已知函数f(x)=x²-2ax+3a+2的值域是[4,+∞),且满足|a|>根号三.问:(1)f(x)的解析式(2)已知m∈R,设P:当x∈[1,2]时,不等式f(x)+3>2x+m 恒成立;Q:g(x)=根号下【f(x)-mx+1】的定义域为R.如果满足P成
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/12 15:56:26
![已知函数f(x)=x²-2ax+3a+2的值域是[4,+∞),且满足|a|>根号三.问:(1)f(x)的解析式(2)已知m∈R,设P:当x∈[1,2]时,不等式f(x)+3>2x+m 恒成立;Q:g(x)=根号下【f(x)-mx+1】的定义域为R.如果满足P成](/uploads/image/z/4336005-21-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dx%26%23178%3B-2ax%2B3a%2B2%E7%9A%84%E5%80%BC%E5%9F%9F%E6%98%AF%5B4%2C%2B%E2%88%9E%29%2C%E4%B8%94%E6%BB%A1%E8%B6%B3%7Ca%7C%3E%E6%A0%B9%E5%8F%B7%E4%B8%89.%E9%97%AE%EF%BC%9A%EF%BC%881%EF%BC%89f%28x%29%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F%EF%BC%882%EF%BC%89%E5%B7%B2%E7%9F%A5m%E2%88%88R%2C%E8%AE%BEP%EF%BC%9A%E5%BD%93x%E2%88%88%5B1%2C2%5D%E6%97%B6%2C%E4%B8%8D%E7%AD%89%E5%BC%8Ff%28x%29%2B3%EF%BC%9E2x%2Bm+%E6%81%92%E6%88%90%E7%AB%8B%EF%BC%9BQ%EF%BC%9Ag%28x%29%3D%E6%A0%B9%E5%8F%B7%E4%B8%8B%E3%80%90f%28x%29-mx%2B1%E3%80%91%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BAR.%E5%A6%82%E6%9E%9C%E6%BB%A1%E8%B6%B3P%E6%88%90)
xRJ@AX3)Z0 T@qH,\T1AYiU|!b}kǤi]X bs=w2z>_xg!6hGĢf))bt(¨Z^;)kw|Zv:%.S}בn BK XKO !atP4-mbzELxw q+c*j%,cr;WN~0#
[AG&"}+h0=pl8ya(`~C|?QIK!5O6B[ l
cWE'Zl2TJ7䱜@]$͋R%d!ȞԵox2j͌fWs9{rLd5bU"**a
nlkvAZ[L&qS>v
已知函数f(x)=x²-2ax+3a+2的值域是[4,+∞),且满足|a|>根号三.问:(1)f(x)的解析式(2)已知m∈R,设P:当x∈[1,2]时,不等式f(x)+3>2x+m 恒成立;Q:g(x)=根号下【f(x)-mx+1】的定义域为R.如果满足P成
已知函数f(x)=x²-2ax+3a+2的值域是[4,+∞),且满足|a|>根号三.
问:(1)f(x)的解析式
(2)已知m∈R,设P:当x∈[1,2]时,不等式f(x)+3>2x+m 恒成立;Q:g(x)=根号下【f(x)-mx+1】的定义域为R.如果满足P成立的m的集合记为A,满足Q成立的m的集合记为B,求CRA∩B
已知函数f(x)=x²-2ax+3a+2的值域是[4,+∞),且满足|a|>根号三.问:(1)f(x)的解析式(2)已知m∈R,设P:当x∈[1,2]时,不等式f(x)+3>2x+m 恒成立;Q:g(x)=根号下【f(x)-mx+1】的定义域为R.如果满足P成
(1)∵f(x)=x²-2ax+3a+2=(x-a)²+3a+2-a²
∴3a+2-a²=4,解得a=2或a=1,
∵|a|>√3,∴a=2,故f(x)=x²-4x+8
(2)∵f(x)=x²-4x+8,∴x²-4x+8+3>2x+m,即(x-3)²+2-m>0
∵1≤x≤2,∴-2≤x-3≤-1,∴1≤(x-3)²≤4,∴3-m≤(x-3)²+2-m≤6-m,
∵当x∈[1,2]时,不等式f(x)+3>2x+m 恒成立,∴3-m>0即m