If function f(x) = a{x^3} + b{x^2} + 11x - 6 satisfies conditions of Rolle's theorem for interva

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5. Continuity and Differentiation

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

$1, -6$

$-2, 1$

$-1$, $\frac{1}{2}$

$-1, 6$

Solution

(a) Given $f(x) = a{x^3} + b{x^2} + 11x – 6$
From $f(1) = f(3)$
$a + b + 11 – 6 = 27a + 9b + 33 – 6$
$13a + 4b = – 11$…..$(i)$
and $f'(x) = 3a{x^2} + 2bx + 11$
$\because$$f'\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = 0$
$\therefore 3a\left( {\frac{{13}}{3} + \frac{4}{{\sqrt 3 }}} \right) + 2b\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = – 11$…..$(ii)$
From $(i)$ and $(ii),$
$13a + 4b = 3\left( {\frac{{13}}{3} + \frac{4}{{\sqrt 3 }}} \right){\rm{ }}a + 2b\left( {2 + \frac{1}{{\sqrt 3 }}} \right)$
$b = – 6a$ ……$(iii)$
From $(i)$ and $(iii),$ $a = 1$ and $b = – 6$.

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