If the function f(x) = ax^3 + bx^2 + 11x - 6 | vedclass

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(A) Given $f(x) = ax^3 + bx^2 + 11x - 6$. Since $f(x)$ satisfies Rolle's theorem on $[1, 3]$,we have $f(1) = f(3)$. $f(1) = a(1)^3 + b(1)^2 + 11(1) -

Vedclass · Accepted Answer

$1, -6$

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(A) Given $f(x) = ax^3 + bx^2 + 11x - 6$. Since $f(x)$ satisfies Rolle's theorem on $[1, 3]$,we have $f(1) = f(3)$. $f(1) = a(1)^3 + b(1)^2 + 11(1) - 6 = a + b + 5$. $f(3) = a(3)^3 + b(3)^2 + 11(3) - 6 = 27a + 9b + 33 - 6 = 27a + 9b + 27$. Equating $f(1) = f(3)$: $a + b + 5 = 27a + 9b + 27 \Rightarrow 26a + 8b = -22 \Rightarrow 13a + 4b = -11$ ... $(i)$. Now,$f'(x) = 3ax^2 + 2bx + 11$. Given $f'\left( 2 + \frac{1}{\sqrt{3}} \right) = 0$: $3a\left( 2 + \frac{1}{\sqrt{3}} \right)^2 + 2b\left( 2 + \frac{1}{\sqrt{3}} \right) + 11 = 0$. $3a\left( 4 + \frac{1}{3} + \frac{4}{\sqrt{3}} \right) + 2b\left( 2 + \frac{1}{\sqrt{3}} \right) + 11 = 0$. $3a\left( \frac{13}{3} + \frac{4}{\sqrt{3}} \right) + 2b\left( 2 + \frac{1}{\sqrt{3}} \right) + 11 = 0$. $(13a + 4\sqrt{3}a) + (4b + \frac{2b}{\sqrt{3}}) + 11 = 0$. Substituting $13a = -11 - 4b$ from $(i)$: $-11 - 4b + 4\sqrt{3}a + 4b + \frac{2b}{\sqrt{3}} + 11 = 0$. $4\sqrt{3}a + \frac{2b}{\sqrt{3}} = 0 \Rightarrow 12a + 2b = 0 \Rightarrow b = -6a$ ... $(ii)$. Substitute $(ii)$ into $(i)$: $13a + 4(-6a) = -11 \Rightarrow 13a - 24a = -11 \Rightarrow -11a = -11 \Rightarrow a = 1$. Then $b = -6(1) = -6$. Thus,the values are $a = 1, b = -6$.

If the function $f(x) = ax^3 + bx^2 + 11x - 6$ 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

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