ \frac {\pi}{8}$，所以$\cos \frac {2\pi}{7} ＜ \cos \frac {\pi}{8}$，故B错误；因为$\cos 226° = \cos (180° + 46°) = -\cos 46° = -\cos (90° - 44°) = -\sin 44°$，$\sin 224° = \sin (180° + 44°) = -\sin 44°$，所以$\cos 226° = \sin 224°$，故C错误；因为$\tan 1600° = \tan (4 × 360° + 160°) = \tan 160° = \tan (180° - 20°) = -\tan 20°$，$\tan 1415° = \tan (4 × 360° - 25°) = -\tan 25°$，又$\tan 25° ＞ \tan 20°$，所以$-\tan 25° ＜ -\tan 20°$，故D正确。故选AD。

5. * (2025·黑龙江哈尔滨高一期末)已知函数$ f(x) = \left(\frac{1}{2}a - \frac{3}{2}\right)\cos x + \left(\frac{\sqrt{3}}{2}a + \frac{\sqrt{3}}{2}\right)\sin x $是偶函数.若将曲线$ y = f(2x) $向左平移$\frac{\pi}{8}$个单位长度后,得到曲线$ y = g(x) $,则函数$ y = g(x) $的单调递增区间是 (C)

A.$\left[k\pi - \frac{7\pi}{12}, k\pi - \frac{\pi}{12}\right](k \in \mathbf{Z})$
B.$\left[k\pi - \frac{\pi}{12}, k\pi + \frac{5\pi}{12}\right](k \in \mathbf{Z})$
C.$\left[k\pi - \frac{\pi}{8}, k\pi + \frac{3\pi}{8}\right](k \in \mathbf{Z})$
D.$\left[k\pi - \frac{\pi}{4}, k\pi + \frac{3\pi}{4}\right](k \in \mathbf{Z})$

答案： 5.C解析：因为$f(x) = \left( \frac {1}{2}a - \frac {3}{2} \right) \cos x + \left( \frac {\sqrt {3}}{2}a + \frac {\sqrt {3}}{2} \right) \sin x$为偶函数，所以$f(-x) = f(x)$，即$\left( \frac {1}{2}a - \frac {3}{2} \right) \cos x + \left( \frac {\sqrt {3}}{2}a + \frac {\sqrt {3}}{2} \right) \sin x = \left( \frac {1}{2}a - \frac {3}{2} \right) \cos x - \left( \frac {\sqrt {3}}{2}a + \frac {\sqrt {3}}{2} \right) \sin x$，所以$\frac {\sqrt {3}}{2}a + \frac {\sqrt {3}}{2} = 0$，解得$a = -1$，所以$f(x) = -2 \cos x$。将曲线$y = f(2x)$向左平移$\frac {\pi}{8}$个单位长度后，得到曲线$g(x) = -2 \cos 2 \left( x + \frac {\pi}{8} \right) = -2 \cos \left( 2x + \frac {\pi}{4} \right)$，函数$y = 2 \cos \left( 2x + \frac {\pi}{4} \right)$的减区间即为函数$g(x)$的增区间。$2k\pi \leqslant 2x + \frac {\pi}{4} \leqslant 2k\pi + \pi$，$k \in \mathbf {Z}$，所以$2k\pi - \frac {\pi}{4} \leqslant 2x \leqslant 2k\pi + \pi - \frac {\pi}{4}$，$k \in \mathbf {Z}$，即$k\pi - \frac {\pi}{8} \leqslant x \leqslant k\pi + \frac {3\pi}{8}$，$k \in \mathbf {Z}$，所以函数$g(x)$的增区间为$\left[ k\pi - \frac {\pi}{8},k\pi + \frac {3\pi}{8} \right] (k \in \mathbf {Z})$。故选C。

6. * (多选)(2025·吉林通化高一月考)已知函数$ f(x) = \cos(\omega x + \varphi) + 1 $(其中$\omega$,$\varphi$均为常数,且$\omega ＞ 0$,$|\varphi| ＜ \pi$)恰能满足下列4个条件中的3个:
①函数$ f(x) $的最小正周期为$\pi$;
②函数$ f(x) $的图象经过点$\left(0, \frac{3}{2}\right)$;
③函数$ f(x) $的图象关于点$\left(\frac{5\pi}{12}, 1\right)$对称;
④函数$ f(x) $的图象关于直线$ x = -\frac{\pi}{6} $对称.
则这3个条件的序号可以是 (AB)

A.①②③
B.①②④
C.①③④
D.②③④

答案： 6.AB解析：若①正确，则$\frac {2\pi}{\omega} = \pi$，解得$\omega = 2$；若②正确，则$f(0) = \cos \varphi + 1 = \frac {3}{2}$，$\cos \varphi = \frac {1}{2}$，$|\varphi| ＜ \pi$，故$\varphi = \pm \frac {\pi}{3}$；若③正确，则$\frac {5\pi\omega}{12} + \varphi = \frac {\pi}{2} + k_1\pi$，$k_1 \in \mathbf {Z}$；若④正确，则$- \frac {\pi\omega}{6} + \varphi = k_2\pi$，$k_2 \in \mathbf {Z}$；对选项A：$\omega = 2$，取$\varphi = - \frac {\pi}{3}$，$\frac {5\pi}{6} - \frac {\pi}{3} = \frac {\pi}{2}$，满足条件，此时④不满足，正确；对选项B：$\omega = 2$，取$\varphi = \frac {\pi}{3}$，$-\frac {\pi}{3} + \frac {\pi}{3} = 0$，满足条件，此时③不满足，正确；对选项C：$\frac {5\pi}{12} + \frac {\pi}{6} = \frac {7\pi}{12} = \left( \frac {1}{4} + \frac {k}{2} \right) T = \left( \frac {1}{4} + \frac {k}{2} \right) \pi$，$k \in \mathbf {N}$，$k = \frac {2}{3}$，不成立，错误；对选项D：相减得到$\frac {5\pi}{12}\omega + \frac {\pi}{6} - (-\frac {\pi}{6}\omega + \varphi) = \frac {\pi}{2} + k_3\pi$，$k_3 \in \mathbf {Z}$，则$\omega = \frac {12}{7} \left( \frac {1}{2} + k_3 \right)$，$k_3 \in \mathbf {Z}$，此时$-\frac {\pi}{6}\omega + \varphi = - \frac {\pi}{6} × \frac {12}{7} \left( \frac {1}{2} + k_3 \right) + \varphi = \frac {-2}{7} \left( \frac {1}{2} + k_3 \right) \pi + \varphi = k_2\pi$，整理得$7\varphi = (7k_2 + 2k_3) \pi + \pi$，$k_2$，$k_3 \in \mathbf {Z}$，而$\varphi = \pm \frac {\pi}{3}$，故不成立，错误。故选AB。

7. * 若$ f(x) = 2\sin\left(2x + \frac{\pi}{6}\right) - 1 $,且$ f(x_1) · f(x_2) = -3 $,则$|x_1 - x_2|$的最小值为 (B)

A.$\pi$
B.$\frac{\pi}{2}$
C.$2\pi$
D.$\frac{\pi}{4}$

答案： 7.B解析：$f(x) = 2 \sin \left( 2x + \frac {\pi}{6} \right) - 1$周期为$T = \pi$，值域为$[-3,1]$，即$f(x)_{\max} = 1$，$f(x)_{\min} = -3$，则$\begin{cases}-3 \leqslant f(x_1) \leqslant 1 \\-3 \leqslant f(x_2) \leqslant 1 \\-3 \leqslant f(x_1)f(x_2) \leqslant 9\end{cases}$，由$f(x_1)f(x_2) = -3$，得$\begin{cases}f(x_1) = 1 \\f(x_2) = -3\end{cases}$或$\begin{cases}f(x_2) = 1 \\f(x_1) = -3\end{cases}$，即$f(x_1)$，$f(x_2)$为函数的最大值与最小值，或最小值与最大值，当$x_1$，$x_2$对应$f(x)$图象上相邻两最值点时，$|x_1 - x_2|$的值最小，故$|x_1 - x_2|_{\min} = \frac {T}{2} = \frac {\pi}{2}$。故选B。