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

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(B) $1$. For Rolle's theorem on $[1, 3]$, $f(1) = f(3)$.$f(1) = a(1)^3 + b(1)^2 + 11(1) - 6 = a + b + 5$.$f(3) = a(3)^3 + b(3)^2 + 11(3) - 6 = 27a + 9

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$a = 1, b = -6$

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(B) $1$. For Rolle's theorem on $[1, 3]$, $f(1) = f(3)$.$f(1) = a(1)^3 + b(1)^2 + 11(1) - 6 = a + b + 5$.$f(3) = a(3)^3 + b(3)^2 + 11(3) - 6 = 27a + 9b + 27$.Equating $f(1) = f(3)$: $a + b + 5 = 27a + 9b + 27 \implies 26a + 8b = -22 \implies 13a + 4b = -11$ (Equation $1$).$2$. Rolle's theorem states $f'(c) = 0$ for some $c \in (1, 3)$.$f'(x) = 3ax^2 + 2bx + 11$.Given $c = 2 + \frac{1}{\sqrt{3}}$, $f'(c) = 3a(2 + \frac{1}{\sqrt{3}})^2 + 2b(2 + \frac{1}{\sqrt{3}}) + 11 = 0$.$3a(4 + \frac{4}{\sqrt{3}} + \frac{1}{3}) + 4b + \frac{2b}{\sqrt{3}} + 11 = 0$.$3a(\frac{13}{3} + \frac{4}{\sqrt{3}}) + 4b + \frac{2b}{\sqrt{3}} + 11 = 0 \implies 13a + 4\sqrt{3}a + 4b + \frac{2b}{\sqrt{3}} + 11 = 0$.Group rational and irrational parts: $(13a + 4b + 11) + \frac{1}{\sqrt{3}}(4a + 2b) = 0$.Since $a, b$ are constants, $4a + 2b = 0 \implies b = -2a$.$3$. Substitute $b = -2a$ into Equation $1$: $13a + 4(-2a) = -11 \implies 13a - 8a = -11 \implies 5a = -11$ (Note: Re-evaluating calculation).Correction: $f(1) = a+b+5$, $f(3) = 27a+9b+27$. $26a+8b = -22 \implies 13a+4b = -11$. $f'(c)=0 \implies 12a+4b+11 + \frac{4a+2b}{\sqrt{3}} = 0$. Thus $4a+2b=0 \implies b=-2a$. $13a+4(-2a)=-11 \implies 5a=-11$. Given the options, let's re-check $f(3)$. $27a+9b+33-6 = 27a+9b+27$. Correct. $a+b+11-6 = a+b+5$. Correct. $26a+8b = -22$. Correct. $f'(x) = 3ax^2+2bx+11$. $f'(2+1/\sqrt{3}) = 3a(4+4/\sqrt{3}+1/3) + 2b(2+1/\sqrt{3}) + 11 = 12a + 4\sqrt{3}a + a + 4b + 2b/\sqrt{3} + 11 = 13a+4b+11 + \sqrt{3}(4a + 2b/3) = 0$. Thus $4a+2b/3=0 \implies b=-6a$. $13a+4(-6a)=-11 \implies -11a=-11 \implies a=1, b=-6$.

If the function $f(x) = ax^3 + bx^2 + 11x - 6$, defined on $[1, 3]$, satisfies all the conditions of Rolle's theorem for $c = 2 + \frac{1}{\sqrt{3}}$, then

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