Let f:[0,3]→ A be defined by f(x)=2x3−15x2+36x+7 and g:[0,∞)→B be defined by g(x)=x2025+1x2025, If both the functions are onto and S={x\in Z;x\in A or x\in B}, then n(S) is equal to :
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Official solution
as f(x) is onto hence A is range of f(x)f′(x)=6x2−30x+36=6(x−2)(x−3)f(2)=16−60+72+7=35f(3)=54−135+108+7=34f(0)=7
hence range ∈[7,35]=A
also for range of g(x)g(x)=1−x2025+11∈[0,1)=Bs=0,7,8,…..35n(s)=30