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Let f(x)=⎧⎨⎩b3+b−2b2−2b2+5b+6−x2 ;0≤x<1 3x−4 ;1≤x≤3 where b∈R. If f(x) has minimum value at x=1, then the least integral value of b is

Answer» Let f(x)=⎧⎨⎩b3+b−2b2−2b2+5b+6−x2 ;0≤x<1 3x−4 ;1≤x≤3
where b∈R. If f(x) has minimum value at x=1, then the least integral value of b is


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