f′(x)=x−22−2x+a≥0f′′(x)=(x−2)2−2−2<0f′(x)↓f′(3)≥02−6+a≥0a≥4amin=4 g(x)=(x−1)3(x+2−a)2 g(x)=(x−1)3(x−2)2 g′(x)=(x−1)32(x−2)+(x−2)23(x−1)2=(x−1)2(x−2)(2x−2+3x−6)=(x−1)2(x−2)(5x−8)<0x∈(58,2)100(a+b−c)=100(4+58−2)=360
Let (2,3) be the largest open interval in which the function f(x)=2loge(x−2)−x2+ax+1 is strictly increasing and (b, c) be the largest open interval, in which the function g(x)=(x−1)3(x+2−a)2 is strictly decreasing. Then 100(a+b−c) is equal to :
Held on 24 Jan 2025 · Verified 6 Jul 2026.
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