If Rolle's theorem holds for the function f(x | vedclass

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(A) For Rolle's theorem to hold on $[1, 2]$,we must have $f(1) = f(2)$.$f(1) = 1 - a + b - 4 = -a + b - 3$$f(2) = 8 - 4a + 2b - 4 = -4a + 2b + 4$Equat

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$(5, 8)$

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(A) For Rolle's theorem to hold on $[1, 2]$,we must have $f(1) = f(2)$.$f(1) = 1 - a + b - 4 = -a + b - 3$$f(2) = 8 - 4a + 2b - 4 = -4a + 2b + 4$Equating them: $-a + b - 3 = -4a + 2b + 4 \Rightarrow 3a - b = 7$ $.......(1)$Given $f^{\prime}(x) = 3x^{2} - 2ax + b$.Since $f^{\prime}\left(\frac{4}{3}\right) = 0$,we have $3\left(\frac{4}{3}\right)^{2} - 2a\left(\frac{4}{3}\right) + b = 0$.$3\left(\frac{16}{9}\right) - \frac{8a}{3} + b = 0 \Rightarrow \frac{16}{3} - \frac{8a}{3} + b = 0$.Multiplying by $3$,we get $16 - 8a + 3b = 0 \Rightarrow 8a - 3b = 16$ $.......(2)$From $(1)$,$b = 3a - 7$. Substituting into $(2)$:$8a - 3(3a - 7) = 16 \Rightarrow 8a - 9a + 21 = 16 \Rightarrow -a = -5 \Rightarrow a = 5$.Then $b = 3(5) - 7 = 15 - 7 = 8$.Thus,$(a, b) = (5, 8)$.

If Rolle's theorem holds for the function $f(x)=x^{3}-ax^{2}+bx-4$ on the interval $x \in [1, 2]$ with $f^{\prime}\left(\frac{4}{3}\right)=0$,then the ordered pair $(a, b)$ is equal to

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