f:[1,3] rightarrow R is a function defined as | vedclass

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(B) Given, $f(x)=x^3+a x^2+b x$.We are given $f(1)-f(3)=0$, which implies $f(1)=f(3)$.Substituting the values into the function:$1+a+b = 27+9a+3b$$-26

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$-17$

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(B) Given, $f(x)=x^3+a x^2+b x$.We are given $f(1)-f(3)=0$, which implies $f(1)=f(3)$.Substituting the values into the function:$1+a+b = 27+9a+3b$$-26 = 8a+2b$Dividing by $2$, we get $4a+b=-13$ ... $(i)$.Now, find the derivative $f^{\prime}(x) = 3x^2+2ax+b$.Given $f^{\prime}\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}ight)=0$, let $x = \frac{2 \sqrt{3}+1}{\sqrt{3}} = 2 + \frac{1}{\sqrt{3}}$.$3\left(2+\frac{1}{\sqrt{3}}ight)^2 + 2a\left(2+\frac{1}{\sqrt{3}}ight) + b = 0$$3\left(4 + \frac{4}{\sqrt{3}} + \frac{1}{3}ight) + 4a + \frac{2a}{\sqrt{3}} + b = 0$$12 + 4\sqrt{3} + 1 + 4a + \frac{2a}{\sqrt{3}} + b = 0$$13 + 4\sqrt{3} + 4a + \frac{2a}{\sqrt{3}} + b = 0$Substitute $b = -13-4a$ from $(i)$:$13 + 4\sqrt{3} + 4a + \frac{2a}{\sqrt{3}} - 13 - 4a = 0$$4\sqrt{3} + \frac{2a}{\sqrt{3}} = 0$$\frac{2a}{\sqrt{3}} = -4\sqrt{3}$$2a = -4 	imes 3 = -12$$a = -6$.Using $a=-6$ in $(i)$, $b = -13 - 4(-6) = -13 + 24 = 11$.Therefore, $a-b = -6 - 11 = -17$.

$f:[1,3] \rightarrow R$ is a function defined as $f(x)=x^3+a x^2+b x$. If $f(1)-f(3)=0$ and $f^{\prime}\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0$, then $a-b$ is equal to

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