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Question

If the function f(x)=ax3−11x2b+11x−6 satisfies conditions of Rolle's theorem in [1,3] and f′(2)=0, then find a+b.

A
176
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B
116
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C
76
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D
4
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Solution

The correct option is D 4
If given function f(x) satisfies Rolle's theorem on interval [1,3], then it will satisfy following conditions:
1) f(x) will be continuous on closed interval [1,3]
2) f(x) will be differentiable on open interval (1,3)
3) f(1)=f(3)
And there will be some point c in (1,3) such that f(c)=0.
Using 3rd condition we will get,
a11b+116=27a99b+33626a88b+22=0(i)
As given f(2)=012a44b+116=0 (ii)
Solving equation (i) & (ii), we will get
a=0 & b=4
a+b=0+4=4

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