(1)∵f(x)=f(x-1), 设x∈[1,2],则0≤x-1≤1, ∴f(x)=f(x-1)=(x-1)2(2-x). (2)设x∈[n,n+1],则0≤x-n≤1, f(x-n)=27(x-n)(n+1-x), ∴f(x)=f(x-1)=(x-2)=(x-3)=…=(x-n)=(x-n)2(n+1-x), ∴y=f(x),x∈[0,+∞]. f(x)=(x-n)2(n+1-x),x∈[n,n+1],n∈N. ∴f′(x)=[2(x-n)(n+1-x)-(x-n)2] =-[3x2-2(3n+1)x+n(3n+2)] =-[x2-2(n+)x+n(n+)] =-(x-n)[x-(n+)], ∴问题转化为判断关于x的方程-(x-n)[x-(n+)]=-1在[n,n+1],n∈N内是否有解, 即(x-n)[x-(n+)]=-1在[n,n+1],n∈N内是否有解, 令g(x)=(x-n)[x-(n+)]-=xn-x+-, 函数y=g(x)的图象是开口向上的抛物线, 其对称轴是直线x=n+∈[n,n+1], 判别式△=(-)2-4(-)=+>0, 且g(n)=-<0,g(n+1)=-=. ①当0≤n≤4,n∈N时,∵g(n+1)>0, ∴方程(x-n)[x-(n+)]=-1分别在区间[0,1],[1,2],[2,3],[3,4],[4,5]上各有一解, 即存在5个满足题意的点P. ②当n≥5(n∈N)时,∵g(n+1)<0, ∴方程(x-n)[x-(n+)]=-1在区间[n,n+1],n∈N,n≥5上无解. 综上所述,满足题意的点P有5个. (3)由(2)知f′(x)=-(x-n)[x-(n+)], ∴当x∈(n,n+)时,f′(x)>0,f(x)在(n+,n+1)上递减, ∴当x∈[n,n+1],n∈N时,f(x)max=f(n+)=, 又f(x)≥f(n)=f(n+1)=0, ∴对任意的n∈N*,当xn∈[n,n+1]时,都有0≤f(xn)≤, ∴Sn=f(x1)+f(x2)+…+f(xn) ≤++++…+ =4-<4, ∴0≤Sn<4. |