2道高中函数题1、函数f(x)=loga|x-b|(以a为底|x-b|的对数)(a>0,a≠1)是偶函数,且在(0,+∞)上单调递减,则A.f(a-3)≤f(b-2) B.f(a-3)>f(b-2) C.f(a-3)≥f(b-2) D.f(a-3)<f(b-2)2、f(x)=b[1-2/(1+2^x)]+asinx+3(a,b
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/03 12:07:39
![2道高中函数题1、函数f(x)=loga|x-b|(以a为底|x-b|的对数)(a>0,a≠1)是偶函数,且在(0,+∞)上单调递减,则A.f(a-3)≤f(b-2) B.f(a-3)>f(b-2) C.f(a-3)≥f(b-2) D.f(a-3)<f(b-2)2、f(x)=b[1-2/(1+2^x)]+asinx+3(a,b](/uploads/image/z/9935784-0-4.jpg?t=2%E9%81%93%E9%AB%98%E4%B8%AD%E5%87%BD%E6%95%B0%E9%A2%981%E3%80%81%E5%87%BD%E6%95%B0f%28x%29%3Dloga%7Cx-b%7C%EF%BC%88%E4%BB%A5a%E4%B8%BA%E5%BA%95%7Cx-b%7C%E7%9A%84%E5%AF%B9%E6%95%B0%EF%BC%89%EF%BC%88a%3E0%2Ca%E2%89%A01%EF%BC%89%E6%98%AF%E5%81%B6%E5%87%BD%E6%95%B0%2C%E4%B8%94%E5%9C%A8%EF%BC%880%2C%2B%E2%88%9E%EF%BC%89%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%87%8F%2C%E5%88%99A.f%28a-3%29%E2%89%A4f%28b-2%29++B.f%28a-3%29%EF%BC%9Ef%28b-2%29+C.f%28a-3%29%E2%89%A5f%28b-2%29++D.f%28a-3%29%EF%BC%9Cf%28b-2%292%E3%80%81f%28x%29%3Db%5B1-2%2F%281%2B2%5Ex%29%5D%2Basinx%2B3%EF%BC%88a%2Cb)
2道高中函数题1、函数f(x)=loga|x-b|(以a为底|x-b|的对数)(a>0,a≠1)是偶函数,且在(0,+∞)上单调递减,则A.f(a-3)≤f(b-2) B.f(a-3)>f(b-2) C.f(a-3)≥f(b-2) D.f(a-3)<f(b-2)2、f(x)=b[1-2/(1+2^x)]+asinx+3(a,b
2道高中函数题
1、函数f(x)=loga|x-b|(以a为底|x-b|的对数)(a>0,a≠1)是偶函数,且在(0,+∞)上单调递减,则
A.f(a-3)≤f(b-2) B.f(a-3)>f(b-2)
C.f(a-3)≥f(b-2) D.f(a-3)<f(b-2)
2、f(x)=b[1-2/(1+2^x)]+asinx+3(a,b为常数),若f(x)在(0,+∞)上有最大值10,则在(-∞,0)上有
A.最大值10 B.最小值-5 C.最小值-4 D.最大值13
2道高中函数题1、函数f(x)=loga|x-b|(以a为底|x-b|的对数)(a>0,a≠1)是偶函数,且在(0,+∞)上单调递减,则A.f(a-3)≤f(b-2) B.f(a-3)>f(b-2) C.f(a-3)≥f(b-2) D.f(a-3)<f(b-2)2、f(x)=b[1-2/(1+2^x)]+asinx+3(a,b
1,答案选D
因为单调递减,所以0 (x+b)^2=(x-b)^2 => 4xb=0 => b=0
f(a-3)=loga|a-b-3|=loga|3-a|
f(b-2)=loga2
因为0
1.由偶函数 得b=0
由(0,+∞)单减可知 0f(b-2)=log(a)2
f(a-3)=log(a)|a-b-3|=log(a)|a-3|
2<|a-3|<3
f(b-2)
2.f(x)=b*[2^(-x)-1]/[2^(-x)+1]+asinx+3
设f(x0)=7+3 ...
全部展开
1.由偶函数 得b=0
由(0,+∞)单减可知 0f(b-2)=log(a)2
f(a-3)=log(a)|a-b-3|=log(a)|a-3|
2<|a-3|<3
f(b-2)
2.f(x)=b*[2^(-x)-1]/[2^(-x)+1]+asinx+3
设f(x0)=7+3 x0>0 且为最大值点
f(-x)=-[b*[2^(-x)-1]/[2^(-x)+1]+asinx]+3
f(-x0)=-7+3=-4
最小值为-4
选C
收起
12345