(文)若f(x)=atan(x+)+btan(x-)(ab≠0)是偶函数,则有序实数对(a,b)可以是_____

(文)若f(x)=atan(x+)+btan(x-)(ab≠0)是偶函数,则有序实数对(a,b)可以是_________正确的一组数字即可)

答案

答案:

(文)(1,-1)(答案不唯一)  ∵f(x)是偶函数,∴f(-x)=f(x),即atan(x+)+btan(x)=atan(-x+)+btan(-x).∴(a+b)[tan(x+)+tan(x)]=0.∴a+b=0.例如a=1,b=-1.

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