1号卷·A10联盟2025届高三一轮复习试卷(二)2试题(数学)

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1号卷·A10联盟2025届高三一轮复习试卷(二)2试题(数学)试卷答案

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分析(1)由题意和三角函数公式化简可得f(x)=4sin(2x-$\frac{π}{3}$)+1,由x的范围可得;
(2)解绝对值不等式可得m-2<f(x)<m+2,由p是q的充分条件可得$\left\{\begin{array}{l}{m+2>5}\\{m-2<3}\end{array}\right.$,解不等式组可得.

解答解:(1)由题意和三角函数公式化简可得
f(x)=$4×\frac{1-cos(\frac{π}{2}+2x)}{2}$-2$\sqrt{3}$cos2x-1
=-2cos($\frac{π}{2}$+2x)-2$\sqrt{3}$cos2x+1
=2sin2x-2$\sqrt{3}$cos2x+1
=4sin(2x-$\frac{π}{3}$)+1,
∵$\frac{π}{4}≤x≤\frac{π}{2}$,∴$\frac{π}{6}≤2x-\frac{π}{3}≤\frac{2π}{3}$,
由三角函数的最值可得
当2x-$\frac{π}{3}$=$\frac{π}{2}$时,f(x)max=5,
当2x-$\frac{π}{3}$=$\frac{π}{6}$时,f(x)min=3;
(2)∵|f(x)-m|<2,∴m-2<f(x)<m+2,
又∵p是q的充分条件,∴$\left\{\begin{array}{l}{m+2>5}\\{m-2<3}\end{array}\right.$,
解得3<m<5

点评本题考查三角函数恒等变换以及最值,涉及简易逻辑的应用,属基础题.