已知f(x)=ax²+bx+3a+b是偶函数,定义域为〔a-1,2a〕求a,b的值
人气:464 ℃ 时间:2019-08-21 06:30:28
解答
∵定义域应关于原点对称,
故有a-1=-2a,
得a=1/3
又∵f(-x)=f(x)恒成立,
即:ax²+bx+3a+b=ax²-bx+3a+b
∴b=0.
故答案为:a=1/3b=0
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