If f is defined in [1,3] by f(x)=x^3+b x^2+a | vedclass

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(C) Given $f(x) = x^3 + bx^2 + ax$. Since $f(1) = f(3)$,we have $1 + b + a = 27 + 9b + 3a$. This simplifies to $8b + 2a = -26$,or $4b + a = -13$ (Equa

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$(11,-6)$

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(C) Given $f(x) = x^3 + bx^2 + ax$. Since $f(1) = f(3)$,we have $1 + b + a = 27 + 9b + 3a$. This simplifies to $8b + 2a = -26$,or $4b + a = -13$ (Equation $1$). The derivative is $f^{\prime}(x) = 3x^2 + 2bx + a$. Given $f^{\prime}(c) = 0$ at $c = 2 + \frac{1}{\sqrt{3}}$,we have $3(2 + \frac{1}{\sqrt{3}})^2 + 2b(2 + \frac{1}{\sqrt{3}}) + a = 0$. Expanding this: $3(4 + \frac{4}{\sqrt{3}} + \frac{1}{3}) + 2b(2 + \frac{1}{\sqrt{3}}) + a = 0$. $12 + 4\sqrt{3} + 1 + 2b(2 + \frac{1}{\sqrt{3}}) + a = 0$. $13 + 4\sqrt{3} + 2b(2 + \frac{1}{\sqrt{3}}) + a = 0$. Substitute $a = -13 - 4b$ into the equation: $13 + 4\sqrt{3} + 4b + \frac{2b}{\sqrt{3}} - 13 - 4b = 0$. $4\sqrt{3} + \frac{2b}{\sqrt{3}} = 0$. $4\sqrt{3} = -\frac{2b}{\sqrt{3}} \implies 12 = -2b \implies b = -6$. Using $a = -13 - 4b$,we get $a = -13 - 4(-6) = -13 + 24 = 11$. Thus,$(a, b) = (11, -6)$.

If $f$ is defined in $[1,3]$ by $f(x)=x^3+b x^2+a x$,such that $f(1)-f(3)=0$ and $f^{\prime}(c)=0$,where $c=2+\frac{1}{\sqrt{3}}$,then $(a, b)$ is equal to

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