(Ⅰ) , ![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018065235-52043.gif) 令 得 ,解得![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018065235-68504.gif) 故 的增区间 和![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018065234-50329.gif) (Ⅱ) (x)=![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018065236-42675.gif) 当x∈[-1,1]时,恒有| (x)|≤ . 故有 ≤ (1)≤ , ≤ (-1)≤ , 及 ≤ (0)≤ , 即 ①+②,得 ≤ ≤ , 又由③,得 = ,将上式代回①和②,得 故![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018065237-78731.gif) . (Ⅲ)假设 ⊥ ,即![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018065233-42444.gif) = 故(s-a)(s-b)(t-a)(t-b)="-1 " [st-(s+t)a+a2][st-(s+t)b+b2]=-1, 由s,t为 (x)=0的两根可得,s+t= (a+b), st=![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018065238-10268.gif) , (0<a<b) 从而有ab(a-b)2=9. 这样![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018065238-94027.gif) 即 ≥2 ,这与 <2 矛盾. 故 与 不可能垂直. |