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Question

If the function $$f(x)=ax^3 -\displaystyle \frac{11x^2}{b}+ 11x - 6$$ 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 $$3^{rd}$$ condition we will get,
$$ a- \dfrac{11}{b} + 11 - 6 = 27a - \dfrac{99}{b} + 33 -6 \implies  26a -\dfrac{88}{b} + 22 = 0\>\>\>\rightarrow $$(i)
As given $$f'(2) = 0 \implies 12a - \dfrac{44}{b} + 11 -6 = 0\>\>\>\rightarrow $$ (ii)
Solving equation (i) & (ii), we will get 
$$a=0$$  &  $$ b = 4 $$
$$\therefore   a + b = 0+4 =4$$

Mathematics

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