解:(Ⅰ)(1)g(x)=a(x﹣1)2+1+b﹣a 当a>0时,g(x)在[2,3]上为增函数 故![](http://img.shitiku.com.cn/uploads/allimg/20190819/20190819150327-30599.png) 当a<0时,g(x)在[2,3]上为减函数 故![](http://img.shitiku.com.cn/uploads/allimg/20190819/20190819150327-43039.png) ∵b<1 ∴a=1,b=0 (Ⅱ)由(Ⅰ)即g(x)=x2﹣2x+1.
. 方程f(2x)﹣k 2x≥0化为![](http://img.shitiku.com.cn/uploads/allimg/20190819/20190819150328-58830.png) , 令 ,k≤t2﹣2t+1 ∵x∈[﹣1,1] ∴![](http://img.shitiku.com.cn/uploads/allimg/20190819/20190819150328-75637.png) 记φ(t)=t2﹣2t+1 ∴φ(t)min=0 ∴k≤0 (Ⅲ)方程 化为![](http://img.shitiku.com.cn/uploads/allimg/20190819/20190819150329-80199.png) |2x﹣1|2﹣(2+3k)|2x﹣1|+(1+2k)=0,|2x﹣1|≠0 令|2x﹣1|=t,则方程化为t2﹣(2+3k)t+(1+2k)=0(t≠0) ∵方程 有三个不同的实数解, ∴由t=|2x﹣1|的图象知,t2﹣(2+3k)t+(1+2k)=0有两个根t1、t2,且0<t1<1<t2或0<t1<1,t2=1 记Φ(t)=t2﹣(2+3k)t+(1+2k)则
或![](http://img.shitiku.com.cn/uploads/allimg/20190819/20190819150330-67410.png) ∴k>0.
![](http://img.shitiku.com.cn/uploads/allimg/20190819/20190819150330-58150.png) |