设函数f(x)=(a×2^x-1)(1+2^x)是R上的奇函数,求(1)a的值,,(2)若k∈R,解不等式log2^(1+x)\(1-x)>log2^(1+x)\k
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![设函数f(x)=(a×2^x-1)(1+2^x)是R上的奇函数,求(1)a的值,,(2)若k∈R,解不等式log2^(1+x)\(1-x)>log2^(1+x)\k](/uploads/image/z/13352840-8-0.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D%28a%C3%972%5Ex-1%29%281%2B2%5Ex%29%E6%98%AFR%E4%B8%8A%E7%9A%84%E5%A5%87%E5%87%BD%E6%95%B0%2C%E6%B1%82%281%29a%E7%9A%84%E5%80%BC%2C%2C%282%29%E8%8B%A5k%E2%88%88R%2C%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Flog2%5E%281%2Bx%29%5C%281-x%29%3Elog2%5E%281%2Bx%29%5Ck)
设函数f(x)=(a×2^x-1)(1+2^x)是R上的奇函数,求(1)a的值,,(2)若k∈R,解不等式log2^(1+x)\(1-x)>log2^(1+x)\k
设函数f(x)=(a×2^x-1)(1+2^x)是R上的奇函数,求(1)a的值,
,(2)若k∈R,解不等式log2^(1+x)\(1-x)>log2^(1+x)\k
设函数f(x)=(a×2^x-1)(1+2^x)是R上的奇函数,求(1)a的值,,(2)若k∈R,解不等式log2^(1+x)\(1-x)>log2^(1+x)\k
(1)因为f(x)是R上的奇函数,所以f(0)=-f(-0),f(0)=0,所以(a*1-1)/(1+1)=0,a=1
(2)有函数单调性知,原不等式同解于1+x/1-x > 1+x/k (a) 且1+x/1-x >0 (b),1+x/k>0 (c)
由(b)得-10 x-1>0 1/1-x > 1/k 1-x < k x>1-k
k>=2时 -1
是大多数都是