Solution :
We are given that f(1) = f(3) = 0.
∴ 13–6×1+a+b = 33–6×32+3a+b = 0
⟹ a + b = 5 and 3a + b = 27
Solving these two equations for a and b, f'(c) is zero or not.
We have,
f(x) = x3–6x2+ax+b
⟹ f(x) = x3–6x2+11x–6
⟹ f'(x) = 3x2–12x+11
∴ f'(c) = 3c2–12c+11 = 3(2+1√3)2 – 12(2+1√3) + 11
= 12 + 12√3 + 1 – 24 – 12√3 + 11 = 0
Hence, a = 11 and b = -6.
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