已知函数y=f(x)对任意x∈(-π/2,π/2)满足f′(x)cosx+f(x)sinx>0则下列不等式成立的是1,根号2f(-π/3)<f(-π/4)2,根号2f(π/3)<f(π/4)3,f(0)小于2f(-π3)4,f(0)>根号2f(π/4)

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/05 21:37:53
已知函数y=f(x)对任意x∈(-π/2,π/2)满足f′(x)cosx+f(x)sinx>0则下列不等式成立的是1,根号2f(-π/3)<f(-π/4)2,根号2f(π/3)<f(π/4)3,f(0)小于2f(-π3)4,f(0)>根号2f(π/4)

已知函数y=f(x)对任意x∈(-π/2,π/2)满足f′(x)cosx+f(x)sinx>0则下列不等式成立的是1,根号2f(-π/3)<f(-π/4)2,根号2f(π/3)<f(π/4)3,f(0)小于2f(-π3)4,f(0)>根号2f(π/4)
已知函数y=f(x)对任意x∈(-π/2,π/2)满足f′(x)cosx+f(x)sinx>0
则下列不等式成立的是
1,根号2f(-π/3)<f(-π/4)
2,根号2f(π/3)<f(π/4)
3,f(0)小于2f(-π3)
4,f(0)>根号2f(π/4)

已知函数y=f(x)对任意x∈(-π/2,π/2)满足f′(x)cosx+f(x)sinx>0则下列不等式成立的是1,根号2f(-π/3)<f(-π/4)2,根号2f(π/3)<f(π/4)3,f(0)小于2f(-π3)4,f(0)>根号2f(π/4)
∵对任意x∈(-π/2,π/2)满足f′(x)cosx+f(x)sinx>0
∴[f(x)/cosx]'=[f′(x)cosx+f(x)sinx]/cos²x>0
∴f(x)/cosx是(-π/2,π/2)上的增函数
∴f(0)/cos0即f(0)< f(π/4)/(√2/2)∴f(0)<√2 f(π/4)<2f(π/3)
B,C,D均错
又f(-π/3)/cos(-π/3)∴2f(-π/3)<√2f(π/4)
即.√2f(-π/3)选A

已知函数f(x)对任意实数x,y属于R,总有f(x+y)=f(x)+f(y)已知函数f(x)对任意实数x,y∈R,总有f(x+y)=f(x)+f(y),且x>0时,f(x)<0,f(-1)=2(1)求证:f(-x)=-f(x)(2)求证:f(x)为减函数(3)求函数f(x) 已知f(x)是定义在R上的函数,对任意的x,y∈R,都有f(x+y)+f(x-y)=2f(x)f(y),已知f(x)是定义在R上的函数,对任意的x,y∈R,都有f(x+y)+f(x-y)=2f(x)f(y),且f(0)≠0(1):f(0)=1(2):判断函数的奇偶性 已知二次函数f(x)对任意x、y∈R都有f(x+y)=f(x)+f(y),且x>0时,f(x) 已知二次函数f(x)对任意x、y∈R都有f(x+y)=f(x)+f(y),且x>0时,f(x) 已知二次函数f[x]对任意想,x,y∈R总有飞f[x]+f[y]=f[x+y],且当X>0时,f[x] 已知函数f(x)对任意x,y∈R,总有f(x)+f(y)=f(x+y),且当x>0时,f(x) 已知函数f(x)对任意x,y属于R,总有f(x)+f(y)=f(x+y),且当x>0时,f(x) 已知函数fx对任意xy∈R满足f(x+y)=f(x)+f(y)求 1 f(0)的值2 f(x)为奇函数 已知函数y=f(x)(x∈R)对任意函数x,y,有f(x)+f(y)=2f(x+y/2)*f(x-y/2)恒成立,且f(0)≠0,(1.)求f(0)的值;(2)试判断函数y=f(x)(x∈R)的奇偶性 已知函数f(x)的定义域为R且对任意x,y∈R,有fx+y)=f(x)+f(y)+2, 已知函数对任意x,y∈R,都有f(xy)=f(x)+f(y),且f(2)=3,求f(8)的值 已知函数对任意x,y∈R,都有f(xy)=f(x)+f(y),且f(2)=3,求f(8)的值. 已知函数f(x)对任意x,y∈R,满足条件f(x)+f(y)=2+f(x+y),且当x>0时,f(x)>2,f(3)=5,求f(a^2-2a-2) 已知不恒为0的函数f(x)对任意实数x,y满足f(x+y)+f(x-y)=2【f(x)+f(y)],则f(x)的奇偶性是 函数的解析式的求法已知对任意的x,y,f(x)满足f(x)+f(y)=1/2f(x+y)求f(2) 已知函数满足对任意xy属于R都有f(x+y)=f(x)*f(y)-f(x)-f(y)+2成立,且x2,证明x 已知定义域在R上的函数f(x)对任意实数x,y,恒有f(x+y)+f(x-y)=2f(x)f(y),且f(0)不等于0.求证f(0)=1 已知f(x)是定义在R上的函数,对任意的x,y∈R,都有f(x+y)=f(x)+f(y),且x>0时有f(x)>0 ⑴判断函数奇偶性已知f(x)是定义在R上的函数,对任意的x,y∈R,都有f(x+y)=f(x)+f(y),且x>0时有f(x)>0⑴判断函数