If the function $f(x) = ax^{3} + bx^{2} + 11x - 6$ satisfies the condition of Rolle’s theorem in [1, 3] and $f'\left( 2 + \frac{1}{\sqrt{3}} \right) = 0$, then the values of $a,\ b$ are respectively

Question

Dr. Abubacker · Accepted Answer

(c)
By Rolle’s theorem, $f(1) = f(3)$
$\Rightarrow \ \ \ a + b + 11 - 6 = 27a + 9b + 33 - 6$
$\Rightarrow \ \ 13a + 4b + 11 = 0$ ...(i)
Now, $f^{'}(x) = 3ax^{2} + 2bx + 11$
$\Rightarrow \ \ f^{'}\left( 2 + \frac{1}{\sqrt{3}} \right) = 3a\left( 2 + \frac{1}{\sqrt{3}} \right) + 2b\left( 2 + \frac{1}{\sqrt{3}} \right) + 11$
$\Rightarrow \ \ \ \ \ 0 = 3a\left( 4 + \frac{1}{3} + \frac{4}{\sqrt{3}} \right) + 4b + \frac{2b}{\sqrt{3}} + 11$
$\Rightarrow \ \ \ \ 13a + 4b + \frac{12a}{\sqrt{3}} + \frac{2b}{\sqrt{3}} + 11 = 0$
$\Rightarrow \ \ \ \  - 11 + \frac{12a}{\sqrt{3}} + \frac{2b}{\sqrt{3}} + 11 = 0$ …..[from Eq.(i)]
$\Rightarrow \ \ \ 6a + b = 0$ …(ii)
From Eq. (i) and (ii),
$a = 1,\ \ \ b = - 6$

Q. If the function $f(x) = ax^{3} + bx^{2} + 11x - 6$ satisfies the condition of Rolle’s theorem in [1, 3] and $f'\left( 2 + \frac{1}{\sqrt{3}} \right) = 0$, then the values of $a,\ b$ are respectively

Explanation

Reviewed by Dr. Abubacker

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