函数竞赛题20分定义域为R的f(x)满足 ①x>0时,f(x)>1;②x≠y时,f(x)≠f(y);③f(x+y)=f(x)f(y)对任意x,y都成立.1.讨论f(x)单调性2.若集合M={(x,y)|f(x²)*f(y²)<f(1)}与P
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![函数竞赛题20分定义域为R的f(x)满足 ①x>0时,f(x)>1;②x≠y时,f(x)≠f(y);③f(x+y)=f(x)f(y)对任意x,y都成立.1.讨论f(x)单调性2.若集合M={(x,y)|f(x²)*f(y²)<f(1)}与P](/uploads/image/z/13846882-58-2.jpg?t=%E5%87%BD%E6%95%B0%E7%AB%9E%E8%B5%9B%E9%A2%9820%E5%88%86%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BAR%E7%9A%84f%EF%BC%88x%EF%BC%89%E6%BB%A1%E8%B6%B3+%E2%91%A0x%EF%BC%9E0%E6%97%B6%2Cf%EF%BC%88x%EF%BC%89%EF%BC%9E1%EF%BC%9B%E2%91%A1x%E2%89%A0y%E6%97%B6%2Cf%EF%BC%88x%EF%BC%89%E2%89%A0f%EF%BC%88y%EF%BC%89%3B%E2%91%A2f%28x%2By%29%3Df%28x%29f%28y%29%E5%AF%B9%E4%BB%BB%E6%84%8Fx%2Cy%E9%83%BD%E6%88%90%E7%AB%8B.1.%E8%AE%A8%E8%AE%BAf%EF%BC%88x%EF%BC%89%E5%8D%95%E8%B0%83%E6%80%A72.%E8%8B%A5%E9%9B%86%E5%90%88M%3D%EF%BD%9B%EF%BC%88x%2Cy%EF%BC%89%EF%BD%9Cf%EF%BC%88x%26sup2%3B%EF%BC%89%2Af%EF%BC%88y%26sup2%3B%EF%BC%89%EF%BC%9Cf%EF%BC%881%EF%BC%89%EF%BD%9D%E4%B8%8EP)
函数竞赛题20分定义域为R的f(x)满足 ①x>0时,f(x)>1;②x≠y时,f(x)≠f(y);③f(x+y)=f(x)f(y)对任意x,y都成立.1.讨论f(x)单调性2.若集合M={(x,y)|f(x²)*f(y²)<f(1)}与P
函数竞赛题20分
定义域为R的f(x)满足 ①x>0时,f(x)>1;②x≠y时,f(x)≠f(y);③f(x+y)=f(x)f(y)对任意x,y都成立.
1.讨论f(x)单调性
2.若集合M={(x,y)|f(x²)*f(y²)<f(1)}与P={(x,y)|f(ax+by+c)=1,a≠0}满足M∩P=空集,求a,b,c应满足的条件
函数竞赛题20分定义域为R的f(x)满足 ①x>0时,f(x)>1;②x≠y时,f(x)≠f(y);③f(x+y)=f(x)f(y)对任意x,y都成立.1.讨论f(x)单调性2.若集合M={(x,y)|f(x²)*f(y²)<f(1)}与P
1. 因为f(x+y)=f(x)f(y),那么f(x)=f(x)f(0) 得f(0)=1
所以f(0)=f(x)f(-x) 得f(-x)=1/f(x)
又因为x>0时,f(x)>1,所以x<0时,0
则f(x+t)-f(x)=f(x)[f(t)-1]>0
所以f(x)单调递增
2. M={(x,y)|f(x²)*f(y²)<f(1)}
则f(x^2+y^2)
则f(ax+by+c)=f(0) 即ax+by+c=0
令y=(-c-ax)/b代入x^2+y^2<1
得x^2+[(-c-ax)/b]^2<1 化简得(a^2+b^2)x^2+2acx+c^2-b^2<0
要使M∩P=空集,那么上述不等式无解
则△=4a^2c^2-4(a^2+b^2)(c^2-b^2)=4a^2b^2-4b^2c^2+4b^4≤0
那么b=0或a^2+b^2-c^2≤0