证明:(1)∵平面AEFD⊥平面EBCF, ∵ , ∴AE⊥EF, ∴AE⊥平面EBCF,AE⊥EF,AE⊥BE, 又BE⊥EF,故可如图建立空间坐标系E﹣xyz. ∵EA=2,∴EB=2, 又∵G为BC的中点,BC=4,∴BG=2. 则A(0,0,2),B(2,0,0),G(2,2,0),D(0,2,2),E(0,0,0), ∴ =(﹣2,2,2), =(2,2,0), =(﹣2,2,2)(2,2,0)=0, ∴BD⊥EG. 解:(2)∵AD∥面BFC,所以 f(x)=V D﹣BCF=V A﹣BFC= =![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021232444-67485.png) = , 即x=2时f(x)有最大值为 . (3)设平面DBF的法向量为 , ∵AE=2,B(2,0,0),D(0,2,2),F(0,3,0), ∴ , =(﹣2,2,2),则
,即 ,![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021232446-77506.png) 取x=3,y=2,z=1, ∴![](http://img.shitiku.com.cn/uploads/allimg/20191021/20191021232446-12206.png) ∵AE⊥面BCF, ∴面BCF一个法向量为 ,则 cos< >= , 由于所求二面角D﹣BF﹣C的平面角为钝角,所以此二面角的余弦值为﹣ .
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