已知定义域为R的函数f(x)=-2的x次方+b/2的x+1次方+a是奇函数(1)定义域为单调减(2)若对任意的t属于R,不等式f(t的平方-2t)+f(2t平方-k)
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![已知定义域为R的函数f(x)=-2的x次方+b/2的x+1次方+a是奇函数(1)定义域为单调减(2)若对任意的t属于R,不等式f(t的平方-2t)+f(2t平方-k)](/uploads/image/z/134842-58-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BAR%E7%9A%84%E5%87%BD%E6%95%B0f%28x%29%3D-2%E7%9A%84x%E6%AC%A1%E6%96%B9%2Bb%2F2%E7%9A%84x%2B1%E6%AC%A1%E6%96%B9%2Ba%E6%98%AF%E5%A5%87%E5%87%BD%E6%95%B0%EF%BC%881%EF%BC%89%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BA%E5%8D%95%E8%B0%83%E5%87%8F%282%EF%BC%89%E8%8B%A5%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84t%E5%B1%9E%E4%BA%8ER%2C%E4%B8%8D%E7%AD%89%E5%BC%8Ff%28t%E7%9A%84%E5%B9%B3%E6%96%B9-2t%EF%BC%89%2Bf%282t%E5%B9%B3%E6%96%B9-k%29)
已知定义域为R的函数f(x)=-2的x次方+b/2的x+1次方+a是奇函数(1)定义域为单调减(2)若对任意的t属于R,不等式f(t的平方-2t)+f(2t平方-k)
已知定义域为R的函数f(x)=-2的x次方+b/2的x+1次方+a是奇函数
(1)定义域为单调减(2)若对任意的t属于R,不等式f(t的平方-2t)+f(2t平方-k)
已知定义域为R的函数f(x)=-2的x次方+b/2的x+1次方+a是奇函数(1)定义域为单调减(2)若对任意的t属于R,不等式f(t的平方-2t)+f(2t平方-k)
∵f(x) = { -2^x + b } / { 2^(x+1) + a } 是奇函数∴f(0) = 0,且f(-x)=-f(x)
根据f(0)= 0,{-2^0 + b } / { 2^(0+1) + a},{-1+b} / {2+a},∴b=1,且a≠-2
根据f(-x)=-f(x)
{ -2^(-x)+1} / { 2^(-x+1)+a } = - { -2^x+1} / {2^(x+1)+a }
左边分子分母同乘以2^x:
{ -1+2^x } / { 2+a * 2^x } = { 2^x-1} / { 2^(x+1)+a}
{ -1+ 2^x} * {2^(x+1)+a} = {2+a * 2^x} * {2^x-1}
-2^(x+1) - a + 2^x * 2^(x+1) + a * 2^x = 2 * 2^x - 2 + a * 2^x * 2^x - a * 2^x
-2 * 2^x - a + 2 * 2^(2x) + a * 2^x = 2 * 2^x - 2 + a * 2^(2x) - a * 2^x
(2-a) * 2^(2x) - (2-a) * 2^x + (2-a) = 0
(2-a) * { 2^(2x) - 2^x + 1 } = 0
(2-a) * { (2^x - 1/2)^2 +3/4 } = 0
∵(2^x - 1/2)^2 +3/4 > 0
∴2-a=0
∴a=2
∴f(x) = { -2^x + 1 } / { 2^(x+1) + 2 }
= -(2^x - 1) / (2 * 2^x + 2)
= -1/2 (2^x - 1) / (2^x + 1)
= -1/2 (2^x + 1 - 2) / (2^x + 1)
= -1/2 + 1 / (2^x + 1)
∵2^x在定义域上单调增;∴2^x + 1单调增;∴1 / (2^x + 1)单调减;∴ -1/2 + 1 / (2^x + 1)单调减
∴f(x)在定义域上单调减.
f(t^2-2t) + f(2t^2-k) < 0
-1/2 + 1 / {2^(t^2-2t) + 1} -1/2 + 1 / {2^(t^2-k) + 1} < 0
-1 + 1 / {2^(t^2-2t) + 1} + 1 / {2^(t^2-k) + 1} < 0
∵2^(t^2-2t) + 1>1,2^(t^2-k) + 1 >1
∴两边同乘以 {2^(t^2-2t) + 1} {2^(t^2-k) + 1}不等式不变号
∴- {2^(t^2-2t) + 1} {2^(t^2-k) + 1} + {2^(t^2-k) + 1} + {2^(t^2-2t) +1} < 0
- 2^(t^2-2t) * 2^(t^2-k) - 2^(t^2-2t) - 1 - 2^(t^2-k)+ 2^(t^2-k) + 1 + 2^(t^2-2t) +1 < 0
- 2^(t^2-2t) * 2^(t^2-k) +1 < 0
2^(t^2-2t+t^2-k) >1
2(2t^2-2t-k) > 1
2t^2 - 2t - k > 0
f(t) = 2t^2 - 2t - k 开口向上,必须判别式<0时才能与x轴无交点,f(t)=2t^2 - 2t - k 恒大于0
∴△ = (-2)^2-4*2*(-k) < 0
4+8k<0
k<-1/2
已知定义域为R的函数F(x)=-2的x次+b/2的(x+1)的次方+a是奇函数,求a,b的值?〔过程详细,拜托各位大哥了。〕 a=0 b=2 由奇函数性质