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It is given that for the function f(x) = x36x2+ax+b on [1, 3], Rolles’s theorem holds with c = 2+13. Find the values of a and b, if f(1) = f(3) = 0.

Solution :

We are given that f(1) = f(3) = 0.

  136×1+a+b = 336×32+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) = x36x2+ax+b

  f(x) = x36x2+11x6

  f'(x)  = 3x212x+11

  f'(c) = 3c212c+11 = 3(2+13)2 – 12(2+13) + 11

= 12 + 123 + 1 – 24 – 123 + 11 = 0

Hence, a = 11 and b = -6.


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