题目内容
已知 为偶函数,则a+b= ( )
A.-6 | B.-12 | C.4 | D.-4 |
A
解:∵∫-11(xcosx+3a-b)dx=2a+6,即(xsinx+cosx+3ax-bx)|-11=2a+6,
6a-2b=2a+6,⇒2a-b=3,①又f(t)=∫0t(x3+ax+5a-b)dx即:f(t)="(1/" 4 x4+1 /2 ax 2+5ax-bx)| t0 ="1/" 4 t4+1 /2 at 2+5at-bt因为偶函数,∴5a-b=0,②
由①②得:a=-1,b=-5.则a+b=-6.故选A.
6a-2b=2a+6,⇒2a-b=3,①又f(t)=∫0t(x3+ax+5a-b)dx即:f(t)="(1/" 4 x4+1 /2 ax 2+5ax-bx)| t0 ="1/" 4 t4+1 /2 at 2+5at-bt因为偶函数,∴5a-b=0,②
由①②得:a=-1,b=-5.则a+b=-6.故选A.
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