If the function f(x) = ax^3 + bx^2 + 26x - 24 | vedclass

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(A) $f(x) = ax^3 + bx^2 + 26x - 24$ ...$(i)$ on $[2, 4]$.Since $f(x)$ satisfies Rolle's theorem,$f(2) = f(4)$.$f(2) = a(8) + b(4) + 26(2) - 24 = 8a +

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$-9$

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(A) $f(x) = ax^3 + bx^2 + 26x - 24$ ...$(i)$ on $[2, 4]$.Since $f(x)$ satisfies Rolle's theorem,$f(2) = f(4)$.$f(2) = a(8) + b(4) + 26(2) - 24 = 8a + 4b + 28$.$f(4) = a(64) + b(16) + 26(4) - 24 = 64a + 16b + 80$.Equating $f(2) = f(4)$: $8a + 4b + 28 = 64a + 16b + 80 \Rightarrow 56a + 12b + 52 = 0 \Rightarrow 14a + 3b + 13 = 0$ ...(ii).Differentiating $f(x)$ w.r.t. $x$: $f^{\prime}(x) = 3ax^2 + 2bx + 26$.Given $f^{\prime}\left(3 + \frac{1}{\sqrt{3}}ight) = 0$:$3a\left(3 + \frac{1}{\sqrt{3}}ight)^2 + 2b\left(3 + \frac{1}{\sqrt{3}}ight) + 26 = 0$.$3a\left(9 + \frac{1}{3} + 2\sqrt{3}ight) + 6b + \frac{2b}{\sqrt{3}} + 26 = 0$.$3a\left(\frac{28}{3} + 2\sqrt{3}ight) + 6b + \frac{2b}{\sqrt{3}} + 26 = 0$.$28a + 6\sqrt{3}a + 6b + \frac{2b}{\sqrt{3}} + 26 = 0$.$(28a + 6b + 26) + \frac{1}{\sqrt{3}}(18a + 2b) = 0$.Comparing rational and irrational parts: $28a + 6b + 26 = 0$ and $18a + 2b = 0$.From $18a + 2b = 0$,$b = -9a$.Substitute into $28a + 6(-9a) + 26 = 0 \Rightarrow 28a - 54a + 26 = 0 \Rightarrow -26a = -26 \Rightarrow a = 1$.Then $b = -9(1) = -9$.Therefore,$ab = 1 	imes (-9) = -9$.

If the function $f(x) = ax^3 + bx^2 + 26x - 24$ satisfies the conditions of Rolle's theorem in $[2, 4]$ and $f^{\prime}\left(3 + \frac{1}{\sqrt{3}}\right) = 0$,then the value of $ab$ is equal to

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